A-level Math Paper 1 | Complete Revision Workshop | MathPrep with Ahmad Bilal

追加: 2026-04-29

[00:00:00]Any quick questions in the beginning before we start?

[00:00:05]Okay. So again, it's a revision session.

[00:00:07]So I'll be quickly going over all the all the major concepts and the common types of questions that you get on these topics. All right, we're starting with binomial first of all. So let's begin with this.

[00:00:22]All right. So what's binomial expansion first of all? Binomial expansion is when you have two terms being added like this and you have a power on top of that. If you have a plus b whole per square one for instance that's fairly straightforward that's a plus b but you've learned in in polles or ICS for instance what a plus b whole square expands to it's a square + 2 a + b square but if you have to expand higher powers like a plus b whole raised to power 3 four five it becomes very tedious to do it manually like this we need some general formula to be able to do that right and in binomial expansion formula we what we can do is using this binomial expansion formula we can expand find any integer power for an expression like this. So we have two terms added two terms adding in a bracket like this and we have a power on top of that like this. We can use this formula to expand that expression to uh whatever is the number of terms that it has. Now there's a condition for using this formula and that condition is that the power has to be an a positive integer. Now in paper one you will always get get questions in which power is power is a positive integer. If it's not a positive integer you will not get get that in paper one.

[00:01:33]You have another binomial binomial formula that's that's given in the formula sheet. Uh since you get the same formula sheet in uh the exam. Just be careful that the other formula that looks something like this that's not uh what you end up using here because that's not relevant for this 1 + nx something. This is the formula that you'll be using a power n plus n c1 into a rais^ n minus 1 b and so on. All you have to do is just input values in the formula. Whatever is the first term that's going to be a whatever is the second term that's going to be b. For instance, if you want to expand a plus b whole raised to power 4. What you do is you you basically figure out what is in place of the in place of a in the formula. You've got a here. What's in in place of b? In place of b you've got b.

[00:02:21]In this case there's nothing nothing different inside the inside the bracket but the power is four.

[00:02:27]All you have to do is in the formula replace n with four. Right? That's all that you're doing. Just replacing n with four. And now how exactly do you evaluate this? This is straightforward a^ 4 - 1 into b a^ 4 - 2 b ^2 and so on.

[00:02:43]But what about 4 c1 4 c2 4 c3. It's a function that you have in your calculator. If you look at the division function, division button on your calculator, on top of this division button, you have this button in yellow. You can use that to evaluate these numbers 4c2. So what you do is, for example, if you have to evaluate 4c1 on your calculator, you go like this.

[00:03:07]You press four. You press the NCR button and then you press one. And that gives you the result. That's 4C1. 4C1 is four.

[00:03:14]Similarly, if you want to evaluate 6 C2, you press six here. Then you have 2 6 C2. That gives you 15. So evaluating this is going to be fairly straightforward. After that, once you've input values in the formula, you just evaluate this on your calculator. 4 C1 4 C2 and so on. And then you just multiply those numbers like this. Now that's a fairly straightforward case. What if you have something else like this? For example, if you have x - 2 y^ 8 x - 2 y^ 8. If you have an expression like this, what about this? Now in place of a we've got x. In place of b we've got minus 2 y and the power is 8. Okay.

[00:03:57]Now all you have to do is in the formula wherever you've got a you replace that with x. In the formula, wherever you've got B, you replace that with minus 2 Y, right? And wherever in the formula you've got 8, you replace that. Wherever in the formula you've got N, you replace that with 8. And that's it. Then you just simplify this and you get the result. In this case, you have X to power 8 plus this plus this. Just be careful with the negative signs here because when there is a negative here that is going to be part of the value for b because the formula has a + b whole raised to power n. If the second term for instance is negative that means the value for b is negative that's minus 2 y. Okay. So when you input the value here make sure that you put that inside brackets like this. A lot of careless errors happen happen in this type. Just make sure you write values inside brackets so that you do not make those sign errors. And then you can just evaluate this and see what the result comes out to be. If it was something more complicated, sometimes you can have fractions like this as well.

[00:05:04]x^2 + 1 / 2x. Now in place of a you've got x^2. In place of b you have 1 / 2x and in place of n you have 7. All you have to do is in your for in your calculator in in the formula you replace a with x^2 b with 1 / 2x and n with 7 and just simplify everything. Now how are you going to evaluate this on your calculator?

[00:05:32]Don't try to do it manually even if your mental math is good. Yes, you can think about the result on um in in your head as well. However, the actual calculation you can you should be doing on your calculator. For instance, if you have to evaluate this, what you would do would be on your calculator, you'll input it like input it like this. You'll say it's 7 C1. Now, this thing should be put inside brackets otherwise you can end up getting errors sometimes. So, you have to be careful about that. You input this inside brackets. 7 C1 7 C1 and now you've got X squ. Now, you can't put X in your calculator like that. It will it will not give you any result for that.

[00:06:14]You ignore x, think of x as one, and just input all the numbers instead. What are the numbers? We've got this number left, 1 and two in the denominator. So now you put 1 / two in the denominator like this and that will give the result 7 over2.

[00:06:41]Now that that that was fairly straightforward, but the next one is going to be much more useful. Now if you use the calculator directly, we could skip this particular step that we're doing and we could directly write value from the calculator. For instance, you can say this is 7 C2 and then inside the bracket what do you have there?

[00:07:10]One, that's one number. Two, that's another number. In your calculator, ignore X.

[00:07:17]Input this number 1 /2. Let's input that 1 /2. You'll get that number directly.

[00:07:25]The coefficient of that term that's 21 / 4. So this coefficient and this coefficient, you can get that directly from the calculator. And now just do this in this manually. the powers of x power of x here x x raised to power 2 into 6 that's 12 in the denominator you've got x you subtract one from that and you get x^ 11 from that so you just have to calculate the powers of x yourself everything else you just put on your calculator okay don't do it manually that will make you do careless errors just do put put the numbers in your calculator like that okay now in your in the formula sheet that you get in the exam the formula is going to be written in a slightly different form. So you have to be familiar with that. Yes, you should remember this formula anyway. But in case you forget in case you forget some particular term in this in the formula in the formula sheet, you've got the formula written in this form. Okay. Now it's not the form that we normally use because the calculator uh has the NCR button, this button. So normally we use NCR in our uh in the formula that we write in the formula sheet when it says n_sub_1 like this that basically is equivalent to nc1. In the formula sheet when it says n_sub_2 that's equivalent to n c2 and the rest of the formula is as it is. So in case you forget the formula look at the formula sheet just replace this expression with ncr and you would get that formula. For example, this is NC1. This is NC2, this is NC3, right? Okay. So, it's in the formula sheet. It's written like this. You should be familiar with that. Okay. Now, that's the basic idea of this how to use the binomial expansion formula to expand something. Now, let's look at some different types of questions that you will see on this. Now, one thing is that you expand the uh expand an expression using the formula. Sometimes what they do is they give you the final result that you would get from that from that expansion and they want you to find some missing values in in that expansion.

[00:09:30]Some missing uh letters, some missing variables from that expansion. So in this case for instance, we've got this constant a that we do not know what it is. We have this power n again we do not know what that power n is.

[00:09:46]How do you find the values of a and n?

[00:09:48]They have given us that the expansion of this is 32 - 40 + bx^ 2. Right? They've given the expansion to us. Now how will we find the values of a and a and n from this? We will do the expansion ourselves. We've got 2 + ax^ n. We apply the binomial expansion formula on this and see what we get from that. 2^ n plus n c1 into this plus nc2 into this. And now we simplify that and we end up getting an expression that looks something like this. 30 - 40x plus u sorry we we get this expansion for this and they gave us in the question that the expansion is 32 - 40x + bx^2 right in the question they said that this is the expansion. All you have to do now is compare these terms. The first term that you're getting here the constant term you getting 2^ n. They said in the question that this term is 32. So these should be equal to each other. 2^ n should be equal to 32. The second term that they gave us was - 40x. What we're getting is n c1 into 2^ nus 1 into ax.

[00:10:55]Right? So these should be equal. And similarly the third term the x² term they should be equal. We make three equations from this and we solve those three equations simultaneously to find the values of those unknowns. For example, in this case, when you solve the first equation, we get n equal to 5 from that. When we solve the second equation, we get this expression. We put the value of n in that that we find there. The value of n is five. And then simplify that. That gives us the value for a. And similarly, from the last one, then we can find the value of b as well.

[00:11:23]Now, sometimes in questions like these, you can end up getting an expression like this. NC 0 or NC or sometimes NC1.

[00:11:37]And if you do not know the value for N in this, you should be able to figure out what they evaluate to without even knowing the value for N. For example, if you have n C 0, you you need to remember what that evaluates to. For example, if you try on your calculator 6 C 0, what's that? That is one. If you try 8 C 0, what's that? That's one. any number c 0 that always is equal to one. So in case you get something like this n0 you should know that you can that you can write that as one directly n0 is always equal to one. Similarly when the number on top of this function and the number at the bottom of this function when they're both the same in that case as well it always evaluates to one. So in in case you end up getting n somewhere you should know if you do not know the value for n that means you cannot input that in your calculator you should know that ncn should always evaluate to one because 6 c6 is 1 7 c7 is 1 10 c 10 is one so we can say in general ncn is always equal to one and if you have n c1 that always evaluates to n okay so sometimes this number on top might be unknown you should be able to simplify that like is n c1 simplifies to just n.

[00:13:01]Okay, because 9 c1 is 9, 15 c1 is 15, 4 c1 is 4. So whatever is the number on top, if you have one at at the bottom, it always evaluates to the number on top. Okay, this is going to be helpful in some in some problems. There are very few of them in past papers, maybe just a couple, but you should know how to simplify this. Okay. Now there's another complication that comes sometimes that instead of two terms you have three terms inside the bracket. Okay. So for example you have you have something like this 2 - 2x - x^ 2 and then whole raised to power 5. Now binomial expansion only works for two terms at a time. In this case you can use a substitution sometimes. Now in most questions they will they will ask you to expand something in the first part and then you have to use that to do this expansion of three terms. Now for instance in this case they asked us to expand 2 - y^ 5 and that expanded to this thing and we got 30 minus 32 - 80 y + 80 y 2 at the end from that. Now using that what if we had to expand this 2 minus this whole thing.

[00:14:10]Now you compare these two and you see in place of y you getting 2x - x square here right? So in that expression inside the bracket what has changed is that in place of y now you have 2x - x^2. So all you have to do is in the expansion that you've done in the first part, you just replace y with 2x - x^ 2. Simplify that and you get your result from that. In this case, they were asking for the coefficient of x². So we simplify that and we see what term in what's the term in x square that we're getting and we write down the coefficient. That's 400 in this case. All right. So you should know how to expand three terms inside the bracket like this. it'll be done using a substitution. They might in more difficult questions in in recent papers they they have started increasing the difficulty level sometimes as well. They might just skip the first part and they might directly ask you to expand this 2 - 2x - x^2 all^ 5. Now how would you do that?

[00:15:11]You would use substitution yourself. You would say let's substitute this thing with something else. So you'll say let u or any other letter that = 2x - x^2 and then you will write that expansion like this 2 - u^ 5 you will do this expansion and then once you have done that expansion in that expansion you'll replace u with 2x - x^ 2 and then simplify that so that you get that that expansion right so three terms as well sometimes they can give you something like this to expand out all right let's go one now In questions you will see sometimes they use this this terminology that there's no term in a particular power.

[00:15:52]For example, no term in x square, no term in x cube, no term in x^ 5. What does that mean? If there's no term in x cube, you could have something like this. For example, 5 x^ 4 - 2x^2 + 7 x - 3. So this is an expression that does not have any term in x cub. Now when they say there's no term in x cub, what that actually means is the coefficient of x cub is supposed to be zero. Because when the coefficient of x cube is 0 then 0 * x cube it's going to get eliminated and we would get an expression like this that does not have any x cub term in it.

[00:16:24]So whenever in in a question they use this terminology there's no term in a particular power of x what that means is the coefficient of that particular power of x that's going to be that's going to be zero. So you'll find that term put the coefficient of that equal to zero.

[00:16:40]Then there's another terminology that that you'll see being used in questions like these and that's that find the term independent of X. Term independent of X.

[00:16:49]What does that mean? In this expression that I wrote above this term is independent of X. That means it does not have any X on it. It does not depend on X. Right? It's not X is not multiplying with this. Now what what exact term is that? That's the constant term or you can think of it as the X power zero term. So when they ask you for the term that's independent of x that means it's a constant and the power of x in that case is equal to zero.

[00:17:17]Okay.

[00:17:19]All right. Now let's have a look at a couple of questions on this.

[00:17:24]So in this in this question they asked you to expand this in ascending powers of x first of all. So let's say you've done that. Now they're saying find the value for k for which this expression has no term in x squ. Now what does that mean? If there's no term in x square the coefficient of x² that you get at the end that should be zero. Okay. So what we do is in the first part we've done this expansion 2 - x^ 6. In the second part they ask us to expand this 1 + kx into 2x 2 - x^ 6. What we do is we will put that expansion in the second part 64 - 192x + 240x^ 2. Simplify this. Find the coefficient of x². Now in this case it's not necessary to do this whole multiplication as we will see uh in a few minutes. You could just you know do these two multiplications as well. But once you have found the coefficient of x² what you do at that point is you put that coefficient equal to zero. So what's the coefficient that we're getting? We're getting two x square terms. One is 240x^2 and the other is - 192 kx^2.

[00:18:32]We write both of them down. take x square common that that gives us that the coefficient of x square is therefore 240 minus 192k.

[00:18:41]Now since we want no term in x square that coefficient should be put equal to zero and we find the value of k from that. Okay. So whenever they say there's no term in a particular power of x you put that coefficient equal to zero.

[00:18:56]Now the next major thing that you need to know is sometimes they ask you to find a specific term in an expansion.

[00:19:01]Right? So simple expansions are fairly straightforward. You just apply the formula and you get the result. Now sometimes they don't ask you for a complete expansion. They will ask you to find the coefficient of a particular power. So in this case they ask you for the coefficient of x square for instance. How do you find a particular power or a particular term in an expansion? Now whenever you have binomial expansion, whenever you have binomial expansion, let me show you uh these uh examples above.

[00:19:34]The powers of x that you get at the end.

[00:19:36]For example, this was one expansion that we did.

[00:19:41]This one x^2 + 1 / 2x^ 7. You see what the powers of x turned out to be. It was 14 11 8. Do you see any pattern in these powers? All of them have a constant difference. It started from 14. Then the next was 11. The next was 8. So it's a linear sequence. It's an arithmetic sequence. The next power necessarily is going to be x^ 5. The next after that is going to be x^2. And the next after that is going to be x^ minus one and so on.

[00:20:09]It's always in an arithmetic sequence.

[00:20:12]Okay.

[00:20:13]Now that is something that can help us find a specific term in an expansion. So for so in this case for instance we want to find the coefficient of x square in this expansion. Yes there's a general term formula that exists but this is much easier much quicker and less prone to errors. What you do is you just do the expansion the normal way. Okay. So x + 2 / x^ 6. You just expand it like you normally would. The first term turns out to be x^ 6. The second term turns out to be this thing.

[00:20:46]We just note the coefficient and the power of x that we would get from that.

[00:20:50]The first term is x^ 6. The power from the second term is going to be x^ 4.

[00:20:54]How? Because we've got x^ 5 in the numerator. We've got x^ 1 in the denominator.

[00:20:59]That simplifies to x^ 4. Right? Now, these two terms are going to give us a pattern. The first power is six. The second power is four. What's the next one going to be? x^2. And the next one x^0. And the next one x^ minus 2. Which one? Which which which specific power do we need? we needed x². So we know that the next term after this after x^ 4 is going to be x² and that's it. We don't have to expand any further. We just simplify this particular term and and that will give us the result that we want. Okay. So we just simplify that and we get that result. Now this was relatively simpler because we got that coefficient. We got that specific power of x in the very beginning just in in the third term. Sometimes what can happen is it might be further along uh the way on the right side after maybe seven iter. What do you do in those cases?

[00:21:48]Let's take this example. So we want to find the term independent of x in this expansion. Okay, independent of x means that's the term that has x raised to power 0. That's the term that we're looking for in this expansion. X^0 term.

[00:22:02]Now we do it do do it the regular way.

[00:22:05]We say let's use the expansion formula on this expression and see what we get.

[00:22:10]Now we expand the first term and that turns out to be 4xq raised to power 8.

[00:22:16]Okay, what's the power of x that we're getting? 3 * 8 that's 24. Okay, what about the next one? Now when we expand this we get x^ 21 and then there's an x in the denominator as well. So 21 - 1 that's x^ 20 and that gives us the pattern. The first term is 24. The next power is 20. Now we know the next is going to be 16 and then 12 and then 8 and then 4 and then zero. And this is the term that we are looking for. Okay, that's the term that we're looking for.

[00:22:48]Now how do we know what that term is? Do we write all of these values? This one, this one, this one, this one, and this one. And then we reach that final final term.

[00:22:58]Now for this, you can remember this general term formula. Now, where does this general term formula come from?

[00:23:06]Uh, where is the formula? Let me just go back to that quickly.

[00:23:12]Should have copied that beforehand.

[00:23:16]Okay, that was the formula that we had right in the NCR form. Now, look at the pattern that you see in this. So, that's a way to remember the generator term formula. That's why I'm just going through that quickly. This is the binomial expansion formula, right? Look at the terms.

[00:23:33]Look at this term. What's that? n c1 into a ra to power 1. A^ n - 1 * b and so on. You see in every term the only thing that's changing is this number.

[00:23:46]This is one and this term. This is two.

[00:23:50]This is three. That's the only thing that's changing. Right? So from this you get get a general term formula. This is the only number that's changing. Right?

[00:24:00]So you can write this expression like this. You can say this is n c something and then a raised to power n minus something and then b raised to power something. That's that same number. We call this number r. That's ncr into a power n minus r into b raised to power r. That's the general term formula that we have. Okay. So if you forget that you can make it yourself from the expansion formula. Just replace these numbers with r and you get that formula.

[00:24:31]So every term is going to be of this format. Okay, every single term in this expansion is of this format and it starts from r equal to0. The first term is when r is equal to 0. The second term term is when r is equal to 1. In this case r is equal to 3. Right? This is three sorry 2.

[00:24:50]Then the next one is when r is equal to 3 and so on. So we have a general term formula that looks like this. n c r into a power n - r into b power i. So from this pattern what we can do is we can figure out what the value of r is going to be and that's it.

[00:25:08]So let me explain what that would be. So we started from this x^ 24 r is equal to 0 here x^ 20 r will be equal to 1 in the next case and then we are looking for this particular term. What's the value of r going to be there? That's all that we need to figure out.

[00:25:26]2 3 4 5 6. This is what we get. R is equal to six. There now you should now you would remember this general term formula. NCR into a power n - r into b^ just input that value of r here and simplify that you'll get that result. So in in this case a is = 4x cub b is equal to 1 1 / 2x n is = 8. Just input those values and the value for r is equal to three. Input those values, simplify that and you would get that term that you're looking for. Okay, so this pattern is very helpful. You can find any specific term that you want from this expansion like this, right? Just find the value for r and input that value of r in this general term formula.

[00:26:16]Now sometimes when they ask you for specific terms they have products like this.

[00:26:24]Okay.

[00:26:26]Now in the first part of this question 1 + 2x all^ 6. They want us to to find the coefficient of x cube in this expansion.

[00:26:33]And that's straightforward. We know that the first term is x^0 then the next is x^ 1 and then the next is x^2. This is the term that we're looking for. We simplify that and that turns out to be 160. Now in the second part they say find the coefficient of x cub in this expansion. Now we've got two things multiplying and we want to find the coefficient of x cube from that product. How do you find the coefficient of x cube from this product? The first thing that you should always do in questions like these is think about what terms you would end up needing from this expansion. Right? because in the first part they just made you find the coefficient of x cq and that was this. Now that's not the only term that you would need for the second part. You might need something else as well. So before you do anything, think about what terms you are going to need from that expansion.

[00:27:33]How do how do you figure that out? X cube is what you want. You've got one here. What do you multiply one by to get x cub? You multiply one by x cub term to get x cub x^0 will have to be multiplied by x cube. So that's one term that you need from that expansion. You need x cube from that 3x. What do you multiply 3x to get x cub? That's x^ 1. You have to multiply this by x² to get x cube. So this will have to be multiplied by x^2.

[00:27:58]So from this expansion we need two terms. We don't don't just need x cq. We also need x². Now this is where expanding it like this without the general term formula becomes so much more helpful because we've got both of those terms there already. We don't have to do any additional work. Just simplify this term as well and that gives you 60x^2. Now these are the two terms that you need for this for the next part. You say 60x^2 + 160x cq.

[00:28:31]These are the two terms that you need from this expansion. You just write those two terms. Ignore everything else.

[00:28:37]You could write that but it's not required. It's much more efficient to just use those specific terms from this.

[00:28:42]Now since you're looking for x cube, you just do the multiplications that this case x cube would be when we multiply 1 by 160 x cq and that gives us this number. And the other x cube would come when we multiply - 3x by 60x^2 and that gives us this other number. We could just ignore everything else. Just do these two multiplications and that would give us the result. Simplify that and get the coefficient of x cq from that. So when you have a product like this and they want you to find a specific term, you first should think about what terms you would need from that expansion so that you use those terms in the expansion that you've already done in the previous part or if if you've not done that you expand that expression up to uh that many terms at least. Let's take one more example of this. Okay, it's taking time too slow today. Okay, find the coefficient of x square in this expansion.

[00:29:44]Again, what do we do for this? We use that pattern. Expand a few terms and you get the coefficient of x square. The first term is x^ 6. This is x^ 4. And the next one has to be x^2. That's that's that's what we need. Okay.

[00:29:59]All right. Now, what about this the second part? You want the coefficient of x² in this expansion. Again the first thing have have to do is you have to think about what terms you are going to need from that expansion. You want x squ at the end 1 + x^2. Think about what terms you need from that expansion. What do you multiply one by to get x square?

[00:30:20]You multiply 1 by x² to get x square.

[00:30:22]What do you multiply x square by to get x square? You multiply that by a constant or x^0 term. That means from that expansion you will need these two terms x^2 and x^0.

[00:30:33]Okay.

[00:30:37]So you just you know simplify those two terms from that expansion the x^ 2 term and then you have to expand one more term that's x^0 you bring those two terms down here 60 x^2 from here and then - 20 from here and then just do the multiplication that you need to do multiply 1 by 60 x^ 2 * x^ 2 by -20 20 and then simplify that you get that coefficient. Okay? So you see in the second part now you just write down one more term and you get that next term as well. If you were to use that general term formula you'll have to use that all over again and that will be like double working. So if you were to use general term formula, yes, you'll have to find this coefficient separately first and then for the next part since you need x^0 as well, you apply that again on this to get minus 20. If you just do the expansion like this normally and look at the pattern, you get both of those terms from this expansion directly. Okay. All right. Let's move on.

[00:31:51]Now this is something that you will see in some recent papers as well. They ask you for coefficient of 1 /x 1 /x^2 1 /x cub something like this. Now don't get confused by that 1 /x what's that 1 /x is actually the same thing as x^ minus one right 1x^2 is just x^ -2 1 / x cub is just x^ minus 3. Okay. So what they're asking asking you for is the coefficient of these powers x^ -2 -3. So when you have x in the power you just bring that to the numerator make the power negative and you find that particular coefficient from that.

[00:32:28]In this case now it's a relatively more complicated example that you will see in past papers.

[00:32:34]They're saying the coefficient of x in this expansion is p. The coefficient of 1 /x that means x^ minus one in the second expansion that is q. And then they're saying p= 6 * q. Find the possible values of k. Now, how do you think about this? First, first of all, you find the value of P and the value for Q. How do you do that? They have told you how to find that coefficient of X. That's going to be the value for P.

[00:32:58]You find the coefficient of X. You put, you say that that is equal to P. Then you're given that P is equal to 6 * Q.

[00:33:04]You find the value of Q from that. And the coefficient from here, the coefficient of X^ minus one, that is going to be equal to Q. So, you find that coefficient, put that equal to Q, and you get the result. Okay? You can have a look at this working in your own time. But this is one relatively more complicated example that you'll see in recent papers. It's nothing uh very complicated as far as the method is concerned, but they just asked uh about a specific term in a slightly different way than than the other questions that you will see in past papers.

[00:33:38]This is very important. Now look at this question. It says it's given that in the expansion of this the coefficient of x² is k. It's also given that there's only one value of a which leads to this value of k. So sometimes now in recent papers you'll see that they've started combining binomial combining quadratics with a number of topics. You'll see questions where they where they require you to use quadratics as in b square minus 4 a c in equations of circle in this case you have to use it along with binomial expansion and there are other topics as well in which you have to use that b square minus 4 a c thing. So think about this they're saying that in the expansion of this the coefficient of x² is equal to k. Okay.

[00:34:26]So first of all what we do is we find the coefficient of x square in this expansion. How do we do that? First of all, we have to think about which specific terms we need from this expansion. How do you figure that out?

[00:34:37]What do you multiply four by? What do you multiply 4 by to get uh x²?

[00:34:45]You multiply you you multiply this by x square to get x square. What do you multiply 2x to get x square? You multiply 2x to get x square. So these are the two terms that you need from that expansion. x and x². You do that expansion. You find those two terms. do the multiplication and you end up getting the coefficient of x square that looks like this. Now they said that the coefficient of x square is supposed to be equal to k. So we put that equal to k like this. And that gives us an equation like this. It's a quadratic equation. And then they're saying there's only one value of a. Look at the question. They said there is only one value of a which leads to this value of a.

[00:35:27]What does that mean?

[00:35:30]This equation has only one solution. It has only one solution. It's a quadratic equation. When does a quadratic equation have only one solution? When b² - 4 a c is equal to 0. So you convert it to to the to the standard form. A x square + b x + c. You say that a is equal to 320, b is - 160 c is minus k. Put b square - 4 a c=0. That gives you the value for k.

[00:35:55]Once you have the value for k, you can solve that equation then to find the corresponding value for a as well. So when they say there's only one solution and you have a quadratic equation that should remind you about the discriminant that b² minus 4 a c should be equal to zero there. Okay, it's a binomial question. Sometimes you don't even think of quadratics in questions like these.

[00:36:17]But that's that that's what it is about the recent papers. They combine a lot of things in a single question. Okay? So you have to have your concepts really strong.

[00:36:26]Okay, so that's it with binomial. If you guys have any quick questions on this, let me know. Otherwise, we're going to move on to coordinate geometry next. After this, which paper was the previous question from uh I'm not sure. You'll have to search for that. Uh you can take a picture. You'll have these notes available as well. You can maybe search for that. So the way to search for a particular question on Google is that you you know find some unique words in that and put them on Google in inverted commas along with 9709 math and generally it pops up as the first two three inside the first two three results. In this case though there does not look seem to be any uh unique words as such. Maybe you can try this equation. P= 6 Q. Input this in inverted commas and possible values of K in inverted commas and then put 9709 and then hopefully find that close.

[00:37:27]Okay, Abdi. That's okay. That that's perfectly okay. Okay. So, whatever method you generally learn first, you tend to find that easier. So, it's okay. Whatever method you prefer, that's perfectly okay.

[00:37:51]Yeah, you can Google scan now as well. I have not used that, but yeah, that's a good idea.

[00:38:13]That's not something that is uh always going to be true. So the question is uh so let me know if I understand your question correctly. You're saying a + b raised to power n. If you have an expression like this, you're saying if a was a fraction, then in the formula a^ n + n c1 into a power n -1 into b + n c2 into a power n - 2 b ^ 2 and so on.

[00:38:46]You are saying we apply the power r and not n minus r.

[00:38:57]So you you're saying you switch this you have this power one and this power n minus one and this power two and this power n minus 2. Is that what you're saying?

[00:39:19]that would not make sense. Uh there's no reason to do that. Uh if you want to do that, you would actually have to actually have to switch these terms. You would have to write B first and A second, right? And then you can treat the first term as B and this this term as A. Otherwise, you would not do that.

[00:39:39]That does not make sense otherwise.

[00:39:45]So whatever is the first term, whatever is the first term, you take that as a.

[00:39:48]Whatever is the second term, you take that as b. Okay. So I I think I understand where you're coming from. So sometimes uh in fact in in questions sometimes they say expand something in ascending powers of x and sometimes they say expand something in descending powers of x. Now in all the past paper questions that you will find whatever order they ask for it automatically comes from the expansion from the expression that that they've given. For example, if you have 1 over or maybe 5 x 5 x^2 + 1 /x and you want to expand this.

[00:40:37]If you expand this, it'll automatically turn out to be in descending powers.

[00:40:41]It'll automatically turn out to be in descending powers. The first power would be 5x square power.

[00:40:47]Let's put seven here. This power 7. The next one would be n c or 7 c1 into 5x^ 2 raised to power 6 into 1 /x^ 1. It's going to be in descending powers. They will say in the question expand this in descending powers. It will automatically come in that order. If it happens, it hasn't happened until now, but there's always a first time for everything. In the recent papers, Cambridge has proved that there's always a first time for, you know, doing giving you something totally uh new that you haven't seen in in the other past papers. You might get this expression.

[00:41:26]You might get this expression and they might ask you to expand this in ascending powers of x. So if they ask you to expand this in ascending powers of x for instance.

[00:41:38]Now if you do the expansion the normal way you're getting it in descending powers. How do you convert this to ascending powers? In that case you would switch these two. If you had to find this in ascending powers you would switch these terms. You would say let's write this as 1 /x + 5x^2 raised to power 7 and then you treat this as a and this is b and now it'll come in ascending powers. Okay. So if you want to switch descending to ascending you just you know switch these two terms.

[00:42:12]That is one case in which you might have to do it. But I haven't seen any question like this in past papers in which the order that they ask for it doesn't come automatically. If it does then you just switch to those terms. You consider the second term as A and the first term as B. That's what you can do.

[00:42:30]We normally do ascending. No, whatever they say. So sometimes they say descending in question. Sometimes they say ascending in question. You generally don't worry about that because the expansion that you do it automatically comes in that order. So if they say ascending just expand normally it turns out to be ascending. If they say descending just expand normally and it turns out to be descending. Okay. If it doesn't happen to be that way again it hasn't happened until now in past papers but if it happens I told you what to do in that case switch A and B right the second term first first term second and the order will switch Okay.

[00:43:21]You you get the grade for all the components combined. That's always an overall grade. Weightage of this topic, it's normally four to six marks. If they combine some other topic, for example, in this case, they combine quadratics with this. In that case the number of marks slightly increases but generally it's four to five marks sometimes six.

[00:43:40]Uh it depends on the question.

[00:43:44]All right so I think we can move on. Uh there's going to be more time soon as well.

[00:43:52]Okay. So I think I think we might actually need to extend our classes a bit. Uh I hope you uh don't have any problems with that. if anyone has any classes because in case we end end up having to extend our class time a bit uh you will have the recording available of course I'll try to be quick now but in case you have to attend another class or you have something more important to do uh which uh I don't know why you would have that but if you do then you can watch the recording later on as well okay just one last Okay, in some recent uh questions you will see that do the complete expansions. For example, they might ask you to do something like this 2x + 1 /x^2 raised to power 4. And generally in in the earlier past papers they always say expand the first three terms first four terms maximum.

[00:44:53]Sometimes they say in the recent past papers find the complete expansion.

[00:45:00]Find the complete expansion of this thing. Now what does complete expansion mean? You expand all the terms of the sequence of of this expansion.

[00:45:12]All terms. Now all terms what does that mean? Just input values in the formula.

[00:45:16]It will be 2x^ 4 + 4 C1 into 2x^ 3 into 1 /x^ 2 power 1 + 4 c2 into 2x^ 2 into 1x^2 power 2 and so on. The next term is going to be 4 c3 into 2x^ 3 power 1 into 1 /x^2 power 3. You keep going on like this. But how do we know when to stop?

[00:45:46]When exactly do we stop?

[00:45:52]We stop when the numbers on top of C and at the bottom of C they become equal to each other. When these two numbers are the same, we stop at that point because we can't go any further than this. The next term on this pattern would be 4 C5.

[00:46:07]That's uh not something that's valid. It gives you an error. You can try that on your calculator as well. If you've done S1, it's saying that select five objects out of four. Doesn't make sense. If you put that in your calculator, 4C5, that gives you an error, right? So, you stop when the number at the bottom becomes bigger than the number on top. You can't go any further than this. When in fact, you stop when they become equal, right?

[00:46:34]You can't you can't go beyond that. And that's your complete expansion. You will simplify this and you'll get the result.

[00:46:39]Okay? So, that's binomial expansion.

[00:46:42]I hope that was helpful. Let's move on to the next one now. Yes, the fifth term would be when n is equal to four. Okay.

[00:46:51]Now the next thing is arithmetic and geometric progressions.

[00:46:59]Okay. Let's start to this now.

[00:47:00]Arithmetic and geometric progressions.

[00:47:02]Now what are arithmetic and geometric progressions? This is just another name for sequence, progression, sequence, series, right? These are all the same thing for our purposes. series does have some detail that you will study in university later on. But uh for our purposes, we treat all of these three different letters, three different words the same way, right? They all mean the same thing. Now, there are two sequences in your syllabus. In O levels, you used to have, you know, a number of different sequences. In A levels, you can only have two types of sequences. It's either going to be an arithmetic progression or it's going to be a geometric progression. There can't be anything else apart from these two. Okay.

[00:47:39]What are arithmetic progressions? This is just another name for linear sequence. If you have a sequence like this in which the difference between consecutive terms is constant, we call this an arithmetic sequence. An arithmetic sequence is described by two things. The first term and the common difference. To be able to uh to write down the nth term of an arithmetic sequence, we need these two things. The first term and the common difference.

[00:48:00]Now the first term is obvious. Whatever is the uh term that's coming in the first place, that's the first term. What about the difference? Difference is calculated like this. next term minus the previous term. Next term minus the previous term. Now you have to be a bit careful when the values are decreasing.

[00:48:16]For example, if you have 5 2 - 1 - 4 - 7, if you have a sequence that looks like this in this case, don't just say 5 - 2 the difference is three. No, you have to subtract the first term from the second term. So 2 - 5 that will be the difference. Min - 1 - 2 that will be the difference.

[00:48:34]Right? So next term minus the previous term that's the difference. So when the terms are decreasing the difference is going to be negative.

[00:48:42]Now we need to know the nthterm formula for an arithmetic progression and the sum formula. This is what the nth term formula for an a for an arithmetic progression looks like. A + n minus1 * d. So if you have a sequence that looks like this 5 2 -1 -4 - 7 and so on and you have to find a particular term of the sequence for example the 50th term.

[00:49:04]What would we do in that case? We will say the value for a is five. We input that here. The value of n is 50 in this case. We input that here. The value of d is minus 3. We input that here. And that gives us that specific term. And we get that result. If you had to write down the nthterm formula for the sequence, how would you do that? Sometimes they ask you to find to write down the nth term formula as well. What's the nth term formula of the sequence?

[00:49:26]You would write that in terms of n, right? So you don't put any value of n there. What's the value of a? a is equal to 5 here. and then n minus one times difference which is minus 3. You simplify that and you get that nth term.

[00:49:40]You can see what that turns out to be. 5 - 3 n plus uh 5 - 3 n + 3 and that gives you 8 - 3 n. That's the nth term formula of the sequence. 8 - 3 n. You put n= 1 into this you get the first term. You put n equal to 5 you get the fifth term and so on.

[00:49:59]Now in the formula sheet you have this formula given but it's written as un we normally use tn that's I think what most of you would have would have used in the formula sheet it says un so just understand that's the same thing as uh the nth term right tn or a un they mean the same thing now let's have a look at this example so you you could have something as simple as this they are saying that there's a sequence that starts from five and The 30th term is 55. What does that mean? The value of the 30th term t30 that is equal to 55. And what we have to do is we have to find the common difference. Now how can you do that? You can just input values in this t30. What would that be using the nth term formula? The nth term formula looks like this. A + n minus 1 into d. You input the value for a in this. Input the value of n that is 30. In this case d is missing. And then you just replace t30 with 155. we find the value of d from that. So you might have a specific term given and the first term or the difference one of those two things might be missing. You can find that like this.

[00:51:08]Now another thing that you need to know about this is how to find sum of a specific number of terms. Now for this we have a sum formula we call that sn that gives us sum of the first n terms of a sequence. For the for an arithmetic sequence, the sum formula looks like this. Half n into 2 a + n minus 1 * d.

[00:51:29]That's the more common formula that we use. There's another as well. We'll talk about that. What does it stand for? Sn.

[00:51:36]It's the sum of the first n terms starting from the first term. So for example, if you have s, that means sum of the terms from 1 2 3 4 all the way up to 10. Okay?

[00:51:46]So for instance, in this case, s3 would be these three terms added. But if you had to find for example S10, you can use the formula just inert values n a and d in this you would get the result for uh that for the sum of those number of terms. That's one formula. There's another one that's written like this in the formula sheet half n into a + l/ n into a + l.

[00:52:11]Now in this formula a is the first term l is the last term. Now this there's a specific meaning that we have for this letter L here the lost term. Let's suppose we have a we have a sequence that looks like this 4 10 16 22 and so on. It goes on like this and we want to find the sum of the first 10 terms of the sequence. Now if we want to use this formula the value for n would be 10. The value for a would be four. That's the first term. But what about L? There's no end to the sequence. It it just continues like this. What's the last term going to be? The last term here basically means out of the terms that you want to sum for, what's the last term. So how many terms do you want to sum for? You want to sum for the first 10 terms. These are the first 10 terms. What's the last term in these 10 terms? That's 58. So the 10th term in this case when you're finding sum of the first 10 terms that's considered the last term. So we often write this formula in a slightly different way. In the formula it's written like this in order to save yourself save yourself from confusion.

[00:53:26]You should write it like this half n into a + tn. So this l basically stands for the nth term. Right? So if it's s10 then this is t the nth term. Okay. And that's how you can use this formula.

[00:53:43]Mostly you would be using the first formula. That's what works in most cases. This formula. Sometimes the second formula is going to be more efficient when you have the last term given or when you have to find the last term. In those cases, this other formula is going to be more efficient. But otherwise, uh the first formula works every time as well. Uh you can use that as well. Okay.

[00:54:07]Now I will let you think about this example. There's a line going on here.

[00:54:11]We have one merry and a zero.

[00:55:54]Okay, let's just we'll have a break from in about 15 20 minutes in just after a APGP. We'll have a break. Uh we'll have a longer break about 25 30 minutes. 25 minutes not 30. Uh but you'll have to wait for that a bit. Okay.

[00:56:16]All right. Let's move on. Now let's see uh this example. It says find the sum of all the terms between the 15th term and the 25th terms both inclusive. So sometimes you have to find the sum of some terms in the middle of the sequence. Now the formula that we've got is for the first n terms. Sn is for the first n terms. How do you find the sum of these terms that are in the middle somewhere? So what exactly are we looking for? We've got the sequence. We know the first term of the sequence. We know the difference of the sequence. If you can find that we want the sum of 25th no 15th to 25th term right 15th term all the way up to 25th term. Now how can you use the sum formula to find this specific sum? You can think of it like this. You can say let's find the sum of all of these 25 terms. So I've written down the terms here like this t1 t2 t3 t4 and so on all the way up to t25.

[00:57:19]What you can do is one idea is you find the sum of these 25 terms using the formula. You can do that right.

[00:57:28]You can find the sum of the first 25 terms using this formula.

[00:57:33]And then since you only want the sum of 15th to 25th term. Okay, you only want sum of the of 15th to 25th term. You do not want the first 14 terms, right? You want the 15th term that has to be included in the sum. But what you don't want is the first 14 terms. So what you do is you subtract from S25 S14.

[00:58:03]So S25 gave you the sum of all the terms from 1 to 25.

[00:58:10]If you subtract the sum of 1 to 14, what are you left with? you're left with 15 to 25 and that's what you want. So if you want the sum of all the trends from 15 to 25, you can use the sum formula like this. You can say find S25 subtract from that S14 not 15 S14 and that will give you the result. After that it's fairly straightforward. You just input values in this and you get the result from that.

[00:58:38]Okay. All right. Let's go one.

[00:58:41]Now uh this is uh relatively straightforward. If you have three consecutive terms that are given for an arithmetic progression, what does that mean? How do you find the value for x here? One possibility is that you say find the difference from these two terms and find the difference from these two terms. Since the difference has to remain constant x - 30 should be the same as 72 - x.

[00:59:06]x - 30 should be the same as 72 - x. So you make an equation like this. Find the value of x from that. In arithmetic progression, there's another way as well and that is the middle term is always going to be the average of the terms on the sides. So if you have one term here 30, another term on the side that's 72.

[00:59:24]The middle term is always going to be average of those two terms. Right? So you can say x is equal to 72 + 30 / 2.

[00:59:33]Okay? Or you can say that the difference remains constant.

[00:59:38]Now let's move on. Let's move on to geometric sequence. Now this is a geometric sequence.

[00:59:47]Geometric progression or geometric sequence.

[00:59:52]Now what about a geometric sequence? A geometric sequence is in which the ratio between consecutive terms remains constant. Right? So you multiply by the same number to get the next number. In uh APG in arithmetic progression we used to either add or subtract the same number to get the next number. Right? In this case we multiply or divide by the same number to get the next to get the next number. For example 3 * 4 is 12, 12 * 4 is 48 and so on. So the ratio in this case is four. Right? In the next case, 10 / 2 is 5. 5 over 2 is 2.5. 2.5 divided by 2 is 1.25. Or in other words, we can say 10 * 0.5 is 5. 5 * 0.5 is 2.5. And so on like this. Okay.

[01:00:45]Now we write this in product form generally because the ratio is considered to be this number that you have to multiply by to get the next number. A geometric progression is described by two things. the first term and the ratio. For an AB, it was first term and difference. In this case, it's the first term and the ratio.

[01:01:07]How to find ratio in this case? Ratio is next term divided divided by the previous term. So for any two consecutive terms, if you divide the next term by the previous term, you get the ratio from that. For example, in this sequence, you could do 20 minus 20 divided by 80 and that will give you the ratio. That's 1 / 4. You could do 5 over 20 or you could do 1.25 over five. Any of them is going to give you the ratio of the sequence. Now this is the n term formula for a gp a r raised to power n minus one. That means the first term for example is going to be uh the first term is just a but then the second term is going to be a r a r² a r c and so on. Right? And then we have the sum formula. It looks like this. A into 1 - r^ n / 1 - r. These are both given in the formula sheet. just need to be able to use them. Again, this formula is for sum of the first n terms.

[01:01:59]Okay, for the first n terms.

[01:02:02]Okay, now let's consider an example.

[01:02:04]Let's say we've got a sequence like this.

[01:02:07]The first term of the sequence is two.

[01:02:09]How do you find the ratio? Take any two terms, divide the next term by the previous term. So - 6 divided by two, what's that? That's minus 3. That's the ratio of the sequence. Okay. Now, if you want to find the 15th term, for instance, just use the formula. What's the formula? The nth term formula is a ris power n minus 1. We just input values in that t15. We say a is 2, r i is - 3, n is 15 and that simplifies to this number. The sum formula is like this. A into 1 - r^ n / 1 - r. Just input values in that and that gives you the result. In your calculator, when you input this, be careful about negative numbers. When the ratio is negative, make sure to always put that inside brackets. So this number for example in the in the power this is r raised to power n. Right? So ratio is -3 -3 that whole thing is going to have this power 10. Okay? So you have to write the negative inside the bracket like this.

[01:03:09]And now you can just put this whole thing on your calculator directly. So when you put this on your calculator be careful put it exactly like this. Use the fraction button 2 * 1 - and then inside bracket minus 3 raised to power 10 and then close that bracket outside and then in the denominator you have 1 - 3 that becomes 1 + 3 and that gives you the result. Okay. So put it exactly like this in your calculator making sure you put the negative values inside uh the brackets. Okay. All right.

[01:03:46]Now you can write the nth term formula as well. If you have a sequence like this minus 100 minus 50 and so on.

[01:03:51]What's the nth term formula for this going to be? What's what's the first term here? The first term is minus 100.

[01:03:57]What about r? Ratio is not negative here. The ratio is going to be positive.

[01:04:01]When you calculate it -50 over - 100 that's 0.5. You have the value for a.

[01:04:06]You got the value for r. If you want to wanted to find the nthterm formula in this formula, just input the value of a and the value for r. You will get that nth term formula. Okay. All right. Let's move on.

[01:04:19]Now you will get questions in which uh they'll have some scenario given and a lot in a lot of those questions they will have something increasing by a certain percentage something something decreasing by by a certain percentage.

[01:04:32]Whenever there is a percentage increase when there or there's a percentage decrease remember that's always going to be a geometric progression. This is always going to be a GP. Okay. you'll be multiplying by some number to get the next number. Now, how exactly do you get that ratio? Let's suppose they say it's there's a 5% increase. 5% increase. Now, you know percentage change for every next number you're increasing the number by 5%.

[01:04:59]That has to be a geometric progression.

[01:05:00]Now, how do you find the ratio of this geometric progression? What's the way to do that? If it's a 5% increase, you say the original value was would be 100%.

[01:05:10]When you add 5% to that, what do you get from that? 105%.

[01:05:16]Write 105% in decimals. That will be your ratio. Okay? 100% plus 5% that gives you 105%, write that in decimals, that's 1.05. That's your ratio. Let's have a look at some more examples of this. If you are increasing something by 15%, what does that mean? 100% plus 15%.

[01:05:38]That's 115%, what's the decimal for that? 115 over 100. Ratio is 1.15. If something is decreasing by 8%, that means it's 100% minus 8% that gives you 92%. What's 92% in decimal? 0.92. If something is increasing by 140%, what does that mean?

[01:06:02]100% plus 140% that becomes 240%. The ratio becomes 2.4. four. Okay. So you can find the ratio like this. If you have the percentage increase or percentage decrease uh given, you see how much change is going to be uh from 100% and then you find the ratio according to that.

[01:06:21]Now this is something that you'll see a lot in the recent papers. You'll have three consecutive terms given for a GP and you have to find something that's unknown. Now this is a simplified version of that. If you have one number that's 15, another number that's x, another number that's 135, these are three consecutive terms of a geometric progression. Okay. If you had to find the value for x in this, that's the term in the middle. How do you do that? Now, one possible possible way of doing that would be using the nth term formula. But that becomes very complicated in some cases. I would not recommend that you do that. So that would be like this that you say t2 is equal to u. So first of all, you find the ratio. You find the ratio like this. You say ratio is x over 15. And then you say that the second term is going to be a which is 15 into r which is x over 15 in this case. Now that equals the second term. In this case that does not give us anything anyway. Uh you can find the ratio from the other other one 135x. So basically the idea is that you in this example probably that will not work very well that explanation but when you have three consecutive terms like this a lot of people what they try to do is they find the second term using a * r they try to find the third term using a r 2 equate a r with x yeah so that's what they would try to do a r equal to x and then they would say a r 2 that is equal to 135 and then use that to find the missing you.

[01:07:52]Now, in the more difficult questions, it'll become very very complicated and it's very unlikely that you would end up getting the correct answer at the end because they end up being too many variables and uh you almost certainly end up doing some careless errors there.

[01:08:08]The more efficient way of doing problems like these is when you have three consecutive terms, three consecutive terms given in a geometric progression.

[01:08:16]You say the ratio can be calculated as this x over 15. That's one way to calculate the ratio. The ratio can also be calculated as 135 /x. That's another way to calculate the ratio. These two are supposed to be equal to each other.

[01:08:29]So we make an equation like this. We say it's x over 15= 135 /x. We solve this equation. Find the value of x from that. So in this case we get two possible values of x which are plus - 45. Both of these are geometric regressions. We could have the middle term of 45. It would be minus 45. If it is 45, the ratio would be three, right? You're multiplying by 3. If it is minus 45, that means the ratio is minus3. This is another another important point.

[01:09:00]Whenever the ratio is negative in a geometric sequence, you will find that the terms alternate. One term will be positive, the next will be negative, then positive, then negative. They will always be alternating like this.

[01:09:11]Whenever the ratio is negative, because when you multiply by a negative number, the sign changes, right? it changes in every single term. So you'll have plus minus plus minus and so on like this. So that's how you find the missing term in problems like these. Okay. So that's the general rule that we have. If you have three consecutive terms for a GP, you say B / A that equals C over B.

[01:09:32]Okay. Now that was about geometric progressions.

[01:09:42]Already done that. that ended up being copied twice. Okay. Now, what about sum to infinity?

[01:09:50]That's the last major concept that you have in APG. Finding the sum to infinity of a geometric progression. Now, what is sum to infinity? It's basically sum of all the terms of a neverending sequence.

[01:10:03]So, if you have a geometric progression that looks like this, 2 3 4.5 7.5 and so on and it never ends. It continues like this. There's no end to it. Infinity means a very large number. So sum to infinity means sum of all of these terms, sum of these infinite terms. Now that looks like it doesn't make any sense because if you were to add an infinite number of terms, the result should turn out to be infinite. That is true in some cases because in this case for instance if the GP is two then uh if the first term is true and the ratio is 1.5 what happens is the numbers keep increasing 2 3 4.5 7.5 and it continues like this. If you were to add all of these terms you will see that the sum turns out to be infinity and it'll it'll turn out to be infinity because the numbers keep getting bigger these numbers are going towards infinity. So if you add all of them up the sum is also going to be infinity. However, if you have a GP that looks like this in which the first term whatever it is, the next term is smaller than that in numerical terms. The next term is smaller. Then what happens is these values keep getting smaller and smaller and eventually these values converge to zero, right? They approach zero. They become almost equal to zero.

[01:11:26]So when you have a GP that looks like this in which the in which the next terms are becoming smaller and smaller until you get zero at the end in that case the sum of all of these terms is actually a finite number that's not infinite. Now there's a whole background to this how this formula comes but now this is given in the formula sheet a over a over 1 minus r this is the sum to infinity formula when does this sum to infinity formula work only when the GP is like this that the terms are becoming smaller the terms are becoming smaller when does that happen when do terms become smaller when the ratio is between minus1 and one the ratio is between minus1 and one because when you have a number that's less than one in numerical terms when you multiply by that the value becomes smaller right so if you have a GP that looks like this in which the ratio is between -1 - 1 and 1 the sum to infinity is given by a / 1 - If I is greater than one, the sum formula is R before one. Uh you could do that. U there's there's no uh as in it's not required that you write it like that that way. There's only one formula. You could write it two ways.

[01:12:59]This formula can be written in two ways.

[01:13:01]You can write it in this way or you could also write it in this way. A into R^ N -1 / R minus one. Both of these formulas are correct for any R. Okay, they are both correct for any value. It doesn't make any difference. No, it doesn't make any difference. They're both correct. They're both equivalent.

[01:13:17]They work for any R. Negative, positive, it doesn't matter. So sometimes uh some teachers what they do is they ask you to use this when uh the ratio is uh less than one and they ask you to use this when the ratio is greater than one because they don't trust that you will be dealing uh fine with the negative numbers because when you have um r less than less than one here you would end up getting or in the first formula when r is greater than one you would end up getting a negative number inside and that may uh make some people do careless errors but it's not necessary. Both of these are correct.

[01:13:58]You can write this formula like this or this. They're both the same. There's no difference between these formulas. They work for R for any value for R. Apart from one of course in that case you get zero in the denominator. That doesn't work. Uh but otherwise they work for any value of R. Okay. All right. Now this is the sum to infinity formula. A over 1 - r. This is the sum to infinity formula.

[01:14:21]Sum to infinity only exists when r is between minus1 and one. Okay. Now this is another thing that you will find very few questions for but it's important.

[01:14:31]You you will find it more in the recent papers.

[01:14:34]Convergence of a GP.

[01:14:37]When does a sequence converge? Converge means that it approaches one fixed value and it doesn't change after that.

[01:14:47]If you have a GP like this in which it's becoming smaller and smaller that means the ratio is between minus one and minus one and one. What happens is the values keep getting smaller and smaller and eventually they approach zero every time.

[01:15:04]We call this in mathematical terms that the sequence converges to zero. Okay, this sequence is basically converging. It's approaching.

[01:15:19]It's becoming stable at it's not it's becoming almost stable at zero, right? It'll approach zero.

[01:15:29]Okay, that's what it means. It converges to that value.

[01:15:33]When does a geometric progression converge? The condition for that is the same as the condition for the sum to infinity that the ratio is between minus one and one. So whenever they say the question given that a sequence converges and then they want you to find something from that you find the ratio you put that ratio between minus1 and + one and you solve that inequality. Okay and see you see whatever you get from that. So whenever a series a geometric progression converges ratio is between minus1 and one. That's the condition that we required for that.

[01:16:10]This is how the formulas are written in the formula sheet. You should be familiar with them. You should of course when when you're done now practice you will start remembering the formulas anyway. But the for but the uh formulas are written in the formula sheet in this format. Okay. The nth term is a + n minus 1 * d for it. And these are the sum formulas and in in place of tn they write un here. Just uh understand that.

[01:16:38]Okay. Uh just give me a minute. I'll be back.

[01:17:35]Okay. So this is how the formula this is how the formulas are given in the formula sheet. Just be familiar with that that in place of tn they write un everything else is the same. And also for the sum to infinity uh they write it in this form. Modulus of r is less than one. That's basically just the same thing as r being between minus1 and + one. Okay. So that that condition is the same as this.

[01:17:58]This I would recommend that you remember because I think many of you are not familiar with this function the modus function. So just remember it like this that r is between minus1 and + one.

[01:18:09]Okay. All right.

[01:18:11]Let's move on. Now let's have a look at some questions quickly late for. Yeah.

[01:18:17]Okay. In an AP the sum Sn of the n or the first n terms is given by this. Now we have the sum of the first n terms formula. So sometimes what they do is they give you the formula themselves in the question and they ask you to find for example in this case the first term and the common difference. Now in this case one quick way of doing this would be that this is the sum formula 2 n square + 8 n right what you do is you say you are looking for the value for a and the value for d right these are two values that we want to find out using the sum formula we can say let's find the sum of uh the first uh one term first two term first three terms and use that to find the values of a and For example, in this case, if I input n= 1, that should give me the sum of the first one term. That's basically that's the first term. So if I input n equal to 1, I get 10 from that.

[01:19:17]So that means that the value for a is equal to 10. If I input n equal to two in this, what does that give me? That gives me 24. That's the sum of the first two terms, right? S2. Now if I know the first term is 10 and the sum of the first two terms is 24 that means the first term which is 10 plus the second term that should equal 24 and we can find the second term from that as well and then once we have the first term we have the second term we can find the difference that turns out to be four.

[01:19:46]This is one way to do problems like these. So when you have a formula given, you input some values of n from that n in that and use those values that you get to figure out whatever they've asked for. Now in more complicated examples like this, you'll have to use another method as well. For example, in this case, they gave us that 2 n square + 8 n. That's the sum formula. They might have some constant there. Sometimes they might give you something like something like this 2n + 8 k * n. And then you have to find the value of k there as well. somehow in those problems you should do it another way and that's like this in the question they gave us that for this arithmetic progression it's an arithmetic progression right it's an AB for this AB the sum formula is 2 n square + 8 n right what's the general sum formula for an AP that looks like this n into 2 + n - 1 into d this is the formula that we know this is the formula that they gave us for that particular sequence. They're both supposed to be the same, right? Because they're both sum of the first n terms of the AP formula.

[01:20:57]What we can do is we can compare both of them. How do you compare them? For comparison, we need both of them to be in the same format, right? Both of them have to be in the same format. Now, how are they going to be in the same format?

[01:21:10]We would have to simplify this. So, we simplify this. At the end we get a form that looks like this. And then we say let's look at the coefficient of n squ that we have here. That's half d. That's the coefficient of x n^2.

[01:21:26]And look at the coefficient of n that we have here. That's a minus/ d. And do a comparison. They gave us that it's 2 n + a 10. We are getting this. That means the coefficient of n² that we have and what they have they should be the same.

[01:21:41]and the coefficient of n that we have and what they have they should also be the same. So we compare them and that gives us the values of a and d like this. Okay. Now in this case it was much quicker and much easier to use this method. But in some problems when it's not possible to find s1 s2 like this because it ends up being in terms of some other constant they might have an extra constant here in this formula. In that case you can use the comparison method as well. You look at you use the standard formula that's given in the formula sheet.

[01:22:15]Compare that to the formula that they give us there.

[01:22:20]Okay. So you'll often see examples see questions in which you have uh something increasing or decreasing by a percentage but you'll have scenarios for that.

[01:22:30]You'll have to interpret them. In this case for instance there's there's a prize money that's increasing by $1,000 each day. What does that mean? it's increasing by the same distance, same uh value. So if it was 100 1,000 initially, then it becomes 2,000 3,000. That's an AP with a difference of 1,000.

[01:22:49]If the prize money is increasing by 10% each day, what does that mean? 100%, it's a GP, right? First of all, it's an increase or decrease by a percentage.

[01:22:59]What's the ratio going to be? We say 100% plus 10%, that gives us 110%. in decimals the ratio will be 1.1. So you should be able to interpret from a scenario like this whether it's an AP or it's a GP if it's a percentage increase or a decrease. Remember it's always going to be a geometric progression. Now this is a very common type of question that you will see in past papers and that is when they give you two sequences and they say some specific terms from one sequence are equal to some other specific terms from the other sequence and then they want to find something that's missing. Now in this case they tell us that for an arithmetic progression the first term is eight and the first term of the AB and the first term of a GP they are both the same.

[01:23:59]So, what are they saying? The first term of the AP and the first term of the GP, they're both the same. And then they're saying fifth term of the AP and the second term of the GP, they're both the same. And then the eighth term of the AP and the third term of the GP, they're also both the same, right?

[01:24:18]We need to write down two equations to find the values of R and D somehow. Now, what you how you would approach these questions would be you would write down both the sequences in two rows like this. You write down the AB and you say okay the first term is eight. Fifth term we'll figure that out and then you have the eighth term. These are the three terms that they're talking about. In the next row you write down the three terms of the GP that you need first, second and third. Now in this case you knew that the first term is eight for the AP.

[01:24:50]Since the GP is the same that means that is also going to be eight.

[01:24:55]Right? How do you write down the next terms? Now you have the nth term formulas for AP and GP. So we have to compare the fifth term of the AP and the second term of the GP. What we can do is we can use the nth term formula of the AP and GP to write down those terms. For example, the fifth term here, what would that be? A which is 8 + n minus 1 that means 4 times the difference which is unknown. So that's 8 + 4. Similarly, this turns out to be 8 + 7. And then you write the nth terms of the second and third terms of the GP as well. The formula is a ris n minus one. If you know the n if you know the first term the first term is 8. This would be 8 * r. This would be 8 r². And now just compare them. Make equations. You say that these two should be equal to each other and these two should be equal to each other. You get two equations. Solve them simultaneously and you get the result from that. This is something that happens a lot in recent papers. Yeah.

[01:25:48]I'm sorry we're getting late. Just one last question.

[01:25:51]Uh so you just compare them and you get the result. Okay. Uh just one one last question then we end.

[01:26:00]Okay. So you just compare them and you make equations and solve them simultaneously. You get the result. So you'll see these questions like these in recent papers a lot. Now one last thing and that is sometimes you will see questions involving circles and they have angles there. Remember, whenever they have a circle and they're talking something about angles, we'll always have one equation that basically says that the sum of all angles sum of all angles is equal to 360° or if it's if it's in radians 2 pi radians. It sometimes is you know something that you don't think of in the in the exam because of you know the stress but the given information in the question that that will tell you uh that will give you some information to make uh one equation but then since in this case they're talking about angles remember whenever they talk about angles in a circle you can make an equation like this that sum of all the angles that should be 360. So in this case for instance we had six angles. So we can say s6 that should be equal to 360. Uh I just wanted to point this out separately because you you see a number of questions a number of questions uh like this in recent papers when they have circles like this. Okay. Sum of all angles should be equal to 360. And then finally we talked about this already.

[01:27:34]they have these three consecutive terms of a GP. If you wanted to find the value for K in this, the most efficient way of doing this would be you say the second term divided by the first term, the third term divided by the second term, they should be equal. So we get k + 6 over 2k + 3 and k over k + 6 they should be equal to each other and then you solve that to find the result. Okay, so this is very important in this case. If you use the nth term formulas that makes it much more complicated when you have three consecutive terms of a GP given you make equations like these you say second over first should equal third over second. All right I hope that was helpful. I'll take your questions after the break now on AGP. We will stop here for uh for Namas and just for a general break as well. We'll come back in about 25 minutes in at 7:40. Okay, 740. You can let me know if you have any questions at that point as well on APGP.

[01:28:33]We're going to be doing coordinate geometry next and uh that including equation of circle. That's going to be an important part and then hopefully uh we can do quadratics as well. Let's see.

[01:28:45]Okay.

[01:28:48]Okay. So, let's quickly have a look at this. It says a circle is divided into six sectors in such a way that the angles of the sectors are in arithmetic progression.

[01:28:56]Okay.

[01:28:58]So angles are in arithmetic progression.

[01:29:00]What that means is we can say if the first angle is a then the other five angles they are going to be in an arithmetic position.

[01:29:11]They have also given that the angle of the largest sector is four times the angle of the smallest sector.

[01:29:17]So if the first angle is a then the lost is supposed to be four times a right. If if we have written them in increasing order, if it's an arithmetic sequence and we're saying the smallest angle is the first one, then it's increasing like this and the last one will be four times a. Now they're saying given that the radius of the circle is 5 cm. Find the perimeter of the smallest crow. We'll talk about that later. First of all, can you find the value of a from this?

[01:29:43]We've got this AP. That's what we know.

[01:29:45]We know that the sum of all of these six terms. What is that supposed to be?

[01:29:49]Since it's a circle, it's part of a circle, right? A circle is divided into six sectors.

[01:29:55]Sum of all angles in a sector in a circle is always 360° or you can say 2 pi radians.

[01:30:02]That means the sum should be 360. And now since in this question we know the last term that is 4 a and we know the first term. This is that type of question in which using that second formula for sum of an a is going to be more efficient where we say half n into a + l. So in place of l i said we will be writing pn. So using that formula becomes much more efficient. Here we input values in this. We uh put s6 equal to 360. On the right side we put the value for n that is six. The value of the first term and the last term that gives us the value for a. Once we've got the value for a, we know the know the angle for the first sector and that's the sector for which we want to find the parimeter for. Right? We've got the radius of the sector. We found the angle. we can find the parameter using uh s= r theta. So this is actually from a time when we had not done circular measure yet. So that's why I'm using the whole level formulas here. But you can use uh s= r theta as well or use uh these formulas. Both of them are fine and that will give the result. Okay. All right. Let's move on now. Okay. So next we've got coordinate geometry.

[01:31:15]Coordinate geometry. This is now a relatively uh straightforward one. So I'll try to be quick. So what do you need to know about coordinate geometry?

[01:31:22]Most of the stuff is O level stuff.

[01:31:24]There's nothing uh that's very new. Uh so you need to be able to able to find distance between two points. You need to remember this formula. It's not given in the formula sheet. X2 - X1 square + Y2 - Y1 square root of that. There's a plus in the middle. Do not forget that. And these are minus signs. Okay? Some people end up mixing these up. Just make sure that you get these signs right. X2 - X1 square + Y2 - 51 whole square and square root of that. That gives us distance between two points. Midpoint formula x1 + x2 over2 plus y1 + y2 over 2. That's what you need to remember. Gradient of a line y2 - y1 / x2 - x1. Fairly straightforward. That's basically change in y over change in x. Now how do you interpret the gradient? What does it what does it gradient mean? If a line has little 1.25, how do you interpret that? That basically means if on the x-axis something increases by one unit, how much does y change? That's what the gradient represents. So if it's 1.25, what that means is if x if the x coordinate increases by one unit, the ycoordinate increases by 1.25 units.

[01:32:31]That's the interpretation of this gradient. Okay.

[01:32:36]If the gradient is negative, for instance, minus 3, what does that mean?

[01:32:39]If x increases by one unit, that means y would decrease by three units. So it's going to be a line like this that as you go one unit to the right, you go three units down, right? That's what a gradient of neg3 means. So gradient tells you for one unit increase in x, how much does y change? Right? That's the interpretation. That's the understanding of gradient. Okay? Now lines like this have positive gradient.

[01:33:06]Right? upward sloping. These are called upward sloping line. As you go from left to right, if the line is going in the upwards direction, that's positive gradient. As you go from left to right, if you're going downwards, that's negative gradient. If it's a horizontal line, the gradient of this line is zero because there's no change in change in y in this case. So, the numerator is going to be zero. The gradient turns out to be zero in this case. For vertical lines, gradient is undefined, right? Because if you find the var of any vertical line what happens is the change in x there's no change in x in the denominator you end up getting zero and anything over zero is what that's undefined. So for vertical lines we say that the is undefined. Now some people say it's infinity at this level you can say that although this is more mathematically correct but uh you can also think of it for now don't quote me when you go to go to university that I taught you wrong that it's not infinity but for now you can say that this is infinity okay that's okay for for this for this level that's okay all right now parallel lines have equal gradient parallel lines have equal gradient so if one line has gradient m1 and the other line has gradient m_sub_2 then we'll say they're both going to be equal if they're parallel perpendicular lines the product of gradients of perpendicular lines that's there that's always uh equal to negative one so if you have one gradient the other is going to be negative reciprocal of that okay so for instance if you know that the gradient of one line is four then the other is going to be - 1 / 4 so what you do is you change the sign and you take reciprocal the gradient of one line is - For the other it's going to be + 1 / 7 and so on. If for one line it is 5 / 6 for the other it's going to be -6 over 5 you change the sign and take reciprocal.

[01:34:53]That's what gives you the gradient of the other line if the lines are perpendicular. Okay. So gradients of perpendicular lines they are negative reciprocals of each other. Now you need to be able to find equations of straight lines.

[01:35:08]So let's say you have this straight line that you see on the right side. We want to find the equation of this line.

[01:35:14]Equation means a line that represents every single point uh an an equation that represents every single point on the line. So for instance, if we say that this is the equation of the line, what that means is every single point on this line, it satisfies the equation of this line. So if you have this point 03, if you input that point here, it will satisfy satisfy uh the equation. every other point. If you take any other point on the line, it will always satisfy this equation. And in other words, you can also say you can find every single point on the line using this equation. How? If you input x equals 0, you get one point.

[01:35:50]If you input x equals 1, you get another uh value for y. If you input x= to 10, you get another value of y. If you input x= - 51, you get another value for y. So you put different values of x, you get different values of y. All of those points are going to lie on that line.

[01:36:06]Okay? So that's the connection between that line and this equation. How do you find this equation?

[01:36:12]This is the standard form for a straight line equation y = mx + c. m represents the gradient here. C represents the y intercept. If you want to find equation of any straight line, all you have to do is you need uh the gradient of that line and the y intercept. If you want to use this form, this is one possible form of uh straight lines. Let's take an example. If you want to find the gradient of a line that passes through 0 - 3. Now that's the y intercept, right?

[01:36:38]Because the x coordinate is 0. When x coordinate is 0, that's the y intercept, right? Gradient is five. What's the equation of the sign going to be? y = 5x - 3 because the gradient is five. We place uh we put five in place of m and in place of c we put minus 3. That's the equation of this line.

[01:37:01]Now you won't always have the y intercept given already. They could they could say something like this. You have a point and it has a point of the line and the gradient of the line is 1 /2.

[01:37:11]Now this point is not the y intercept.

[01:37:13]So what do you do in this case? You say y is equal to mx + c and then gradient is 1 /2. You input that point in this find the value of c from that and that gives you the equation. It turns out to be this. Okay.

[01:37:28]Then you could also have the gradient not given as well. For example, they could give you two points. What do you do in this case? How do you find equation? You find the gradient of the line first. Then you put that gradient in the line y in the equation y= mx + c.

[01:37:41]Find the value of c by inputting any one of these two points in that equation.

[01:37:46]Once you have the value for c, just put that in y= mx plus c form and you get that equation. Now this is one way to find equation of straight lines y mx plus c. A lot of times in A levels the more efficient uh method is going to be the other one. This other form that's called the gradient point form.

[01:38:04]It looks like this. Y - Y1 = M into X - X1. If if somebody's interested in where this comes from. So you can just have a look at this later on. This is where that form comes from. But that's the other form that you y - y1= m into x - x1.

[01:38:21]you uh have this other form that is more efficient whenever the y intercept is not given. How do you use this? You remember this? First of all, y - y1= m into x - x1. M in this case as well is also is the gradient and uh x1 y1 these are basically the coordinates of any specific point any given point on the line. So if you want to find equation of a line that passes through this point and has a gradient to this, we did this using the other method earlier. But now if you use this equation y - this form y - y1= m into x - x1. You can write down the equation directly. You'll say y - - 5 since this is y this is y1 this is x1 and this is m. Just input their values in that in this form and you get the equation. It turns out to be y - - 5 =/ into x - 2.

[01:39:15]Now in a lot of questions this is going to be acceptable as the final answer right they they won't uh require you to convert it necessarily to another form.

[01:39:26]Uh so if the person is silent about it they don't say anything about this final form that they want. You can actually leave your answer in this form and that will be acceptable. So that saves your time as well. If you want to convert it to the to the other form you can do that. If you had two points again you find the gradient. First of all, gradient is required in this form as well. So you have to find that find that gradient. But once you have the gradient, what you can do is you can input that in this form along with any one of these two points. You can you can call any of them x1 y1. You can call this x1 y1. You can call this x1 y1.

[01:39:57]Just input them in this form and you get the equation of the line. Okay, that's pretty basic stuff. So how do you decide which form to use? This is what I would recommend. If you have the y intercept given, uh it's better to use y= mx plus c. If you don't have the y intercept given, it's more efficient to use y - y1= m into x - x1. Both of them work in every case. It's just that in some cases, one one is more efficient. In other cases, the other is more efficient. Now, then the next thing that you need to know about in this topic is solving simultaneous equations. Now, there are multiple ways that you have have for solving simultaneous equations.

[01:40:31]You have elimination, you've got substitution, and you've got two different ways of sub doing substitution. You can solve these equations using any method that you prefer. You can use elimination if you want. Uh you've done that in O levels. A lot of times we uh just don't do it uh because uh for some reason students find it difficult. Uh but if you want to do it, the idea is that you make one of x or y terms the same in both equations.

[01:40:57]Right? So for instance, in this case, the first equation has x x + 6 y. The second has - 6 - 4x + 3 y. If we make for instance the y terms the same in them, how do we do how do we do that? We can multiply the second equation by 2.

[01:41:11]When we do that, that becomes - 8x + 6x + 6 y = 66. Then we can just subtract these two and that will give us an equation in terms of x. Find that value of x from that. Put that in the other equation. You get that result. You could do that with y. You could do that with x as well. We could say let's multiply x with minus4. In fact, this first equation by minus4 or plus4, doesn't matter. You can multiply it by +4 and that gives you 4x + 24 y = 44. And then you just add those two equations to eliminate x. That's one possible method.

[01:41:42]But mostly we will be using substitution. How does substitution work? If you have two equations like this, one equation is this. Another equation is this. Substitution is that you make one variable the subject in one equation. Put that in the other equation. Generally it's easier to make that variable the subject that does not have any coefficient with it. So for example, if you have these two equations in the second equation, you see with x, I don't have any coefficient. So it's going to be more efficient and easier if you make x the subject from the second equation. Put that in the first equation. Why? Because that will avoid fractions. When you don't have any coefficient with a variable. When you make that the subject, you would not have any fractions. That will make it simpler for you to solve that equation.

[01:42:26]Right? So that's one way to do substitution. You make x the subject here. Put that in the other equation.

[01:42:30]Get the value of y from that. Put that in the other equation back and get the corresponding value of x. And another way of doing of doing substitution is that we do a lot of times as well that we make the same variable the subject in both equation. Right? That's also possible. For example, one way to do solve this simultaneous simultaneous equations could be that we make y the subject in both equations. So we make y the subject in the first equation. We make y the subject in the second equation and then we just equate both of them. Right? This is also a substitution but a slightly different way of doing sub substitution because we make the same variable the subject in both equations. first of all and now we could we are basically saying let's input this y value in place of this y here. So it turns out that we are just equating the two uh equations essentially this is also substitution but a slightly different way of doing this. You find the value of x from this and then you find the corresponding value of y from that. That's how you solve simultaneous equations. Now whenever you have two functions whenever you have any two functions any two graphs and you have to find out their points of intersection any two random graphs how do you find points of intersection you always find points of intersection of any two graphs by solving those functions simultaneously right so if you have any two graphs and you want to find the points of intersection what you do is you solve their equations simultaneously and that gives you the result okay now sometimes what you what you have to do is that that you're given the equation of a straight line and you have to find the gradient of the line using that.

[01:43:57]Okay, so you're given the equation, you have to find the gradient. How do you do that? If it's an equation in this form y - 5x + 3, then uh the standard form is y= mx + c, you compare with that and you say in place of m you've got - 5. So that's the gradient. But if it is not in that form, for example, if it's - 2 y = -= 7 x - 4. In this case, if you want to compare it with y, y = mx + c, then you first have to make y the subject. So you make y the subject and that gives you - 7 /2x + 2. Now after you make y the subject, now you can you can compare with the standard form. In place of m, you've got - 7 /2. So that's the gradient of this line. Similarly here you have - 5 s + 2 y. If you want to find the gradient of this line, what you do is you make y the subject compare with y = mx + c. So the gradient turns out to be 5 /2. You can also compare with the other form. If you have an equation like this, for instance, y + 3 into -2 into x - 5. If you have this equation, now converting it to y = - y = mx + c might take a couple of steps. You could just directly just directly uh compare it with y - y1= mx - x1 as well.

[01:45:06]And that gives you the gradient as well.

[01:45:07]Although this works every time y= mx + c. So if you understand this that's going to be sufficient as well. Okay. Then you have to find equations of perpendicular bis sectors. Now what is a perpendicular bis sector of uh any two points. Perpendicular bis sector is that it's a line that passes through the midpoint of those two points. So if I if I have two points like this there is one point a here there's another point b. perpendicular bis sector would be a line that passes through the midpoint of A and B and is perpendicular to that. So perpendicular means the angle is 90 and bis sector means that it cuts in half. Okay. So you want to find the equation of a line that cuts AB in half and makes a 90° with that. How do you do that for any equation for any straight line equation?

[01:45:52]You need two things. One gradient in the other uh other point. Okay, I think I'm getting tired now.

[01:46:01]So, am I speaking too fast by the way or is it okay?

[01:46:07]Anyone having a problem understanding me?

[01:46:12]It's okay. All right.

[01:46:15]All right. Find the perpendicular bis sector of the sign joining B such that the coordinates are 1 2 and - 5 7. Now, perpendicular bis sector is perpendicular bis sector is when you have a line that makes a 90° with with that original line and it cuts that in half.

[01:46:37]So, my chain of thought broke there.

[01:46:41]Yeah. So, you want the equation of the perpendicular bis sector. What you do is you want the midpoint of that line. So, you use the midpoint formula to find that and you want the gradient of that line. Now since these lines are perpendicular since these lines lines are perpendicular their gradients are supposed to be negative as approval of each other. So what we can do is we can find the gradient of gradient of AB gradient of AB turns out to be minus 5 / 6. So the perpendicular bis sector that's that gradient is going to be negative reciprocal of this. So that's 6 over 5. Once you have these two things the midpoint that's a point on that perpendicular bis sector and you've got the gradient of the perpendicular bis sector. What you can do now is you can just use the form y - y1 = m into x - x1 and that will give you that equation that you're looking for. Now a lot of times people end up using point a or point b uh in the equation of the perpendicular bis sector. You cannot do that. Perpendicular bis sector is not passing through b or a. It's passing through the midpoint. So you have to use the midpoint of uh those two points for this equation. Okay. Now this is important using vector method to find an missing point. So sometimes what happens is you are given a line and on this line let's say you've got two points. One is A here and another is B. The point B is known but the point A is not known. You have to find that point. But what you also know is the midpoint. You know the midpoint of the line. The midpoint is 2 and four. Now one possible way of finding the coordinates of A is that use the midpoint formula. That works. You can do that but that is slightly longer. So you would say that okay if the coordinates of a are x and y then x + 5 / 2 that's the uh x coordinate of the midpoint. y - c /2 that's the y coordinate of the midpoint. You equate that with 2 and four. You get two equations from that and you solve those two equations to find the values of x and y. That's one possibility. You can do that and that will work.

[01:48:36]But a more efficient way of doing this is you have this point B and you've got the midpoint and you have to find the point A. Now how can you do that? You think of this in terms of vectors. You say okay this is the x coordinate of this point uh that's five. The midpoint has an x coordinate of two.

[01:48:58]How many units do you have to go in order to go from five to two? You have to go three units to the left. three units to the left. Now, this point is the midpoint. How far would you go further to the left to get to the point A? You'll go three units more. So, in order to go from B to M, you have to go three units. So, how do you go from M to A? Then you go three units further.

[01:49:25]So, 2 minus 3 that gives you minus one.

[01:49:28]Similarly, for the ycoordinate, you can think of it like this. You can say the ycoordinate of this point is minus3.

[01:49:38]ycoordinate of the red point is four.

[01:49:40]How many units do you have to go in order to go from minus3 to 4? 7 units.

[01:49:45]You go 7 units more to get to the ycoordinate of the point a. What would that be? 4 + 7. That turns out to be 11.

[01:49:54]Now, how will you show this working in the exam? You'll do something like this.

[01:49:59]You write down one coordinate B that the one that you know that's 5 - 3. Okay.

[01:50:06]And then you write down the midpoint.

[01:50:08]You write down the midpoint. Midpoint is 2 and four. And what you're looking for is the point on the other side. How do you find that point? You see, okay, from 5 to 2, you have to go minus three. From for in order to go from minus 3 to four, you have to go + 7. You go minus 3 more.

[01:50:23]2 - 3 that gives you 1. You go + 7 more for y. 4 + 7 that gives you 11. So this is often used when you have the midpoint given and the point on one side given and you have to find the point on the other side. It's called the vector method. You can do it like this.

[01:50:38]Another uh case where you will have to use u this method um apart from the midpoint is when you have uh figures like this a triangle a rhombus a parallelogram something like this in which the opposite sides are parallel to each other. Now in this case if you have one point missing, one vertex missing and you have the other vertices given for example you have A given, you have B given and you have C given but B is unknown. How do you find the coordinates to point B? You could say whatever is the number of units have to go from in go in order to go from B to C that's the same number of units right left and up down that you will have to go in order to go from A to D. So from B to C how do you go from B to C? The xcoordinate of B is 12. The x coordinate of C is 14. So from 12 to 14, we have to go plus two units. So in order to go from A to D as well, we say the xcoordinate of A is 5. We do plus two with that and that gives us seven. That should be the X coordinate of 7. All right. Similarly, in order to go from B to C, how many units do you have to go on the Y-axis? 1 to 6. that is five units. You go the same number of units on the y-axis in order to go from a to d as well. So you go you go 3 + 5 and that gives you the y-coordinate of the point d and you can find it like this without any working and that makes it much more efficient.

[01:52:11]Okay. So that's called the vector method. You can use it like this. Then there are some properties of diagonals of some quadrilaterals that you need to remember. These are very important. You need to memorize these uh because they are tested in in the questions that you get.

[01:52:28]There are three things that you need need to know about all of these quadilaterals. Number one, are the diagonals equal to each other?

[01:52:35]Do they meet at midpoints? As in uh this point where they intersect is that the midpoint of the diagonals and the third thing are they perpendicular to each other or not? Now for a square for a square all of these properties are true.

[01:52:49]For a square the diagonals are equal.

[01:52:52]The diagonals meet at midpoint. That means this point m that's the midpoint of both the diagonals. And also they meet at right angles. This is very important.

[01:53:02]Diagonals of a square they make right angles with each other. Okay. So the point of intersection is the midpoint.

[01:53:08]The angle between the diagonals is 90° and the diagonals are also equal. For a square all three of these things are true. Now we need to know about these for all of these quadrilaterals. for after so after a square we've got rhombus. For a rhombus the diagonals are not equal. They're not equal.

[01:53:26]They're not equal. However, they meet at midpoints and they're perpendicular. Okay, they're not equal, but they meet at midpoints and they're perpendicular. Now, for both square and rhombus, for both a square and a rhombus, the diagonals meet at midpoints and they are perpendicular. Does that remind you of something? Titles are perpendicular and they meet at midpoints. That's the definition of a perpendicular bis sector. So for a square and a rhombus, remember this. Diagonals are perpendicular bis sectors of each other.

[01:54:02]So if you know one, you can find the other. Just find the perpendicular bis sector. You'll get the other diagonal from that. So diagonals of a square and the diagonals of a rhombus, they are perpendicular bis sectors, right? One diagonal is perpendic one diagonal is perpendicular bis sector of the other diagonal.

[01:54:20]That's true for both a square and a rhombus. Okay? But rhombus does not have equal diagonals. That's important. For a rectangle, the diagonals are equal. They meet at midpoints, but they are not perpendicular. They're not perpendicular. These are used in coordinate geometry questions that come with circles as well. In circles, you get kite, for instance. You sometimes get square, you get rhombus. uh they they're used there these properties.

[01:54:46]Okay.

[01:54:49]So diagonals are equal for a rectangle. Diagonals meet at midpoints for a rectangle but they're not perpendicular for a parallelogram.

[01:54:59]For a parallelogram they're not equal. They meet at midpoints for so in the previous example that we saw here it was a parallelogram.

[01:55:10]since they meet at midpoints. If you knew the diagonals and you were given the midpoint for instance, you see some such questions in past papers. If you have the midpoint and you have this point B, you should know that this point point of intersection of the diagonals, that's the midpoint of the diagonals.

[01:55:26]Okay, that's that's an important property to remember. So for a parallelogram, the diagonals meet at midpoints. However, they're not perpendicular and they are not equal as well. Now, this is very important. The kite in circle questions you often have this kite when you have a tangent and uh you join that point with the radius.

[01:55:45]When you have two tangents coming from a point outside the circle you always get the kite shape. For a kite the diagonals are always perpendicular to each other.

[01:55:54]Diagonals are perpendicular to each other. They're not equal. They're not equal.

[01:56:00]What about meeting and midpoints?

[01:56:04]The point of intersection. This point of intersection is the midpoint of one diagonal but it's not the midpoint of the other diagonal. So the the point X here the point of intersection of this of these two diagonals it's the midpoint of A C but it's not the midpoint of BD.

[01:56:20]So the point of intersection in case of a rumble in case of a kite is midpoint of one diagonal but not the midpoint of the other diagonal. Okay. But they are perpendicular. This propert is very important that they are perpendicular to each other. For a trapezium, none of these properties is true that the diagons are not equal. They don't meet at red points and they are not perpendicular. Now these properties are used a lot for a parallelogram that the diagons are equal for a kite that they are perpendicular and they meet at midpoints. So all the tick marks that you see here all of them are used in past papers. You need to remember these properties.

[01:56:59]Okay.

[01:57:01]All right. Uh so this is going on here.

[01:57:03]We'll have a 1 minute break. So Okay, let's resume. So you you do get questions like on on this in recent past papers since the focus is more on circle equations. You don't get a direct question on this in recent past papers.

[01:59:11]But when you have uh in circles when you have tangents and they're joined with the radius you sometimes do get figures like rhombus kite. uh you will see that uh in the recent past papers but in the earlier past papers you will see a lot of questions like these uh where you have these shapes you know trapezium parallelogram where you have one some point given and the and another point is missing so you get a number of questions like this in past papers where you have where you have to use those properties okay uh what what I can do is I can share this so you'll find this in any topical worksheet if you are interested I post in the WhatsApp group and I'll share this with you through some one of my students can probably share this with you and uh you can have a look at those questions. Okay.

[01:59:58]All right. Let's move on. Now this is another thing that you won't find a lot in past papers but there is going to be uh some questions on this in the past papers that you will see but since it's not very frequent you might have missed it as well. So let's understand this as well. Sometimes what happens is they give you coordinates of a figure and they want you to find the area of that figure using those coordinates. For example, if you have a triangle that looks like this, it has these three points 2 5 2 - 3 and - 42. These are the vertices of this triangle. And we have to find the area of this triangle ABC. We have to find the area of this triangle ABC. Now, how do you find that? using coordinates.

[02:00:47]Now, in 99% of the questions that you will see on this, especially when you have a triangle, you will have one side of the triangle, one side of the triangle that is either horizontal or vertical. Either horizontal or vertical.

[02:01:04]What you do is whatever side is horizontal or vertical, you take that as the base. For example, in this triangle, we have one side that's vertical. this side AB this is vertical you take that side as the base okay when you take that side as is the base height is always perpendicular to the base so height is going to be perpendicular to that like this this is going to be height and then you can use the area formula that gives you the result that gives you the area of that triangle so whichever side is horizontal or vertical you keep that side as the base okay whichever side is horizontal or vertical take that as the base and then the height height is going to be perpendicular to that. And then what you can do is you can find the base and the height using coordinates. How do you do that? You've got the coordinates of a.

[02:01:53]They're 2 and 5. This is 2 and minus 3.

[02:01:55]So what's the vertical distance from -3 to 5? That's 8 units, right? So this distance is 8.

[02:02:03]Similarly, since we know that this x coordinate is 2 at at this point and this x coordinate is minus4, what's the height?

[02:02:15]-4 to 2 that's 6 units. So it can be done in one single line. You say half into base and height base is 8 in this case. Height is six and you get the area using the half into base into height formula. All right. Mostly in 99% of questions that you get get like this, you can find it directly like this.

[02:02:34]So if you have this shape for for instance you have a base that is horizontal you have one side that is horizontal you take that as the base that's the base height is going to be perpendicular to that in this case the height is going to be outside the triangle right this is perpendicular so you say the height that's from 5 to 11 that's 6 units because this ycoordinate is was five this is 11 so the height becomes six and the base from this point to this point that's from minus 3 to 2 that's 5 units.

[02:03:07]So base is 5 units height is six just use the formula half into base is height and that will work. So whenever you have a triangle in which one side is either horizontal or vertical you take that side as the base and then the height would be perpendicular to that.

[02:03:20]Sometimes, very rarely, you could get a question in which none of the sides is either horizontal, either horizontal or vertical. Either horizontal or vertical.

[02:03:34]In that case, you use this method that we call the sholess formula. I would not recommend using it in simple cases like this. That's so much extra work. If you just use some brain, you can just find the uh base and the height directly and then just just use the formula. Curious formula ends up uh making you do a lot of careless errors sometimes.

[02:03:59]But uh this is an this is a method that you can use when none of the sides none of the sides is perpendicular. None of the sides is sides horizontal or vertical. Okay. of the sides is horizontal or vertical. In that case, if you have three points A, B and C, A, B and C, then what you do is you have this formula that we call the shoest formula. This is the actual formula that you see here.

[02:04:30]But to remember this, we use this form.

[02:04:33]And the idea is we say these three points, they have coordinates x1, y1, x2, y2, x3, y3. We write down those coordinates like this. X1 Y1 X2 Y2 X3 Y3 and then at the end the first point is repeated. At the end the first the first point is repeated outside you've got 1 /2 multiplied.

[02:04:59]Okay. And then what you do is you multiply these coordinates like this and add them up together. X1 Y2 X2 Y3 X X X X3 Y1 this is what you get and then you multiply the other diagonals these and you subtract them in the formula. So that gives you a formula that looks something like this.

[02:05:26]Then you just input those values. You have those coordinates just input those values this formula x1 y1 x2 y2 x3 y3 and then just simplify this and you get the result. The you have these vertical lines on the sides. This is basically the modulus function. What it does is whatever is the number inside if the number inside turns out to be negative you will just make that positive. Right?

[02:05:47]That's what that's all that this notation means. These vertical lines that if the number inside is negative you'll make that positive because area cannot be negative. Then you can use this like this.

[02:05:59]Do we multiply by 1 /2? Uh even if it's not a triangle. So this method is actually a general method that works for any figure. Uh for a quadilateral, for pentagon and so on. This works for any figure.

[02:06:12]So for a quadilateral for instance you would you could extend like this x1 y1 x2 y2 x3 y3 x4 y4 and then at the end this first point is repeated.

[02:06:23]The number outside that remains the same. That remains 1 /2. Yes, that does not change. Okay. All right. So uh if you want to understand this in detail uh can someone please share the sholess formula videos somewhere maybe in the group or in the chat share the link for that because I think a couple of people some people do not know about that. So you can go through that.

[02:06:49]Could could we draw perpendicular bict from P to height? Okay let me actually come back to the question later on. Then we have area of rhombus and a kite. area of a rhombus or a kite.

[02:07:01]Area of a rhombus or a kite that's given by this half into d1 into d2. That's the formula that we've got for area of a rhombus or area of a kite where d1 and d2 are the diagonals of that shape. For both rhombus and kite, this formula works.

[02:07:19]One way to do this would be of course that you break the triangle into two parts. You make the uh you break the kite or rhombus into two parts. Make it make two triangles. For example, if you break it like this, you make two triangles and then you can uh then you can uh find the area of one triangle, multiply that by two and you'll get the area of the complete figure. Okay, that's one one possibility. But from that we actually get this direct formula as well that you can use if you want. half into d1 into d2. So if you know the diagonals the area of a character around this is given by this half into d1 into d2. All right. Now this is another another important thing that a lot of people forget because again it's tested in just a few questions uh in the past papers maybe five six area between a line and x-axis or area between two lines. Area between any line Area between any line and x-axis it is given by this formula tan theta equals the gradient tan theta equals gradient.

[02:08:27]If you want to find angle between a line, angle between a line and the x-axis, angle between a line and the x-axis, how do you find that?

[02:08:39]You have the you would need the gradient of that line. For example, if you have a line that has gradient 2, you can say tan theta= 2, that gives you theta= 63.4. That's the angle between that line and the x-axis. Sometimes they say find the obuse angle then you would just use the formula find the angle that you get. If it turns out to be acute you do 180 minus that and you get the corresponding obuse angle from that as well. All right.

[02:09:09]So that's the formula that we can use to find gradient as the angle between a line and x-axis tan the leg is gradient.

[02:09:16]Okay. Now from that method we actually can derive another method that's called that's basically angle between any two lines.

[02:09:24]We can generalize that and the formula that for that that we have is tan theta equals m_sub_1 - m2 / 1 + m2. If you want to find angle between any two lines any two lines and you have their gradients m_sub_1 and m_sub_2 then you can use this formula to find the angle between those two lines. Just input one gradient in place of m_sub_1 the other gradient in place of m_sub_2 and that will give you the angle that you're looking for. You need to remember these formulas for angle between line and x-axis tan theta equals gradient for angle between any two lines. This is the general method that works for any two lines where one line does not necessarily need to be the x-axis. It could be any two lines. You say tan theta is equal to m1 - m2 / 1 + m1 m2. All right.

[02:10:14]In case you forget this formula, in the formula sheet, in the formula sheet, you have uh an identity written that looks something like this. Tan of a minus b.

[02:10:25]Now, if you have not done paper 3, you might not be familiar with this, but at least you can read that that you have this identity given in the formula sheet. tan a minus tan b / 1 + tan a tan b.

[02:10:39]Just replace tan with m_sub_1 m2. So tan a put m1 in place of that. Tan b put m2 in place of that and you get that formula. Okay. So in case you forget in the formula sheet you will find this identity given it actually comes from this identity. So if you forget the formula a lot of times people end up mixing the sign mixing up the signs.

[02:11:00]They forget that this is minus and this is plus.

[02:11:03]How do you remember this? In case you forget how do you remind yourself about it? Look at this identity in the formula sheet. It's written in the trigonomet on the trigonometry page.

[02:11:13]Just replace tan a with m1 and replace this with m2 and you get that formula tan theta= equals that. Okay. All right.

[02:11:22]You can use this to find angle between any two lines. For example, if you have gradient of one line that's 2 over 3, gradient of another line that's minus one. Then tan theta equals m1 - m2 / 1 + m1 m2.

[02:11:33]And that will give you the result. Okay, so these are these are the two formulas.

[02:11:39]tan theta= gradient tan theta= gradient.

[02:11:43]It's used for angle between any two lines. Okay, I'm sorry. Angle between a line and x-axis tan theta equals gradient. The first formula and the second formula tan for any two lines.

[02:11:54]Okay, where does this come from? It comes from trigonometry. Uh again that's out of the scope of this revision video. uh you can find the detailed video where I will explain that but just remember these formulas tan theta equals gradient of line when you have to find uh uh the angle between a line and x-axis and for angle between any two lines you use this other formula okay that's golden geometry all right that's what you know need to know about this apart from equation of circle we move on to move move on to equation of circle in a Any quick questions before we move on to the next one?

[02:12:39]Any quick questions?

[02:12:43]I think everyone is getting tired. It's okay. It's just a month.

[02:12:50]app will be free in okay all right good uh there was a question there in the middle that I just quickly respond to and perpendicular bis sector from P to find height and okay Okay. So, one of you suggesting that we find the perp in this triangle for instance, we find the perpendicular bis sector of uh QR and uh that will give us this height number one that will become quite long.

[02:13:46]That's one that's one uh problem with this. And the other problem is that it's not correct. It's wrong. Right? That's actually the more major problem. Why are you assuming that this is the midpoint?

[02:13:57]You can't assume that that uh where you make a 90° here that's going to be the midpoint. You can't make assumptions, right? So no, that would not work.

[02:14:09]But that's going to take too long anyway. Even if it was the midpoint, that'll, you know, require you to find that midpoint first of all. In fact, uh, no, that's not going to take that long.

[02:14:19]You just find the midpoint. Then you find the distance between these two lines and then you also find the distance for find the distance qi using the distance formula.

[02:14:28]Still much longer than this and also incorrect. Again, that's the major problem.

[02:14:33]Okay. All right.

[02:14:36]So we just we're just saying we're just saying uh any side any side that is either horizontal or vertical in the triangle you take that as the base. Whatever side is either horizontal or vertical you take that side as the base of the triangle.

[02:14:56]Then height is always perpendicular to that. Height is always perpendicular to the to the base. So in this case base is this yellow line. What would the height be? the perpendicular coming from the third point of the triangle to that base. So that's in this case is going to be outside the triangle. So that's this length here. That will be the height. And you can find this height. Now how do you do that? You know that the ycoordinate at this point is five. The ycoordinate at this point is 11. What's the difference between them?

[02:15:24]That's six. The x coordinate at this point is minus3. The x coordinate at this point is is two. What's the difference between them? Five. So that's the base. Base is five. Height is six and then you just input those values in the formula. If you have any side in the triangle that is vertical, you take that as the base and the height would be perpendicular to that like that. Is that okay?

[02:15:49]We take it we can take any side as the base.

[02:15:54]Logically we can take any side as the base. The formula would work. But if you take any other side as the base finding the height would be very difficult.

[02:16:02]Right? Because if you take this side as the base, finding this height is very difficult. How do you how do you find this height? That's the problem. So we take the horizontal side as the base because then if we know the coordinates, you just have to do plus minus to find the length for base and height. So that makes it much more quicker, much more much more easier. Make sense?

[02:16:26]Okay. All right. Let's move on. Let's move on to the next point of coordinate geometry and that's equation.

[02:16:52]Please don't pray for anything like this. Cancellation of exams, delay of exams, you will be regret regretting this later. Every year this this has been happening for the last three few years and your grades get get affected by that because the expected grades that you get on those missed papers they're not very good generally.

[02:17:13]Yes.

[02:17:15]Okay. Let's move on to the next one.

[02:17:22]Equation of circle. Okay. So this is also part of coordinate geometry but this is relatively more difficult and the concepts are not that difficult but the questions that you get in this they are sometimes quite tricky. All right so uh what do we need to know about this?

[02:17:36]First of all remember what the equation of a line represents. If you have an equation of a line like this y= 2x + 5 y what does that equation represent? This line this equation represents every single point on this line. Now we want something similar for a circle as well.

[02:17:49]So let's say we have a circle that looks like this.

[02:17:52]We want to find equation of the circle such that that equation represents every single every single point on the circumference of the circle. Okay. So the equation that that we're going to find it should represent every single point on the circumference of the circle. It would not represent points inside the circle or outside the circle. only points only points on the circumference of the circle. Now what's that equation? It looks like this. X - a^ 2 + Y - B²= R². That's the standard form for the equation of a circle. You need to remember this coordinates of the center are a and b and r is the radius of the circle. Just like for straight line we have y = mx + c. We have y - 51 = 7 x - x1. For a circle, we've got this standard form x - a square + y - b square= r². In order to write down the equation of a circle, you need two things. Center and radius.

[02:19:00]For straight line, you need gradient and a point. For a circle, you need center and the radius. All right?

[02:19:09]And this equation represents every point on the circumference of the circle. on the circumference of the circle. Let's take an example. Let's say we want to find the equation of a circle that has radius 7 and its center is at 4 and minus 5. Now 4 and minus 5 these are going to be values of a and b. a is equal to 4 and b is equal to minus 5 and radius is 7. So r is equal to 7. Right?

[02:19:32]All you have to do is just input values in this form. Input a input four in place of a input five in place of b and input uh seven in place of r. You get that equation. This is the equation of the circle. Now this is one form for the equation of a circle. This is the completed square form. You can also write the equation in the expanded form.

[02:19:52]So if you expand this whole thing, you get this for this equation. This is another way to write down write down the equation of the circle. They can sometimes specify that they want the equation in this form. In that case you would write the equation first in the standard form then simplify that to get this final form. Okay? So you should be familiar with both of these forms. So that's one thing that given the center and the radius, you should be able to write down the equation of a circle.

[02:20:17]This is not a formula. This is the standard form that you that you need to know about. Just like y= mx + mx + c or y - y1= m into x - x1. This is not given in the formula sheet. You need to memorize this. Find the radius. So sometimes you have to do the other do it the other way around that you're given the equation and you have to find the radius of the circle and the coordinates of the center of the circle. How do you find that? Let's say we've got this equation. We want to find the equation of the we want to find the radius of the circle and the coordinates of the center. We compare this with the standard form x - a² + y - b² that equ= r².

[02:20:57]Compare with that - a= 5 that means a= - 5 - b= - 6 that means b is equal to 6 and r² = 121. So r = 11. What happens is you'll always just flip this sign. So this is + 5. You flip this to get - 5.

[02:21:16]That's the x coordinate of the center.

[02:21:18]This is minus 6. You flip this sign.

[02:21:19]That becomes + 6. That's the ycoordinate of the center. So the coordinates of the center therefore are -5 and + 6.

[02:21:27]All right. And then the radius. What's that? Radius is going to be equal to 11.

[02:21:32]That's the square root of this number.

[02:21:35]Okay. Now, what if you have this expanded form? So, sometimes they can do something like this as well that they give you this expanded form and then they ask you to find the radius of the circle and the center of the circle. How do you do that? Now, you remember the standard form. The standard form looks like this. X - a whole square + Y - B²= R². That's the standard form. So what you do is you complete the square for this equation. So you complete the square for the x term separately. So x^2 - 18x then you do + 9^ 2 - 9^ 2 complete the square for this. Similarly you complete the square for the y term separately and then you convert them to the standard form and then you you can compare to find the coordinates of the center and the radius. Right? That's one possible way of doing this. Another possible way of finding the radius and the center of this equation is that you do the other way around. You've got this equation in expanded form. So what you do is you have the equation in this form. That's what you remember.

[02:22:39]Instead of completing the square for this for the for the equation that they gave, what you do is you simplify the standard form. You expand that and you get something like this from that. x^2 + y^2 - 2 ax - 2 b y + a square + b square - r square = 0.

[02:23:00]Look at the coefficients. Coefficient of x² that's just one. So that does not give us anything interesting. But coefficient of x for instance you have - 2 a here in the equation that they gave that that was - 18. We compare them you get a= to 9 from this. Compare the coefficient of y that gives you b= - x.

[02:23:17]Compare the constant and that gives you uh the radius as well. Now some people uh memorize this as well. If you think your memory is really good, you can do that if you want. That makes things quicker.

[02:23:32]But yeah, so so this is what we can do. Uh Dina, let me answer that at the end. Okay. So it basically basically comes from Kbase. R is equal to R is equal to or A square + B square - R square that equals this constant term that you've got here. Okay, you can do something like this. Then you rearrange this you will get that formula that that you're talking about. It will always work if you use it correctly. That will always work. uh but I don't recommend memorizing a lot of a lot of stuff uh because that ends up you know uh getting you end up getting something wrong in the exam you end up forgetting some plus minus if you're very sure about your memory that you will not forget that then you can try to memorize those complicated formulas but otherwise I think this is not that complicated you can just you know either complete the square for that or expand the standard form and then do comparison what you can remember is apart from the radius you can write down the coordinates of the center relatively easily. Uh and that would be like this.

[02:24:44]If you have this expanded form, if you have this expanded form, if you divide the coefficient of x by minus2, that gives you the x coordinate of the center. If you divide the coefficient of y by minus2, that gives you the y-coordinate of the center.

[02:25:04]So finding the coordinates of the center is pretty straightforward. You can do that directly without completing the square. You just divide these numbers, the coefficients of x and y by minus2 and you'll get the coordinates of the center directly. For radius, it's relatively more complicated.

[02:25:21]You'll have to do something like this.

[02:25:22]So in uh this equation for instance, you would first of all divide this number by two.

[02:25:31]Square that that's 9 squ plus this number. So essentially the coordinates of the center you have r 2 = a 2 + b 2 minus this constant right whatever you want to call that um you can call that k or something so minus that constant and then you take square root of that to get the radius.

[02:25:58]If you want to remember this, you can do that. U otherwise just do it to use in the ways that we've just discussed.

[02:26:06]Is that okay?

[02:26:10]All right. Can I move on?

[02:26:15]Okay. Then uh so these are the core concepts that the key concepts that we have in equation circuit. Okay. But along with this you will need to uh know some other things as well.

[02:26:27]For example, you you will need to be able to use the angle properties of circle that you've done in O levels. All of these properties, if you remember the circle theorems, angle angles in the same segment are equal to each other.

[02:26:44]Angle made by diameter is 90.

[02:26:47]If a line coming from the center, it cuts a cord at midpoint. The angle made made is 90. Here this property where you have two tangents coming from outside outside a circle. How the two tangents are equal and some other properties that you have along with that angle at center is twice the angle of circumference. Cyclic quadilateral and tangent is rad. Tangent is perpendicular to the radius. All of these properties you need to be able to use them. Now I'll specifically some spend some time on the properties that are used more often.

[02:27:23]This is the one of the more common commonly used properties.

[02:27:28]Angle made by diameter is equal to 90.

[02:27:30]Angle made by the diameter. Angle made by diameter on the circumference of the circle that's always 90. So every angle, so if this yellow line is the diameter, every angle that it makes at the circumference is 90.

[02:27:43]Not just this, the only chord that can make a 90° angle on the circumference is diameter. So if you know there's any chord in a circle, there's any chord in a circle, any random chord. So let's say I've got this chord and I don't know if if it's the diameter or not. If I know that it makes 90° at the circumference, then we know for sure that this line has to be the diameter. This is used a lot in circle uh equation questions. If you know that the angle at the circumference is 90 then this line definitely has to be diameter because no other chord apart from the diameter can make 90. Okay, that's one thing. Then so angle made by diameter on the circumference is 90.

[02:28:28]Angle made by the diameter inside the circle. So if you have an if you have a point inside the circle, let's call this alpha. This angle is always going to be greater than 90. Angle made by diameter inside the circle that's greater than 90.

[02:28:42]Angle made by diameter outside the circle that is less than 90. So if this is beta this angle is less than 90 on the circumference the diameter makes 90.

[02:28:54]Inside the inside the circle it makes greater than 90. Outside the circle less than 90. Okay. All right. So that's one property that's used a lot. This one chord coming from the center of the circle.

[02:29:07]sorry line coming from the center of the circle to the midpoint of a chord, it always makes a 90°. And also true, it's also true the other way around. If the angle that a chord is making at uh if the if a line is making if a line coming from the center of the circle to the middle point of a chord.

[02:29:30]Okay, let me repeat. So it it can be understood two ways. One, if you have a line coming from the center of the circle to the midpoint of a chord, the angle that it makes is is 90. That's one thing. And the other thing is if you know that there's a line coming from the center of the circle and the angle that it's making is 90, then that means that this point is the midpoint. So it works both ways. If you know that point is the midpoint, the angle is 90. And if the angle is 90, that point is the midpoint.

[02:29:58]Right? So one nec one basically requires the other to be true as well. Okay. The angle is 90 point is midpoint. If the point is midpoint then the angle is 90.

[02:30:10]Now this is again used a lot. From this property we actually get something very useful in circle equations and that is for every chord for any chord that you have in a circle. For example, this chord, this chord, if you were to draw perpendicular bis sector of this chord, perpendicular bis sector of the chord, these perpendicular bis sectors will always pass through the center of the circle. Perpendicular bis sector of every single chord in a circle always passes through the center. Okay, it always passes through the center.

[02:30:46]That's a property that you need to remember as well. perpendicular bis sector of every chord in a circle passes through the center of the circle. So sometimes when you have to find when you have to find the when you have to find the center of the circle to find the equation of a circle for that you often have to do this that you use perpendicular bis sector of a chord right and you say that it it has to pass through the center so you use that to find the center somehow.

[02:31:16]This is the next property that's used a lot in equation of circle problems. If you have a tangent coming from a point outside the circle, two tangents coming from a point outside the circle. So let's say you've got these two tangents coming. I'll just get rid of that so that it makes it easier to understand.

[02:31:34]We have two tents coming. So we have we have a point P outside and from that P from that point P, we have two tangents coming uh that are touching the circle at two points. Let's say that they are uh Q and R. Okay.

[02:31:50]Now these two tangents are always going to be equal to each other. PQ is going to be equal to P R. These tangents are going to be equal to each other.

[02:32:02]PQ is going to be equal to P R. But not just that this a lot of other things follow from this property as well. What are those things? Whenever you have two tangents like this, you will always make a kite figure like this. So you will join the radi like radi uh here as well.

[02:32:21]So you'll join that point with the center. You'll join this point with the center and you'll get a figure like this. Since tangent and radius they are perpendicular to each other. You've got a 90° angle here. You got 90° angle here. And that means this figure that you get oppite because these are two radius.

[02:32:41]These these rad are equal to each other.

[02:32:45]The tangents are equal and that's the definition of a kite. The adjacent sides are equal. It's a quadilateral having equal adjacent side. So that's a kite.

[02:32:55]Now all the properties of kite apply here as well. What are those properties?

[02:32:59]If I were to divide this into two parts like this.

[02:33:06]These two triangles are congruent.

[02:33:08]And the angles that we get here, if this angle is X, then this angle is also X.

[02:33:14]If this angle is Y, then this angle is also Y. These angles are going to be equal to each other. These are going to be congruent triangles. And if I make the other diagonal of this kite like this, then it's a kite. Diagonals of a kite meet at 90°. This angle here is also going to be 90°.

[02:33:31]Okay? This angle here is also going to be 90°. So all of these things follow from this one property. All right, this is something that you have to use sometimes as well. And then finally, this is another commonly used one.

[02:33:44]Radius perpendicular to tangent. That's pretty straightforward.

[02:33:48]Radius and tangent, they are perpendicular to each other. Okay.

[02:33:53]Now, if you have three points given on the circumference of a circle. So, let's say we have the circle and we know three points on the circumference of the circle. What are those three points?

[02:34:09]Points A, B and C. A, B and C. And we want to figure out the equation of the circle. Now, how do you find the equation of the circle? Equation of of the of a circle needs two things. Center and radius. Right? These are the two things that we need to be able to write down the equation of a circle. Center and the radius.

[02:34:31]Now, how do you find the center? If you if you can find the center, you can find the radius easily. to that for finding the center we use that property we say let's make any two chords let's make chord AB center should uh should lie on the perpendicular bis sector of this so center should lie on the perpendicular bis sector of this chord make another chord you can make AC or BC let's make BC the perpendicular bis sector of BC is also going to pass through the center of the circle Okay.

[02:35:05]And now what you can do is once you find these two perpendicular bis sectors, you can solve them simultaneously and that is going to give you that is going to give you the coordinates of the center.

[02:35:17]Okay. Once you have the center, then you can find the distance of the center from either A or B or C and that will give you the radius and you can write down the equation of the circle. So if you have points given on the circumference and you want to find out the equation of the circle what you do is you make any two chords find the perpendicular bis sectors equations solve them simultaneously and that will give you the coordinates of the center. All right. So this is the summary of that. You can have a look at that. Take a picture of this. We'll have this in the notes as well. If you want to find the center, find perpendicular bisector of a perpendicular bis sector of VC. Since we know that perpendicular bisector of every chord passes through the center, it passes through the center. Then that is going to be uh the point of intersection of these two perpendicular bis sectors. We solve these two equations simultaneously and that's that that gives us the center. Once we have the center, we can find the radius by finding distance of O from any of those points. Okay, can't we can't assume that midpoint of any of these chords is uh the center. These are not diameters.

[02:36:23]Okay. All right. Let's take some examples of this. Now, now these are difficult questions. Some of these not all the point A has coordinates 15 and the line L has a gradient -2 over 3 and it passes through the point A. There's a circle that has center at 511 and its radius is under root of 52. So we have a point A with these coordinates. We have a point uh we we have the line L that is this and it passes through A and we're given some some information about the center and the radius. We need to show that line L is a tangent to the circle.

[02:36:58]How do we show that line L? How do we show that line L is a tangent to the circle? In order to show in order to prove that a line is a tangent to a circle we need to prove two things we need to prove two things. One that that line is perpendicular to the radius at that point and it touches the circle at some point.

[02:37:26]Okay. Now let me repeat the question says show that L is the tangent to the circle at the point A. The point A is 15. Now how do we prove that? We have to prove two things here. Number one, the point A lies on the circle. First of all, we don't know that it's not given that point A lies on the circle. So if line L is a tangent at the point A, we we will need to prove that A lies on the circumference. How can we do that? We can input that in the equation of the circle. If the equation equation of the circle is satisfied, that means the point A lies on the circumference.

[02:38:05]That's one condition for it being a tangent.

[02:38:08]The second thing is The second thing is that the line O A the line O A this radius this radius is perpendicular to the tangent. Right? Radius is perpendicular to the tangent.

[02:38:29]How do you do that? You find the gradient of OA. You find the gradient of this line L. If they are negative reciprocal of each other that necessarily means that they are perpendicular to each other. When both of these conditions are met since tangent is perpendicular to radius we can say line L is a tangent to the circle. So in order to prove that line is a tangent to a circle we have to show two things. One the point lies on the midpoint and the point lies on the circumference and number two that it makes a 90° angle there. Yes. So that's another idea.

[02:39:03]Another idea could be that you find the equation of the line. Find the equation of the line. We have a point. We have the gradient. We have the we can write down the equation of the circle as well.

[02:39:13]If the line is a tangent to the circle, you can find the discriminant. Find the discriminant.

[02:39:19]You'll have to write down the equations of both of them both of them first. Then start solving them simultaneously. Make put uh the y value from the line in this equation. So for example in this case the equation of the line is going to be y - 5 = -2 over 3 into x -1. You make y the subject from this y = -2 over 3x + 2 over 3 + 5 that will I think be 67 / 3.

[02:39:47]You put this value in this equation of circle and then you say let's find b square - 4 a c. If b square - 4 a c turns out to be equal to zero then that means that the line is a tangent to the circle. that could also be used here although that will be slightly longer but uh you can do that as well. Okay, so that's one thing. How do you prove that a line is a tangent circle? This is these are the two conditions that have to be met. The point has to lie on the circumference and it should be perpendicular. Both of them have to be shown in order to uh in order for this to be a complete proof. Then it says find the equation of the other circle that has radius under root 52 for which line L is also a tangent at that same point. So we have another line L. We have the same line L. And now they're saying this line L is also a tangent to another circle. Where would that other circle be? That other circle is going to be on the other side of the line like this. Then this would be a tangent. How do you find the equation of the circle?

[02:40:44]Now the question said it's the same radius.

[02:40:51]Okay, same radius. So it's the same size circle. We know the radius. How do you find the center? Now you can use the vector method. You can say we know the coordinates of the center of the first circle. We know this point A. The point A will be the midpoint of these two circles. We can use the vector method to find the center of the other circle and then use the standard form for the equation of a circle to write down our final equation. Okay, this is the same method that we use when sometimes they say find the equation of another circle that is the reflection of the first circle in the line L. So if you had to find the equation of a circle that's a reflection of that first circle in the line L, we do exactly the same thing for that. If L is the line of reflection, we say the radius is going to be the same.

[02:41:42]Radius radius is going to be the same.

[02:41:44]And then since a is going to be the midpoint of the centers, we can use the vector method or the midpoint formula to find the center of the other circle.

[02:41:54]Right? And then you can write down the equation. All right? Now, sometimes you have to prove that there's a point that is outside the circle. How do you do that? So there's this point. We have to prove that this point lies outside the circle.

[02:42:09]This is the equation of the circle. How can we do that? We know that the radius of the circle is 10 square root of 100. Find the distance from the center to this point.

[02:42:23]From the center of the circle to the points, what's the center of the circle?

[02:42:26]Center of the circle is 8 and 4. If the distance from the center of the circle is more than the radius, then that necessarily means that point is outside the circle. Right? Because if you have a circle then if the radius is 10 if some point is more than 10 units away from the rad from the center then it has to be outside the circle. Right? So that's a common question that you see in past papers as well to prove that a point lies outside the circle or inside the circle. That's what you do. You find the radius and find the distance of that point from the center and compare them.

[02:42:59]If the distance is less than the radius it lies inside. If the distance is more than the radius the point lies outside.

[02:43:04]If the distance is equal to the radius then that point lies on the circumference. All right, this is a good question. You can look at the working for the rest of the question here but let's move on to this now.

[02:43:24]So we are given three points that lie on the circumference of a circle. Right? We have three points 7 1 7 9 and 1 9.

[02:43:34]And we want to find the equation of the circle three points on the circumference. How do you find the equation of that? Using that method that we learned, we could do that that we find the perpendicular bis sectors of any two chords. We can make chords ourselves, right? Even if they're not given there, we can make chords ourselves, right? So if we have a point A and we have a point B, we can make a chord. Can just join those points to make a chord.

[02:43:59]Make two chords, find perpendicular bis sectors of each of them, solve them simultaneously, you'll get the center.

[02:44:04]However, in a lot of questions, a lot of questions, it's often going to be much easier than that. How? This is where you'll understand why it is important to draw good diagrams in the circle equation questions. Always draw diagrams whenever you have a circle equation question. In this case, if you were to draw a circle and mark these three points, how will these three points look like?

[02:44:34]One point is 71. Another point is 79.

[02:44:36]Another point is 1 N 7 and 7. These two points have the same x coordinate. That means they're going to be on the same vertical line here, right? Because the x coordinates are the same. So if the x coordinate is seven here if if point A is here and point B is going to be here on the same vertical line.

[02:44:56]Similarly the point C and point B they have the same y-coordinates this is 9 and this is also 9 that means if there's there's some point at which the ycoordinate is 9 then B and C are on the same horizontal line.

[02:45:14]What does that tell us? That tells us that in this case the angle that we're getting at the point P, what's that angle? That angle is 90°. So this angle here is 90°. Why? Because one line is horizontal and the other is vertical.

[02:45:28]Between horizontal and vertical, we've got a right angle. Now if that's a right angle, does that remind you of something? 90° on the circumference.

[02:45:36]What does that mean? That necessarily means that this line here, the line AC, that is going to be the diameter. That has to be the diameter.

[02:45:45]Okay. Now if that's the diameter finding the finding the center becomes so much easier. Just find the red point of A and C and that will give you the diameter that will give you the centers coordinates. Now that only works in this case because we have figured out that AC is the diameter. If it was not the diameter then we could not do that right because center is not the midpoint of every chord. It's only the midpoint of diameter. So since in this case we know that this is going to be diameter because that angle is 90 and only diameter can make 90° on the circumference that means that the line AC is the diameter and the midpoint of that is going to give us the coordinates of the center and then once we have the center we can find the radius and write down the equation of the circle from that. Okay. All right. No, which part do I want to cover for this?

[02:46:48]The last part. Yeah, in this case. So, we have the situation. So, I'm not going to go through the whole question. You can have a look at this working later on in your own time. This last part, it says point R lies on the positive y-axis. the point R this is some point R so okay so let me give you some context so we have an equation of a circle that's given so that's circle looks like this we have found the centers coordinates and the radius coordinates all right and then in the second part we have showed that PQ is the diameter of Z so we have shown that PQ P and Q this is point P here and this is point Q if you join P and Q that's the diameter of the circle okay now they're saying the point R lies on the positive yaxis. The point R lies in the positive y ais and the angle P RQ is equal to 90°. The angle P RQ is equal to 90°.

[02:47:43]Okay, find the coordinates of that point R. How can we do that? How can you find the coordinates of the point R?

[02:47:51]This is the only information that we're given. The point R lies on the positive yaxis and the angle P RQ is equal to 90°.

[02:48:01]Okay. Now it lies on the y-axis. So it has to be on this y-axis.

[02:48:08]But it lies on the positive y ais that means it has to be on the positive side.

[02:48:13]That means only in this part it has to be somewhere here. Now where exactly is it going to be? How do you find that point out?

[02:48:23]We are given that angle P RQ is 90°.

[02:48:28]Now the line PQ the line PQ we know that that is the diameter of the circle. This line is the diameter of the circle. We have 2 the previous part. Now if R if the angle at R that this line is making is equal to 90° and this line is diameter then that necessarily means that the point R has to be on the circumference. Earlier we said it could be anywhere on this y-axis.

[02:48:58]But then since the angle is 90 and PQ is diameter that point R necessarily has to be this point here because only then the angle can be equal to 90° using that property that only diameter can make a 90° on the circumference. So that this point has to lie on this on this on the circumference. So this is that point R.

[02:49:22]Now we can find this point R because now we know that it lies on the y-axis that means it's x coordinate is zero and it also lies on the circumference of the circle. So all we have to do is in the equation of the circle in the equation of the circle put x equal to 0 and that gives us two y values. We consider the one that's positive because it's above the above the x-axis and that's the point that we were looking for y= 2. So if you know that a a diameter is making 90° at with some point that point necessarily has to be on the circumference it can't be outside or inside. All right. So using that we've spotted that point out and then we found its coordinates.

[02:50:02]Now this is something that has happened the first time in a very recent paper. I think this is March 2025. We were doing it the other day. Uh so you have to we normally study this in complex numbers in paper three uh but it turns out that you need need to know this here as well. Now if you have a circle if you have a circle and it has some center and there's a point outside the circle.

[02:50:29]Let's say there's a point here. Let's let's call that point X. If you want to find out what's the greatest what's the smallest distance or the greatest distance from the point x to any point on the circumference.

[02:50:43]How do you find that? So the point x it's shortest distance to any point on the circle.

[02:50:51]Is this the shortest distance? It's not.

[02:50:53]Where is the shortest distance going to be? Somewhere around here.

[02:50:58]Right? Somewhere around here. But what exact point is this? This point is if you were to extend this line up to the center then the point where this line cuts the circle this is the shortest distance between this point and the circumference. Okay. And if they ask you for the longest distance for the maximum possible distance what happens in that case?

[02:51:30]The maximum possible distance from that point to any point on the circumference that is going to be when you go on the other side of the circle. Now where exactly that would be when you pass through the center of the circle and reach the other side of the circle and that will give you the greatest possible distance.

[02:51:54]Let me show that to you here. The greatest possible distance would be If you were to pass through the center of the circle and reach a point on the other side of the circle, this distance is going to be the greatest distance between the point and the circumference of the circle. Okay?

[02:52:21]So from point to a circle the shortest distance and the longest distance they always come when you join that point with the center where it cuts the circle first that's the shortest distance when it cuts the f circle on the other side that's the greatest distance okay this is a concept that's used in a recent paper uh the most recent paper March 2025 uh you can have a look at this working of the the working of this question it it uses that concept okay so that's something that you need to know and then finally The concept of discriminant.

[02:52:53]The concept of discriminant.

[02:52:58]The concept of discriminant is also used in circle equation.

[02:53:04]Sometimes how exactly?

[02:53:07]You have a circle and you have a line and they say the line is a tangent to the circle at the point P. Okay.

[02:53:15]Line is a tangent to the circle.

[02:53:19]If you were to solve these equations simultaneously, you would get a quadratic equation from that. Since you can get a quadratic equation from that and you are given that the line is a tangent of the circle, that means b² - 4 a c that is going to be equal to zero.

[02:53:33]So what you do here is you equate these two equations or you know start solving them simultaneously. This is the equation of the circle. This is the equation of the line. We put y =/ x + 6 in the equation of the circle.

[02:53:48]Simplify this. You get a quadratic equation from that. And then since we have a tangent, that means they are only going to have one point of intersection, one point where they meet each other.

[02:53:59]When we convert it to a standard form, ax² + bx + c equal to 0.

[02:54:04]Then we put b square - 4 a c= 0. So discriminant can be used with circles sometimes as well. Just be uh wary of that. Be aware of that. And then you can find the values of a from that.

[02:54:18]All right. So that's how you would uh deal with circle equation problems. It's questions are difficult require a lot of practice. Make sure that you do that practice because in terms of concepts the standard concepts there isn't much but in questions they make you think a lot. Now what's the logic of this? Why the logic of the shortest distance? Uh remember uh shortest distance between uh a point and a line. What's that? That's always the perpendicular distance, right? So here as well, we basically are looking for the perpendicular distance from this point to the circle. Now, where will the perpendicular be made?

[02:54:59]This is the radius. If you draw a tangent here, that would look like this.

[02:55:05]That means this line and this line, they're perpendicular to each other. So the direction of the circle and the direction of this line, they are perpendicular to each other. So we're kind of thinking of the perpendicular distance, right? that that's where the it comes from.

[02:55:18]Okay. So that's the biote equation of circle. If you have any questions now, I will answer the questions quickly and then we have a short break. I'm sorry for uh making this too long for you. I I'm I'm very aware that your generation unfortunately has very short attention spans and I've made you sit here for like so long. It's the same with with us as well by the way by now. But uh yeah, I'm sorry about that. We don't have a choice. Unfortunately, we'll have to extend it. So, we have the recording available as well. If you if you uh are not able to focus, then you can you know keep attending and try to understand as much as you're able to but then you can know the recording as well.

[02:56:09]Okay. All right. Uh it's the normal to the center. Uh normal to the center. Yeah, you can normal not not normal to the circle. Yes, you can say that. Normal to the circle that at that point no irre irrelevant please. We're already late. We don't want to indulge in that.

[02:56:32]Uh okay, let's move on.

[02:56:36]Discriminant. Okay, we already answered that. Betman formula. Yesum.

[02:56:41]Uh, so I'm a I answered that as well. I think that if it does not have a chord, we can make a chord ourselves, right?

[02:56:50]Uh, Abdul Hadi, did you understand that when I explained that? I think that was right here.

[02:57:03]Quadratics. Quadratics. We'll just do quadratics and that's it for today.

[02:57:08]Then the rest tomorrow where where I will again make you sit for long hours.

[02:57:18]Okay, good. All right. So let's have recording uh for today's session. It will not be posted uh right now after tomorrow's class. by tomorrow night in by tomorrow night.

[02:57:37]Tonight is going to be difficult. I will try if I can do that. I will try if I can do that.

[02:57:44]Okay. What I what what I what I do is I will I will make the recording available. I will make the recording available. You need to have your email addresses. Notes I'll share right away.

[02:57:53]You need to have your email addresses working. the ones that you have in the uh registration form, they need to be working and you will get access to that.

[02:58:13]In fact, for now, I will just give you access and then the email will have to be working after tomorrow at least.

[02:58:20]I might not restrict that for now for you.

[02:58:26]All right. So I think we're done with the question file.

[02:58:32]Okay. This point. So we basically had to find the center of the circle. We learned above that center of a circle.

[02:58:44]It lies on perpendicular bis sector of every chord. So any chord that you have in a circle for example you have chord AB.

[02:58:52]this chord AB. If you draw a perpendicular bis sector of that, it has to pass through the center. You have another chord BC, if you draw a perpendicular bis sector of that, that also has to pass through the center.

[02:59:03]Now, both of these lines are passing through the center.

[02:59:07]So, that means their point of intersection is going to be the be the center. So, you just solve them simultaneously and you get the the coordinates of the center from that.

[02:59:17]Does that make sense?

[02:59:19]All right. So uh I need some break now uh maybe around uh 13 minutes. Let's resume around uh 9:30 in sh Okay everyone. So let's begin with this quadratics. This is the next one that we're doing now. Uh let's begin with this. So first of all there are some things there some concepts that you already should know from O levels. Uh some very basic stuff like solving quadratic equations. So you can solve quadratic equations using factorization using uh that's that's the middle term breaking thing right or the quadratic formula solving equations using completing square that's not necessary for you to know but quadratic formula and factorization you should know both of these methods so you I think you already know how factorization works and I not going to spend time spend time on that for instance in this case 2 into minus4 that gives you minus 8. If you think you think about the factors of eight that when subtracted give you seven somehow.

[03:00:17]So in this case that's 8 and minus one and that gives you seven. You break the middle term like this. This middle term basically gets split into two parts. So we break the middle term and then we factoriize whatever we can from the first two terms and from the second two terms and we get the result from that.

[03:00:33]Now at this point this is something that you must understand. You have two things multiplying. When you have two things multiplying and the result is equal to Z, that means necessarily either the first thing is equal to Z or the second thing is equal to Z. When two when you have a product, the result can only be zero when at least one of those expressions is equal to Z. Right? So either the first is 0 or the second is zero. Compare that with an equation like this 2x - 1 into x + 4= 2. Now this is not true. This is not correct. If you do it like this that you say either 2x - 1 is equal to 2 or x + 4 is equal to that would be incorrect. that not that would not make sense. It only makes sense when you have zero on the right side because two could come from any other uh combination as well. It's not necessary that if the product is equal to two then 2x - 1 is equal to 2 or x + 4 is equal to 2. Not necessarily because we could have two when we multiply for instance 0.5 and 4 that also gives us two. It's not necessary that any of them has to be equal to two. When it's zero on the other side then it's necessary. You can't get a product of zero when at least one of those numbers is equal to zero. Right? So that's why we put each expression equal to zero in those questions uh separately like this. How do you solve an equation like this? You would have to simplify this. Take two to the to the other side solve it as a normal quadratic and that would work.

[03:01:51]All right. Now when you are factorizing be careful. Sometimes what happens is when you take wrong factors inside the bracket you end up getting uh different signs. Don't forcefully make them the same that okay let's just make it make the second one x plus2 as well. Don't do that. That that basically is a sign that you're doing something wrong. If the if the uh expressions inside bracket they turn out to be uh different. Okay. So how do you solve this equation? Then if factorization does not work you have another method that's quadratic formula.

[03:02:22]You take everything to one side and then if something cannot be factorized in this case for instance you cannot factoriize this you use the quadratic formula. This is what the quadratic formula looks like. It's given in the formula sheet as well. Minus b + minus square root of these commands 4 a c divided by 2 a. You know how this works.

[03:02:38]The first term is a. This is b. This is c. Just input values and you get the result. Now when you're solving equation using uh when you're solving any quadratic equations, it's necessary to show the working for that. You cannot write down answers directly. You will not get marks for that. You will not even get marks for the answer even if that's right if you're if you're not showing the method. Okay? So showing the method is important. If you're using the quadratic formula, at least show at least show this particular step where you input the values in the quadratic formula. Then you can write down the answer after that. But if you write down the answer without showing this step, you get zero marks for that part. Okay?

[03:03:13]You don't you don't get any marks for that. So make sure that you show at least this step when you are solving using quadratic formula and at least the factoriz factorization part at least uh this step when you're solving something using factorization. Okay, this is important. Okay, so that's how we solve quadratic equations. Now, what what exactly do these solutions represent?

[03:03:34]So, we solved this equation 2x2 + 7 x - 6 equal to 0. We got two values of x. x= 712x equals this. What do these values represent? These values represent the x intercept. These values represent x intercept of x intercepts of this function. Right?

[03:03:51]The values where the graph of this function, the function on the left side that cuts the x-axis. Why is that? is this graph that I have this graph is y equals this right y equals this when we input y equal to 0 we get this on the right side now when is y is zero when y is zero we have the x-axis right that's the equation for for for the x-axis y= 0 so when we have an equation like this when we solve this the solutions of that equation that represent the x intercepts of this function that is there on the left side. Okay. Now you need to know how to sketch graphs of quadratic functions. There are two shapes that you can have. You can either have a U shape or an N shape. The first shape or the second shape smiley face and side face.

[03:04:41]However you want to remember that. How do you get each of these shapes? If the coefficient of X squ is positive, if it's positive, you get the first shape.

[03:04:51]If the coefficient of x square is negative, you get this second shape, the n shape. Okay. Now, there are some properties of the of this uh graph that you need to know about. It's it's called parabola. You should know know that name.

[03:05:07]It's symmetric about the center of the curve. So, if you if you draw a line here like this, it's the line of reflection of this curve. So, the graph is symmetric about this vertical line.

[03:05:18]And this point here, this is the turning point of this line. We can call this the minimum point in the first case. And this is the maximum point in the second case. There are multiple names that you have for this point that again you should be familiar with. You can call this point uh turning point. You can call this vertex and in different in differentiation later on you will also be calling the same point stationary point. Right? So these are different names of the same point that you should know about. Now on one side of this quadratic function in one on one side of this quadratic function the function is going to be decreasing and on the other side the function is going to be increasing. For example in the first case if it's a U-shaped graph on the left side you see the graph is going in the downward direction as you go from left to right the graph is going downwards. So if you think about the gradient of this graph gradient of this graph is negative. If the gradient is negative we call this a decreasing function. And if the gradient is positive, we call this an increasing function. Right? Again, we'll do this in differentiation later on as well.

[03:06:18]Sometimes you get this in quadratics too. Uh on the right side, this is an increasing function. Okay? For the other curve, on the left side, the graph is going upwards. So gradient is positive.

[03:06:29]On this side, the curve is increasing.

[03:06:31]On the right side, when it's going down, the curve is decreasing because the gradient is negative on that side.

[03:06:39]So you need to know about these properties of these graphs. Also uh this constant that you have here C that is the y intercept of this graph. That's true for all polomials for cubic as well. Whatever is the constant value that's the y intercept of the graph. How do you find turning point of a quadratic function? If you just need to find the turning point and there's nothing else that you that you need uh not you don't need a specific form as such in that case what you do is you say let's use the formula x= minus b /2. This is the formula that you have to find the xcoordinate of the turning point of a quadratic function x= - b / 2 a. If you say this is a and this is b, you can use this formula minus b over 2 a and that gives you the xcoordinate of the turning point of this function. Okay. How do you find the corresponding y-coordinate? You put this value of x. For example, in this case, we got this. Put this value back in the fun value of x back in the function. That gives the corresponding y-coordinate. And that's the turning point of this function. Okay? It comes from the quadratic formula. If you forget this, just think of the quadratic formula. In the quadratic formula, just get rid of this part.

[03:07:48]And that's minus b / 2 a. That's actually where this comes from. That's the xcoordinate of the turning point of a quadratic function. Right? That's the quickest way of finding this. Apart from differentiation that that also works the same way similar way as in the sim same number of steps. Uh that's how you can do that. Now there's another method that you need to know. Uh that's completing the square. Sometimes they will specifically ask ask you to complete the square for a function. Now how will that work? Now this is the standard method that most of you would probably be familiar with for completing the square.

[03:08:22]Let me go through this first. So one possibility is if you have a function that looks like this and you want to complete the square for this. Complete the square means you want something that looks like this. A into x - h square + k. Now the signs could be different here. It could be plus here. It could be minus here. It could be plus here. It could be minus here. They could use different letters here. They could use you know b and c here in place of h and k. But that's roughly what the form is going to look like. That's that's in fact the standard form for uh the complete this completed square form. Now how do you convert this function to this this form? That's that's the question.

[03:08:56]Right now this is one way to do it.

[03:09:00]Whatever is the coefficient of x square you take that common outside. You have to make that coefficient one. First of all you take that common outside. So you take two common and then what you do is the coefficient of x after you've taken two common outside you divide this number by two 7 /2 divided by two what does that give you that gives you 7 / 4 now you will square that number you square that square that number and both add that inside the bracket and also subtract that inside the bracket okay so this is a fraction that makes it a bit more complicated. If it was for instance inside if it was 8x what you would do would be you would say let's add 4 squ and also subtract 4 square inside brack and that will give the perfect square but a lot of times you get fractions like this as well. You should know how to do that. So yeah if you have 7 /2 you divide that by two square that and you get you add that number here and also subtract that number inside the bracket.

[03:10:04]And now these first three terms they give you a perfect square. From from the first term you take just x from the second term you take this sign and from the third term you take this number that is inside and that gives you this perfect square x + 7 over 4^ 2 and then outside you've got this left you simplify this. Then at the end what you do is you multiply this number two inside the bracket and that uh gives you the final result in that completed square form. This actually comes from these identities. Uh they are they do they I have them here anywhere?

[03:10:40]No. So this process basically comes from those identities that you have a square + 2 a b + b 2 that is a + b square and a square - 2 a + b square that's a minus b square right that that's what it comes from that's the logic behind it but that's what you need to remember. Now if you want to use this method method that's what you do you uh take the coefficient of x for gamma outside and then you uh divide by divide the coefficient of x by two square that both add and subtract that inside the bracket and then you get a perfect square from these first three terms and then you simplify everything and you get that result. Now this is the standard form that we've got. A into x - h² + k right.

[03:11:24]Now in this form h and k are the coordinates of the turning point of the function. So in this case for instance if you have got this completed square form this form here then what's the turning point from this one you would compare with the standard form. In the standard form you've got minus h. You'll need to remember this.

[03:11:44]You should remember this form. X - H square + K. Compare these these numbers.

[03:11:51]This number and this number minus H= 7 over 4. That gives you H= - 7 over4.

[03:11:56]Compare the numbers outside. That's - 818 and plus K. The value of K turns out to be - 818. If it's the standard form like this, all you have to do is change the sign of this number. That becomes - 7 over4. Keep this sign the same. That's - 818. And that gives you that turning point. So using this completed square form what you can do is you can find the turning points as well. However, if you just want to find the turning point and they have not asked you for a specific form in that case minus b /2 is much quicker. But if you have already converted some something to this form then you can write down the write down the turning point directly from this form. Flip the sign and this remains the same. These are going to be the coordinates of the turning point if it is the standard form. If it is if it is the standard form. This is one way of completing the square.

[03:12:43]Another way of completing the square is using this method which I think is uh much easier much quicker when you have more complicated expression like like this for example when you have fractions in these cases this other method is much quicker but since a lot of students have learned that first method at whole levels uh they don't trust me on this for some reason uh and they still stick to that so a lot of people they find you know learning new things difficult But think about this. If you had that same expression 2x2 + 7 x - 4, you want to convert this to the standard form a a into x - h the completed square form. Right? What you can do is you can mix up those two methods that we that we have learned.

[03:13:29]Remember what the standard form is? The standard form of a of a completed completed square function. It's a into x - h square + k a standard completed square form. Now in this form you know that h and k are the turning points the coordinates of the turning point. If you remember this form what you can do is you can find the turning point using minus b / 2 a. So what I've done is I've used the formula minus b / 2 a to find the turning point. It turns out that the x coordinate of the turn coordinate coordinate of the turning point that is - 7 over4 and the ycoordinate is - 818. So now we've got the turning point. Once we have the turning point, we know this is the value for h. This is the value for k. Just input those values here. A is going to be the same as this. So a was two here.

[03:14:24]That remains the same. A is two. And then you just input the values of h and k there and you get that complete completed square square form. And that's it. All right. That's another alternative that you have for this form for converting something to that form that's that you can also use. Now you can use whatever you prefer. This is another possibility. However, all right.

[03:14:43]Okay. Now, if you have the completed square form, if you have the completed square form from that, how do you write down the coordinates of the turning point? We discussed that you flip the sign in the middle. For the x coordinate, you flip the sign. It becomes minus 3. For the y coordinate, you just keep the sign.

[03:15:00]That's - 8. Okay. And you can look at it for the rest of these expressions. If you have x + 11 here and this is four, what is this? What does this become? - 11 and 4.

[03:15:13]Whether it's a maximum or a minimum is decided by the number outside. If this is negative, then that's a maximum because we know that's going to be an n-shaped curve.

[03:15:22]If there's a positive number outside, then that means it's going to be a minimum because that shape is going to be a U-shaped curve like this. Now sometimes you'll end up getting something like this 2 into x - 4 square.

[03:15:34]What about this? What are the turning?

[03:15:36]What's the turning point of this?

[03:15:37]There's no number outside. Since there's no number outside, we assume that that's zero. So the x coordinate is four and the ycoordinate is 0.

[03:15:46]If you have this 5 into x²us 3, what about this? What's the turning point of this function?

[03:15:53]You can think of it like this. You can say this is 5 into x - 0^ 2 - 3. The number inside is 0. So the x coordinate of the turning point that's going to be 0. And the ycoordinate is the number outside that's -3. That's the turning point. Okay? And sometimes you get a slightly different completed square form. That's not the standard one. For example, you'll get something like this 4x - 8^ 2 + 7. Now what's this? What's the turning point of this function? For this function, we are not just going to say that the turning point is 8 and seven. That's not correct because this is not standard completed square form.

[03:16:31]We've got this four inside as well. In the standard form, we can't have any coefficient of x here. So, in this case, the x coordinate is not going to be seven. The x coordinate is going to be found by equating this thing to 0. 4x - 8= 0. And that gives us x= 8 / 4, which is 2. The x coordinate is going to be two here. and the y-coordinate is going to be seven. So the turning point is actually going to be 2 and 7. That's going to be the turning point. 2 and 7.

[03:16:59]Okay. So the general rule is whatever is inside you put that equal to zero. That gives you the x coordinate of the turning point. And the number outside that's going to be the ycoordinate. That works in this case as well. For example, for x + 11, if you put that equal to 0, you get x x= -1. So that's the x coordinate. Right? If you have 4x - 8, you put this equal to 0, you get x= 2. That's the x coordinate coordinate of the turning point. For the ycoordinate, it remains the same as the number outside that remains seven. Okay?

[03:17:29]So that's how you complete the square.

[03:17:30]Now you've got the concept of discriminant.

[03:17:33]So for any quadratic function we have something that we call discriminant b square - 4 a c b square - 4 a c you seeing something okay we'll discuss that that that's coming okay x - 7 x + 12 that let's say we got a function like this If you solve this equation, how many roots do you get from this? You get two different roots. If you solve and now this equation, you solve two equal roots. If you solve another equation, you get no real roots.

[03:18:13]What's the difference between all of these three equations? I had three I got three equations here. These are these are the three functions that that they are the blue one, the green one, green one, and the orange one. I solved both all of them, and it turned out that one of them gave me two different solutions.

[03:18:27]One of one of them gave me two equal solutions and the other one the last one gave no real solutions. What's different in all of them? Look at this working in one of them inside the quad inside the square root I'm getting a positive number. In the second I get I'm getting zero. When I get have when I have a positive number I have I get one value with plus that and another value with minus that. So I get two different roots in this case. when I have 0 inside the square root I get 6 plus 0 and 6 - 0 that does not change anything both these points turn out to be the same when I have negative inside that does not give us anything but the square root of a negative number that is not real so we don't get any solution from that so this number that's inside the square root that's very important what's that number that's basically this discriminant b squareus 4 a c inside the square root you've got this thing called b square this thing b square - 4 a c we call this discriminant that is basically what decides how many root how many roots you're going to get for a particular quadratic equation. Now there are three different cases in fact four different cases three base cases and then one is derived from them.

[03:19:34]One possibility is in the first case for instance when the number inside is positive you get two different solutions. So when b square - 4 a c is greater than 0 you get two real and distinct or different roots or solutions. when b square minus 4 a c is equal to zero you get two real and equal solutions or you can also say that's this one real root because it's the same thing uh coming twice and the last case is that you have no real roots when when does that happen when b square - 4 a c the part inside the square root that's less than zero in that case you get no real roots now graphically what does that mean graphically what that means is the function that we have and the line that we have on the other side in this in this case the line was the x-axis.

[03:20:21]Sometimes it could be some other line as well. The graphically that means that the curve and the line in the first case they're meeting at two different points.

[03:20:30]In the second case they're only touching at one point. The line is a tangent to the curve and in the third case they're not meeting at all. So you should know you should understand both the graphical interpretations and the interpretation using the roots. When m squareus 4 a c is greater than zero graphically the line and the curve they intersect at two different points. When b square - 4 a is equal to zero. They often say it like this. The line is a tangent to the curve or we can say the line only touches the curve or only meets the curve at one point. And then finally in the last case b square - 4 is less than zero. The line and the curve they do not meet each other. Right? So whenever you have any of these statements written in the questions understand what that what that's referring to. At two points means this is greater than zero. It's tangent to the curve. it's equal to 0. They don't meet b square - 4 a c is less than 0. Now this is only for quadratic equations. Right? When you get a quadratic equation after equating the functions, it's only for that case. If you're not getting a quadratic equation, then b square - 4 c is not relevant.

[03:21:34]Then you use differentiation.

[03:21:36]That's not what we're discussing right now. But then we have another case.

[03:21:39]Sometimes what they say is two functions they have real roots or they might say the line and the curve they meet each other without specifying at how many points.

[03:21:52]So if they have real roots they could have you know two different roots. They could have two equal roots, one root, two roots. We we're not sure about that.

[03:22:00]If they say the line and the curve they meet each other now at how many points do they meet? Do they meet once? Do they meet twice? We're not sure about that.

[03:22:08]When they say any of these statements when it's where it's not clear whether it's uh at two points or one point you basically combine the first two cases you say in this case b squareus 4 a c is greater than or equal to zero that's the combination of the first two cases because it could be any of those two right if the line of the curve then meet that means they could meet at one point or two points so we combine these b - 4 a is greater than or equal to zero in that case so these are three base cases that you need to know about and then one case that's derived from the first two cases.

[03:22:44]Okay. Now let's have a look at this example. So we did complete completing square above. Sometimes they give you they ask you to convert something to a form that is not the standard completed square form. For example, they might ask ask you to convert this expression to this form 2x + a whole square + b. Now this is not standard form. In this case you've got 2x inside. In the standard form, you cannot have any number inside the bracket multiplying with x. In this case, they have that they have that.

[03:23:14]Now, how do we deal with problems like these? One way to do this is that you complete the square the normal way.

[03:23:20]Whichever way you prefer, you can do it that way. Right? I've done this using minus b over 2 a method here. You can do it using the other method. Eventually, what happens is you get this standard form. Right? This is something that everyone can do.

[03:23:34]saying you get to this form but now they don't want in the form that they've given they don't want any number outside the bracket they have a number inside the bracket now what do we do for that the number that's outside four we take that inside but we can't take that inside as it is because this bracket has power two and this four only has power one they have different powers in order to in order to multiply four inside we need to make the power the same for the number outside as well how would that work as 2^ And now since the powers are the same that means you can multiply the multiply the bases. So it becomes 2x - 3^ 2 minus sign and that is the form that they wanted there. This is one way to do this that you do it using the standard way and then the number that's outside you write that in square form and then multiply the base inside and that gives you the result. Another possibility is using comparison method.

[03:24:25]This is also something that can be used for uh normal completing square as well.

[03:24:30]Uh it'll work there as well. So what what they did was they gave us this expression and they wanted us to confir convert it to this form. We can compare them. We can say okay this is the original form 4x - 12x and this is the form that they want us to convert it to.

[03:24:48]We can say let's simplify this thing. 2x + a square + b. What does that give us?

[03:24:51]That gives us 4x^2 + 4 a x + a 2 + b.

[03:24:57]And now we could compare the two. We can say the coefficient of x square is four.

[03:25:02]Here this is also four. That does not give us any any interesting anything interesting. Uh the coefficient of x here is -12. This is 4 a. You compare them. The value of a turn will be minus 3. And the constants there's no constant here. This is a square + b. We compare them and uh that gives us a square + b equ= 0. So we get two equations. 4 a= -12. gives us the value for a a square + b equ= 0 and from that we can find the value for b and then once we have those values the value for a and value of b all you have to do is input those values in that stand in that form that they gave and that gives us the final form all right so that's what we can do as well okay let's move on now uh I think around 15 20 more minutes probably let's let's see there is some still some stuff to to to go because We have some difficult questions coming up that might take a bit time.

[03:26:01]Okay. Now, let's look at another example. Let's say we've got a curve that looks like this. y = x^2 - 3x + 4.

[03:26:09]We need to show that the whole of the curve lies above the x-axis. How do we do problems like these? The whole of this function lies above the x-axis.

[03:26:18]Now, what's the shape of this graph?

[03:26:19]It's a U-shaped graph. The coefficient coefficient of x² is positive. Since it's a U-shaped graph, since it's a U-shaped graph, what I do is I just get rid of that part. Yeah. Okay. If it's a U-shaped graph, that means it looks like this.

[03:26:36]Right? Now, if the whole of this curve has to lie above the x-axis, then that means the turning point here, its y-coordinate should be positive. So, how do we prove this? We find the turning point of this function.

[03:26:50]What's the turning point of the function going to be?

[03:26:57]This is the function y = x - 3x + 4.

[03:27:00]Find the turning point minus b / 2 a.

[03:27:02]You get the x coordinate 1.5. The y coordinate is 1 1.75.

[03:27:07]And the ycoordinate turns out to be 1.75. Since that's positive, that means it's above the x-axis. It's a minimum point. The minimum point is above the x-axis. If the minimum point is above the x axis that means the whole curve therefore is going to be above the x-axis right that's one way to prove this another way to prove this could be if the whole curve lies above the x-axis then you could also say that the discriminant is going to be less than zero right so you find b square - 4 s for this that turns out to be negative but that's not enough proof that's not sufficient proof if you just say b square - 4 s is less than z it could be two possibilities it could either be this form or it could be this that the whole curve lies above the xax is if it's a U-shaped curve or it lies below the x-axis if it's an nshaped curve b - 4 a less than 0 could mean either of these. So if you use b square - 4 a method in that case you would say b square - 4 a c is less than zero but at the same time it is a u-shaped curve. So it's this case not this case and these things both of these things together they mean that the whole whole curve lies above the x-axis. That's another method that you can use for this as well. All right.

[03:28:16]So either show that the minimum point point lies above the x-axis or find the discriminant that's negative and also say that it's a U-shaped curve. That is why the whole curve lies above the x-axis. That's what you have to prove there. Second part that's also another important one. Find the set of values of x for which this is the decreasing function of x. Now we talked about increasing and decreasing function before. It's a U-shaped curve. If it's a U-shaped curve, we know this part here is decreasing. The part that has a negative gradient that part is decreasing. Now what's that part? What are the values of x in this part we found the turning point if we did if you did it using the turning point method in the first part you can just use that if not you would have to find the you you would have to find the turning point in this part. Now the turning point was 1.5 and uh 1.75. They wanted us to find the set of values of x for which this is a decreasing function.

[03:29:08]So decreasing function is on the left side of the turning point. What what are the corresponding values of x that you get from this to the left of 1.5? That's x less than 1.5.

[03:29:18]That's the solution that you have there.

[03:29:19]Okay?

[03:29:23]All right. If you have to find point of intersection between a line and a curve, you just solve them simultaneously.

[03:29:27]Right? Just like we do in coordinate geometry. This is a line. This is a curve. If you want to solve if you want to find the point of intersection of them, we solve them simultaneously. You can make y the subject in both of them and then equate them or make y the subject in the linear equation. Put that in the other equation and then you can solve it simultaneously like this. Okay.

[03:29:49]All right. Now let's look at this.

[03:29:54]We have a line that has this equation mx - 2. Okay. So I'm going to skip the first part. That's pretty straightforward. Let's say we have a line mus and they're saying find the two values of m for which the line is a tangent to the curve. This the line is a tangent of the curve. Does that remind you of something? We just discussed if a line is attention to the curve that means b square - 4 a c is supposed to be equal to zero. Right? If line is a tangent of the curve b square - 4 a supposed is supposed to be equal to z.

[03:30:23]So now we have a straight line which is mx - 2 m. have a curve and this line is a tangent to this curve. So that means they are only going to have one point where they meet each other. So b square - 4 a should be equal to zero. Now how do we use b square - 4 a c. The first thing we have to do for using b square - 4 a c is we first of all equate these two functions. We start the process of solving them simultaneously. Either do substitution or make the same variable the subject and equate both of them.

[03:30:54]Once you equate them, you get one single equation like this. Right? Using those two equations, when you start solving them simultaneously, you get you get one equation from that. You convert this equation to the standard form. Ax² + bx + c equal to 0. You bring all terms to one side and leave only zero on the other side. Okay. Once you've done that, the coefficient of x square, that is going to be one. That that's going to be the value for a. The coefficient of x that's going to be the value for b. So if you have multiple x terms or multiple x square terms, you take x square common to uh in this case you take x common and you see what the coefficient of x turns out to be. In this case that's minus into 4 + 10. That whole thing is the coefficient of uh x. So that's the value for b. And then similarly the constant is everything that does not have any x in it. So that's the constant. All right.

[03:31:46]Now this is the constant 5 + 2 m. Since the the line was a tangent of the curve, we put b square - 4 is equal to 0. Solve that and find the values of m that we get from that. Now sometimes what they ask you for for is to find those points at which the line is at tangent to the curve. Once you have found the values of m, now you know what that equation is.

[03:32:08]You know the equation of the line now.

[03:32:10]But it will be two possible cases, right? One possible case when the value for m is 2. In that case you have y = 2x - 4. The other possible case is when m is equal to minus2. In that case you get this equation. And now what you do is you solve these two equations simultaneously to get those points at which the line meets the curve.

[03:32:33]Okay.

[03:32:42]All right. Let's move on.

[03:32:46]Hidden in this guy's quadratic equation.

[03:32:48]Sometimes you have equations that are in this form such that they're not quadratic.

[03:32:54]However, you can convert them to a quadratic form. For example, you could have something like this x^ 4 - 7 x + 12 = 0. This is not a quadratic equation.

[03:33:02]This is a power four equation. Now, we don't have a meth general method that we have in our syllabus to uh solve quic equations. Power four equation. So, how do we solve this equation? Then we can think of think of it as a quadratic equation. It's a hidden hidden quadratic. How do we do that? And how do you figure out if it can be converted to this form? The power should be in the ratio to ratio 1. The power should be in the ratio 2 ratio 1. One power should be twice the other power. So in this case, one power is four, the other power is two.

[03:33:34]One is twice the other. What we do is we use a substitution. We replace the smaller power with another letter. So in this case we say let u = x^2 and then u ^2 will become x^ 4. Solve this equation. Put that substitution back at the end. In place of u you put x square back again and you get the solutions from that. Compare that with this equation. In this case uh substitution does not work because the powers that you've got there are three different powers right you've got x^ 4 x^2 x^ 1.

[03:34:06]So substitution does does not work here.

[03:34:08]That would not work. It's not a hidden quad value equation. If you had something like this 6x into 7 x raised to power 1 /2 + 2. Again in this case that works because you've got one power and the other power is twice that. So it does satisfy that condition. We replace the smaller power with another letter and then solve it as a quadratic and then at the end we put that substitution back. This is another example. You have these two equations and we have to find the points of intersection of uh these two uh curves. This is one curve. This is another curve. How do you find points of intersection? If you were to equate them, you would get an equation like this. X^ 2 over 3 - 1. That equals X^ 1 / 3 + 1. And that gives you X^ 2 over 3 - X^ 1 3 - 2 = 0. Now one power is twice the other. You can see that this is 1 over 3, this is 2 over 3. So what you do is you replace the smaller power with another letter. You say X, U is equal to X^ 1 over 3. And then u ^2 would be x raised ^ 1 uh 2 over 3.

[03:35:10]And this can be written as a quadratic equation. U square minus u - 2 = 0. You solve this quadratic equation at the end. Put that substitution back and you get the result. Next thing you have have to know about is solving quadratic inequalities. How do you solve quadratic inequalities? When you solve equation, what do you get from that? For example, if you solve this quadratic equation x - 7 x + 12 = 0. When you solve this equation, you get two answers from this.

[03:35:35]x= 3 and x= 4. They represent the x intercepts of this function, right? Why?

[03:35:43]Because we've got this function y and we're saying that is equal to zero. When is y equal to 0? y is equal to 0 on the x-axis. Right? So they represent these solutions represent the x intercepts of this function. But what if instead of an equation we have an inequality for example if you have something like this this less than zero. Now you cannot do it like this that you say let's just factoriize this and then say either x - 3 is less than 0 or x - 4 is less than 0. That's not how it works because if both numbers are negative the product is actually going to be positive right. So that it's not going to be uh it's not going to be negative it's going to be positive. So this logic does not work when you have an inequality. So how do you solve inequalities then?

[03:36:29]Whatever equation, whatever inequality it is that you have to solve, what you do is first of all you treat this as an equation. Treat this as an equation and find its x intercepts. So remove the inequality sign. Treat this as an equation. You know how to solve the equation. You can solve it like this.

[03:36:45]Find the values of x that you get from that. Now it's a U-shaped curve. Draw that U-shaped curve and think about it.

[03:36:51]If it's a less than sign, that means y is supposed to be less than zero. You're supposed to be below the x-axis. Which part of the of this graph is below the x-axis? That's this yellow part. What are the corresponding values of x for this part? 3 to 4, right? It's between 3 and four. This part of the graph uh is below the x-axis when x is from 3 to 4.

[03:37:14]So that's what that's what that's what your answer is going to be. You say x is between three and four like this.

[03:37:22]When you write this form, when you have middle region, you always write it like this. Smaller number on the left side, bigger number on the right side and you have less than signs, these signs in the middle. Okay? When whenever you have two-way inequality, inequality bounded bounded both sides, you always write it like this. Smaller on the left, bigger on the right, and the less and less than signs in the middle, middle, right?

[03:37:47]That's the convention that we have for uh writing down this inequality. Now if instead of less than sign you had a greater than sign there.

[03:37:58]What would you do in this case? When is y greater than zero? That's this part or this part. Right? This is where the y value is greater than zero. So what are the corresponding values of x that we get from this? Either to the left of three or to the right of four. Right?

[03:38:11]These are the corresponding values of x that we have. So that means x is either less than three or it's greater than four. Since in this question statement you had an equal to sign as well, you put equal to signs with the solution as well. All right.

[03:38:26]Okay. Uh so this is these are the inequalities that we get.

[03:38:30]This is the solution to this equation.

[03:38:32]Now if this inequality was like this, if it was if the coefficient of x was x square was negative, then you would say it's an nshared curve, right? So you would draw an nshared curve. So if it's a less than sign, what does that mean?

[03:38:45]Where is the where is the graph below the x-axis?

[03:38:48]It's to the left of minus 1 /2 and to the right of four. So you solve it as an equation. First of all, once you solve sol solve the equation, you find the values of x from that and then using the graph, you can write down the final inequalities. Another possib possible way of doing this could be that if you're not comfortable with with a negative sign here, you can just multiply the whole inequality by minus.

[03:39:07]First of all, that will change the inequality sign as well. that will make it positive because whenever you multiply or divide by a negative number the uh inequality sign also changes. So now we've got this inequality. Treat this as an equation. Solve this solve this equation. You get two x intercepts again. But now in this case since we have a greater than sign that means we're looking for the for for the part of the graph that is above the x-axis.

[03:39:28]The corresponding values of x that we get from that is to the right of four or to the left of minus 1 /2. Right? These these are the corresponding values of x that we get from this. So that's the solution. Now just an exam trick that I sure you would like. If you have this inequality and you don't like graphs and you often get this wrong, you can just use a calculator. How do you do that? You show this process. You show this process.

[03:39:53]This is necessary to show. Okay, you have to show this process where you solve it as an equation, right? Treat the inequality as an equation. Find the two x intercepts. Once you have found the two x intercepts, this is what most people get wrong. the final solution after that just put this on your calculator. So you should have a good calculator that that's that really helps with with the exam. So on this version uh I have this inequality solving function. I can just go there. The degree of this polomial is two. Let's put that there. Which which option do we select here? It's a less than sign, right? Less than zero. That's number two in this. input those coefficients minus 2 7 and four and you get that answer. So after you you're done with this part, you found these x intercepts. If you're not sure about the inequality signs, you get them get them from the calculator. X is less than - 1 /2 and x is greater than four and that's your final answer.

[03:40:52]All right?

[03:40:54]So that's it. That that gives you the result. So you don't have to show this graph. There's no there are no marks for showing this graph. you just write down the answer directly. Okay.

[03:41:06]All right.

[03:41:08]Okay. Let's move on. Now, that's pretty much all in terms of concept. Now, let's have a look at some uh tricky questions and then we end the equation x2 + px + q. Equation x2 + p x + q= 0 where p and q are constants. It has roots 3 and minus 3 and five. We need to find the values of p and q. How do you find values of P and Q in this case?

[03:41:33]This is the equation. We have to find the values of P and Q. We given two solutions of this equation. One solution is -3 and one solution is minus3 and the other solution is five. One idea could be that we say since these are solutions to this equation, we can input those solutions in this equation and the equation should satisfy and you get two simultaneous equations from that. You solve those two simultaneous equations and that gives you the result. That's one possibility.

[03:41:59]That's one idea that will work but it takes quite long. Okay.

[03:42:04]But a quick way of doing this could be it's only two marks, right? You have given the solutions - 3 and 5. Think about what happens when you solve a quadratic equation. If you have this x - 7 x + 12= 0. When you solve it, you get a form like this first factorized form and then you say either x - 4 is equal to 0 or x - 3 is equal to 0. and then you get the solution x= 4 and x= 3. If you had the solutions given and you wanted to go back to the original equation, how would you do that? You can just go backwards in this direction. You can say, okay, we've got x= 4, you've got x= 3.

[03:42:38]Let's go one step back. Bring four to the left side. Bring three to the left side. And that gives you this x - 4= 0.

[03:42:44]x - 3= 0. Go one step further back.

[03:42:47]Combine these factors. Multiply them.

[03:42:49]And then you can simplify everything to get back to that original equation.

[03:42:52]That's another idea that we could use here that will be much more efficient in problems like problems like these. So you are given that the solutions are minus3 and 5. Okay. Now if those are the solutions we can say okay x= -3 and x= 5. That gives us x + 3 = 0. x - 5 = 0.

[03:43:10]And now we can multiply these two factors x + 3 into x - 5 that equals 0.

[03:43:14]Simplify this and you get get the equation x^2 - 2x - 15. They told us that the equation is x2 + px + q= 0.

[03:43:22]Compare the two, you get the value of p that's minus2 and you've got the value of q that is min -5.

[03:43:28]That's another way to do this that's much simpler. Now there's a complication here that that you could have. For example, in this case, if they had said if they had said that the original function original function is of this form, it's not x2 + px + q. It's actually - 3x^2 + px + q = 0.

[03:43:48]and the solutions are the same. What happens in that case? How do you find the values of P and Q in that case? Now, when you get to this point, you are getting x^2 - 2x - 15. This is what you're getting. Now, you cannot compare these two equations because the coefficient coefficients of x square, they are different from each other. You cannot do comparison in this case. That would not work.

[03:44:10]In the previous case, they were both the same. So, we could compare them. Now, we can't. So, how do we compare them? We can only do this if we first from this form get to this form. How do we do that? We just multiply the whole equation by minus3. Multiply the whole equation by minus3. So that the coefficient of x square also becomes minus3. When that becomes minus3, these coefficients become the same. And now we can compare p and q. We can we can say the value of p turns out to be six and the value of q turns out to be 45 here. So whatever you know in those equations they those numbers should be equal. So in this case we knew the coefficients of x² so they had to be equal before we could compare these two equations.

[03:44:57]Right? So that's a more complicated example that you would see. All right.

[03:45:02]Now sometimes what they do is they don't tell you that the line of the curve they meet at meet at multiple points. uh they won't tell you that uh they the line is a tangent to the curve or they don't meet each other but they would want they would ask you to show something. So for example in this case look at this question it's a very common one in recent papers now and it's a bit complicated sometimes you've got the equation of a curve that looks like this and you've got the equation of a line that's this. They want us to show that for all values of m that means no matter what the value of m is the line intersects the curve at two distinct points. Now how do we show that the line intersects the curve at two distinct points. What does that mean? B squareus 4 a c is greater than zero. But is that given to us? That's not given. They're not telling us that the line intersects the curve at two distinct points. They want us to show that. So we are not going to start with this assumption that b square - 4 a c is greater than zero.

[03:46:03]We have to show this. So how will this work? We've got the equation of the curve. We've got the equation of the line. First of all we equate both of them.

[03:46:12]Simplify everything. Convert it to the standard form. A x square + b x + c equal to 0. Now from that form we find b square - 4 a c. Now we are not going to say that b square - 4 a c is greater than z. We're not going to say that.

[03:46:25]We're just going to we're just going to simplify b square - 4 a c. So we input the values of b, a and c. simplify b ^ square - 4 a c and that gives us this eventually 4 m ^ 2 - 32 m + 68.

[03:46:44]Okay, this is what we get. Now once we've got this form b - 4 is equal to 0. We have to prove that this is greater than zero. This is always greater than zero. No matter what value of m you put there, this is always greater than zero. How do you do that?

[03:47:06]This is an expression. This is a quadratic expression. It's a quadratic expression. Can you find the minimum point of this of this expression? Not the minimum point of the original function.

[03:47:19]Minimum point of this discriminant.

[03:47:22]Discriminant is a quadratic function here. How do you find minimum point of any quadratic expression?

[03:47:28]You can do that using minus b over 2 a.

[03:47:30]Right? So what's the xcoordinate of this of this turning point going to be minus b over 2 a. We use that. We get a equal to 4 from that. The x coordinate of the turning point that is going to be four.

[03:47:41]Put that back in this function and that gives the corresponding y-coordinate.

[03:47:44]What that means is this curve it's not the original curve it's the discriminant. This is b square minus 4 a c. We are graphing the discriminant that looks like this. The minimum value of the discriminant what's that value of the discriminant is four. That necessarily means it's always greater than zero.

[03:48:07]If it's always greater than zero that means that the line and the and the curve they meet they they intersect at two distinct points no matter what the value of m is going to be does that make sense this is one idea uh the other idea here after this point could be that you complete the square complete the square when you complete the square for this you would get a form like this 4 into m + 4 2 uh 4 into m - 4 2 + 4 when You complete the square for this you'll get something like this and then from this also you can say and this part is always positive or you can say this is always greater than equal to zero right because we have a we have four that's positive this is a square square can never be negative and this is plus four so that means the overall thing is always going to be greater than zero since it's always greater than zero that means the line and the curve they intersect at two different points Okay, that's another idea. You complete the square and argue based on that. What you need to remember is don't put greater than zero in the beginning. Start with just b square - 4 a c. Simplify that and then the expression that you get is somehow prove that that is always positive in this case. All right, another similar question like that. You can do that yourself. You've got this curve, you've got this line. Again they're saying show that show that for all values of K the curve and the line meet.

[03:49:38]What does that mean? Curve and the line meet. At how many points? We're not sure. They've not given us given us how many points they meet at. It could be one point. It could be two points.

[03:49:48]But they meet. Now that means B square - 4 A C is greater than equal to Z.

[03:49:53]Greater than or equal to Z. It could be both possibilities. But is that given to us? No, it's not given to us. We have to show this. How do we show that? We equate both of them. Find B square - 4 A C. Simplify that. And at the end the result that we get, we find the minimum value of that. If it's it's 0 or above, that means B square - 4 A C is greater than or equal to Z. Okay.

[03:50:19]And then finally, we have this function f of x that's defined for all real values of x.

[03:50:26]f of x is equal to x^2 - 6 x + c where c is a constant. It's given that f of x is greater than 2 for all values of x. What does that mean? f ofx is greater than 2 is greater than 2 for all values of x.

[03:50:46]Okay.

[03:50:50]Find the set of possible values of C.

[03:50:52]Now, how do we understand that?

[03:50:55]So, we've got this quadratic curve. It's a U-shaped curve. It looks like this.

[03:50:58]It's a U-shaped curve, right? It'll have some minimum point.

[03:51:04]Can we find that minimum point somehow?

[03:51:06]X - X + C minus B / 2. That gives us X X= 3. That's the X coordinate of the minimum point. Then we find the corresponding Y coordinate. That turns out to be C minus 9.

[03:51:19]So, we've got the turning points coordinates. The X coordinate is three.

[03:51:24]The ycoordinate is 7 is c minus 9. Now what's the range of this function?

[03:51:31]The minimum point is c minus 9. So the range of the function is going to be greater than c - 9. y is supposed to be greater than c - 9. That's the range of this function. Right?

[03:51:46]They have given us that y is greater than 2. We are getting y greater than c minus 9. Compare the two. What do you get from that?

[03:52:00]If if y is greater than 2, then from here we get c - 9, that should be greater than two. So think of it like like this.

[03:52:12]In fact, if they're saying f ofx is greater than two, then the ycoordinate here should be two because that's what the y value is greater than it's always greater than two, right? So, we say c - 9 that should be greater than two and c is greater than 11 from that. Does that make sense? This is one idea. This is one way to do this.

[03:52:41]This is going to make it much more complicated. Let's just stick to this.

[03:52:45]So, y greater than 2. That was what was given to us in the question that for all values of x y is greater than 2. This function is greater than 2. We are getting that the turning point of this function its ycoordinate is c - 9. So, c - 9.

[03:53:08]therefore has to be always greater than two. So we say C minus 9 is greater than 2 that gives us C greater than 11. All right.

[03:53:17]So that's another type of question that you will see in recent papers. Uh sometimes that's it. I hope that was helpful. If you have any questions, you can let me know now and you can finally go and sleep if you want. Uh although even though you're saying you're tired and all you are all going to be watching reals now watching YouTube shorts it's only while studying maths that you get tired ask her sorry that was not a valid question yes let me know if there are any questions Can we have two curves instead of a curve and a line?

[03:54:05]Uh, it's possible. You could have uh a curve like this. y = 2x^2 + 7x - 3.

[03:54:12]Another curve might be y = I don't know 4x^2 - 3. The idea remains the you just equate both of them. If after equating them you get a quadratic equation then yes that's possible right because after equating them b - 4 only works when after you equate both the functions when after you equate both the functions you get a quadratic equation so you could have any two functions if after equating them you get a quadratic equation then yes that is possible you can get that This last question uh it's the same idea that we that we did before. You'll equate both of them. 2x 2 - 3x + 1 = kx + k^ 2. Take all terms to one side.

[03:55:03]Convert it to ax 2 + bx + c form. Once you've done that, find the value of b square - 4 a c. Simplify that.

[03:55:13]Simplify that. the result that you get you have to show that that's greater than or equal to zero. Complete the square for that and that will help you help you understand that the one above. Okay, what about this? Do you have a specific question on this? So let me let me repeat what what I've done. So we had to show that b square - 4 a c is greater than zero. We had to show that right.

[03:55:41]So what we did was we equated both the functions. first of all and found b square - 4 a c. I think you understand until this point at least right we found the value of b square - 4 a c and we got this form and now we need to show now we need to show that this b ^ square - 4 a c is always greater than zero no matter what the value of m is it's always greater than zero now one idea that I explained was that we say it's a quadratic expression 4 m^ 2 - 332 m + 68 that's a quadratic expression m^2.

[03:56:18]You find the minimum value of this using minus b / 2 a minus b over 2 a. That gives you h equal to 4. That means the x coordinate that turns out to be four.

[03:56:27]The corresponding y-coordinate that you get when you input x= 4. Here you get the corresponding ycoordinate.

[03:56:34]So this value is the minimum value of this expression four.

[03:56:41]Right? So b square - 4 a c it minimum value is four. If its minimum value is four that means b square - 4 a c is always greater than zero and that's what you have to prove that v square - 4 a c is greater than zero.

[03:56:55]Okay.

[03:57:00]Yeah. So whatever value of m you have there since you have proved that the minimum value is four it's always going to be greater than that. If the minimum is four it can't go below four. So that that's greater than zero. Okay. If you if you find this confusing just complete the square complete the square and you'll get something like this 4 4 into mus 4^ 2 + 4 that is what you would get from this. Now rather than thinking of the minimum point of that curve because some people may find it confusing that why are we drawing the discriminant there? Just think of this expression.

[03:57:34]This is four. Four is a positive number.

[03:57:37]This is m - 4 square. That's a square. A square can never be negative. It has to be greater than or equal to 0. It cannot be negative. And then you have a plus4 here that's also a positive number. So overall the whole expression is going to be greater than 0. Right? Because we have plus4 even if this becomes 0 0 + 4 that becomes greater than 0. So all you have to write down is you say b ^ square - 4 a c is after this uh after converting it to completed square form you can just write b square - 4 a c is greater than zero. And that shows that the line in the curve they meet for all values of x.

[03:58:13]Is that all right?

[03:58:40]Abdulhadi we have not drawn this curve that's not the graph that you have drawn there this graph is for the discriminant this expression is discriminant the original curve and the line they meet at two points that will be like this curve is like this and the equation of line is like this they are meeting at two points.

[03:59:06]This is not the graph of the curve. This is the graph of discriminant. The discriminant is always greater than zero. If the discriminant is always greater than zero, that means that this curve and the line they always meet each other.

[03:59:21]Okay.

[03:59:25]Okay. So let's start with the next topic. This is functions. First of all, you need to understand what functions are about. what function notation is. So if you have any function that any equation that's of this type for example y= -2x + 5 there's another way to write this and that is you say this is f ofx = - 2x + 5 right that's just another way to write down this equation that was in terms of y earlier you can write that in terms of f ofx like this as well so what what does f of x mean f ofx means this thing is a function of x it's defined in terms of x you need to understand how to uh deal with this notation as well. For example, in this notation, when you say input x=0, you get y= 5 like this. In function notation, the corresponding notation would be that in place of x inside the bracket, you replace zero.

[04:00:17]That means you're inputting x equal to 0 inside this. And then the result that you get from that, this is the value for y. Five is the value for y. Similarly, if you want to input x= -3 in this, you have minus3 in place of the inside the bracket here. And the result that you get from that, that's the value for y, that's 11. All right? Now, it could be something other than numbers like this as well. So, for example, if you have a function that looks like this f ofx= 2x2 + 4x - 3 and you want to evaluate f of x cub, what does that mean? Wherever you've got x in this function, what you do is you replace that with x cub. So let's try doing that. When we replace x with x cub, this is what we get. Then we simplify that and the result is going to be the expression for f of x cub. Okay.

[04:01:07]Similarly, if you wanted to evaluate f of x - 1 in this case, all you have to do is just replace x in this function with x -1 and then you simplify that.

[04:01:16]You get the expression for f of x - one.

[04:01:19]Okay? So that's how function notation works. Now how exactly does a function work? If you have a function that looks like this f(x)= - 2x + 5, what that means is when you when you input x in this, x is the input that goes into the function and it the function gives you an output. The output that you get that's what we call f ofx or it's can it can also be represented by y. So how does a function work? You give you you give this function an input. That's normally the value for x. the function applies some operations and that for example in this case the operation the function is applying is that it's multiplying this value by minus2 and it's adding five to that it's adding five to that that these are two operations that the function function is applying and it gives you an output from that for example if you input minus4 in this minus2 into minus4 that will give that that will become 8 + 5 and you would get 13 as the result now the first major thing that we need to know about functions is the concept of domain and range of functions.

[04:02:21]Okay, I'm sorry. Just give me a second.

[04:02:23]I forgot to send the link to somebody who just joined.

[04:02:30]Just give me a second. It's a scarf.

[04:02:33]Reminder of the front.

[04:03:00]Okay. So the first major thing that you need to know about functions is domain and range of functions. Okay. Now what's the domain of a function? Domain is the set of values of x that that a function can take as an input. And what about the range of the function? Range is the set of values of y that you can get as an output from the function. So values of x they're known as the domain. Values of y we call them the range. Now about domain there are only a few things that you need to know about. And that is for instance if you have a function that looks like this f ofx = -4x + 6. Now does this function have any restrict restriction as far as the number of as far as the values of x that you can input in this is concerned.

[04:03:44]Is there any restriction? It's a straight line function. You can input any value for x in this and that will work. For example, you can put a negative value in this. You can put a positive value in this. You can put zero in this. There's nothing. There's no value for x at which this function is not defined. For a function like this, we often say the domain of the function is all real values. Now you should be famili familiar with this notation. If it says something like this, x belongs to r. What that means is x can be any real number. Right? So in in the straight line there's no restriction.

[04:04:17]The value for x can be anything. You can input any value of x in this. There's no restriction. It could be all real values.

[04:04:23]Compare that with this function f of x= 2 into roo<unk> of xus 3.

[04:04:28]Now you have a square root function here. In a square root function inside the square root you cannot have anything negative right. So whatever is inside the square root you want that thing to be positive. So or it could be zero as well but it can't be negative. So x - 3 that should be greater than or equal to 0. That means that the value for x should be greater than or equal to 3. So the domain of this function will have to be x greater than or equal to 3. Any value that's less than 3 will not work in this function. The function will break because you will end up getting a negative number inside the square root.

[04:05:00]So that's one thing that you need to know need to know about square root functions that inside the square root you can never have something negative.

[04:05:06]Similarly here in this function the part that is inside the square root that has to be greater than or equal to zero. So what's the corresponding domain that you get from that? X greater than or equal to 7.

[04:05:20]And the second thing that you need to know about domain is if you have x in the denominator. So if you have - 3 over 2x - 8 is this function defined for all possible values of x. Well pretty much all values but there's one exception and that is that the denominator cannot be equal to zero because division by 0 is not possible. If you end up getting zero zero in the denominator of a fraction that fraction becomes undefined. So how do you get 0 in the denominator here?

[04:05:49]Put 2x - 8= 0. What's the what's the value of x that you get from that? You get x= 4. That means that's the value at which this function becomes undefined.

[04:05:59]So what would you call what would you say about the domain of this function?

[04:06:02]You would say it's all real values.

[04:06:04]However, there's one exception that x cannot be equal to 4. So for reciprocal functions in which you have x in the denominator, the denominator's value that cannot be equal to zero. So whatever is the corresponding value of x that you get from that that value is not supposed to be equal to zero. Now these are just two restrictions that you need to know for the domain that inside the square root you want something greater than equal to 0 and in the denominator you you want something that is not equal to z. Apart from this they will always tell you what the domain of a function is. So these are two things that you may have to figure out yourself but generally they will tell you the domain of the function themselves if there is some practical consideration. So these are inherent limitations of these functions that square root cannot take negative value. You cannot have something uh you cannot have zero in the denominator. But there are some functions that may not have any inherent limitation on them in in terms of their u in terms of their mathematical evaluation. For example, if it's a linear function, there's no problem inputting. You can input any value for x in this. You can put a negative value, you can put a positive value. It can take any uh value of x and it would still be defined. However, there might be some practical considerations because of which there might be some restriction on x. For example, if x represents price, in that case, price generally cannot be negative. So, you would say that the value for x would have to be greater than zero. If there's any consideration like this, they will always tell you the domain of the function in the question themselves. So that's why mostly when you see a question on functions, they will tell you the domain and what you would have to do generally would be to find the range of that function. Now that's the major part of this topic. Finding the range of different functions. For domain, we only have those two cases.

[04:07:53]But but for the range of functions, we have a number of cases for which we need to a number of different types of functions for which we need to be able to find the range of the function. Let's start with those different types of functions. First of all, we've got linear functions, straight line functions. Let's suppose we've got a function that looks like this f ofx= 2x + 5. And the domain of this function is x greater than or= 3. Now if this is the domain of the function, what's the range of this function? Remember domain is the values of x, range is the values for y.

[04:08:24]If it's if it's x greater than or equal to three, what you can think about is you can think about the graph of this function. Now whenever when whenever you're thinking about range of functions for some specific kinds of functions for which you know how the graph would look like, you should always sketch the graph. Linear function and quadratic functions in particular, you should always first sketch the graph and from the graph try to think about what the range is going to look like. For example, in this case 2x + 5, that's a straight line. It's going to look like this. It's going to be upward sloping line like this from left to right. And now you consider the domain.

[04:09:00]It's x greater than or equal to 3. Now what you do is you find the y value at this point. When x is equal to 3, what's the corresponding value for y that you get? You're getting 11 from this. So this is the point that lies on the line.

[04:09:13]311. This is the point that li that lies on this line. And now you think about which part of this graph satisfies this condition. What's the condition that x is supposed to be greater than or equal to three. Now if you have three here, if you have three here, which side of this function is going to be valid? The right side. So look at the graph.

[04:09:36]This is the part of the function that is valid, the yellow part. So what's the range going to be? Range is the values of y that on the vertical axis. how much you can go up and how much you can go down. On the right side here in the upwards direction you can say the values of y are always greater than or equal to 11. So that's the range of this function. Now you might think that it's it was not really necessary to draw the graph here because we got the value for y that was 11 and the range turned out to be y greater than or equal to 11 because this was a greater than sign and this was a greater than sign. So we could just put that greater than sign without drawing the graph. But that's not going to work every single time. Why is that? Because think about this example f of x f ofx= - 3x + 8. Let's say we've got a function that looks like this. In this function, let's say the domain is given to be x greater than or equal to 4.

[04:10:34]How do you think about the range of this graph? Now you input that value of x in this. The value of x is four. The corresponding value of y that you get is -4. Now what's the range going to be?

[04:10:43]You might think this is greater than or equal to 4. So the range should be y greater than or equal to minus 4. But that's not correct. And that's something that you understand when you think of the graph of this function. It's a downward sloping line. It's - 3x + 8.

[04:10:57]What does the graph of this function look like? It's going to be going in the downward direction. From left to right, the graph goes in this direction like this.

[04:11:08]So what you do is you think about this point. The value for x at this point is four. What's the corresponding value for y? The corresponding value for y is minus4. So this is the point that we're talking about. Which part of this function is valid when x is greater than or equal to 4? Which side of the graph?

[04:11:26]That's the right side of the graph. So from that point, consider the right side of the graph.

[04:11:33]Look at the yellow part. What's the range of the function? It's going to be below minus4. Less than minus4. Remember range tells you about the values of y.

[04:11:44]Values of y means vertically how far can you go in the upwards direction and how far you can go in the downward direction. So in this case, what's the maximum value of the graph? What's the highest value of the graph? That's -4.

[04:11:56]Lowest value, it can go up to negative infinity. So the range of the graph is going to be y less than or equal to minus4. That's the range of the graph.

[04:12:04]You cannot figure that out if you do not think about the graph of this function.

[04:12:08]So thinking about graphs is really important in problems like these. Take another example. Let's say we've got a function like this. F ofx= - 5x + 4.

[04:12:17]What's the domain of this function?

[04:12:18]Domain of the function is from -2 to 3.

[04:12:21]Again, you can think of the graph. It's a it's it's a downward sloping line because the gradient is negative. So, it's going to look like this. That's the shape of the shape of the graph. It's a downward sloping line like this. From -2 to 3, this part of the function is valid. From -2 to + 3. So, we have a point on the graph. At -2, the value for y turns out to be 14. At x= 3 the value for y turns out to be minus 11. So what part of this function is valid? The part between these two points. So on the x-axis now you are saying any value between -2 and 3 that is valid. On the yaxis any value between -1 and 14 that is going to be valid. So what's the range of the graph? The range of the graph is from -1 to 14. But now again you've got to think about the inequality signs as well.

[04:13:18]Minus2 has an equal to sign with this.

[04:13:21]What's the corresponding value for for minus2? That's 14. The corresponding y value for for minus2 that's 14. So on the vertical axis on the vertical axis on the y-axis the range is also going to have an equal to sign with 14. The domain does not have an equal to sign with three there. So the corresponding value for three that's -1 minus 11 is not going to have the equal to sign here as well. Okay. So inequality signs you have to think about them whether you're going to have an equal sign or not. You think about the original x value. If that's included the corresponding y- value will also be included. If the original x value is not included then the corresponding y-value would also not be included. Okay.

[04:14:05]In a linear function if the domain is all real values what about the range?

[04:14:10]The range is also all real values because horizontally the function can go as far as possible on the right side and as far as possible on the left side.

[04:14:20]Vertically as well there's no no restriction. The graph can go up to positive infinity and in the downward direction it can go up to negative infinity. Right? So there's no restriction. There's no restriction in the uh vertical direction as well. You would say the range of this function is all real values. Now that's about linear functions. In linear functions, all we have to think about about is the end points of the domain. Think about what happens at three. Think about what happens at four. Find the corresponding y value and look at the graph and see what the range is going to look like based on that. Similarly, here the domain was from minus 2 to 3. We just looked at the end points and we find the corresponding y values, looked at the graph to figure out the range of the corresponding function.

[04:15:04]Now, the next type of function is when you have reciprocal functions. They're a bit confusing sometimes.

[04:15:10]These are functions when x are x is in the denominator. So there's any x in the denominator. This type of function is called a reciprocal function. For example, you could have something like this 4x - 2 - 8x2 + 1. We call these reciprocal functions because we've got x in the denominator in these functions.

[04:15:29]How do you find range of functions like this?

[04:15:33]Again, you think of those end points.

[04:15:35]However, the graphs of reciprocal functions are a bit confusing. Um, if you know them, you can draw them.

[04:15:42]However, we can do this without the graph as well. This is one function for which we will not be drawing any graphs ourselves. Sometimes they give the graph in the question themselves. We'll have a look at an example like that. But mostly they will not give you any graph for this. And for this since the graph is going to be slightly tricky, we would not look at graph. We will do it without that. Now, how exactly is this going to work? Let's say we want a function like this for reciprocal functions. If you have any function like this 3 over 2x + 1 and you're given the domain of the function that's x greater than 0. We think about the end points of the domain. If x is greater than zero, what does that mean? What's the minimum value for x? The minimum value for x is 0.

[04:16:20]What's the maximum value for x? It's greater than zero. So on the maximum side, it's basically going up to infinity. There's no limit as such. It can go as far as possible. So we consider the end points of x. We say okay one possible value for x is 0 another possible value of x is infinity.

[04:16:37]So what happens when x is zero? What happens when x is infinity? We think about those extreme points. So we input x equal to0 in this function. x=0 gives us three as the result. We we think about what happens if x becomes infinity. Now what happens if x becomes infinity? You have infinity in the denominator. Any number divided by infinity that is going to be extremely small. So it's going to approach zero.

[04:17:02]If you find that hard to think about, another idea about this could be that in your calculator just input that function 3 / 2 into x + 1 in place of x. Put a very large number and think about what the value looks like. For example, I can put 10^ 10 in place of x here. The value turns out to be 1.5 into 10^ - 10. 10^ - 10.

[04:17:27]That's an extremely small value. So that's basically telling you that the that this value is going to be appro approaching zero. It'll be very close to zero. Right? So you can think about this from a calculator as well. But just remember when you have infinity in the denom in the denominator the result is going to be extremely small and it's going to approach zero eventually. So now the end points that we're getting for the y value they are three and zero.

[04:17:53]So what do we say about the range of this function? We say the range of the function is from 0 to 3. Right? So 0 is the minimum value. Three is the maximum value. Now we're not putting equal to sign with any of them. Why is that?

[04:18:05]Because in the domain there was no equal to sign with zero. The corresponding value the corresponding y value for zero was three. So we don't put any equal to sign with three as well. What about zero? Why we not input not putting an equal to sign with this? Because when the denominator is infinity the value is very close to zero but it does not become equal to zero. So we don't put an equal to sign with 0 as well. We say the range is just between 0 and three like this. And that's our final result.

[04:18:36]Similarly if you have had an example like this 1 /x + 5 you want to find the range of this function. The domain of this function is x greater than zero.

[04:18:45]Now again how do you think about that?

[04:18:47]It's greater than zero. So what's the end points of the domain? The minimum value is zero. The maximum value is infinity. Again it can go from 0 to infinity. We think about what happens when x is equal to zero and we think about what happens when x is going towards infinity. Now the problem in this function is we cannot really input the value for x as zero here because that becomes undefined. So how exactly do we think about this? Since it's greater than zero, we can think about a number that's very close to zero. So if you input 1 / 0 in your calculator, that's going to be undefined, right?

[04:19:19]That will not give you any value. So the work around for that could be that you input a very small value 0 0 input a lot of zeros there and and think about what the result that you're getting that's 10 raised to power 23 or whatever how however many zeros that you have there depending on that you'll get a very large number here. So what that tells you is that when x is going toward and when x is close to zero here the value of this whole function that's extremely large so that goes towards infinity or another way to think about this is just remember when you have zero in the denominator the value becomes very large although strictly speaking in mathematical terms we call that we say that this is undefined. However, since this is slightly this is greater than zero, when you have a very small value, the overall result is going to become very large. So that's actually going to approach infinity. So that is what happens when x goes towards zero. The value for y that is going to go towards infinity. So that's one end point that we've got for the for the range. That's infinity. Then think about what happens when x goes towards infinity. when x goes towards infinity. When x is x is a very large number. Now if x is a very large number we get something like this 1 / infinity + 5. Now what's one one over infinity? 1 over an extremely large number that's going to become zero. But then we're adding five to that. Then we're adding five to that.

[04:20:47]So it becomes five and that's the result that we've got. So what's the range then? The range is from five to infinity. Okay. That's the range that we've got. Five to infinity. This is one way to write this that the range is from y 5 to infinity. Another way to write this could be that just y is greater than y is just greater than five because infinity is not a limit as such. It's just y greater than 5. All right.

[04:21:15]Okay. Let's have a look at this example.

[04:21:18]Let's have a look at this example. 1 / x - c².

[04:21:22]The domain is x is greater than or= 5.

[04:21:26]domain is x greater than or equal to 5.

[04:21:28]That's the domain of this function.

[04:21:30]Again, we think of the end points. We think of what happens when x is five and what happens when x goes towards infinity. Right? That's the these are the end points of the domain from 5 to infinity.

[04:21:42]Now, what happens at five? We can just input five in the function and that gives us 1 / 4. What happens at at infinity? When you have infinity in the denominator like this again what what happens is the number is going to approach zero. So you get zero from that. So the end points that you're getting for the range here are from 0 to 1 / 4. So that's the range that we that we write here. Now do we put an equal to sign with any of those numbers? How do we think about that? It's x greater than or equal to 5. So the corresponding value for 5 1 / 4 that is going to have an equal to sign here. But when the denominator is infinity in this case the value is going to be approximately equal to zero. It's not going to be exactly equal to zero. In that case in that case we don't put an equal to sign with this case. Okay. All right. Let's move on. So okay just a reminder just a reminder uh question I'll be taking your questions at the end. Okay. So we otherwise it'll take too long. It's already going to take too long. Uh like four four and a half hours today. Uh so just a reminder what we discussed yesterday that while I'm explaining this I'll be reading some of your questions. If I can incorporate them right there I will do that. Otherwise I'll be answering your questions once we're done with the topic. Okay? Otherwise it'll become too slow and we'll not be able to do many topics with it. Okay. Okay. All right.

[04:23:13]Now the function is defined as follows.

[04:23:14]So keep a note of your questions. You will be answering it. Okay. All right.

[04:23:19]The function f is defined as follows. f ofx is equal to root of x - one for x greater than 1. That's a function that we plot. Now find an expression for f inverse of x. Now that's fairly straightforward. Now let's think about this this part. It says the diagram shows the graph of y = g of x where g of x is equal to this. Now we've got a graph of this function. This is the reciprocal function.

[04:23:46]In this case, we think of the graph to state the range of the function. How do you do that? Look at the graph. Remember what the range represents.

[04:23:55]Range represents what's the maximum value going to be on the y- axis and what what the smallest value is going to be on the y- axis. What's the highest value on the y-axis on this graph? The highest value on the y-axis is 0.5.

[04:24:11]You can find that by inputting x=0 the y intercept of the graph. Put x equal to 0 here. That gives you 0.5. So you say the maximum value for y is 0.5. And what about the minimum value? You can see it's becoming an asmtote here on this side and also on this side. So on the lower side the value is getting very close to zero. You can think about what what would happen if x becomes very large here on this side. In the denominator you would end up getting infinity and the value is approaching zero. So what's the maximum value that you're getting on the graph? 0.5. The minimum value that's zero. So what's the range going to be? It's going to be from 0 to 0.5. We put an equal to sign with 0.5 there because 0.5 is actually part of the graph. But we do not put an equal to sign with zero there because it's not becoming equal to zero. It's only approaching zero on the x-axis. Okay, that's what the range of this function is going to be. So for functions, if you do not have a graph given, I would not recommend that you draw the graph. It's going to take time and it's a bit complicated.

[04:25:18]Just think of the end points. If you have the domain given, just think of the end points. In this case, for example, if x is greater than zero, 0 is one end point, infinity is another end point.

[04:25:28]Just check what happens at those points.

[04:25:32]If you have a graph given, in that case, you think of the graph itself. In that case, think of the graph itself. Think of what the maximum value on the graph is and think of what the minimum value on the graph is. And that will give you the range of that function. Okay? Now so in general when you're finding range of any function these are the important points that you have to consider end points of the domain we we've seen them in the previous examples that we've seen in straight line functions as well and also in uh the reciprocal functions we think think of the end points of the domain.

[04:26:06]However there are some functions that have turning points as well. For example in this reciprocal function that we that that we had we also had to think of this turning point. Now for reciprocal functions they will always give you the graphs uh in case you have to think of the turning point. If if they do not give you the graph then there would not be any turning point and you can just think of the end points. However, there are some functions for which you need to be able to think about the turning points uh yourself. For example, quadratic functions and trigonometric functions. For these two functions, you will have to consider these two sets of points. end points of the domain and the turning points of the function. These are two points that you will always have to consider to understand what the range of the function is going to look like.

[04:26:53]Next, let's have a look at some examples of quadratic functions. Now, let's say we've got a function like this 2x square - 12x + 24 and the domain is all real values. Now for a quadratic function if the domain is all real values the range is not going to be all real values because the quadratic functions look either U-shaped or n-shaped. So they will have some maximum value or some minimum value. So how do you find range of quadratic functions? You always have to find out the turning point of the quadratic function. We know how to do that minus p over2 differentiation completing square. You find the turning point using any method. Let's say in this case the turning point turns out to be 3 and six. Now since x could be any real value, it could be all real values.

[04:27:34]What that means is on this graph, you think about it, what's the lowest point on the graph? The lowest point of the graph is six.

[04:27:41]On the upper side, there is no limit.

[04:27:43]The value of y could be could could go up to infinity. So we say the range of this function is y greater than or equal to 6. That's the range of this function.

[04:27:54]Remember the range is on your graph.

[04:27:57]what's the maximum point and what's the what's the highest point and what's the lowest point that's the range the lowest point in this case is six there's no highest point so we say the range is y greater than or equal to 6 now if they restrict the domain so if it's if it's the same function but they restrict the domain like this they say that x is now less than five x is less than five now what does that mean how do you think about the range of this function now you have to think of the turning point and you have to think of this point five and consider both of them to figure out what the range is going to look like. Find the value for y when x is equal to 5.

[04:28:35]That turns out to be 14. Mark that point on your graph. Now, this is what the graph is going to look like. It's a U-shaped curve. You mark the turning point. Where is this point going to be? x= 5 is going to be on the right side of the turning point.

[04:28:49]That point is 5 and 14. That's this point. Now, which part of the function is valid? X less than five. That means we're looking at the part that is towards the left side of five.

[04:29:00]So all of this part of this graph is valid. Now in this yellow part, in this yellow part, what's the lowest point?

[04:29:08]The lowest point is six. What's the highest point on the left side? You can see the graph is going as far up up as possible. It's going up to infinity. So what's the range of the function then?

[04:29:18]It's y greater than or equal to 6 in this case. Take another example. If you had that same function but the domain was now x greater than or equal to 4.

[04:29:26]What about this? Now you think of the value for y when x= 4 that turns out to be 8. This is the point on the graph when x is equal to 4. What part of this graph is valid? To the right side of four. To the right side of four. This is the part of the graph that is valid.

[04:29:41]What's the corresponding part on the curve? I've shaded that here. So vertically now what's the range? The lowest point on the graph is eight. The highest point on the graph is infinity.

[04:29:53]So the range turns out to be y greater than or equal to 8. So for quadratic functions, you always have to sketch the graph. Mark the part of the graph that is valid for the given domain. And in that part, think about what's the lowest value, what's the lowest value on the graph and what's the highest value on the graph. And that will be the range of the function.

[04:30:15]Let's consider another example. If it was x less than zero, again you think of that point when when the value for x is 0, the y value turns out to be 24. It's the same turning point because it's the same expression as before. 36. Which side of the turning point is this point going to be? X=0. It's on the left side of the turning point. So we mark that point. Now which part of this graph is valid? To the left of zero. So what's the range? This is the highlighted part.

[04:30:43]This is the valid part of the graph. on the graph. Think about what's the lowest point. The lowest point on the on the shaded part on the valid part is 24. The highest point that's infinity. So the range is going to be greater than 24. We don't put an equal sign here in this case because the point zero is not included. So the corresponding value for y 24 that's also not included. We don't put an equal to sign in this case. Okay.

[04:31:06]Let's take one last example in this or the second last example. The r the domain is x between 0 and four. Okay. So now one of these points is on the left side of the turning point zero. The other point is on the right side of the turning point four. So this is one point. This is another point. You find the corresponding values of y at both of them. They turn out to be 24 and 8. And think about what part of this graph is valid. So this yellow part is valid between those two points between 0 and four. This is the part that is valid.

[04:31:38]Now what's the range going to be? Look at the graph. Look at what's the lowest point of the graph. The lowest point of the graph is six and the highest point of the graph is 24. So what's the range going to be? It's going to be between 6 and 24. Now since the point 6 is included because that's in the middle of that valid part 24 is also included.

[04:32:00]Is that included? No, 24 is not included because the value for x here zero that 24 corresponds to that was not included.

[04:32:09]So we don't put an equal to sign of 24 here. So that's the range. It's from 6 to 24. That's the range of this function. If you had something like this, if the domain is from x from minus1 to 2. Now which part of this function is valid? Again, think of the turning point. The turning point is here. Where are these two points going to be? Both of these x coordinates are less than the x coordinate of the turning point. So both of these points will be on the left side. So this is x = 2. This is x= -1. The value for x has to be between these two points. So now this is the valid part of the graph, the shaded part. What's the range? Think of the lowest point in this in this region and the highest point in this region. The lowest point in this region is 8 and the highest point in this region is 38. So that's the range of this function. It's from 8 to 38.

[04:32:58]That's what we write there. We don't we do not include 38 because the minus one was not included. So we don't put an equal to sign there. But there was an equal to sign with two here. So the corresponding value 8 that does have an equal to sign. Okay. So these are all the different cases that you would have in a quadratic function. You always have to sketch the graph, find the turning point, shade the part of the graph that is valid and in that shaded part, think of what's the lowest point on the graph and what's the highest point in the graph and that will give you the range of the function. Next we've got trigonometric functions. How do you find range of trigonometric functions? Let's say we've got a function like this. y = f ofx = 3 - sin x and the domain is from 0 to 360° and we want to find the range of this function. Now whenever a trigonometric function s and cos in particular whenever their cycle is completing okay yeah whenever their cycle is completing so from 0 to 360 sin x completes its one cycle for s and cos in particular the their values are between minus1 and one right the graph of the graph of s and cos both of them their lowest value values are minus one their highest values are + one.

[04:34:14]They're always between minus one and one. So then whenever they're completing a cycle, if you want to find the range of a function that includes one of these functions, all you have to do is consider the extreme values of s and cos. So in this case, we've got sine.

[04:34:28]What's what are the extreme values of s?

[04:34:31]The lowest value of sign could be minus one. The maximum value of sign could be + one. So once we input sin x equal to one in the function, that gives us one value. And then we input sin x equal to minus one in the function. That gives us another value. And we say the range of the function is going to be from 1 to 5.

[04:34:46]And that's our final answer. Take another example. Let's say we've got this function. Now this is tan on tan does that that rule does not work. So this function is actually there's a misprint here. It's 3 - 2 into tan of/ x. That's what the function is. You want to find the range of this function. Now tan does not have its values between minus1 and 1. tan can go from negative infinity to positive infinity. In this case, you just have to check the end points. Check what happens at zero.

[04:35:14]Check what happens at pi. And that is going to be the range of the function. 0 is giving you three. Pi is giving you negative infinity. So you say the range is going to be from minus infinity to 3.

[04:35:26]Or if if you say minus infinity to 3, that's the same thing as saying that it's just y less than or equal to 3. Now for s and cos functions you try one and minus one if the cycle is completing for tan you just check the end points check what happens at zero check what happens at pi and that will give you the result but if you have sin square or cos square what about that look at this example so let's say we have a function that looks like this y = 2 sin^ square x - 3 cos square x and the domain is from 0 to pi okay Now what about this? Whenever you have a square function, now if you have sine or cos, their values are between minus1 and one. But when you square something, you cannot have a negative value from that. So cos square is going to be between 0 and one. Right?

[04:36:21]So these negative values are also going to become positive.

[04:36:26]And so it's the values values for cos x are basically between minus1 and one. So these negative values when you square the function they also become positive. So now the square function only has values between 0 and one. So for sin square and cos square what you would do would be you would consider 0 and one at as the end points.

[04:36:51]But for finding range of these functions first of all you would have to convert them to the same function. For example in this original expression you have both sin square and cos square. In order to find the range of this function you either want to convert both of them to sin square or both of them to cos square. That's why they make you do that in the first part of this question already. They make you convert the function to cos square first of all. Now once you have done that in this expression now 2 - 5 cos square all you have to do is try the extreme possible values of cos square what were the extreme possible values of cos square they were zero and one input zero ones and then you input one the result that you get from that that will be the range you'll say minus 3 is the minimum value for y 2 is the maximum possible value for y and that's the range of this function this is one way of doing this although I now actually recommend using your calculator for this as well. And that makes it even easier. And that is basically this idea that you just put this function on your calculator. Just put this function on your calculator. It's the same for sin square. Yes, for cos square sin square just you just input 0 and one as the as the extreme uh points.

[04:38:08]But there's there's there's another way to do this and that is the way that we use to sketch functions. Use the table function.

[04:38:16]This uh this function is in degrees right now. So what we can do is we can select degrees from here. In the table function input this 3 - 2 sin x. Now the interval that's given is from 0 to 360.

[04:38:30]So start at 0 end at 360. And what about the step? I recommend whatever is the interval that's given you divide that by 8 generally. Right? So if this is the interval that's given 0 to 360 over 8 what does that give you? That gives you 45. Take that as the step and that will give you enough values to basically understand what the maximum value of the of the graph is going to be what the minimum value of graph is going to be. So step 45 just input this on your calculator and see what you get. This is the function that these are the values that you're getting. Just look at the table and see what the maximum value is and what the minimum value is. In these values, you see one is going to be the minimum value. There's no value that's less than one. And five is going to be the maximum y value. And that's what you write there. You write that the range is from 1 to 5. Similarly here again you can just input this function in your calculator directly and see what the highest value for y is in the table function and what the lowest value of y is in the table function and that should work in this case as well for the square function you could just input that directly like this you would say f ofx is equal to 2 sin^ square x sin of x² so in your calculator you have to write down sin square x in this form sin sin square x + sorry - 3 cos² x just input this function on your calculator and then what what's the what's the domain that you're given it's in radian first of all you have to convert the angle unit sorry let me go back There that's the table function 2 into sin of x² + 5 minus 3 into cos of x².

[04:40:45]You can input this form or you could also input this form. Both of them will work because it's the calculator doing it. So you won't have to worry about this. What's the starting value? It's zero. What's the end value? You're ending at pi. And what's the step going to be? Divide that by 8. Pi / 8. And see what that gives you. Look at whatever is the maximum value in this table. So minus 3 is one value that we're getting.

[04:41:08]That's that seems to be the least value.

[04:41:10]Keep going on. What's the maximum value?

[04:41:12]The maximum value that we're getting is two. The maximum value is two. Minimum is minus 3. Maximum is two. That's what you write there. That's the range of this function. So you can use the calc use your calculator to do it as well.

[04:41:30]Okay, that will also work.

[04:41:36]All right. So now sometimes what happens is they give you they give you the range of the function themselves. They give you the range of the function themselves.

[04:41:47]For example, in this function, they tell you that f of x is equal to a minus b cos x. And they're and they're saying that the maximum value of f of x is 10 and the minimum value is minus2. Now, how do you think about this? How do you find the values of a and b in this first part? We have to find the values of a and b. Now, for this, as we'll see later, you can think of the graph.

[04:42:07]That's one idea that the graph of the cost function it it has its maximum value at 10 and its minimum value is at minus2. So where is the baseline going to be? The baseline is going of the graph is going to be exactly in the middle. What's the what's the midpoint of minus2 and 10?

[04:42:26]That is four. That's the midpoint. Four is the midpoint. Now four is the baseline. So the number that's outside here that represents the baseline. As we'll see in function transformations as well, this represents the baseline. So you can write a equal to 4 there. And then this number b that represents the amplitude. What's the amplitude going to be? The maximum vertical distance from the baseline that is 6 units from 4 to 10 or from minus2 to 4. That's the value of v that you get. You say the value of v is going to be equal to 6. That's one idea. Thinking of the graph, think of the baseline and think of the uh amplitude of the graph. The other idea is this is the function that you've got a minus b cos x and you are given the range of the function again it's a cause function it's completing one full cycle you input cos x equal to cos x equal to 1 once and then you input cos x equal to minus one input those two values and see what you get from that one is giving us a minus b minus one is giving us a + b which of these values is bigger a plus b you're adding two numbers up so a plus b has to be bigger so we say that is going to be 10 and a minus b is the smaller value. We say that's going to be minus2 because they told us that the least value of the function is minus2 and the maximum value of the function is 10. So we say a minus b= -2 and a + b= 10. We get we get two simultaneous equations from that. Solve them and you get the result from that. Now this is one idea but thinking of transformations make it makes it much quicker. So when you have the range given like this, you can think of the graph, think of where the baseline is going to be, think of what the amplitude is going to be and based on that you write down the values of those unknowns.

[04:44:12]All right. So this is the next thing that you need to know about the different types of relations two variables can have with each other. So there there are four different types of relations that any two variables can have with each other. One to one, many to one, one to many and many to many.

[04:44:25]Now what's a onetoone relation? One to one relation means that for each value of x we've got only one value for y and for each value of y we've got only one value for x. So for for every x we've got every for each x we have only one y for each y we've got only one x. What kinds of functions are these? For example straight line like this you could have a curve like this. In this curve as well you have one value for x for when every one value of y and one value for y for each value of x. Right?

[04:44:53]The other type of function is many to one. In this in these functions, you can have multiple x that can give you the same value for y. For example, in this case, you've got these two values of x that are giving you the same value of y.

[04:45:06]One very common example of this is quadratic functions. In quadratic functions, for example, if you have y= x^2, -2 gives you 4 + 2 also gives you 4, minus 1 gives you 1 + 1 also gives you one. These are called many to one relations.

[04:45:23]All polomials apart from linear functions they are generally many to one. Cubic function for example it's if it's if it's graph looks like this. This is a many many to one function. Then you have a one to many relation. In one to many relation you can have multiple y values coming from one single x. So one x can give you many y. That's a one to many relation.

[04:45:48]For example, you would have a curve that looks like this. just one value of x that's giving you multiple values of y one value of y here another value of of y here that's that's called a one to many relation for example if you have a function like this y^2= x this is a one to many relation and then you have a many to many relation for example if you have a circle or an oval this is a many to many relation because one x can give you many y one y can also give you many x so on both sides it's many to many this is a Many to many relation for example this value of x that's giving me this y and also this y and this value of y that's giving me this x here and this x here this is called many to many now what's important to know in these relations is I'm sorry yeah so there these are four different types of relations one to one many to one to many and many to many only the first two one to one and many to one. We classify only these first two as functions. The last two one to many and many to many they are not functions.

[04:46:59]They are relations but they are not functions.

[04:47:03]Only one to one and many to one relations are functions. So for a for a relation to be a function the condition is there should only be one output. So it could be one to one or many to one.

[04:47:15]You cannot have multiple outputs. That's not allowed. And out of these functions only one to one functions have inverse. Many to one functions do not have inverse because when you inverse a many to one function that will become one to many and one to many is not a function. So for many for many to one functions the inverse does not exist. Inverse only exists for one to one functions. So you need to know about these relations for understanding inverse functions. Now what are how do you identify if a function is one one to one or not? We have this test for this that we call the horizontal line test and the test basically says can you draw any horizontal line on the graph of the function that cuts the graph at more than one point. If you can do that then it's not a onetoone function. If you cannot draw if you cannot draw any horizontal line on the graph that cuts the graph at more than one point then that would be a onetoone function. For instance let's try that here.

[04:48:17]on this graph in this on the straight line graph if I draw any horizontal line it can only cut the graph at one point so this is one one on this graph as well if I draw if I draw any horizontal line it can only cut the graph at one point so this is a one one function but in this in in these examples if I draw a straight line here a horizontal line here that's cutting the graph at more than one point if there's even one line that I can draw like this on the graph that means It's not one one because there are multiple values of x that are giving us the same value for y. So this is not one one. Similarly here I can draw this line and that that makes it uh not one one. So this is the horizontal line test. Can you draw at least one horizontal line that cuts the graph at more than one point? If the answer to that is yes, then that's not one to one.

[04:49:08]If the answer to that is no, then it is a one one function. And inverse only exists for one toone functions. Okay. So what is an inverse function? You've done this in all levels already for linear functions. How does it work? If you have a if you have a function that looks like this f ofx= - 2x + 5. How does the process work? You replace f ofx with y first of all and that becomes x= f inverse of y like this.

[04:49:36]Or in fact I could just you know that is why because it's not required right now.

[04:49:40]Okay. So you could say let's just say let let y equals f ofx I'll just cut out the detail there.

[04:49:55]Okay so this is this is how the inverse process works. You say let y= f ofx.

[04:50:00]Then once you do that it becomes y= - 2x + 5. And now you make x the subject.

[04:50:05]When you make x the subject, eventually all you have to do is replace x with f inverse of x and replace y with x and you get your inverse function. Now what exactly is an inverse function? Inverse function basically does the opposite of what the original function does. So in this original function for instance, if you were to input x= 1, what would you get from that? You'll get y= 3 from that. In this inverse function, if you input x= 3, you get y = 1. So the input becomes output in the inverse function.

[04:50:39]So it does basic it basically does the opposite of this. So going in this direction from 1 to three we had to apply f. If you wanted one from three we we apply the inverse function. So x and y they basically interchange in the inverse function. Okay. So their domain and range also reverses like this. So think about this.

[04:51:03]If you've got this function f(x)= - 2x + 5, if you input x= 2 in this, you get one from that. And the inverse function when you input x= 1, you get two out of that. Right? So it does the opposite of what the original function does. So when you think about the domain and range of these functions, this is what happens.

[04:51:24]If the original function f ofx has a domain that is x greater than zero, then that becomes the range of the inverse function. That the range of the inverse function that becomes greater than zero.

[04:51:43]So if this is greater than zero, the range of the inverse is greater than zero. If the range of the inver if the range of the original function is less than five that becomes the domain of the inverse function. So that's the relation between the domain and range of a function and its inverse. The domain of original function becomes the range of inverse. The range of the original function becomes the domain of the inverse. So if the domain is x greater than zero for the original function, the range for the inverse function would also be greater than zero. But we have to be careful about the variables. When we're talking about domain, it's x. When we talk about range, it's y. So this was x greater than zero on the left side.

[04:52:29]For the for the range, it becomes for the inverse it becomes y greater than z.

[04:52:34]Similarly, here the range of the original function was y less than five.

[04:52:38]the domain of the of the inverse function becomes x less than five because domain are values of x and range uh is values of y. So if you think about a point any random point on the original function if its coordinates are let's say a and b. For the inverse function what happens is the coordinates get interchanged and they become b and a. So x becomes y and y becomes b the coord coordinates switch like this. Now this is shown in graphs as well. When we draw a function an f ofx function and its inverse when we draw the both of these graphs on the same line on the same graph what happens is they are always reflections of each other in this line y = x. If you remember your olar transformations when we used to reflect anything in y= x line their coordinates used to switch. X used to become Y, Y used to become X. That's exactly what happens in a function and its inverse as well. Whenever you draw these functions on a graph, they're always going to be reflections of each other in the line y= x. So whenever they say draw a function and its inverse and explain or identify the relation between the two graphs, you would always draw this y= x line as well and show or write that they are reflection of each other in the line y = x. Now while you do that you need to make sure that on both the axes you're using the same scale. Whatever you have marked as one unit on the horizontal axis, you should mark as one unit uh the same distance on the vertical axis as well because otherwise your transformation the reflection is not going to look good and it will not actually look like a reflection if the scale on both axis is not the same. So whenever you draw the inverse function in its original graph always draw y= x9 and also use the same scale on both axis.

[04:54:38]Okay, so that we talked about how to find inverse of linear functions.

[04:54:41]Sometimes you'll have to find inverse of trigonometric function as well. Now that's not very complicated. For example, if you have a function like this f ofx= 5 sin x - 2. What is the inverse going to look like? First of all, we say let y= f ofx. Okay, so that becomes y = 5 sin x - 2. Make x the subject. Now when you make x the subject, if you have sin x here, when s goes to the other side, that's going to become sin inverse. So we get x = sin inverse of y + 2 / 5. And then at the end all we have to do is replace x with f inverse and on the right side replace the variable y with x and you get the result from that inverse of of quadratic functions. Now quadratic functions also have inverse.

[04:55:24]Now how does that make sense? Quadratic functions are generally not one to one and inverse only exists for one to one functions. So if you have a graph like this, if you consider this whole graph, this whole graph is not one to one because it fails this horizontal line test. But you can still find the inverse if you restrict the domain of this graph. So for example, if you had this quadratic function x^2 - 2x + 3 and the domain of this function was x less than 0. Now the turning point of this graph is actually when x is equal to 1. So where is this point going to be? x equals 0. This point is on the left side of the turning point. So that point is somewhere here. This is the point 0 and three. So which part of this graph is valid? Only this part to the left of x uh x=0. Now this yellow part if you just focus on this part, this part of this function is indeed one one. So for this yellow part we can find the inverse. Now how will that process work? How will we find the find the inverse of this quadratic function? Whenever you have to find inverse of quadratic functions, the first thing you always have to do is complete the square because otherwise they would not be able to make x the subject. So whenever you have to find inverse of quadratic functions, you always complete the square for that function first of all. So in this case for instance when you replace f of x with y initially then what you have to do is you have to complete the square for that because then you can make x the subject. When you have completed the square and you got this form, you take the constant to the other side. So you get this thing x - 1 square = y - 2 and now you take square root on both sides.

[04:57:02]When you take square root on both sides do not forget the plus minus because whenever we take square root we always have plus minus. But then we have to decide whether to use the plus sign there or the minus sign there. So when we make x the subject in this case we're getting 1 + -<unk> y - 2. That's the result that we're getting. But that's the expression that is basically the inverse of this whole thing. And that graph basically the graph of this thing would look like this. The one that you see at the bottom here. You have one part here. You have another part here.

[04:57:38]And that's not even a function. This is one to many. So plus is giving you one part and minus is giving you another part. You only have to consider one of them. It can either be plus here or minus here. You always have to first write plus minus there and then figure out whether to use plus there or minus there. Now how do you do that? Once you've made x the subject once you've made it the subject you look at the domain that is given for that question.

[04:58:04]The domain is x less than zero in this case. So you say okay the domain is x less than 0 and we are getting 1 + - roo<unk> of y - 2. How can we get a value of x that is less than 0. Do we get less than zero value when we do one plus something or 1 minus something? We can only get less than zero when we do 1 minus something. So in this case we choose the minus sign and we say the inverse therefore is going to be 1 -<unk> xus 2. So using the domain of the function when we've made x the subject we have to decide between the plus and minus in this function and then just write one of them. Sometimes it's going to be negative sometimes it's going to be positive. You have to think about it using the domain like this. Consider this example. Let's say we've got this function x - 2x + 3. How do you find the inverse of this function? Again, you complete the square. First of all, you make x the subject. Eventually, you will get something like this. 1 + minus something. It's actually the same expression as before, but I've changed the domain. Now, the domain is from 1 to 6. Now, what about the inverse of this?

[04:59:08]Now, the domain is from 1 to six. How can you get a number from this expression that is between 1 to 6? What do you have to do? One plus something or one one minus something. If you do 1 minus something, that value will be less than one. That's not what we want. We want something that's greater than one.

[04:59:24]So the final sign that you use here, it's going to be a plus sign. So the inverse function therefore becomes 1 + roo<unk> of x - 2. And that's your final result. So always complete the square and then you make x the subject. Do not forget to put plus minus with the square root there. And then you use the given domain of the function to figure out whether to use plus or minus in your final result. That's how you find inverse of quadratic functions. Let's have a look at a couple of examples in this quickly. Let's say we got this function 2x 2 - 8 x + 11. And the first part is just about computing the square range. I'm just I just want to focus on this part.

[05:00:04]First of all, explain why f does not have an inverse. So this function is defined for all real values of x.

[05:00:12]Why this function does not have an inverse because the graph of this function would actually look like this.

[05:00:17]It's a U-shaped function. It's not one one. So whenever they ask you something like this whether whether a function is in whether a function has an inverse or not. You think about whether that function is one one or not. So in this case this is not one to one because it fails the horizontal line test. That means that the inverse does not exist. So that's what you write there. It's not a one to one function. But then this last part is important. It says there's another function g of x and that is 2x - 8 x + 11. It's the same function as before. Same expression as before, but the domain has changed. And the domain is that x is less than or equal to some value. Now we need to state the largest value of a for which this g of x function would have an inverse. Now what what's the graph of this function going to be like? It's a U-shaped graph like this. And they are saying that the domain of this graph is x less than or equal to a.

[05:01:17]Now we will have some turning point of this graph. We have already found that in the second part for finding the range. The turning point of this graph is 2 and three. That's the turning point of this graph. Now if on the x-axis you have some point a somewhere.

[05:01:32]What's the largest value of a that you can have so that this function would have an inverse? It it becomes a one one function. If I put a here on this side, then this part of the function would be valid to the left of a. This is not one one and that would not work. If I put a on the left side here, then to the left of a this part of the function would be valid and this is one one. This could be one possible value for h. But they are asking us for the largest value for a for which it was it would have an inverse. So how far can a go toward the right side so that it would still have an inverse. So I can start from the left side of this function. I can keep going on. The yellow part still has an inverse. It still has an inverse. I can go up to the turning point. At this point if I were to go any further, the function would not remain one to one. So what's the largest possible value of a that we can have? The largest possible value is when you when we have it when we are at the turning point. So what's the value of x at the turning point that is two the it was the domain of the function right x less than or equal to a. So we say the largest possible value of a here could actually be two and that's the that's the result that we write down there. We say the largest a is therefore two because that is when this function would still have an inverse. It will be the xcoordinate of the turning point.

[05:02:59]Similar to that sometimes they ask you for the smallest value of something for the function to be a 1 to1 function and then the domain has x greater than or equal to something. In the previous case it was x less than or equal to something. In that case we were looking for the maximum value. Now it says x is greater than or equal to k. Now that means in this quadratic function. So you can go through the rest of the parts in your own time later on. But this is the function and the domain is x less x greater than or equal to k. We want to find the smallest possible value for k for which this function would have an inverse. So again think of it the same way it could be at the turning point. If if the value for k could was even slightly to the left here then the function would not not be one to one anymore. So any such question that like this where they ask you for the maximum or smallest value of a constant such that the function remains one one you do is you think of the turning point the turning point is going to be that maximum or minimum value uh the the x coordinate of the turning point that that will be the value that that you write there it's k equal to three in this case okay then you have composite functions composite functions are when you have one function inside another function and you have one function inside another function. Now, what do they look like?

[05:04:18]So, let's say we've got a function f ofx and we have another function g of x.

[05:04:22]They're both defined for all real values of x. f ofx = 4x - 3 and g ofx = x² + 2.

[05:04:28]These functions are defined for all real values of x. How do you evaluate the function f ofx? Now, fg ofx basically means that you want to input g of x inside the function f. Okay, you want to input g of x inside the function f. So the way to the way to evaluate this is that you in introduce an extra bracket here like this. And then you say inside the function f you have to input g.

[05:04:55]Input g there. The function g is x2 + 2.

[05:04:58]And then you're saying f of x2 + 2.

[05:05:01]Inside this function wherever you have x, you just replace that with x2 + 2.

[05:05:05]Simplify this and you get the function f ofx. What about gf of x? g of f gf of x would be that inside the function g you input f ofx. Now what's f? f is 4xus 3.

[05:05:18]So you basically say in inside the function g wherever you've got x you replace that x with 4x - 3 and that gives you the function gf of x. That's how you evaluate fg or gf. Now it's important to remember fg and gf are not equal to each other in general. These are not numbers that are multiplying that uh if you change the order it would remain the same. These are functions. If you input f inside g that's a different function. If you input g inside f that's a different function. They are not equal to each other in general.

[05:05:53]Then you have this notation f square of something or g square of something. What does that mean? F is not the same thing as saying if the function f ofx is 4x - 3. f square of x is not this that you say 4x - 3^ 2 that would be f ofx^2 that's a different thing f of x^ 2 that would be 4x - 3^ 2 but when it says f² of x what that means is you've got the function f coming twice there so it's a composite function such that you're inputting f inside the function f okay so function f is being put inside itself So inside f you're inputting f_x - 3. So wherever in that function you've got x, you replace that with 4x - 3 and you get that result. You simplify that you get that result.

[05:06:45]You could have three functions like this sometimes as well. You've got f here, you've got g here. You may have to evaluate f² g of x. What does that mean?

[05:06:52]You first of all break it down into into parts like this. You say okay f square means f f and then g. And then you put a bracket with the function that is closest to x.

[05:07:05]What's the function that's closest to x?

[05:07:07]That's g. Right? Now, input the value for g there. The value for g is x² + 2.

[05:07:13]You input that here. And now you put another bracket to the next function like this. Evaluate f of x² + 2. So inside the function f, you input x2 + 2.

[05:07:23]That becomes 4 into x2 + 2 - 3. Simplify this. When you simplify this, you get something like this. Now, finally, it means you input 4x2 + 5 inside this expression. So, when you have three functions like this, you go step by step like this. You put this bracket, you evaluate this thing, then you put another bracket, you evaluate this thing, and then finally you will get something like this that you can simplify finally. Okay? So, you can have three functions like this as well in composite functions.

[05:07:58]Okay? Okay. Now sometimes what you what you end up getting in the in these problems is in recent papers you'll find this multiple times. If you have a function f ofx= 4x - 3 and another function g of x = x^2 + 2 and you are given that f h ofx is equal to g ofx and you want to figure out what the function h of x is in terms of x. So we are given this equation f h of x is equal to g of x and what we have to find out is the function h. So we want to make h of x the subject somehow. Now how can we do that? We want to get rid of f from this side. Now the way to do that is we apply the function f inverse on the left and also f inverse on the right side of this equation. So we can apply the same operation on both sides. Since we want to get rid of f from there, we say let's find the f inverse function. First of all, f inverse turns out to be x + 3 over 4. And now we apply the function f inverse on both sides of the equation in the beginning because f is on this side.

[05:09:04]So we have an f inverse introduced here because f inverse and f they're opposite of each other. They will cancel each other's effect. So we will get rid of them on the left side. We're just left just left with h of x there on the right side. Now we have f inverse of g of x and that's something that we can evaluate. We know the function f inverse in the inside the function f inverse.

[05:09:25]All you have to do is input x² + 3 in place of x and you would get that function h of x. Now in this case it was fh equal to g. It could also be reversed. So what if you had h f ofx equals g of x. What do you do in this case? Again, if you had to find the function h of x from this, you want to get rid of f from this side. How do you do that? How do you get rid of this function f? Now the function f is coming on the right side here as in after h. In order to get rid of this, you will have to introduce f inverse on this side. Now rather than in the in the beginning, you have to introduce f inverse afterwards at this point. So you do that do the same thing on both sides of the equation. You have an f inverse here.

[05:10:13]You have an f inverse here. Now on the left side of the equation these two basically cancel each other's effect and we're just left with h of x and that equals g f inverse of x inside the function g. Now you have to input f inverse and that gives you the function h of x. So if you want to make one's function the subject from an expression like this you whatever function it is that you want to eliminate from that side you apply the inverse of that function either in the beginning or at the end to get rid of that function from there.

[05:10:52]Okay. Now this is an important part of the of of this uh of of composite functions and that is if you have a function fg of x that looks like this fg of x how exactly does this function work this function works like this step by step what happens is the input that you give to the function the input that you give to the function that is first going to go to g the function that's closest to that. So input goes to G first of all and then G gives you an output that output is G of X and then what happens is that output that you get that goes as an input to the next function which is f and from that you get the final result which is fg of x. That's how a composite function works. The function that's closest to the input that's applied first and then the next function after that.

[05:11:49]Now this gives us something very important about the domain and range of composite functions. So you'll have problems like these where they say you have a function f ofx that looks like this. You have another function g of x that looks like this. And you want to figure out if the function fg of x can be formed or not. Now in this we consider the function f as the outer function. We consider the function g as the inner function.

[05:12:17]The function that comes first, we call that the outer function. The first function that is close to x that comes after we consider that the inner function. Now there's a condition for the function fg of x to be formed here.

[05:12:31]And this is what that condition looks like. So think about this. The function f can take any positive value. The function g that can take all real values.

[05:12:44]Now remember what's going to happen. The input is going to go to function G and G is going to give you an output and that output is going to go to the outer function and that outer function is going to give you give you give you the final output.

[05:13:00]Now for this to work the output that you're getting from the function G, the output that you're getting from the function G, the range of function G, since that has to go as an input to the function F, it should be possible to input all possible values from here in the function f otherwise the function would break. So the range of the function g range of the function g it should be compatible with the domain of the function f otherwise we would have a problem the function f would not be able to take that as an input and the function would break.

[05:13:46]For instance in this case if you find the range of this function g it's a quadratic function you can find the range of this the range of this function turns out to be y less than or equal to 3. So it can give you any ne any value that's less than or equal to three. So at this step the function g could give you any output that's less than or equal to 3. But the problem is the function f that can only take values that are greater than zero.

[05:14:12]And now we've got a problem because the function g can give any output that is less than or equal to three. Now less than equal to three also means that you could have some negative values. when you have a negative value and that goes inside the function f you would have a problem because the function f cannot accept it and the function would break.

[05:14:29]So from this we actually get this result that this function cannot be formed in this particular example because the whole range of the function G the whole range of the function G all the possible output values from function G they cannot be input in the function F or in other words the whole range of F whole range of G is not part of the domain of F since it's not part of the domain of f the function would break it cannot exist. Now the summary of all of that is this condition that we've got and that is a composite function for example fg of x it can only be formed if the whole range of the inner function in this case the inner function is g is part of the domain of the outer function. If this condition is met then the composite function can be formed. If this condition is not met then the composite function cannot be formed. The whole range of the inner function it should be part of the domain of the outer function. You need to memorize this range of the inner function domain of the outer function. They should be compatible otherwise the function cannot be formed. Inner function is the function that's next to x. Outer function is the fun function that comes first. This condition needs to be met for uh the this condition needs to be met for the function to work. Otherwise it will break. And if a function can be formed, if a function can be formed. If a function can be formed in that case, let's say if the function is fg of x, if the function is fg of x, the domain of this function is going to be the same as the domain of the inner function. domain of the composite function is the same as the domain of the inner function. Why is that? Because remember in fg ofx function the input was first going into g right g was the first function that took an input. So if function g can take cannot take something as an input that means the composite function cannot take that as an input.

[05:16:54]So the domain of the composite function should be the same as the domain of the inner function. This is the summary of all of that. The condition for a composite function to be formed is this.

[05:17:06]Range of the inner function should be part of the domain of the outer function. And if a function can be formed, the domain of the composite function is the same as the domain of the inner function. Let's have a look at this example. Let's say we've got this function f(x)= 2x + 3. x is greater than equal to 0 and we have another function g of x that's a x² + b and x is less than or equal to q. Now in this case in the first part they ask us to find the values of a and b. Let's say we've done that and we found found that the composite function is this. The composite function is fg ofx = 6x^2 - 21. Now the second part says find the greatest possible value for q. Greatest possible value for q. Now what is q? Q is coming in the domain of the function G and the domain of this composite function. Now what does it mean? Find the greatest possible value of Q. You've got a composite function here. What's that composite function? The composite function is FG of X. This is the composite function.

[05:18:09]The inner function in this function in this composite is G. This is the inner function. So G is the inner function and F is the outer function. What's the condition for the composite function to be formed? The condition is that the range of the inner function should be part of the domain of the outer function. What's the domain of the outer function? Domain is x greater than equal to zero.

[05:18:33]So this is the function g. This was the inner function. This is the function f.

[05:18:39]This is the outer function. The domain of the outer function is x greater than or equal to 0. That means the range of the inner function that should also be y greater than or equal to 0. What's the inner function? It's the function g. So we say this y value. The y value of this function all of this it should be greater than or equal to 0. Right? This represents the y- value. This whole thing is y. So this has to be greater than or equal to 0. So we put that greater than equal to 0. We solve that inequality like in like like we solve quadratic inequalities and we get this result from that x is less than or equal to minus2 or x is greater than equal to 2.

[05:19:20]Since the question said x is less than or equal to q here we consider the case that has the less than equal to sign and we say the greatest possible value of q that we can have in this case is this number here which is 2 minus 2.

[05:19:35]That's the value of q. That's the greatest possible value of q. That's how we use that rule. This is this is one way that we use that rule. Then you may have to find the range of the composite function sometimes as well. So in this case they give us the value for Q and the value of Q now they're saying is minus3. We found the greatest possible value for Q. Greatest possible value for Q was minus2. But then they're saying the value of Q is minus3. Now if the value of Q is minus3, they are asking us to find the range of the function FG.

[05:20:03]The range of the function FG. Now how would you do that? How do you find range of a composite function? Range of a composite function is found just like you find the composite function or just like you find the range of any function.

[05:20:14]So since we know that the domain is now x less than or equal to minus 3 and this is a quadratic function. You find the range of this quadratic function like you find the range of any quadratic function. Think about the turning point.

[05:20:26]Think about the graph of the function.

[05:20:28]highlight the valid part of the graph and see what's the lowest point on the graph and what's the highest point on the graph and you get your range from that. So finding the range of composite function is done using the standard way just like it's done for any function.

[05:20:44]Then how do you find fg whole inverse inverse of the composite function inverse of the composite function? Again that's also done the normal way. If this is the composite function, just think all of think of this whole thing as y and then make x the subject and you'll get the inverse function from that eventually. All right? So that it works the normal way as well.

[05:21:09]This is another example. So I'm going to skip the the other parts. Just look at the last part here. This is a function f ofx. f of x is x - 4 / x + 4. The domain is x greater than 2. Now they're saying explain why the composite function ff cannot be formed. How do you think about that?

[05:21:31]In the function ff, what's the inner function f? What's the outer function f?

[05:21:36]So that means the range of f should be part of the domain of f. We know the we know the domain of f that is x greater than two. And the second part they make us find find the range of the function f as well. It's a rest appropriate function. You find the range of the rest appropriate function like we normally find the range of any rest appropriate function. The range turns out to be 0 to one. Now the both uh the function f is both inner and outer function. Now the range of this is from 0 to 1. Domain of this is x greater than 2. Is this whole range part of the domain? It's not. The range is from 0 to one. that these values cannot be input inside uh the domain of of the same function. That is why you can say this function cannot be formed. Okay, let's look at one last example. It's been a long session on functions. Let's have a look at this. We have a function f. We have functions f and g. They're defined for all real values of x. We did a very similar example in quadratics and this as well. So let's think of think let's consider that again just as a reminder. So we have this function f ofx that's a quadratic function and they say that g of x is another function which is 5x - 1. Given that the range of the function gf is greater than or equal to 39. Find the value for k. Now how do we think about this? We're given the range of the composite function. We're given the range of the composite function. Now what's the composite function? You have to find that first of all. How do you find that g f ofx? You you know input inside the function g the function f simplify everything and it turns out you get end up getting something that looks like this 5 into 3x - 2^ 2 + 5k - 1 that's the function that you end up getting. Now they are saying the range of the function is greater than or equal to greater than equal to 39. This is the range of the function greater than equal to 39.

[05:23:40]So what's the range of the function from this the number that's written outside that's 5k minus one that's the y-coordinate of the turning point what's the ycoordinate of the turning point going to be this is the y-coordinate right 5k minus one in in the standard form y = a into x - h² + k so the range of this function would be greater than or equal to 5k minus one so that means 5k minus one that corresponds to 39 9. So 5k - 1 = 39 and that gives you the value for k that is 39. So if they give you the range of a function themselves and there's something that is unknown, what you can do is you can find the range of the function yourself in terms of that unknown uh variable and then you compare both of them and you get the value of that unknown from that.

[05:24:29]All right. Uh the next part is uh something that we've already discussed.

[05:24:35]You can have a look at this. That's it on functions. I hope that was helpful although we ended up taking a lot of time and I was even though I was trying to be very quick as well. I hope that made some sense. If you have any questions you can let me know. Uh and now we can have uh your questions quickly.

[05:24:57]Will we do past papers? Uh it's a revision session. How can we uh do past papers? So uh just revising these concepts just revising these concepts is going to take us like hours and hours.

[05:25:11]Uh doing one pass paper takes like 1 hour 45 to 2 hours. I have a separate past session going on. Uh if you want to join that you can do that but in the revision session it's it's not really possible to uh do past papers as such.

[05:25:42]Okay. So what what we're going to do is we going to have uh we're going to do circular measure and then we're going to have a long break after that. Okay.

[05:25:49]Circular measure is a small thing. It's not not going to take much time and then we have a break after that. Okay.

[05:25:54]Because then trigonometry is going to take longer. So after a circular measure we'll have a longer break. Okay.

[05:26:07]You want to pray before it's time. It's 25 minutes till we can probably get done with circular measuring that time. We We don't We have a lot to do. We have a lot to do. We have like five topics left.

[05:26:28]We've got a lot to do.

[05:26:33]We can have a longer break. We can have a longer break. Yeah, we're starting circular motion. I'm just going to answer very some questions quickly. If there any quick questions in this, let me know. So, you can ask them again if you ask them earlier.

[05:26:47]Let me actually uh allow you to unmute yourself as well.

[05:26:53]Yeah.

[05:27:00]Okay. So, some of you seem to have problems understanding uh the standard methods uh for some of these questions. For that, you need to make sure that you understand the detailed method. Again, uh the point of the session is revising what you already know. uh if you do not understand something at all then you have to uh make sure that you go through a detailed video for that. So for example the reciprocate functions how to find range of those functions. uh you can you know u what I can what I'll do is I'll add to the group description u the P1 playlist and you can watch that particular video range of reciprocal functions from that to understand how how that works because going into a lot of detail right now unfortunately is not going to be possible because there's so much that we still have to cover and we will be otherwise you know pulling an allnighter then you guys already will get tired by 9:00 p.m.

[05:28:07]I think you're already tired uh at this point from the messages that that I'm seeing. It seems like that.

[05:28:15]So if we not start going going to going into too much detail that will uh be very difficult. Okay. So that's one question that I that I can quickly answer about x equals z.

[05:28:29]Uh where is that?

[05:28:33]What is the appropriate functions? Okay, so some of you got confused here.

[05:28:52]Okay. So in this case for instance when you have domain x greater than 0 you cannot input x equal to 0 in this function directly. If you try doing that on your calculator that will give you an error. One way to think about this is you say okay 1 / 0 just remember that's going to be an extremely large value. So you can think of that as infinity. So infinity + 5 is still infinity. So that's going to give you infinity. Or another way to think about this is since you cannot input zero as such, think of a very small value instead. It's x greater than zero. Right? Greater than 0 means any value that's greater than zero. So you want to check what happens at zero. Try a value that's very close to zero like 0.0000 and then put many zeros there just to understand what it is going to look like. Now one over an extremely small number like this. What do you think is going to happen? the overall result is going to be very big.

[05:29:47]Now in that result even if you add five that does not change much. It's still infinity. So that's what you get. You get infinity in this case. So that's one extreme value that you're getting for y.

[05:29:58]The other extreme value that you're getting is five. So you say the range is between five and infinity. Or in other words, it's just greater than five.

[05:30:05]That's the range here. Okay. Graph or progal functions. You would not have to do that. You can do everything without sketching graphs yourself. If you need the graph for example if the graph has a turning point uh in that case this method would not work in that case they will always give you the graph themselves like in this example there only I think two question like this in recent papers in which you have a turning point on the rest appropriate function and both of them they give you the graph themselves you won't have to draw the graph of of appropriate function yourself that's not something that's required in your syllabus Okay, any other quick questions? Okay, let me go back there. So this this is clear now. The five to infinity ones.

[05:31:00]Okay. So uh when we write this 5 to infinity both of these are acceptable.

[05:31:07]Both of these are acceptable. So 5 to infinity means the value has to be greater than five and up to infinity.

[05:31:13]Now infinity is not a number as such. So that's not a limit. So you can also write that as y greater than five. Both of these are acceptable according to the marking schemes. So you can write any of them. You can say y greater than five or or y is between five and infinity. Both are fine.

[05:31:42]Okay. So in this case we're saying it's 1 /x - 3^ 2. So it might have uh a turning point. No it's not a quadratic function. It's quadratic in the denominator. These functions still do not always have uh turning points as such. um even if they have it the domain that they would give the domain that they would give it would always not include that turning point. So we don't we won't have to worry about the graph. Okay. So in reciprocal functions when they do not give you a graph themselves it always means it's going to be a onetoone function. That means it's not going to have any dirty points. So you don't have to worry about the dirty points in that case.

[05:32:31]Okay.

[05:32:35]If there is a turning point, they will always give you the graphs themselves.

[05:32:39]So that's not going to be a problem for you guys to figure out. Okay. All right.

[05:32:44]Uh is that it? Can you can can we move on to circular measure?

[05:32:51]No further questions.

[05:32:56]Okay. Let's do circular measure next.

[05:32:58]Then we have a break after that. Okay.

[05:33:00]Now this is a very short topic.

[05:33:04]Circular measure. Now the first thing we need to know about circular measure is another unit for measuring angles and that's called radian. Now this is one thing for which the syllabus explicitly says that you want to you need to know the definition of radian as well. You need to understand what one radian means. So what is one radian?

[05:33:25]When you have a circle and in that circle there is a sector such that the arc length of that sector is equal to the radius of that sector.

[05:33:37]So in a sector if the arc length and radius are equal if this and this they are equal to each other. These lengths are equal. The angle that you get at the center that angle is one radian.

[05:33:52]Okay, you need to have this understanding of gradient. When the arc length is equal to the radius, the angle at the center that you have that is one radian. All right. Now, if the arc length is r angle is one radian, what's the arc length of the complete circle? For the complete circle, the arc length is 2 pi r.

[05:34:14]So, what is the rad? What is the angle going to be? Multiply one with 2 pi. the angle turns out to be 2 pi. So for a whole circle the total angle in radians is 2 pi. Okay for a complete circle for a full circle the angle is 2 pi. Now this is the exact value. You can also write this in decimals that's 6.28. So you see it's a much smaller number than the number in degrees. The number the angles in degrees that are larger numbers generally 360 that's a large number. 360° corresponds to uh 6.280.

[05:34:49]Okay. So radian is a bigger unit. So its values are smaller. They correspond to uh so 2 pi radians corresponds to 360° in this case. Okay. So now that you understand what 1 radian is and what and what the angle in a full circle is 2 pi radian. You need to be able to convert between degrees and radians. So 360° corresponds to 2 pi radian. That means 180° corresponds to pi radian. And that's the ratio that we've got between degrees and radians. And that's what we can use to convert from degrees to radians. 180° to uh to pi radian. That's the ratio that we have. Sometimes they ask you for one mark to convert a degree angle to radian or or radian angle to degree. This is the ratio that you use for that. And you should also memorize some of these common angles and their corresponding radian values. A lot of people uh cannot uh do well in trigonometry questions that have radians in them because they cannot visualize properly what these angles are going to be. Pi / 2, pi over 3, pi / 4. You need to memorize what each of these corresponds to in terms of degrees because you're you're used to working in degrees since your early years. Radian is the first time this is the first time that you're studying gradient. So you need to be able to connect that to the corresponding degree angle. So you need to memorize this without any calculation. You should you should be able to figure out that p<unk> /2 radian corresponds to 90°, p<unk> / 3 is 60°, p<unk> / 4 is 45, p<unk> / 6 is 30° and 0 is 0. You need to memorize these angles. Okay? P<unk> /2 is 90, P<unk> 3 is 60, and so on. Now this is the ratio that we've got 180 to pi. If you want to convert an angle from degrees to radians, you multiply that with pi over 180. If you want to convert an angle from radians to degrees, you multiply that by 180 over pi or just use ratios and that will give you the result. Now we've got the two formulas mainly in this topic. One for arc length and the other for area of a sector. Now arc length in degrees was given by this theta 360 into 2 pi r in radians. That formula becomes this arclength equals r into theta. S= R theta. S represents S represents the uh arc length and theta is supposed to be an angle that is in radians for this formula to work. Okay, S= R theta. That's one formula that we've got in degrees. The area of sector formula was this. When we convert that to radians, that becomes half r² theta.

[05:37:20]This is the second formula that we've got/ r² theta. So there are two formulas primarily that you need to know about in this topic first of all and they are s= r theta and area of a sector that's equal to half r² theta in both of them the angle theta has to be in radians otherwise these formulas don't work and in circular measure generally you'll be doing everything in radians anyway. Okay so that's these these are two formulas that we've got primarily. Now this is all that you need to know in terms of concepts as far as circular measure is concerned. However, along with these concepts, you have to use some other concepts that you know about already.

[05:37:58]For example, you will need to find areas of triangles. How do you find area of a triangle? There are two formulas that we normally use for that. Half into base into height. For this formula, base can be any of the three sides. Any of the three sides in the triangle. You can take any three any side as the base.

[05:38:14]Okay? You can take this as the base. If you take this as the base, height is going to be perpendicular to this.

[05:38:18]That's the height. If you take this side as the base, height is going to be the perpendicular coming from the third point to that point and so on.

[05:38:27]The other formula that we will be using more because we will not always have the perpendicular height. Sometimes we we will but sometimes most of the times we won't. So we'll be using this other formula that's half a b sin c. What what's half a b sin c? A and B are two sides of the triangle and C is the angle between those two sides. Two sides and the angle between them. We can use this formula then for area of triangle. Now make sure your calculator is in radian mode for this otherwise you will have problems. You get wrong answers. So always doing any questions make sure that you look at make sure that you always look at the top of your calculator whether it's saying R or D. Otherwise you sometimes you end up doing careless errors in the exam. Instead of degrees you you use radians instead of radians you use degrees and that makes you lose a lot of points. So make sure that you're careful about this always when you're doing calculations involving sign cost and look at the top of your calculator whether it says r and d and whether that's correct or not. All right. Now using this area of triangle we can actually find area of segment as well.

[05:39:34]It's not necessary to memorize this formula as such. You can just do sector minus triangle as well. But you have to do this a lot in this topic. Finding areas of segments. So if you want to memorize this formula as well, you can do that. It's not given in the formula sheet. And it works like this. If you have a if you have a sector like this, this green shaded part, this is known as a segment. Remember sector has three sides, two radius and one arc length.

[05:40:00]This whole thing is a sector. And this shaded part that you see there, that's called a segment. that only has two sides, one chord and another arc. That's a segment. How do you find area of a segment?

[05:40:13]How do you find area of a segment? You can say find the area of the sector.

[05:40:16]That's half r² theta minus the area of the triangle. That will be half a sin c.

[05:40:22]Both the both the sides are the same in this case r and r into s of the angle sin theta. If you simplify that, that gives you this formula/ r² into theta minus sin theta. For area of segment, you can use this directly as well. Or you can just use sector minus triangle.

[05:40:38]Do whatever you prefer. Then you'll have to use some trigonometry here as well.

[05:40:43]So you need to remember the basic trigonometry like the ratio s cos tan and what what they mean in in terms of a right angle triangle. So whenever you have a right angle triangle you have these ratios defined like the sign is perpendicular hypotenuse cos is base over hypotenuse. Tan is perpendicular over base. Just be careful. The angle whatever angle it is that you're considering the side opposite to that that's the perpendicular the side adjacent to that that's the base. Okay.

[05:41:10]And then hypotenuse is just the side that is uh opposite to the 90° or pi /2 radian angle. I'm sure you're already familiar with this. If you do not have right angle triangles you only use these ratios when you have a right angle triangle. When you do not have right angle triangles in that case you can either use sign rule sometimes or cosine rule sometimes. Although they're used very rarely uh in in circular images but sometimes you do need to use them. So in triangles that are not right angle you can also use sine rule cosine rule if you want. Okay. Uh okay I I missed that.

[05:41:46]Let me just add that here.

[05:41:49]Okay. So you need to know something about isosis triangles as well. A lot of people end up not doing these questions correct that involve some isocus triangles for some reason.

[05:42:05]So let's say we've got an isocles triangle that looks something like this.

[05:42:12]We've got two triangles. Now these are these are isocus triangles. They have two sides equal. That means these base angles are also going to be equal. Now in isocuses triangles like this which you have two sides equal and you have to do some calculations always think of something like this. Breaking this triangle into half from the middle.

[05:42:35]The side that has the two equal angles in an isocious triangle we call that the base side. So to the base side make up a perpendicular from the third vertex. you will always get a 90° here and this point will always be the midpoint. In the second form on the right side, if if this was an isocles triangle with these two sides equal, you can say this is the base side of the triangle. If you were to break the triangle into two parts like this, the angle here if uh if the angle is 90, if you make a perpendicular here, this point has to be the midpoint.

[05:43:11]This helps a lot. Rather than using cosine rule formulas or sign rule formulas in equ in in triangles like this, always break them into two parts and then use simple sin cos tan ratios in either the triangle on the right side or the triangle on the left side.

[05:43:25]Otherwise, when they ask you to prove something in a question and they want a specific form using sine rule and cosine rule, getting that form that they ask you to to to to show that becomes very very difficult sometimes. So always break is triangles into two parts like this and use s cos tan ratios on them directly. Then there are some angles for which you need to use the exact values in degrees. These angles are 0 30 45 60 90 in radians. Those corresponding angles are going to be 0 pi / 6 pi 4 pi over 3 p<unk> /2. When you have these angles and you have to find sin cos tan of these angles, the question often will ask you to to give your answer in exact form. In these case, in this case, never use decimals. Now, you do not necessarily have to memorize these values. You can just use a calculator for them. You must have a calculator that gives you these exact values. For example, if you try this on your calculator, cause of 3 or pi over 3.

[05:44:29]Pi uh pi over 6. In fact, that should give you under root 3 over2 and not just a decimal value 866. If your calculator is not giving you under root 3 over2, I would highly recommend up upgrade that calculator. You need something that can give you exact values otherwise you'll be in trouble in your in your A level exam. So you need exact values. So for these angles you would often be asked to write exact answers. In that case how should you show your working? If you have to find the area of this triangle for instance and the angle is pi p<unk> / 3 sin p<unk> / 3 you'll evaluate that on your triang on on your calculator and you would first show that the value of sin p<unk> / 3 is under root 3 over2 and then you simplify that afterwards to to give the exact answer. So they would either ask you to give an exact answer or they will say give your answer in terms of pi or they will say give your answer in third form. In all of these cases make sure that you show all the working. Make sure that make sure that show show all working and uh you're not skipping any steps.

[05:45:35]Okay. So do this calculation on a calculator step but show clearly that you're inputting this value and then simplify that afterwards.

[05:45:46]And then these circular circle theorems that we talked about in in in equation of circle as well. Uh you need to know these for uh this topic as well. We talked about an equation of circle as well. Uh you need to know about these circle theorems. All of these properties they are used in this uh topic as well.

[05:46:06]Angles in the same segment are equal.

[05:46:08]Angle made by diameter is 90. If the angle is 90, uh the chord is supposed to be a diameter. It works both ways. And this property line coming from the center to the midpoint of a chord, it always makes a 90. If the angle is 90, then the line cuts at the midpoint. And this property tangents coming from outside a circle. If you were to u find their lens there, they're going to be equal to each other PQ and PR. And if you were to join those points with the center like this, the shape that you get, that's always going to be a kite.

[05:46:42]And all of these angles Y and Y, they're going to be equal. X and X are going to be equal and so on. We talked about this in equation of circle before as well.

[05:46:49]Angle at center is twice the angle of the circumference. Cyclic, quadilateral, and tangent is perpendicular to B. Yes, you still need to need to know all of these properties for circular measure as well. Sometimes you will have to use them. Again, some are used more often than not. uh the same four properties they're they're going to be used more.

[05:47:06]The kite one this is used a lot. Tangent is perpendicular to radius. This is used a lot. And then diameter uh makes a 90° at the circumference. And then finally this one line coming from the center to the midpoint of a cord that's always perpendicular. All right. Now how will the questions look like? You will have to find areas. You'll have to find parimeters arc length. And for areas for instance you would be you would have to figure out what area to subtract from which which area.

[05:47:39]For example, if you have to find this shaded area you you have to think about how exactly you can get get this region.

[05:47:46]For example, one idea could be that you find the area of this rectangle and then from that rectangle you subtract the area of this triangle and then you also subtract the area of this sector. So you'll have to do do this thinking and then just apply the formulas and do that plus minus and get the result. It's fairly straightforward. Generally when they ask you for exact answers, you just need to make sure that you don't put anything in decimals. For example, in this particular question, they're asking that you should give your answer in this form. A square root of 3 minus b pi. Now don't get confused by just reading this form that what form it is that they're asking for. All that all it means is that you should not use decimals and you would automatically get this form as the result as long as you're using exact values. Okay? You should always use radians. You should always use radians.

[05:48:33]You should never use degrees in this topic. Always use radians unless a question specifically asks you should convert an angle in degrees. In that case, you just do that for that part.

[05:48:41]Otherwise, for all the formulas, these formulas only work when the angles are in radians. Otherwise, they don't work.

[05:48:47]Okay? Sometimes what you have to do is you have to add some additional lines u in your diagram as well. So do a when you do a lot of practice you you know understand how to make these extra lines sometimes for example in this part for example in the first part they're asking you for the length of SR. Now how do you find the length SR? You can add another line here and that's XQ. And now you get a right angle triangle that you can actually use to find the length of XQ. XQ will be the same as RS. So keep a look out of look out for this that you may sometimes need to add some extra lines on the diagram to basically make make the problem simpler. All right.

[05:49:31]Okay. There's a going on here. We just have a one one minute break and then we'll resume She Hold.

[05:51:01]Okay. So, sometimes you may have to add extra lines to your diagram to make your problem simpler. So make sure that you uh practice a lot so that you can recognize problems like these. In this case as well again you have to find the value of theta. This is the value of theta.

[05:51:24]This value now in this triangle on the right side there isn't much that you know you only know this hypotenuse.

[05:51:37]If you make this triangle on the left side, you have more information on the left side here. You know the hypotenuse.

[05:51:42]You also know this angle. So using that other triangle on the left side, you can find this length and this length would be equal to this length. And then you use trigonometry and the triangle on the right side to find that value of theta.

[05:51:56]Again, in the original figure, you did not have these lines and you have to think about the think about these yourself. In this question, in fact, they do give you a hint that you consider the perpendicular distances from uh each of those points to uh the base.

[05:52:11]But sometimes they might not give you the scent as well. You might have to think about this yourself like this.

[05:52:15]Okay. So have a look at the solution.

[05:52:18]These are fairly straightforward problems. Generally when once you figure them out, figure out what to do, the actual working after that is fairly straightforward for parimeter. Remember perimeter means the length of the boundary, right? So perimeter does not have any standard formula. As such, whatever figure it is that you have, all you have to do is just sum up all of those sides. For example, if this is the shaded region, you want to find the perimeter of that.

[05:52:46]You find this length. However you can, you can use the arclength formula for this. You find this length, that's dx.

[05:52:52]Use the arclength formula in the other sector. and DE find that length and in this case that is basically found by considering this complete length and subtracting from from that these two separate lengths and that that's that gives you this this value of YZ in the middle that's the same thing as DE and that gives you that value then you can find the overall perimeter from that now it changes from every from question to question you have to figure that out yourself and the questions you do a lot lot of lot of practice for that and that helps you do That again this is another example in which you have to draw some lines yourself to uh figure out what what whatever you have to how do you find the stated region for instance it's not some standard shape that you have the direct formula for this. So one idea that we could use here is that we break this figure into two parts like this and now this becomes a segment. So we can find the area of the segment by using this sector for instance and then just multiply that by two right find the area of the sector multiply that by two of the segment sorry uh and then multiply that by two that will give the complete shaded area okay so think about how you can add any additional lines to make the problem relatively easier.

[05:54:08]Sometimes you will get areas that are a bit hard to spot as in how exactly to calculate that will require more working. You need to break them down into smaller chunks. You need to break them down down into smaller chunks. How do you do that? For example, in this case, this is not any standard shape. So how do you think about it? You do it step by step. So this is the figure.

[05:54:29]You can think of it like this. First of all, you say okay, we've got the semicircle and in the semicircle we have this shaded part from which from from the semicircle we are subtracting this part to get that shaded part. Okay. So area of semicircle we can find that somehow. Now the problem would be how to find the area of the se segment. So think about this separately.

[05:54:55]For finding the area of that segment you can add these lines here.

[05:55:00]Find the area of this sector. Find the area of the sector and then subtract from that area of this triangle.

[05:55:08]This triangle and that will give you the area of the segment. Once you get that area of segment, you had the area of semicircle earlier. Subtract from that that area and you'll get the result. So sometimes it it's not very easy to spot.

[05:55:20]You have to break that down in in parts.

[05:55:23]So in this case what you can do is find the area of the semicircle. As I said, find the area of that segment separately. How do you do that?

[05:55:34]Area of segment formula or if you want area of sector minus triangle, you get the area of that segment and then eventually you find the difference and that gives you the result. Again, you have to do practice for that.

[05:55:50]And in question sometimes the difficult part is where in the first parts they ask you to show something. They ask you to prove something. They will give you some information in the question. For example, in this case they're saying in this diagram AC divides the sector into two regions of equal area. So we have two regions here. One region is this.

[05:56:10]This is one area. This is another area.

[05:56:13]It's dividing this region into two areas like this.

[05:56:16]We need to show we need to show that sin alpha into cos alpha is equal to half alpha. Now don't worry about how exactly this form is going to come. Just use the information that's given in the question. They are saying AC divides the sector into two regions of equal area.

[05:56:32]This is one area. This is another area.

[05:56:35]Just make an equation out of that.

[05:56:37]Somehow you can find the area of this triangle.

[05:56:41]Find the area of the region on the side and then equate them somehow. Now in this case finding the area of this region is going to require some work right you'll have to find the area of the whole sector from that subtract the area of the triangle. So an easy way of thinking about this could be you could say if this is area A this is another area A that means the total area of this whole sector what would that be that will be 2 * A. So this statement that they've written here this could be simplified to a form like this. You could say area of this complete sector area of the complete sector the sector is A O B that would be equal to 2 * the area of the triangle O A you make you make an equation using that information and then just simplify that it will automatically turn out to be something like this. You just keep simplifying and keep the end in mind that this is what you want and you will get that eventually. Don't start thinking about this that okay how will you get that form just use that given information simplify that and you get the result.

[05:57:44]For example in this case when we do that area of sector O that's equal to 2 * area of that smaller triangle. Simplify that we automatically get that form that we wanted there.

[05:57:56]All right.

[05:57:58]Okay. And then the rest you can figure out in the questions. That's the key.

[05:58:03]These are the key concepts that we've got in circular measure. The rest is all practice.

[05:58:08]Can we pitch in? So I think I've got most people right now. Okay. Let's start with trigonometry now. So in trigonometry again first of all we've got to know these basics uh the definitions of s cause and tan. How they're defined in terms of a right angle triangle. S is perpendicular over hypotenuse. Cause is base over hypotenuse. understand is opposite perpendicular over the base.

[05:58:31]Uh you need to make sure that you understand that the side that's adjacent that's base side that's opposite to any angle that's the perpendicular with respect to that. For example, with respect to theta_2 x is perpendicular and y is the base. If you don't have a right angle triangle in that case we use we can use s rule and cosine rule.

[05:58:52]Sometimes that also works. Now this is what you know already. Then we talked about these exact values already in circular measure as well that uh when you have these these specific angles you need to be able to write their exact values when required and use them in your working and how to show that in your working as well. We also talked about that. Okay. Now uh this is the new stuff.

[05:59:19]So until O levels you saw that whenever you had to find whenever whenever you had to find sin cos tan you basically thought of a right angle triangle and you defined sin cost co with respect to to the right angle triangle like this which is correct. Uh you do it like this but that was a restricted definition in in in the sense that you could only have acute angles uh in a in a right angle right angle triangle. You cannot have any angle that's greater than 90° in a right angle triangle. So these definitions only used to work for acute angles, right? Angles that were less than 90°.

[05:59:58]But what if you had some bigger angles?

[06:00:01]So 150, 225, 420, -60. Sign cost 10 exist for all of these values. If you try these on a calculator, you get value. You get results for these values as well. Now what are these angles? How are s cost 10 defined for this? That's the first thing that you understand in trigonometry. First of all, how do we understand angles?

[06:00:22]Where do angles greater than 360 come from? Where where do angles in the negative direction come from?

[06:00:31]If you have quadrants like this, if you have x and y axis like this, let's say this is x axis, this is y axis. Remember angle is always going to be measured from the positive x-axis. Angle is always measured from this side. From this yellow line, it's measured from this line. And if we go in the anticlockwise direction, we consider that angle to be positive. If we go in the negative direction, if we go in the ant, if if we go in the clockwise direction, we consider that angle to be negative. Okay? Anticlockwise positive, clockwise, negative. So if you had a line, this blue line here, for instance, this line that you see here, if this line makes an angle of 30° with this horizontal, then we could say this line has an angle this, which is 30°. So the angle of that line is 30°. But that's one way to describe the angle angle of this line. We could also say that we can go one full circle and come back to that line and that would become 30 + 360 which is 390. So that same line can have an angle of 390 as well. Right? And similarly you can keep going on in circles keep adding 360 230 and you get more and more angles 390 750 and so on. All of these angles are equivalent angles, right? All of them have the same signs, the same cause, the same time. These are equivalent angles.

[06:01:49]So that's where the angles greater than 360 come from that we can go beyond 360 and we get to the same same point.

[06:01:54]Again, what about negative angles? If you go in the clockwise direction, that's where negative angles come from.

[06:01:59]For example, in this uh case, the complete angle in a circle is 360.

[06:02:04]Subtract 30 from that, that's 330. But since since we're going in the clockwise direction, we say this negative 330. So that same blue line can also be said to have an angle that's -330. This angle is also equivalent to this to all of the other angles here. So this is how we measure angles anticlockwise or clockwise. We can go one full circle and go beyond that to find more angles as well. Now in in this quot in this region on the left side if this sign was uh to have an angle of 45° with the horizontal we could say one way to measure that angle is that this angle is 180 minus 45 and that's 135 that's that's this angle. So we can say this line has has an angle of 135. But then we could also say let's go one full circle come back to that point again.

[06:02:48]This line could also have an angle of 495.

[06:02:52]Similarly we could also say let's go in the clockwise direction. We can also say this line has an angle ofus 225°. So there are different ways that we can measure the angle of that same line. All of them all of them are equivalent. All of these have the same sign cause and time. Now the region on the right side here on top that's the first quadrant.

[06:03:09]That's that's how we name these coordinates. We number these quadrants.

[06:03:12]This is quadrant one this region. Then on the left side this is quadrant 2.

[06:03:17]This is quadrant three. This is quadrant four. Now in each of these quadrants you can measure angles on the same format like this. You start start merging from this positive x-axis and then you either go anticlockwise or you go clockwise to find that particular angle.

[06:03:32]Anticlockwise is positive, clockwise is negative. For example, this line at the bottom here, we could say it has an angle of 240.

[06:03:42]600 - 120. All of these are the same angles because this is 240.

[06:03:48]180 + 60. This is 180 - 60. 120. But since you're going clockwise that's 120.

[06:03:54]So you should be able to measure these angles. That's where these angles come from. Now what happens is in the first quadrant since the values of uh in fact should should we go into this detail? Okay. So I'll go through this very quickly. So the definition of s cos and tan they're basically extended like this that instead of saying base hypotenuse and perpendicular we say instead of base and perpendicular we will say sign is defined as y / r where y is the y-coordinate of that point and r is the distance from the origin. If we have a have a circle like this that on the circle we've got points like this then we have the definitions of s and cos like this sine is y / r cos as x / r and tan is y /x and from that we basically get that since at this point the xy coordinates are both positive the value of sign is going to turn out to be positive here the value of co is also positive the value of tan is also positive but I think it's going to take too much time to explain that in our revision videos. But I in the revision class what I'll do is I'll just get to the result right away. So all of you know this already. I'll just uh move on from this. This is actually the logic behind where where you get to it uh how you get uh this asc. But now for the exam purpose this is all that you need to know. So you have these four coordinates. You have these four coordinates and in these quadrants what basically happens is in the first quadrant all ratios are positive. In the second quadrant only sign is positive. In the third quadrant only tan is positive. In the fourth quadrant only cause is positive. You need to remember this a cc. In the first quadrant all ratios are positive. In the second sign is positive. In the third tan is positive. In the fourth cause is positive. You need to memorize this.

[06:05:49]There are different ways of memorizing this. Add sugar to coffee and some others. You need to remember this. Now that's one thing that you need to understand to be able to solve equations that in which quadrant which ratio is positive. The second thing is the concept of a basic angle. Basic angle is defined as the acute angle that any line makes with the horizontal axis. So for example, if you have a line like this that makes an angle of 30° to the horizontal here.

[06:06:19]The basic angle for this line is 30. If you have this line that's 150° angle because this angle here is 150. The angle that it's making with the horizontal there on the left side that is also 30. So this line that has an angle of 150 the basic angle for that is also 30°. Similarly in in the third quadrant below if we have this line that has an angle of 210° the basic angle here that's also 30°. And we have another line here that's 330°. The basic angle of this sign is also 30°. So this is what we call a basic angle. The angle that a line makes with the x-axis.

[06:07:01]This is another very important angle to be able to solve equations. Because what happens is whenever whenever the basic angle of two angles is the same, their sign values are going to be the same.

[06:07:13]Their cause values are going to be the same and their tan values are also going to be the same. So for instance all of these angles 30 150 210 330 all of them had the same basic angle. If you were to find their sign values sin 30 sin 150 sin 210 sin 330 you would see that all of them turn out to be 1 /2.

[06:07:35]All of them have the same magnitude.

[06:07:37]They're all 1 /2. The only difference is that some are positive, some are negative. The first two are positive.

[06:07:42]The next two are negative. And where do why why is that? Because the first two angles are in the first and second quadrants. And we know that sign is positive in the first and second quadrants. These values are positive.

[06:07:55]And the third and fourth values, they're in third and fourth quadrants. Those values are negative. But the magnitudes are all the same. They're all 1 /2. So that basic angle that basically tells you what the magnitude of the sign cause and tan ratio is going to be. So in this case, the magnitude is 1 /2. For cos for instance cos 30 cos 150 cos 210 cos c30 all of them have the same magnitude. Why is that? Because all of them have the same basic angle. So whenever the basic angle is the same the magnitude of these ratios that you get there that's always going to be the same. The only difference is going to be the plus minus and that is decided by quadrants. So these two things together they will help us find solutions to trigonometric equations. In which quadrant what ratio is positive and the basic angle because basic angle tells you this number the magnitude and the quadrant tell you plus minus. Okay. Now how are they going to be used to solve equations? This is the process for that. So let's say you have you have to solve solve an equation like this. Sine of x= 0.5 and the interval that you're given is from 0 to 360. How do you solve this equation? How do you find all possible solutions of this equation? The way to do this is you first of all find the basic angle. How do you find the basic angle? This is the process that you follow for that. Replace X with alpha.

[06:09:18]So we denote basic angle with alpha.

[06:09:22]Okay. So this is this is what you'll always do. Whatever is the equation that you're given what you what you start with would be you would replace x with alpha and on the right side of the equation whatever is the number that you've got you will make that positive.

[06:09:39]Now in this case that was al already positive so so you would not have to do anything but in case it was not positive you will make it positive. So it becomes sin alpha= 0.5. you get an equation like this. And now you put this on your calculator, evaluate sin inverse of 0.5 and see what that gives you. And that will give you 30° in this case. So the basic angle turns out to be 30°. Now if that's the basic angle, what you do now is you use quadrants to figure out what the solutions of this equation are going to be within the given interval. Now how do you do that? The basic angle is 30.

[06:10:13]The basic angle 30 could be in four different positions. It could be here.

[06:10:18]It could be here. All of these are 30 30°, right? All of these basic angles.

[06:10:23]Basic angle is the angle with the x-axis. All of these are basic angles.

[06:10:27]But now we know that s is positive.

[06:10:29]Where is s positive? We know that a s t c sign is positive in the first two quadrants. So we only consider the first two quadrants. We ignore the ones below.

[06:10:38]And now we write down all possible angles in these first two quadrants inside the given interval. The interval is from 0 0 to 360. All you have to do is this is the first angle that's 30°.

[06:10:48]This is the other angle that's 150° that's 180 - 30. And that's how you solve any trigonometric equation. Let's look look at some examples of this.

[06:10:56]Let's say you have cos theta= -0.3.

[06:10:59]How do you start with this? You say the basic angle is alpha. You replace the left side with the theta on the left side with alpha with alpha. And that gives you cos alpha= 0.3. Find the basic angle. The basic angle turns out to be 72.5°. And now you make coordinates. The coordinates look like this. Since cause is negative, you think about where the basic angle can be. It's ac tells you that cause is positive in the first quadrant and in the fourth quadrant. But cause is supposed to be negative here. So which coordinate should we consider? The second and third in those coordinates. Now we write down all the possible angles inside the given interval. The interval is from 0 to 360.

[06:11:39]That means you have to consider angles in the positive direction in the clock in the anticlockwise direction. So this is the first angle that you get 180us 72.5 and the second angle is 180 + 72.5 because this is 180 + 72.5 that gives you the angle in the third quadrant.

[06:12:00]Similarly you could do it for other angles as well. They could change this interval. So if you have an interval that has some negative values as well.

[06:12:06]What do you do in that case? So if you have tan theta= 0.3 the first thing you do is you first of all replace theta with alpha. Make the right side positive. Find the basic angle. See what the basic angle turns out to be. In this case, the basic angle turns out to be 16.7. Now, you draw that basic angle in the coordinates that are valid. Since tan is positive here, tan is positive in the first quadrant and in the third quadrant. So, it's either this or this.

[06:12:30]Now, you consider the basic angle. But now look at the interval. The interval is from minus 360 to + 360. How do we write down all the possible angles in this? This is the first angle that we get 16.7°.

[06:12:41]That's straightforward. This is the other one that we get 180 + 16.7. Okay.

[06:12:46]Now, in the positive direction, we have found the two angles that we needed.

[06:12:49]However, the interval is also going in the negative direction. So, we also have to check if there's any possible angle in the negative direction that also falls inside the given interval. So, how do you go in the negative direction?

[06:12:58]Now, you go clockwise like this. This is the first angle that you have. It's 180 - 16.7. However, since we're going clockwise, you put a negative sign with that. And that turns out to be minus 163.3. Similarly, we we go beyond this and we see is there any any other angle here in inside the interval. This is another angle that we're getting 360us 16.7 put a negative sign with that and that gives you minus 30 - 343.3 and that's another angle that you are getting. Uh that's inside the interval.

[06:13:27]So it has four possible solutions that look like this. Whenever you have have an interval that has some negative values as well, always check if you can find any possible angles in the clockwise direction. In this case, we can. So they're also your solutions.

[06:13:48]Okay, let's consider this now in gradients. Let's say we have sin theta= 0.7. Now sin theta is positive here.

[06:13:54]Again the first thing you do is you find the basic angle. Sin alpha= 0.7. Find the basic angle from that. Now in this case the angle is in radian. So make sure your calculator is in gradient mode for this. Find the basic angle and see what that turns out to be. The basic angle turns out to be this.

[06:14:09]Now you have to write down all the possible angles with this basic angle in which in which two quadrants? S is positive. S is positive in the first quadrant and in the second quadrant.

[06:14:17]That means it's either this quadrant or it's this quadrant. It's one of these two quadrants. Now what angles are going to be possible here? In the first quadrant you get this angle first. In the positive side that's 7754. You get another angle that's this.

[06:14:34]What's that going to be?

[06:14:36]This is zero.

[06:14:38]180° corresponds to pi radians. So in the second quadrant now what you'll do would be you would say the angle is pi minus 0.7754.

[06:14:47]Pi is the complete angle. You subtract from that 0.7754 and that gives you that result 2.37. Now these are two angles that we're getting in the positive direction. However, look at the range.

[06:14:56]Look at the interval that's given. It's also going on the negative side. So we also have to check if there's any possible angle that we can get.

[06:15:04]any possible angle that we can get in the clockwise direction here. What are the possible angles that you can get in in the clockwise direction? This is one angle. That's 180 plus 77 54. That's pi + 7754. But since since we're going clockwise, we put a negative sign with that as well. And then we have another possible angle that we're getting here. That's this angle here.

[06:15:28]What's that going to be? 2 pi. Complete circle is 2 pi. Subtract from that 7754.

[06:15:34]Since that that's in the negative direction, we put a negative sign with that. That's minus 5 5.51. These are four possible solutions that we get in this. These are these are our final answers.

[06:15:48]Okay. Anticlockwise was was uh done there as well. So depends on the intervals in the interval. If you have both anticlockwise and clockwise, if you have both positive and negative values, you have to check in both directions anticlockwise and clockwise. So in this case we had these angles in both directions anticlockwise and clockwise.

[06:16:08]Now these are some cases uh that are relatively special. For example if you have sin theta= minus one. Now what happens in this case sin theta= minus 1.

[06:16:16]You when you try finding the basic angle in this case it turns out to be 90°. So you replace theta with alpha make the right side positive. It becomes sin alpha= 1. Alpha turns out to be 90°. Now where is 90° going to be?

[06:16:31]90 is the basic angle. What does that mean? It could either be here on the vertical axis above or it could be here on the vertical ais down. Now which one do we consider from that? S is negative.

[06:16:43]Now 90° is not falling in any quadrants, right? It's it's on the yaxis. It's not in any quadrants. It's on the boundary.

[06:16:51]Now we know that s is positive. S is positive in these two quadrants. and sign is negative in these two coordinates. So the angle that we're getting on the upper side that is not going to be possible because we have positive here, positive here. We want sign to be negative. So it can't be there. So when it's 90° with a negative sign, what we do is we consider only the part that is on the lower side. That's the angle that we consider here. And we say inside the given interval, this angle turns out to be 270° like this. All right. Let's take another example of a similar. situation.

[06:17:26]Let's say we have sin theta equal to 0.

[06:17:28]What about this case? Find the basic angle from this. What does the basic angle turn out to be? The basic angle turns out turns out to be 0 in this case.

[06:17:36]Where is zero going to be? Now 0 can either be on the right side here or it can be on the left side here. Which one do we consider in inside this interval from 0 to 360? Now we know that s is positive in the first and second quadrants. It's negative in in the third and fourth quadrants. This is zero. Is 0 positive or negative? 0 is neither positive nor negative. 0 is actually between plus and minus. Okay, that's that's interesting. Between plus and minus. What does that mean? The angle that we're getting on the left side here, this one, this is between plus and minus. That means that works. The one that we're getting on the right side, this is also between plus and minus.

[06:18:10]That also works. So, we consider both of these angles in the first uh on the right side here, this angle is zero. So, that's one possible answer.

[06:18:18]On the left side, this angle is 180. That's another possible answer. However, the interval is going until 360. So, we can also go further and we can say when we complete one full circle, this angle here, when we come come back to the x-axis, that's another possible angle. So, sin theta equals 0, it actually ends up having three solutions inside this interval. 0, 180, and 360. All of them are possible solutions of this equation. Okay, let's take another example. If you have cos theta= minus1, what do you do with this?

[06:18:46]You find the basic angle. Cause alpha= 1. The basic angle turns out to be zero in this case. Again, like what does that mean? We have to consider this side on the right right side and this side on the left side. That's when the basic angle is right on the x-axis.

[06:19:01]Now, which one of those two is going to be possible? Cause is positive in the first quadrant and in the fourth quadrant. It's negative in the second quadrant or in the third quadrant. So, which one do we consider? It has to be negative. Negative means it has to be on the left side, right? So, we only consider the part on the left side.

[06:19:20]That's this part here. What's the angle there? that turns out to be 180° and that's a final answer. It has to be 180 180° in the given interval. All right.

[06:19:28]So that's how we solve equations that look like this using quadrants and uh basic angles.

[06:19:42]Now for these special angles when you have a basic angle of 90 0 or 0 in this case you can also think of the graph when you understand the graphs when you remember the shapes of the graphs sign we know that looks like this. So you can just think of it in your head as well the sign is negative 1 at this point this is 270°. So without any working as well you can write down the answers in this case that will turn out to be 270.

[06:20:05]If you want to use coordinates you can do it like this. But if the basic angle is either zero or 90, you can also use your graph. Think about the graph in your head. Maybe do a rough sketch of that and based on that you can write down what the angle is going to be. In this case, for instance, sine is zero.

[06:20:19]In the second case, in this example, s is 0. Think about the graph. Where is sin 0? This is what the graph looks like. Sine is 0 here. S is 0 again here.

[06:20:27]And sine is 0 again here. These are the three points where where s value is equal to 0. What are those three three points? 0, 180, and 360. You could write down those answers without any working as well if you remember your sign graph well. Cos theta is minus one. Again this is a special case because the basic angle turns out to be zero. So when the basic angle is either 0 or 90 you can just think of the graph and write down the answer directly from that. So in this case it's co what does the graph of cause look like? The cost graph looks like this. Now on this graph we know that it's equal to one. The value is equal to 1 at x= 0. Then it becomes uh x is equal to 1 when x is equal to 0. y is equal to 1 when x= 0. And when it's 90, the y value is 0. When it's 180, the y value is minus1. So just think of the graph and you can write down directly that the solution in this interval for this equation would be that theta is equal to minus uh theta is equal to 180 because that is when the value for the function becomes equal to minus1. So for these special angles when the basic angle is 0 or 90 you can think about the graph for the rest of the angles you make quadrants like this find the basic angle mark the coordinates and write down all possible angles in the given interval. If it's positive angles, you go anticlockwise. If it's negative angles, you go, you go in the clockwise direction. Sometimes the ang the interval that's given might go beyond 360 as well. For example, if this interval goes beyond 360 and it goes up to like 720. What we'll do in that case, we'll complete one full circle and then come back to those points again to see if there's any other point that also falls in the interval. So you have to consider three possibilities. One, it's just 0 to 360. That's just one full circle in the positive direction. Then if it's negative like in this case, you also have to go in the clockwise direction.

[06:22:22]And number three, if sometimes the interval ends up going beyond 360, let's say it goes up to 720, then you have to complete one full circle and come back to those lines again in case you find any other angle in the interval as well.

[06:22:36]Okay. So that's how you find that's how we solve these basic equations. But mostly you'll get something more complicated than this. For that you have to use these identities. There are two identities that you have in your have in your syllabus for paper one trigonometry.

[06:22:49]Before we do them you need to remember these notations. So whenever you have sin theta square this is the actual function sin theta square. There's a short form for writing this and that is sin square theta. Similarly if you have cos theta whole cube there's another way of writing that and that's cos cube theta. So whenever you have square like this or cube like this sin square cos cube remember that the correct form of writing that is sin theta square the problem mathematical form is this is a short way of writing that so you need to understand that this is how this notation works it does not work for negative angles so if you have sin theta^ minus 1 it does not work for negative powers not angles does not work for negative powers so if you have sin theta^ minus1 this is 1 / sin theta Sin inverse theta. This is a different function. These are not equal to each other. So this is a very common error that happens here. Sin theta minus 1.

[06:23:46]This is 1 / sin theta.

[06:23:49]Sin inverse theta is a different function. That's the inverse function of s. Inverse is different from reciprocal.

[06:23:55]Okay. So in negative powers we cannot use this notation to mean that it's power minus one. Power minus one is different. Sin inverse is a is a different thing. In positive powers we say that these are equivalent.

[06:24:09]This is just a short way of writing that positive power. Now we have two identities that basic that we basically have here. Sin square theta + cos square theta that's identical to 1 and tan theta is identical to sin theta cos theta. If we rearrange these we get something like this. Sin square theta can be written as 1 - cos square theta.

[06:24:27]So a lot of times what we have to do is we have to convert sin square to cos square or cos square to sin square. We can use these identities to do that conversion. If you have sin square, we can write that as 1 - cos square. If you have cos square, we can write that as 1 - sin square.

[06:24:42]In this other identity, tan theta= sin theta cos theta. This is written in the formula sheet directly. Sometimes you have tan square. What do you do in that case? Tan square can be written as sin square theta cos square theta. Okay. So that's also some something that you have to use. Sometimes make sure that you remember this as well. If tan is s cos then tan square theta is equal to sin square theta cos square theta. However, these are not correct. For this identity identity to work, you need to have squares. You can't have single powers.

[06:25:11]Sin theta is not 1 minus cos theta.

[06:25:13]That's wrong. cos theta is not 1 minus sin theta. That's also wrong. Indices rules do not work when you have plus minus uh like this. They don't those rules don't apply in this case. They apply only in multiplication division.

[06:25:27]So this works for both power one and power two. But this only works for squares. It does not work for this.

[06:25:34]Okay. And it this is also incorrect. Sin theta + cos theta square that's also not equal to one. Okay. Only sin square theta + cos square theta that is equal to 1.

[06:25:48]Now why are those identities important?

[06:25:50]Because you get equations that look something like this. You'll have multiple functions inside uh an equation. So you we know how to how to solve for example sin theta= equal to something cos theta equal to something.

[06:26:00]We know how to solve equations like those. But what if we have something like this? When where we have multiple functions s cos we whenever you have to solve any equation involving trigonometric functions we have to convert that equation to a single function. So in this case we've got multiple functions s and cos. How do we convert them to a single function? In this case when you have sine and cos a lot of times what you do is you simplify everything and then divide cause on the other side to convert it to tan. That's something that that will happen a lot.

[06:26:27]Keep that in mind. When you have s and cos you try try try try to convert them to tan somehow when and they have single powers right when you have powers one then you try to divide cos on the other side and make it sign over cos it can be converted to time and then once it converts to tan then it's fairly straightforward to solve after that you can just find the basic angle find coordinates and that equation can be solved like that another idea is that you could just divide that whole equation by cos theta so one idea is that you take sign to one side cost to cos to another side and then you divide cos like this that you take it to the denominator on the other side. This is another idea that you think directly like this. If you have an equation like this 3 sin theta minus 3 cos theta minus sin theta divide the whole equation by cos what happens in that case is cos and cos they get cancelled sin over cos that becomes tan and then you can solve that equation further.

[06:27:20]If you have squares like this for example 3 sin square minus 2 cos theta.

[06:27:24]Now this is where that sin square theta + cos square theta= 1. That identity is going to be useful. You can convert sin square to cos square so that it becomes a quadratic equation like this and then you solve it like a quadratic find the root find the possible roots from that use coordinates and find the basic angle.

[06:27:42]Now this is something that can happen as well sometimes that you get disguised quadratic equations in trigonometry. 2 sin raised to power 4 theta plus sin square theta - 1= 0. So they want you to convert this form to this form. First of all, that's straightforward that you convert tan square to sin square over sin cos square and then you simplify that and that gives you this form. How do you solve this equation further? It's a disguised quadratic. We've got powers that are in the ratio to ratio 1. One power is two times the other. Right? So this is something that can be solved as a disguised quadratic.

[06:28:16]What we do is we substitute the smaller power with another letter. So we can say let let u equ= sin square theta and then you solve it like a quadratic. At the end you put that substitution back and you get this result. Now in this case you're getting sin square theta= minus 1. That's not possible because sin square cannot be negative. But from this you get sin theta = + roo<unk> 1 /2. Now do not forget plus minus in such questions we're getting plus minus. That means sign could either be positive it could be negative. What does that mean? It could be in any one of the four quadrants because it's plus minus sign is positive in the first and second quadrant. It's negative in the third and fourth quadrants. We have to consider all possible quadrants. So we find the basic angle. The basic angle turns out to be 45° here. And then we say it could be in all four all four quadrants. So we write down possible angles in all four quadrants like that. Then you have to prove identities somehow as well. What does it mean to prove identities? You have two sides of this. What you have to do is you have to go start from one side and prove that equal to the other side.

[06:29:21]Okay, how would you do that? These are two identities that you can use to prove an prove an identity. Sin square theta + cos square theta is 1 and tan is equal to s over cos. So you start from this and you think about okay what's the final result that you want and the final result that you want you don't have any any sign in that. So you have to eliminate sign somehow from this also on the left side you have multiple fractions on the right side you only have one fraction. So what you want to do is you want to convert it to a single fraction. For that you take LCM on this side. First of all when you take LCM you simplify that try to use those identities and eventually get this form that you want on the right side. Now while you're proving identities these are some things that you need to keep in mind. You start from the more complex side generally. So the side that looks more complex you start simplifying that and prove that to be equal to the other side. So in this case for instance the left side is more complex because it has two different functions. It has s and cos. It also has two different fractions. The right side only has one fraction. So how do we identify which side is more complex? The side that has more different types of functions. For example, if one side has sign and co and the other side only has sign, the other side only has cos then the one that has more types of functions that's more complicated. And then if one side is a fraction and the the other side is not a fraction, you start from the side that that that is a fraction because that will be easier to work with. And then if the first two rules they you cannot differentiate them based on these first two rules. So think about which side has more number of terms. If so one side has two terms and the other side just has one term. it'll be easier to convert these two terms to a single term rather than breaking this single term down to two terms. Right? So these are some things that you need to keep in mind to decide which side to start from. Now in most of the question 90 95% of the questions you will be starting from the left side. So they give you identities in a form like this that the left side is generally more complex. But in some rare case when uh it looks when when you figure out that the right side is more complex you can also start from that and prove that equal to the left hand side and how do you decide that based on these rules that I've given you so there are three ways of proving identities one you start from the left hand side prove that equal to the right hand side you start from the right hand side and prove that equal to the left hand side both of them are acceptable if both sides look a bit complicated this is another way that you can do you can say let's simplify the left hand side a simplify the right hand side and then somewhere in the middle they will both become equal to each other and then you say that the identity is proved. That's also possible. This is also allowed.

[06:32:00]However, while you do that you cannot take terms from one side to the other.

[06:32:04]You can't say let's take one take one side from one side and take one term from one side and minus that on the other side and divide that on the other side. That's not allowed. the ter the both both sides of the identity they should be uh dealt with separately. You can't take terms from one side to to the other while you're proving identities that's not allowed. Okay, you can simplify both sides separately and then meet somewhere in the middle. That's okay. But don't take str from one side to the other. Now these are some things that will help you prove identities more efficiently. When you have fractions, when you have any fractions and there's some denominator somewhere, do not simplify that denominator. Why is that important?

[06:32:45]Because a lot of times what happens is you end up having an opportunity to cancel something in a later step. So if you have the denominator in a factorized form like this sin theta into 1 plus cos theta, it'll be easier to cancel something something there. For example, it might be that the numerator ends up having 1 plus cos theta and you can cancel it out or you might be able to cancel sin theta somewhere. If you multiply that, if you multiply it like this and write it sin theta plus sin theta cos theta, yes, you can factoriize that again if you want to convert it to that form. But it's very likely that you would not get that idea in the exam that you can you know factoriize and then do cancellation. So whenever you have in the denominator something like this that's factorized form always consider the possibility of a cancellation at a later step and do not simplify this.

[06:33:32]Keep it in this in this form. When you have figured out that there's no cancellation possible then yes you can see if you need to simplify that further. This is an identity that's used a lot while proving identities. Keep that in mind as well. A squareus b square that's a minus b into a + b. So often in this direction for instance you might have something like this 1 - tan square theta and to simplify that you may be able to write it in a + b into a minus b form. How do you do that? You can write 1 as 1 2 so it becomes a square - b square then and then you write this as a + 1 + tan theta and 1 - tan theta. If it was a fraction and you had something in the denominator as well, it it's very likely that the denominator would have either this term or this term and you can do some cancellation there. So this identity is used a lot uh this algebraic identity.

[06:34:25]Whenever you have a square minus v square, you can factoriize it like this.

[06:34:30]This is another thing that you'll find very helpful when googling identities.

[06:34:33]When you have tan in any identity, I'll show you some examples of this as well.

[06:34:37]When you have tan on one side of an identity that you're starting from and there's no tan on the other side, it's mostly more efficient to convert that tan to sign over cause right away in the very beginning. And that will help you prove that identity more efficiently than if you were to not convert it in the beginning and just simplify it as it is and then at a later step convert it to sign or because it'll make your working much longer. So if there's an there's there's an identity in which on one side you have a tan term on the other side there's no tan for example you might have sign there or might have co there you know that tan has to be eliminated eventually because if you want to convert to sine or cos somehow rather than waiting for later you would right away replace it with sin theta or cos theta and that will simplify your working in most cases you can try that while we're proving identities Now tan can be written as s over cos. When you have cos over s somewhere, you can also write that as 1 / tan. Okay? And then you can sorry if you have 1 / tan somewhere, you can also write that as cos / sin. And that will simplify your working. Let's have a look at this example. So you have 1 - 1 sin theta - 1 / tan theta squ. And you have to prove that this is identical to 1 - cos theta over 1 + cos theta. Now one idea could be that we take LCM here first of all and then we square that and then we somehow you know simplify everything and then convert tan to s over cos compare that with replacing 1 / tan with co over s at the very beginning. What happens when you do that do that is the denominators become the same and the simplification becomes much easier much quicker right it becomes much quicker and you are able to prove the identity much faster so whenever you have tan in the in an identity and you do not want tan in the in the final result that you got convert tan to s over cause right away or one over 10 to cause over sign right away okay and that will help you do this better do this quicker and then a lot of times what happens is they make you prove an identity in one part and then they make you use that in the next part. For example, they ask you to prove this identity in the first part and then they ask you to solve this equation. Okay.

[06:36:55]Now, how do you solve this equation? You say you proved in the first part that this thing is equal to this thing. So, you replace it with that.

[06:37:06]So, the equation becomes 1 - cos theta / 1 + cos theta = 2. And now that's straightforward to solve. you make cos theta theta the subject and then you find use quadrants and basic angle to find out the final answer from that. Now this is uh a relatively complicated part and that is okay. So I'll answer that later. Solving equations having angles that are not just x or theta, not just x or theta.

[06:37:47]Now what does that mean? So until now the equation that we have seen look like this. For example, cos theta= c 7. We know how to solve that. You could have x here or theta here.

[06:37:59]However, what if you have something like this sine of theta + 60. How do you solve this equation? Now this is what I recommend that you do for this. Let's say you've got this interval given theta from 0 to 360. That's the interval that's given, right? What you do is you say whatever is inside the bracket you substitute that with another letter. So you say a = theta + 60. You do substitution like this. Whatever is written inside the bracket you substitute that with another letter. So in this case a becomes theta + 60. So now that equation is reduced to sin a= 0.5 and that becomes much simpler and you can solve it using the previous way.

[06:38:35]However there's a catch that is the interval that was given that was for theta. The angle now is a that's not theta. So you also have to adjust the interval. How do you do that? Theta is from 0 to 360. And you are saying that a is equal to theta + 60.

[06:38:53]How do you find the corresponding interval for a? You input 0 here in place of theta. What's the corresponding value of a that you get? That's 60°. So the minimum value for a is going to be 60°.

[06:39:04]Input 360 here in place of theta. What's the corresponding value of a that you get from that? That's 420°. So the interval that you've got for a that's from 60° to 420°. So now when you solve this equation, you have to solve this within this interval. So you do it the normal way. Now find the basic angle.

[06:39:21]The basic angle turns out to be 30°. And now sign is positive. S is positive in the first quadrant or in the second quadrant. Since the basic angle is 30, the first angle that you get here is this angle. This is the first angle in the first quadrant. That's 30°. However, if you notice that falls outside the interval that's not in the interval. So, you do not consider that. Okay? Because 30 does not fall in the interval. You you cannot consider that. So, you go beyond that. This is the next angle that you get. That's 150. 150 does fall in that interval. So, that's your first angle. That's your first angle. Okay?

[06:39:55]And then since the interval is going beyond 360, that means you have to go in one full circle. go one full circle and see if there's any other possible angle that you can get inside that interval.

[06:40:06]So you can go for one full circle that's 360 come back here that's 390. 390 falls in this interval. You will consider that another possible angle. You'll say the value for a could be either 150 or it could be 390. And then you put that substitution back. Theta + 60 is equal to 150. Theta + 60 is equal to 390. And that gives you the corresponding values of theta. If you had not adjusted that interval, you would miss out on this possible answer 360 + 30. So adjusting this interval is very important. Whenever you do a substitution like this, you always have to adjust the interval as well.

[06:40:43]Let's take another example of this.

[06:40:44]There's actually a going on here. We'll have a one minute break and then we can have a look at this example. We'll discuss it then.

[06:42:02]Look Okay, before we resume any uh one having problem understanding what I'm saying?

[06:42:11]Am I am I speaking too fast or is it okay? Cuz I I'm feeling that I'm sometimes speaking so fast that I can't understand what I'm saying myself. Is it okay?

[06:42:27]Does everyone understand what I'm saying? It's okay. Of course, if you've done these concepts, then um if you're if you're if you're seeing something for the first time, then uh you have to cover that concept in detail. That's that's going to be important.

[06:42:43]Okay, got let's we're doing more examples of that. We're doing we're doing more more examples of that. Okay.

[06:43:03]Okay, let's assume sine of 2x + 3 cos 2x. Again you see the angle is not x or theta is 2 * x. So what do you do in this case? You do use a substitution.

[06:43:12]You say let a or any other letter equals 2x. Before that we have convert that converted that to one function. This was sin 2x= - 3 cos 2x.

[06:43:24]So we divide cos 2x on the other side it becomes tan and we have tan 2x=us 3. And now since the angle is not x, it's 2x.

[06:43:32]We use a substitution. We say let a = 2 * x. So whatever is the angle, you substitute that with another letter. In this case, we say a is equal to 2x. Now the interval for x goes from 0 to 180.

[06:43:44]What's the interval for a going to be?

[06:43:45]You input 0 in place of x here that becomes 0. Input 180 in place of x here that becomes 360. That means the interval for a therefore is from 0 to 360. So now when you solve this equation tan a= minus 3 find the basic angle the basic basic angle turns out to be this when you uh are looking at the quadrants now this is the interval that you have to consider from 0 to 360. Now tan is negative where is tan negative tan is negative in the in the second quadrant or in the fourth quadrant. Find out the angles within this interval. So in the second quadrant quadrant it'll be 180 minus this. In the third in the fourth quadrant it will be 760 360 minus this.

[06:44:25]and you get two angles from that. Put that substitution back at the end. Put 2x= equal to this. 2x equal to this. And you find the corresponding values of x from that. The final answer that you get when you follow this method correctly.

[06:44:38]The final answers are automatically going to fall in the given interval that you started from. So the value values of x that we had. Okay. Now when it comes in radians, it becomes a bit too complicated. For example, if you had something like something like this sine of 3 theta minus<unk> /4. Let me actually show you this example first.

[06:44:57]We have to solve this equation. Sine of 2 theta +<unk> / 3 = 1 /2 where theta is between pi 0 and pi.

[06:45:07]That's this. This is this is the equation. Angle is between 0 and pi.

[06:45:10]What do we do? The angle is not just x or theta. It's something multiplying with theta and something added adding as well. How do we deal with this? Whatever is inside the bracket, you substitute that with another letter. So you say let a= 2 theta +<unk> / 3. Okay. Now you have to find the interval for a as well.

[06:45:28]How do you do that? Input 0 in place of theta that gives you p<unk> / 3. Input pi in place of theta and that gives you 7 p<unk> / 3. So now when you solve this equation sin a= 1 /2. You have to find all possible angles in this interval.

[06:45:42]That's<unk> / 3 to 7 / 3 *<unk>. Now you solve it the normal way. You find the basic angle. The basic angle turns out to be 6. And now s is positive. Where is s positive? sign is positive in the first coordinate and it's positive in the second coordinate. Okay, in the first quadrant the angle turns out to be p<unk> / 6. So in the first coordinate we've got p<unk> / 6. Now is that in the interval? The lowest value that you have here is pi over 3. Pi / 6 is smaller than that. So you cannot consider that it does not fall in the interval. Even though in the original interval the smaller value was zero but we're not solving for theta here. We're actually finding the value for a. The value for a has to be between p<unk> / 3 and 7 p<unk> / 3. So the first angle that we're getting here p<unk> / 6 we cannot consider that that's outside this interval.

[06:46:34]Then what about the next angle? This is this is the next angle that we're getting. This is going to be p<unk> -<unk> / 6 and that gives us another value. Uh this is in inside the interval. This is 5 pi / 6. You can compare that with pi over 3. That isn't the interval. So you consider that. Then you go further. You go one full circle.

[06:46:52]Since the interval is 7 pi over 3, 7 p<unk> over 3 is greater than 2 pi.

[06:46:57]Right? Complete circle is 2 pi. 7 p<unk> over 3 is 2.33 pi. So that's more than one full circle. That means it may be possible that you go one full circle and come back to get another possible angle.

[06:47:08]So if you go one full circle, that's 2 pi plus that basic angle pi / 6 that's 13 p<unk> / 6. Is 13 p<unk> / 6 in the interval? Yes, 13 p<unk> / 6 is in the interval. So you just equate that to 2 theta + p<unk> / 3.

[06:47:24]Find the value of theta from that.

[06:47:25]That's another possible angle that you get. If you are finding it difficult difficult to compare these two, just put them in your calculator. 13 / 6 and 7 over 3. 7 over 3 is bigger. 13 / 6 is smaller. That means it does fall in the interval. So you get two possible solutions from this. These are the two values of theta that you get. Now sometimes you end up getting something something more complicated. It's it it happens more in paper three later on um less in paper one but it can happen that you end up getting four answers like this. So you can have a look at this example. So basically this is this is the angle. We replace that with another letter. We say let a equals this. We adjust the interval the same way that turns out from minus<unk> / 4 to 11 p<unk> /4. And then since the interval is going going in the negative direction, we also have to consider angles in the negative direction. So we have to have to check if there's anything in in the in the interval in the negative direction. And the upper limit is going beyond 2 pi. So that means we also have to check if there any other any angles in the positive direction after completing one full circle. So one full circle and then you can come back here to get more angles.

[06:48:30]So in this case you end up getting four possible angles all of them in the positive direction, none in the negative direction. You can check that. Have a look at this working. go through this.

[06:48:39]This is a good example. Okay. Now, sometimes you end up getting equations that look like this where you have sin inverse where you have sin inverse of something. Sin inverse of x - 1 =<unk> / 3. Now, how do you solve this equation?

[06:48:52]Sin inverse of x squareus 1. That's not something that something that you can evaluate on your calculator because this is x here. Now, on the other side, you've got pi / 3. What you can do is you can take sin inverse to the other side and that gives you sin pi / 3.

[06:49:07]And now it becomes a simple quadratic equation. You've got x square and you've got an a number on the other side. Just solve it like you would normally solve any quadratic and you can solve it and you can get the value of x from that. So when you have sin inverse in an equation given consider if so sometimes sometimes you can have something like this as well sin inverse of maybe under root 3 over2.

[06:49:28]If you have some some number inside just evaluate that on your calculator and that will work. if they have some variable inside like in this case they have a variable inside in that case this is something that you cannot evaluate from a calculator directly. So what you would consider doing is you would consider taking this to the other side.

[06:49:43]So in this case when you take s to the other side that becomes sin pi / 3 and then you can solve that equation like this.

[06:49:49]All right let's move on. Now these are some identities that you need to know about when when the question does not specify if it's if it's if it has to be a fraction or decimals it doesn't matter.

[06:50:01]You can write any of them and that will that will be fine. If the question specifies then you have to follow that instruction. Okay. Complimentary angle identities. Now we've got some identities that are not given in the formula she C that you need to know about. Complimentary angles are those angles that add up to 90. Right? So for example 40 and 50 these are complimentary angles because they add up to 90. We have these identities that for example if you have sine of 25 that is equal to cos of 65. If there are two angles that add up to 90, sine of one is going to be cause of the other.

[06:50:36]So for example, if you have sine of 60, sine of uh 70 that's the same thing as cos of 20 because they add up to 90. Sine of one is going to be equal to cause of the other. Now there's a logic behind it.

[06:50:56]You can have a look at this. But that's the identity that we have. Sin of 90 - theta that's cos theta cos of 90 - theta is sin theta and tan of 90 - theta is 1 / tan theta. These are three identities that you need to remember. These will be helpful. These will actually be needed in some uh trigonometry questions. Sin 90 - theta is cos theta. Cos 90us theta is sin theta and tan of 90us theta is 1 / tan theta. In radians those same identities become this sin of pi minus theta<unk> / 2 - cos theta. cos of P<unk> /2us theta is sin sin theta and tan of P<unk> /2us theta is 1 / tan theta you need to remember these identities they're not given in the formula sheet you need to use them sometimes let's have a look at this example so they have given you that cos of X= P where X is an acute angle in degrees and we need to find the value of the value of sin X tan X and tan of 90 - X in terms of P now one way to do this is using identities you can say okay if cos X is equal to P we know that sin square x + cos square x= 1. Just input the value of cos x in that. Find the value of s from that. So s turns out to be plus minus roo<unk> of 1 - p^². Now do we use the plus sign or the minus sign? Since the question says it's an acute angle that means it has to be in the first quadrant. In the first quadrant sign is positive. So we only consider the positive value here. We say sin of x=<unk> 1 - v^2. Now in the second part they say find the value for tan x. tan is equal to sin / cos. We already are given the value for cos. We have found the value of s sign s in the first part. So we can say tan x= sinx over cos x. And that gives us the result. And then they say find tan of 90 - x. How do you do that? We just did did that identity. Tan of 90 - x is the same thing as 1 / tan x. If we know tan x, we just you just take reciprocal of that and that will give the value for tan of 90 - x. That's one way to do this. You can do it like this. But sometimes there will be some questions that you will not be able to do using only identities. And for that there's another method that you also need to know and that is by making right angle triangles. So if they're saying cos of x is equal to p and x is an acute angle. What we can do is we can say x is an acute angle that's in the first quadrant. We we can make a triangle like this in the first quadrant in the first quadrant such that this is the angle x right? If this is the angle X and and we're given that the value of cos of X is equal to P. Now we can consider P as P over one. That's the same thing, right? Cause X is equal to P. That's what the question said. We could say that's P / 1. Now cos is defined as base over hypotenuse. So we can say if the base is P, then the hypotenuse is going to be one. So we can say base is equal to P, hypotenuse is one. Can we find the perpendicular from that using Pythagoras? We can find the value of perpendicular from that. And in this triangle now we know that the base is P. The perpendicular is square root of 1 - P^ square and this hypotenuse is equal to 1. So now once we have this we can write down the values of sin cos 10 directly from that triangle. S is perpendicular over hypotenuse.

[06:54:00]Perpendicular is 1 - P^ square over the hypotenuse that's 1. That gives you the value for S. Tan is perpendicular over the base. perpendicular over base that gives you tan tan of 90 - x. If this angle is x, the angle in top here is 90 - x. So if you find tan of that angle, that turns out to be opposite over adjacent. That's p /<unk> 1 - p^ 2. Okay. Now in this particular question, you could use identities as well and they will work just as well. However, you'll find some examples in which you would not be able to use identities or you'll have to remember too many identities in that case like there like 15 20 of them. If you want to memorize all of them, you can do that. But uh that that's too much that's it'll be very difficult to remember all of them. So understand the logic behind it. Acute angle x is such that tan of xal is equal to k where k is a positive constant express in terms of k tan of pi minus x. Now we don't have any identity that we've learned for this tan of minus x. How do you figure this out? You make a right angle triangle for this. What right angle triangle?

[06:55:05]The angle X is acute. Okay. So, where is X going to be? X is going to be in the first quadrant. So, let's make a triangle in the first quadrant like this. First of all, such that this is this is the acute angle. That's X. Now, we are given that tan of X is equal to K. So, we can say tan X is equal to K.

[06:55:25]Right? If tan is equal to K, we can say that's K over 1. That's the same thing as that tan is defined as perpendicular over the base. So if perpendicular if perpendicular is equal to k then base is equal to 1. Use pythagoras to find hypotenuse and that turns out to be the square root of 1 + k 2. So now that tells us that this is k this is 1 and this is square root of 1 + k square.

[06:55:51]Okay. Now this will help you solve some of these parts. For example, in the last part, that's the two marks, two mark part, they're asking you for sinx. Okay, what is sin x? Sin x is perpendicular over the hypotenuse.

[06:56:05]Perpendicular is k in this case.

[06:56:07]Hypotenuse is this. So the value of sign turns out to be this k over 1 + k. Now for the other parts for the first for the second part as well, let's do that as well. So tan of p<unk> / 2 - x<unk> /2 is 90°. We can say if this angle is X then this is going to be P<unk> /2 - X right because this is P<unk> / 2 so the remaining angle is P<unk> / 2 - X find the tan of that and that tan of that turns out to be opposite over the base that's 1 / k but the problem is what about - tan - x now where is that going to be if you think about it x is an acute angle x is an acute angle if you do pi minus that you think about an acute angle in degrees 40° Pi is 180°. So 180 - 40, what is that going to be? 140° that's an obtuse angle. So if X was an acute angle, p<unk> - X is an obtuse angle, right?

[06:57:03]That will go in the second quadrant. So now we've got this triangle in the second quadrant.

[06:57:10]This one such that the angle here, this angle is pi - x. Now what's the basic angle in that case? If the straight line is pi, then sorry about that.

[06:57:45]We were here, right? Yeah. So this is the triangle in the second quadrant and this angle is p<unk> - x. Now we know that the straight line is pi radian. The straight line is pi radian.

[06:58:03]This straight line is pi radian. If this is p<unk> - x that means this angle here is going to be x. So the basic angle is x. If the basic angle is the same that means the triangle here that we would get it would have it would have the same length as that first triangle the perpendicular is K.

[06:58:20]The high part is square root of 1 + K 2 and the base is going to be 1. So that means whatever was the tan whatever was the tan value earlier K the value of tan here is also going to be K k over one.

[06:58:35]However, since we're in the second quadrant. In the second quadrant, we know that tan is negative, we put a negative sign with that. So, the value turns out to be negative. Okay? So, you need to understand how to work with these triangles. I'm going to show you some more examples that you can have a look at in your own time. Have a look at this working.

[06:58:53]Then have a look at this working as well.

[06:59:00]Okay. Then the next thing that you have is graphs. So you can have a look at those examples later on in your own time. So trigonometric graphs that's the next major concept that you have here.

[06:59:13]You need you need to be able to sketch these graphs sin x cos x and tan x and any of their transformations. You might have something multiplying with them something adding subtracting with them.

[06:59:23]You need to be able to sketch these graphs.

[06:59:26]Now you need to remember the shapes of the graph. Sinx looks like this.

[06:59:30]Its shape is this. Okay, you need to remember the shape.

[06:59:36]The cause graph looks like this. You need to remember this shape. And the tan graph is in multiple parts like this.

[06:59:46]And it has asmmptotes at different points as well. How would you actually draw these graphs? Whatever function is given whether it's just sin x or is it or if it's sin of 2x or if it's 2 sin x - 4 whatever is given all you have to do is put that in the table function on your calculator put that in in the table function on your calculator and you get some values of y from that for example if you want to draw the graph from 0 to 360 all you have to do is put this in the table function on your calculator go to menu table in this case the angle unit is degrees.

[07:00:22]So we we choose degrees here. f ofx equals sin of x.

[07:00:28]Put that on your calculator.

[07:00:30]Start from zero. End at 360.

[07:00:34]Take a step so that we have enough values. You can take 45°.

[07:00:37]If you know the shape um dividing the interval by 4 is generally going to be enough. So you divide 360 by 4 and that's 90. You have these values. plot these values and through those values just draw the sign graph and that's it.

[07:00:52]So that's all that you have to do.

[07:00:54]Whatever transformation it is if it's multiplying by something if if something is being added or subtracted you don't care about that. Just put that in the table function. Plot those points join those points and you get your graph. All you have to do is remember the shape of the graph. Sign looks something like this. It could either be like this or some reflection of this. Cause looks like this and the tan shape as well. You need to remember those shapes. As long as you remember the shapes, you will plot some points. You will rec recognize what the shape looks like and that's it.

[07:01:20]You'll get your graph. Okay? So, you just use a stable function for sketching any of these graphs.

[07:01:26]Now, we have some things that are defined for these graphs that that you need to know about. First of all, we've got period. Period is the interval in which the graph completes one full cycle. For s and cos, the period of the graph is 360° or 2 pi radians. For tan the period is 180 degrees of pi radian.

[07:01:45]Then you have you've got the baseline of the graph. Baseline is the line about which the graph moves up or down. For example in all of these graphs in the standard graphs the x-axis is the baseline. This is the line in the middle of the maximum point and the and the min minimum point. It's exactly in the middle. The graph is oscillating about this line like this. This is the baseline the x-axis and the standard graph for all of these functions for s co and also tan.

[07:02:14]Now for s and cost you'll be able to to figure out the baseline by finding the midpoint of the maximum and minimum point. So if there's some transformation we'll see some some examples in which you have to figure out what graph it is or what function it is. You can find the baseline by just finding finding finding the midpoint of the maximum point and the minimum point for tan. However, since no maximum or minimum exists, it goes from negative infinity to positive infinity. For recognizing what the baseline in a tan graph is, you basically identify where the graph is changing its direction. So it it goes in one direction first and then goes in the other direction like this. So when it changes direction, that's the point where the baseline is going to be for a tan function. Okay. Now that's the baseline. Amplitude is the maximum distance the graph can go uh from the baseline in the vertical direction. In s and cause the standard amplitude is amplitude is one. For tan there is no limit. So this is the summary. For s and cos the amplitudes are one. For tan it's infinity or you can say it's undefined.

[07:03:22]Periods for s and cos are 360 and 2 pi.

[07:03:24]for co for tan it's 180 180 degrees or pi radians. Now you you need to understand what happens when there's some transformations of these graphs.

[07:03:35]So when when there's something multiplied outside or some some some number multiplied outside uh some a sign multiplied outside plus minus or something being some not plus minus just minus multiplied outside or something is being added or subtracted outside. What happens in each of these each of these cases with this function? When you multiply by a negative sign that's a reflection in x-axis when you multiply by a number that's a stretch with scale factor a. But in terms of the sign function you can also understand that this number represents the amplitude of the graph. For both the sign and cos functions if you have a number multiplying outside that represents the amplitude of the graph. For tan the amplitude does not exist. So for tan that's not the case. You just think of that as stretch. But for sign and cause you understand this number as the amplitude of the graph and the number that's adding or subtracting outside that translates the graph up or down generally. But in terms of the trigonometric function you can also understand that it represents the baseline of the graph. So the number that's outside that's the baseline of the graph. So if you have sine of x + 3 the baseline of that graph is y= 3. This horizontal line. If you have a function that's 2 sin x, the amplitude of this graph is equal to two. So this is what these numbers represent. So in vertical transformations, something is happening outside the whole function, right? Number multiplying or minus multiplying or something adding or subtracting.

[07:05:07]Only the ycoordinates change because of this, right? Then you have horizontal transformations. In horizontal transformation, what happens is you have some operations applied inside the function with with x. X could be multiplied by negative X could be multiplied by a number or this there could be something adding or subtracting to X like this.

[07:05:26]Now when when you have minus multiplying here that means it's a reflection in the line yaxis.

[07:05:33]When you have a number multiplying that's a stretch with scale factor 1 / as we'll do in as we'll see in function transformations as well. But in terms of trigonometric functions, this number that's multiplying inside it also means that the period is going to change because of that. So if you have tan of 3x for instance, tan of 3x. Now this basically tells you that the period of the graph is going to get divided by three. So what's the original period for tan? Original period for tan is 180°.

[07:06:09]When you have 3 * x, the period gets divided by three. And now tan completes one full cycle in 60°.

[07:06:16]If you had sine of half x, what about the period? Its original period is 360 or in radians is 2 pi.

[07:06:27]That gets multiplied by 2 because 360 over 1 /2. What's that? That's 360 * 2 that becomes 720. So because of this number that's multiplying, what happens is the period changes. weed gets divided by that number. Okay? And then the number that's adding or subtracting here, there's no special interpretation of this. It's just translation. If you have plus here, that means you're going left. If you have minus here, that means you're going right. So, for example, if you have tan of x - 90, what that means is you're going 90 units to the right side.

[07:07:00]Let's have a look at some quick examples and then we uh finish this and I'll take your questions if there are there if there are any. It says sketch on a single diagram the graphs of y= cos 2 theta and y = 1 /2. How do you do this?

[07:07:14]You put this on your calculator. y= cos 2 theta. Make sure your calculator is in radian mode.

[07:07:20]Take an interval of maybe 2<unk>i / 4.

[07:07:23]That's p<unk> /2.

[07:07:25]Or if you think you need more values, you take an interval of pi /4. Take a step of pi over4. Plot those values.

[07:07:32]Join those values. You will get the cause graph. Okay? There's no thinking that's required here. Just plot values and draw the graph. What you do need to remember is the shape of the graph so that you can draw that correct shape through those points. Y= 1 /2 is just a horizontal line. So we've drawn those two lines here. Now they will ask some questions based on this. They're saying where do the number of roots of this equation? Remember graphs of functions and O levels. You want the function that you have sketched on one side of this equation. How do you do that? This is 2 cos 2 theta minus one.

[07:08:05]So you rearrange rearrange this in such a way so that on one side you get cos 2 theta. Now on one side you get cos 2 theta on the other side in this case you end up getting 1 /2. So that means it becomes cos 2 theta= 1 /2. The solutions of this equation are going to be the same as the points of intersection of these two graphs cos 2 theta and 1 /2.

[07:08:26]And these are the same two graphs that you've drawn in the first part. Look at the points of intersection. There are four points of intersection. 1 2 3 four.

[07:08:33]So you say that's the number of solutions. That's four. They will not ask you to find the exact solutions because that's not possible from a sketch. They will only ask you for the number of solutions. Just look at the number of number of points of intersection and you write down the result. Now there could be something more complicated like this as well.

[07:08:50]They're saying reduce the number of groups of that same equation in the interval from 10 pi to 20 pi. Now from 0 to 2 pi there were four solutions. Now since the graph is symmetric and it's going to keep repeating like this. It's a cyclic graph. It's it's going to keep repeating like this. In an interval of 2 pi we're getting four solutions. Using ratios we can say if from 0 to 2 pi we're getting we're getting four solutions. Then from 10 pi to 20 pi that's a gap of 10 pi radians. In that interval it's going to be 20 solutions because 2 * 5 is 10 pi. So 4 * 5 that gives us 20. That's the number of solutions that would have that we would have in that case.

[07:09:28]You can they sometimes ask you to solve inequalities using graphs like this as well. So in the first part they just ask you to solve this equation in this in this question and then they ask you to sketch two graphs. Let's say you've done that again using the table function on your calculator. You just sketch these two graphs. But then in the last part they're saying use your answer to part one and two to find the set of values of x for which x is between 0 and 360 for which 2 cos x plus 3 sin x is greater than zero. Now how do you make sense of that?

[07:09:59]You have solved this equation in the first part. 2 cos x + 3 sin x. We're almost done. We're almost done with trigonometry.

[07:10:08]2 cos x + 3 sin x = 0. We've got this equation. Right? Now we solve this equation. We found two solutions from that. One solution was this and the other solution is this.

[07:10:19]And then they want us to solve this inequality. Now what did what did the makers do in the second part? They made us sketch these two functions 2 cos x and - 3 sin x.

[07:10:34]Now if 2 cos x + 3 sin x is greater than zero. That's the inequality that we want to solve. If we rearrange this, what do we get from that? We get 2 cos x greater than - 3 sin x. Now these are the two functions that we have drawn. We have drawn 2 cos x. That's one function that we run and - 3 sin x and the inequality is saying 2 cos x is greater than - 3 sin x. 2 cos x is greater than - 3 sin x. In what region on the graph is the 2 cos x graph above the minus 3 sin x graph.

[07:11:13]So I represented the two cos x graph in orange and the other one in green. Where is the orange graph above the green graph in these highlighted parts? This part and this part. That is where the orange graph is above the green graph.

[07:11:28]So that those are the going to be the values of x that satisfy this inequality. Now what are those values of x? We have found in the first part the points of intersection. So we know that this point of intersection is 146.3 and we know that this point of intersection is 326.3. So what are the solutions going to be? X between 0 and 146.3. this left side and x between 326.3 and 360 that's this part of the graph. These are the solutions of this inequality. That's why we can do that using this graph.

[07:12:00]Okay. So sometimes they can ask you to solve inequalities like this as well from the graph. One last example.

[07:12:08]So you've got the sketch y= 2 sin x.

[07:12:10]Again you do that using the table function in a calculator.

[07:12:14]And then they ask you to add a suitable straight line to your sketch. Again, graphs the function O levels to determine the number of real roots of this equation. 2 pi sinx=<unk> - 6. Now, how does this connect to the first part?

[07:12:31]This is the graph that we have drawn. We rearrange this equation in such a way that on one side of this equation, we get 2 sin x somehow. How do we do that?

[07:12:37]We divide pi on the other side. And then on the left side, we're just left with this 2 sin x. And now we simplify this and that means this is one function y= 2 sinx. We already know the graph for this. And the other function is y = 1 - x / pi. We would have to draw this graph and then look at the points of intersection of these two functions. Now this is a straight line although it looks a bit weird but it is a straight line because it's mx plus c form.

[07:13:06]Minus 1 / pi is just a constant.

[07:13:09]So that's the gradient of the line. X plus the Y intercept. So what you do is you draw this graph. Take two points on this. Draw that graph. Look at the points of intersection. There are three points of intersection.

[07:13:23]So that's the number of real roots. And what's the equation of the straight line? We already wrote that there.

[07:13:28]That's y = - 1<unk> * x + 1. That's the equation of that straight line. And the points of intersection are three. So there are three solutions because of that.

[07:13:39]Okay.

[07:13:41]I hope that makes sense. Uh if you have any okay so let let me know if you have any questions now I I'll be answering those questions. Uh before that just a small pointer you have questions on function transformations. You have function and trigonometry transformation in which they give you graphs and they give you a graph and based on that they ask you to figure out what the equation of that function would be. Right now for that uh we will uh those questions we'll actually discuss when we do uh function transformations. That's the next thing that that we'll be doing. In that uh in that topic we will also discuss the transformations of trigonometric graphs.

[07:14:20]We would have graphs given and using those graphs we have to find out what the function is going to be.

[07:14:27]Okay. So let me know if there any any questions that you guys have.

[07:14:36]Okay, how do you determine the period for graph using calculator is generally given? So this is 0 to 2 pi. So that's start start is zero end at 2 pi.

[07:14:51]What do you mean by period? The starting the starting point and the end point open this mic as well.

[07:15:02]Okay, you can do that period for sketching graph the graph using calculator. What do you mean by period and the step? Okay, so the step uh you're talking about talking about the step is the interval that you want the gap between the gap that you want between values. So if if the graph is from 0 to 2 pi, how much gap do you want between each value? For example, in this case, you have a graph of 2 sin x, right? I generally recommend that you divide this interval into eight parts. Whatever the interval is, divide that by 8. So 2 pi over 8, that gives you pi / 4 in this case.

[07:15:50]However, if you remember the shapes of the graphs, well, taking four points is also going to be enough in a lot of cases. So if you think you remember your graphs well you can get away with just dividing by four as well. So take an interval of pi / 2. So in radians you take an interval of either pi /2 or pi / 4. In degrees you either take an interval of 45° or 90°. Generally that's the interval that you take either 90° or 45°. It depends on you depends on the number of points that you need. Okay.

[07:16:44]So divide the interval by eight. That will give you enough values. If you know your graphs well, uh you can also just divide by four and then if you remember the shape, you can draw that shape through those points and that will also be fine.

[07:17:06]So for figuring out this straight line remember graphs how do we uh solve equations graphically?

[07:17:15]We have one function that we have sketched already that's 2 sin x and the equation that we want to solve is 2<unk>i sin x =<unk> - x. We want on one side of this equation this function 2 sin x. So we want to get rid of this pi from there. How do you do that? You divide pi on the other side. When you divide pi on the other side, this is what you get. Pi / pi - x over pi. That becomes 1 - x over pi. That's the equation. I've just rearranged that here. It's not required to do that. You can just leave it like this as well.

[07:17:48]That's the equation.

[07:17:50]And then how do you sketch that? Take any two points. It's a straight line.

[07:17:54]Take put one value of x, you get some corresponding value provide. Put another value of x to get another value of y.

[07:18:00]And then draw a straight line through those points.

[07:18:06]If your calculator, if your calculator is not giving you exact values, I would highly recommend that you change that calculator. So if you have this version and that's not giving you exact values, then that's not original. That's some copy and that can cause some other problems as well. It might give you some wrong results for some for some uh expressions. It it can happen. It's it's a risk. I would not recommend that you take that risk. Change that calculator and upgrade that.

[07:18:40]Yeah, if if you take any copy that would not uh that would not be um reliable.

[07:18:48]You you can end up getting wrong answers from that as well. Any reliable shop. So if you are I I only know about Lahar. If you're in Laor you can go to main market. You've got those big bookshops like uh Anise I think and uh there's another in front of that uh ikabal book corner I think you've got readings get them get it from anywhere there it's expensive it is expensive uh there is another brand uh that has an has an exact uh same version and that's relatively more reliable as compared to the fake copies that you have in the Ed uh I'm forgetting the name for that. I think people from Karachi would be familiar with that uh because I think it's very common. There's some teachers there also recommend that. I've I've seen Shasad I think recommend that.

[07:19:40]Maybe not Shasad. So Sman probably I'm forgetting that name unfortunately.

[07:19:47]So that's basically another brand that makes copies of this as well but that's a relatively reliable brand as compared to the the copies that you have in the market generally. But I I don't remember the name for that right now. Let me think about it and get back to get back to you.

[07:20:04]Have I told about that calculator about that brand before? Uh Bazar has this calculator for 1,00. It's very unlikely that all the functions work on this.

[07:20:19]Make sure that you check that thoroughly. Make that Make sure that check that thoroughly that everything that you are doing it's giving. Yeah, Abaku. Abaku calculators. Thank you for reminding about about that. Abaku, that's another brand that's relatively cheaper. It has the exact same version.

[07:20:34]Uh, and that's relatively reliable. I think that's half the price of this. If you can get that and get that in Lahar, that would be a better choice.

[07:20:45]Online, I think it might have a website that you can order from. You can check online. I think it does have a website.

[07:21:01]What about this?

[07:21:15]Okay. So, if you're sure that it's working fine for you, then then that's okay.

[07:21:28]In the second part, we just sketch these these graphs or the third part. Are you asking about the second part or the third part?

[07:21:41]Can you just ask verbally? That will be easier. That quicker.

[07:21:46]Okay. Third part. Okay. 2 cos x + 3 sin x. You're saying this is the inequality that that you need to solve. We rearrange that and we get this 2 cos x is greater than - 3 sin x.

[07:21:58]What does that mean? On your graph, you've got two functions. One is 2 cos x and the other is - 3 sin x. The inequality is saying 2 cos x should be bigger than - 3 sin x. That means the value of 2 cos x should be greater than the value of - 3 sin x. So look at the graph and see the region in which the 2 cos x graph is above the minus 3 sin x graph. So that's in two regions. That's this region. This highlighted region that's where the orange graph is above the green graph and the region on the side that is on the orange graph is above the green green graph. So what's the interval that you've got for that?

[07:22:36]You know the points of intersection? You know this point. You know this point from the first part. So it's either between 0 and 146.3 or it's between 326.3 and 360. So those are your values of x. Okay. All right.

[07:22:52]Let's begin function transformation. So this is the next con the the next topic that we're doing function transformations.

[07:23:04]So there are three types of types of transformation that you need to know about in A levels. Reflection, translation and stretch. Reflection and stretch. Uh sorry, reflection and translation I've done in O levels as well. Stretch is something new for you.

[07:23:17]Now there are only going to be two types of reflections. Reflection in X-axis and reflection in Y-axis. Okay. So if you have this triangle here on the left side, it can either be reflection in X axis like this or Y axis like this. How would you describe this? You would say either reflection in x-axis so sorry in y-axis if it's in y axis or you would say reflection with yaxis as the line of reflection do not write reflection along y-axis that's not acceptable so you can either write in the short form in the short form you would use the word in okay reflection in y-axis or reflection in x-axis or to be safer you can write it full like this reflection with y-axis line of reflection or reflection with x-axis as the line of reflection. Now when it's reflection in yaxis what happens is if you've got this original figure and that gets reflected to this the movement of the points is horizontal right the movement is horizontal if the movement is horizontal we call this horizontal reflection when it's reflection in x-axis the movement of these points is vertical so we say this is a vertical reflection so reflection in yaxis we will be calling this horizontal reflection.

[07:24:45]Reflection in X-axis we'll be calling this vertical reflection because horizontal reflection means the direction in which you've got the points moving. Okay. Now that's something that you understand in this y-axis when that's the line of reflection it's horizontal reflection. When x axis the line of reflection that's vertical reflection.

[07:25:05]These are two types of reflection that you need to know about. Then you got translation. There are two types of translation that you need to know about.

[07:25:12]horizontal translation and vertical translation. In this case, you will always have to write down vector.

[07:25:18]So if you have a point, if you have a triangle that that's going from ABC to this transformed figure A dash B dash C dash, what's the translation vector for that? It's going one unit left and uh 1 2 3 4 units up. The translation vector is going to be min -14. Minus one means you're going one unit to the left. Four means you're going four units up. Four units up.

[07:25:45]Now for describing a translation, you always write down a vector. You cannot write this in words. You always have to write down a vector. That's necessary.

[07:25:55]Now these are two transformations that you already know about from before from O levels as well. Then there's one new trans transformation that's stretch. How does stretch work? There are two types of stretch that you have in syllables.

[07:26:06]Vertical stretch and horizontal stretch.

[07:26:08]In vertical stretch, what happens is you see the distance from the x-axis.

[07:26:17]If the scale factor is three, for instance, the stretch works like this. Every point, whatever it distance was before, it becomes three times that distance.

[07:26:27]For example, if you have this point C, this point C has a distance of one unit from the X-axis. After transformation, what happens is it becomes three units from the X-axis. That's a stretch with scale factor three. For B, the distance was 2 units from the x-axis. After transformation, what happens is it becomes 2 * 3 and that becomes six units. That's a vertical stretch. When you uh when the y-coordinates get multiplied by the scale factor. If it's scale factor of three, then you say whatever was the y-coordinate before, it gets multiplied by the scale factor. You can have a horizontal stretch as well.

[07:27:01]In horizontal stretch you think about the distance from the vertical axis distance from the y-axis.

[07:27:11]So for example if ABC this was your original figure and you want to stretch this by a scale factor of 1 /2 what what would you do for that? You would say B point is 3 units away. You multiply that 3 by - 1 /2 and that becomes - 3 /2.

[07:27:31]That's where the point B is going to be.

[07:27:33]Right? Similarly for A this point is at a distance of one unit to the right side. If it's stretched by a scale factor of minus 1 /2, what happens is the x coordinate 1, it gets multiplied by minus 1 /2 and it goes here. And that's how stretch works. So all you have to do is in case of vertical stretch multiply the y-coordinate by that number. In case of horizontal stretch multiply the x coordinate by that number. And that's how the graph moves. Your your figure becomes bigger in one direction. If it's a vertical stretch you will see that the height of the figure will change. If it's a horizontal stretch you would see that the base that the width of the figure that's going to change. So it was three earlier it becomes 1.5 like this. Right?

[07:28:19]So if it's a vertical stretch, the height changes. If it's a horizontal stretch, the width of the graph changes.

[07:28:26]Now these are three transformation that you have. How do you describe these transformations? You have to describe these in exactly one of these ways.

[07:28:33]Exactly one of these ways. So there are three types of transformations. First of all, you've got reflection. You ei talked about this before as well. Either say reflection in x-axis or reflection of the x-axis of the line of reflection.

[07:28:44]Similarly for the y-axis, this is a vertical reflection because the movement of the graph is vertical. Then you've got translation, you have to write down the translation vector for this. If it's a vertical translation, you have something in the y component here. If it's a horizontal transformation, you have something in in the x component here.

[07:29:02]For stretch, you always describe it like this. If it's a vertical stretch, you either say it's a vertical stretch with scale factor something. If it's a horizontal stretch, you say horizontal stretch with some scale factor. Another way to describe this is you say for vertical stretch, it's a stretch along x-axis. Stretch along x-axis with that scale factor. And similarly for the horizontal stretch or you can say stretch parallel to y-axis for vertical and stretch parall to x-axis for horizontal. These are the only three descriptions that you can have for stretch. Don't write anything apart from these three. Okay? Just one of these three. Don't write multiple. If one of them is wrong and the others other is correct, that will be marked wrong. So just write one of these whichever you're most most sure about. You can write just vertical or horizontal and that will be fine as well. Or you say either stretch along that axis or parallel to that axis. In reflection do not use along. If you want to write it in short form, use in or just say it's a reflection with that line as the line of reflection.

[07:30:10]Okay, and that's how you describe these transformations. So you've got six transformations in total here. Three vertical, three horizontal and vertical y-coordinates change and horizontal x coordinates change. Now how is that relevant to functions, right? These are transformations. How how do they relate to functions? So basically what can happen is you may have a function f ofx and it can have something multiplied or a minus sign multiplied or something adding or sub something added or subtracted outside. When you have something like this something adding or subtracting outside minus multiplying outside or a number multiplying outside these are vertical transformations. the y-coordinate changes because of that because you're multiplying something with f of x, right? The y changes.

[07:30:57]In vertical transformations, the graph moves in the vertical direction. Only the ycoordinates change and the operations, some operations are applied outside the function either plus minus multiplication with negative or multiplication with a number.

[07:31:14]These are three things that can happen when you multiply by the negative sign.

[07:31:18]That means it's a reflection with x-axis as the line of reflection.

[07:31:23]If you had points x and y, since the movement of the points is vertical, the point that was above that goes down. The point that was down that goes up. So the ycoordinate changes plus y, y becomes minus y, right? The sign of the ycoordinate changes. X coordinate remains the same. What about this number that's multiplying? When you have a number multiplying outside the function that mean it's a stretch along y-axis or you can say vertical stretch the scale factor that number which is a. So if you have x y is the original point that becomes x a y the ycoordinate gets multiplied by that scale factor. And if you have something adding or subtracting outside if you have plus or minus something that's a translation that's a vertical translation. Plus means you're going up minus means you're going down. Okay. And if you have a point xy that becomes x y + minus b the y-coordinate has b added or subtracted like this. Then you have some horizontal transformation. In horizontal transformation what happens is operations are applied inside the function and the graph moves horizontally and only the x coordinates change instead of outside the function.

[07:32:34]You you've got something happening to x inside x plus minus something or minus into x or a into x. Right? Now these are the three transformation that you've got. You can have something you can have minus multiplied outside minus multiplied with x. What does that mean? It's a reflection with y-axis the line of reflection. It's horizontal reflection. So the x coordinate changes.

[07:32:59]x becomes min - x. Y remains the same.

[07:33:02]when you multiply by a here you multiply by a number that's a stretch that's a horizontal stretch or you can say stretch along x-axis but scale factor reciprocal of this number so in horizontal transformation things work a bit counterintuitive when you multiply by a number it's basically a compression so the scale factor of this x is going to be reciprocal of that that's 1 / a so what happens to the coordinates if the original coordinate was x y it becomes x / a x is divided by uh that scale factor. X is divided by that scale factor. And then for a translation, if you have plus minus V, if you have plus minus V, what happens in this case is when you have a plus in the function, you go left. When you have a minus in the function, you go right. Okay. Now the translation vector, what is that going to look like?

[07:33:54]If you have plus here in the translation vector, you would have minus.

[07:33:59]And when you have minus here in the translation vector, you would have plus because plus there plus in the function means you're going left. Minus in the function means you're right. You're going right. So for example, if you have something like this f of x + 3, that means it's a translation three units to the left. The vector is going to be minus 3 0 for this like this. Okay, that's that's what happens in horizontal transformation.

[07:34:23]When you add or subtract something, that's translation.

[07:34:28]Outside the function, vertical. Inside the function, horizontal. When you multiply, divide by a number, that's a stretch. If you do outside the function, that's vertical. Inside the function, that's horizontal. When you multiply by negative 1, that's a reflection. When you do outside the function, that's vertical reflection. When you do inside the function, that's horizontal reflection. These are six transformations that you've got. Now, this is very important. Order of transformations. So sometimes you can have multiple transformations happening at the same time. So when you have only one transformation describing that is straightforward. Okay. But when you have multiple transformations when you have multiple transformations in that case the order is important. You have to apply those transformations in the correct order. Now how do you figure out what the order is?

[07:35:14]When you have horizontal and vertical transformation some horizontal and some vertical. In that case the order does not matter. You can do horizontal first.

[07:35:21]You can do vertical first. It does not matter. Both of them have the because the horizontal transformations are only going to change the x x coordinates.

[07:35:31]Vertical transformation is only going to change the ycoordinate. So they don't have anything to do with each other. You can do any of them for us and that will be fine. When you have multiple vertical transformation, for example, when you have something like this minus2 of f of x + 7 or or let's just skip this for now. minus minus 2 into f of x plus 7.

[07:35:51]What you do in vertical transformations is you follow bm mass. Whatever b mass says that's the order that you that you follow for transformations. Now bmass says you do multiplication first. So that means you do stretch or reflection first right minus2 that's multiplied. So you do this first and then you do translation. So you do plus 7 later on after that. However, if it looks like this, when they have an extra bracket sometimes like this and that happens a lot in papers when they have extra brackets like this, what that means is now BMA says bracket gets to gets evaluated first. So plus 7 has to happen first. So that means in this case you do translation first. It's a translation 07 first and then number two you do the stretch after that and reflection after that. Right? Stretch and reflection they have the same priority between them. It doesn't matter which one you do first.

[07:36:41]But translation in the second case happens first because that's what B mass is. And in the first case, translation happens second because that's what B masses. In vertical transformation, you just have to follow B mass.

[07:36:51]In horizontal. However, you just do the opposite of that. You do opposite of B mass. Whatever B mass says, you do the opposite of that. So for example, in this case, you have f of f of - 5x + 3.

[07:37:01]Now what does b mass say? Bmass says that you would multiply by minus 5 first and then add three. In horizontal, you do the opposite. to say no you'll do this first that means you apply that translation first and then after that you do stretch okay if you have something like this that's the most complicated uh the most confusing one so you have f of minus2 into x + 4 what do you do in this case again this is horizontal transformation because something is happening inside the function you're you're doing something to x now what does bm mass say because of this extra bracket Botm is that you would do +4 and then you would multiply by minus2. But this is horizontal transformation. You'll do the opposite of that. You'll say multiply by minus2 first. That means do stretch reflection first and then do a translation. So in horizontal transformations you just have to do opposite to whatever bot mass is.

[07:37:56]That's the order. That's the way you uh figure out the order of transformations.

[07:38:01]Okay. Now for instance if you had 2 into f of 3x look at look at the first example you've got one vertical one horizontal it does not matter which one you do first you could do this first you could do this first it does not matter because between them we don't care which one we do first because horizontal only change x vertical only change y if you have multiple vertical transformations like this minus4 and then plus one in this case you do stretch and reflection first or you can just say it's a stretch with scale factor factor minus 4 that's also fine and then you say it's a translation after that that happens second in this case the order matters that's important when you have horizontal transformations multiple f of - x - 2 - 6 is -2 you do the opposite of what mass that means you have to do transformation first and then you do uh stretch or reflection after that that happens second because that's the opposite of part mass sometimes you can have all of these happening in the same time like this you've All vertical transformations, all horizontal transformations. What's the order going to be? You can start with either horizontal or vertical. That would not matter. You can start with either horizontal or vertical. It would not matter.

[07:39:15]So let's start with vertical for instance. In vertical you have to do stretch first in this case because that's what BMA says. You do stretch first and then you do this translation that's 02 and then you can start with horizontal. In horizontal you do translation pro in this case because opposite to body mass that's one zero translation and then stretch scale factor minus 1 over4. Now when you have multiple transformations like this and you have two translations coming one after the other two translations coming one after the other. It's also possible to combine them as a single transformation and single translation and you can say two and three could be combined to give one two and that will also be acceptable.

[07:39:55]Okay. However, if you had another transformation, another transformation between these two translations, then you could not combine them. Then that would not be allowed. But if two translations are coming one after the other, then it's possible to combine them and write them as a single translation as well.

[07:40:09]And that will be fine. y= x reflection in line y= x, we do that when we have a function and its inverse. But we don't consider that consider that in function transformations. The only uh place where that is relevant is when we have a function and its inverse function. So we don't study that in function transformations here.

[07:40:32]Let's look at this example. If you have a function f of x f ofx= x^2 and it's being transformed to this function g ofx and we have to describe the transformations that are happening. How do how do we describe these transformations? So we've got two vertical transformations.

[07:40:48]X square was the function right outside the square we've got two things happening we've got two multiplying and outside the square we've got three adding which one do we do first we apply B mass board mass says because this is vertical transformation right we apply B mass here so B mass says you do plus three first so you uh you do translation vertical translation in this case I've written horizontal first it doesn't matter between horizontal and vertical doesn't matter as let's start with horizontal You this can come at any place it would not it would it would not matter because there's only one horizontal transformation and that's this minus 2 that that's a translation 2 0 okay so that's one translation that's happening but then vertical transformations are two here this is one one vertical transformation this is another so in this case between those two the order does matter so you do translation first in this case because what that's what BMA says 03 and then you do vertical stretch scale factor 2 now this is one way of describing is you could also say this is an actually act actually exam question u this particular function another way to think about this could be that you simplify this expression first of all and you say let's simplify g of x in a way that this two gets multiplied with this and this so that the bracket opens and you write it like this and now if you do it like this you can say it's a horizontal translation 2 0 that's the same as before but vertically Now since there's no extra bracket there you do multiplication here first and then you do then you do addition. So that means now you can say it's a vertical stretch scale factor two and then it's a vertical translation that's 06 and that's also acceptable. These two are equivalent to each other even though these vertical trans translations that you have now they are different but since the order is changing that takes care of that. So these two are equivalent. You can either do it the way on the right side or on the on the left side. Both of them are equivalent to each other. Now when you have quadratic functions, so this is how you apply transformation, right? This is how the order of transformation works. This is the easier part that you are given some transformation and you have to describe that. Okay, that's that's relatively easier to do. The difficult thing normally is that you have to identify the transformations yourself.

[07:43:13]In this case though uh we're doing that as well. So let me show you that example.

[07:43:27]Okay. So this is what I'm talking about.

[07:43:28]So sometimes what they do is they give you something like this. They'll give you equations like this. y= x + 2x - 5 and they say it's transformed to this.

[07:43:36]Figure out the transformation. So this is normally the tricky part. How how do you how do you understand how to go from this to this? So normally when they give you quadratic functions when they give you quadratic functions you should always complete the square for for those functions first that will make it easier for you to spot those transformations.

[07:43:58]Okay. So when you have a whenever you have a quadratic function and you need to describe the transformations as in identify what transformations are happening to go from one function to another function first complete the square and that will help you identify what transformations are happening there. Okay. So that's another important thing. Now have a look at let's have a look at some examples because that's where the that's where people get confused. The concepts are not that difficult but it's questions are quite tricky sometimes. Let's have a look at this. This is a very straightforward example. First up, we've got this function f ofx and this is transformed to this. What kind of what kind of transformation is that?

[07:44:32]That's a reflection in the line yaxis.

[07:44:35]Now, you have to be careful. Is it is it a horizontal reflection or vertical reflection? It's a horizontal reflection because the movement of the graph is horizontal.

[07:44:43]We want to write down the function f uh the the equation of this graph in terms of the function f. F was the original function. What's the equation of this transformed graph? This is horizontal reflection. That means inside the function we will multiply with minus. So that's f of min - x like this.

[07:45:04]What about this? Now you've got this original graph and that's been transformed to this. Now stretch is the easiest to spot because not the easiest but it's like very easy to stop spot because you would always have the size of the figure changing. Right? In this case you can see vertically the height of the figure is changing. It was two earlier now it's becoming four. Right? Horizontally however there's no size change in size. There's only one uh it was two earlier after that as well it's two. So it's a vertical stretch. It's a vertical stretch by what scale factor? Height was two earlier then it becomes four. So it's multiplying by two. So what's the function that we have here? We say this is 2 * f ofx. Now there's no translation here. When you do stretch the points automatically move up because of that because this was one unit after stretch it becomes two units so it moves up.

[07:45:59]This point was two units up after it stretched it becomes four units up. So there's no extra translation that's happening. It's the stretch that is moving this graph vertically upwards as well because the y-coordinates are getting multiplied by two. So this is 2 * f of x. That's a vertical stretch but scaled back to two. Think about this.

[07:46:17]You've got this original function. It's being transformed to this. Now that's clearly a cross translation. The graph is shifting from one point to another.

[07:46:25]How many units to the left? One, two, three. Three units uh and then four units left. So that's min -4 and then 1 2 3. Three units down. So the translation vector for this is going to be -4 and minus 3. Four left and four down. However, we are not looking for translation vector. We want the equation of this function in terms of f of something.

[07:46:51]How do we write that? Now translation is when you have plus minus something. Now minus3 is a vertical translation.

[07:47:00]So we have minus3 outside the function.

[07:47:04]Horizontally the graph is going four units left. Left means inside the function you would have plus. So inside the function it'll be f ofx + 4 right in horizontal things works a bit counterintuitive plus means left minus means right. Okay so when you're doing when you're going left in the inside the function notation you'll say it's plus4 and that's what the function becomes in this case let's look at this now you have to describe the transformation that takes you from this graph to this graph.

[07:47:36]Now whenever you have to whenever you have to identify transformations always think of stretch first. Always think of identifying the stretch first because that's going to be very clear to spot in that case as in you can't get you can't be wrong with that because you know exactly what stretch is happening just by looking at the figure. Look at the width that's not changing. So there's no horizontal stretch that's happening. But vertically the height was one earlier afterwards it becomes three. So you know for sure it's a stretch by scale factor three. So when you have to identify transformations from a diagram always identify the stretch first and clearly show that on the diagram and then think of the next transformation. So in this case we have figured out that okay this is this has to be a vertical stretch scale factor 3 stretch along y-axis scale factor 3 that is definitely there. Okay. Now you don't think of the next transformation until you have sketched the result that you would get from this first transformation. So when you stretch this figure by a factor of three in the vertical direction, this point goes here. Sorry, this point goes here. One goes to three. The point at two goes to six. So that that's here.

[07:48:50]And this point at two goes here. And after that stretch, the figure comes here. Now when it comes here now you think of what further you have to do to this to take it to this graph. Now now that's a translation just four units to the right and that's what you write down it's a stretch along y-axis scale factor 3 and a translation of 4 0. If you do not draw this intermediate intermediate mediary step this step you would wrongly think that there's a vertical translation happening as well. That's not happening. The vertical movement is because of the stretch itself. So when you show what happens after that stretch, it becomes easier to spot what the next transformation is going to be after that. That's just that's that's just a horizontal translation. There's no vertical translation after that.

[07:49:36]Okay? And then you have to write down that function in terms of f ofx. So that's what it becomes. Three, that's a vertical stretch. And then minus4 inside because it's four, right? That means inside the function you do minus4.

[07:49:50]Another very similar example you have to spot this transformation. Uh this is the original graph. Let me share that for you.

[07:50:04]This is the original graph the highlighted one and this is going to become this dotted graph. Now how do you figure out what transformations are happening? Again the first thing you should you should think about is what stretch is happening? Then think about any other transformation. What stretch is happening? Is there any horizontal stretch? Horizontally it's four units afterwards and it's four units before there's no horizontal stretch.

[07:50:34]Vertically however it's two units before and it becomes four units afterwards. So it's a vertical stretch scale factor two.

[07:50:42]That's what you write down first. You say it's a vertical stretch scale factor two. First of all, do that stretch. When you apply that stretch, you get a figure like this.

[07:50:53]So, this was the original graph. The green one is the original graph. After applying stretch, you get this yellow graph. After applying the stretch, you get this yellow graph. Now that you've done that, now you think of what further you have to do to this yellow graph to get to that dotted graph. Okay? Now, there could be multiple ways of thinking about this. Now, you could say, let's go one unit to the right. There are three transformations, right? So you can say go one unit right and then do a reflection in y-axis or you can say do a reflection first so that you get this green graph and then you go one unit left. So after you've applied that stretch then identifying the rest the next transformations becomes quite straightforward. So this is what you get. This is one way to describe this.

[07:51:41]There are other possibilities as well.

[07:51:43]Stretch is definitely going to be this but apart from stretch the other transformations they could be described in other uh ways as well. This is one way to do this. You say you reflect in y-axis and then you have a translation minus 1 0 and then you have to express g of x in terms of f ofx as in the transformed function was g of x. Express that in terms of the function f the original function. How do you do that?

[07:52:05]It's a stretch in the y direction scale factor two. So you multiply by two outside reflection in y-axis that's a horizontal reflection when you reflect in y ais that's a horizontal reflection.

[07:52:17]So you multiply by -x inside that's happening first right. So you apply you multiply minus x first inside the function and then once you have that you do this translation. Translation is min - 1 0. What does that mean? You replace x with x + one. put that inside bracket because you have a negative negative outside in inside bracket you would say x gets replaced with x + one because that's what this translation means then you simplify that that's the result that you get 2 f of - x -1 that's the that's the function g of x in this form okay this is a good example have a look at this uh later on as well try to think about this again let's move on let's have a look at this now find a translation that transforms the graph this to this how do we Think about this now. So you've got 5x - 3x and it's transformed to this graph. Now just by looking at it, you might get the wrong idea first that it's a horizontal translation of 2 units, right? Because it's going from f ofx to f of xus 2. So you say it's a translation of 2 units, right?

[07:53:29]And there's a plus 10 outside. So that's 210 translation, right? That's what you would think of at first if you do it in a hurry. You would get 210 from this.

[07:53:40]But that's not correct. That's not correct because even though this x is going and becoming x - 2, what about the other x?

[07:53:52]If you if you go from f of x to f of x - 2, every x should be replaced with x - 2. The other x is staying x. That's not changing.

[07:54:02]So again that's why it's important to do it step by step. What's the transformation that you're very sure about? X is becoming X -2 inside the square. We can say that okay this is becoming X -2. Apply that transformation first that okay f of X is going to f of X -2. Apply that transformation and that gives you this. So you replace X with X -2 and you get a function like this.

[07:54:29]Replace X with X -2 here. replace x with x - 2 here. Now you simplify this. After simplification, you already have a plus 6 outside like this.

[07:54:38]After doing that horizontal translation, you already have a plus 6 outside. And then you think of how much more do we need to add here to get to that point 10. You have to add plus 4. And that gives you a vertical translation of only four and not 10. And the translation vector turns out to be 24. So when you have to identify transformations, do it step by step. Write down the transformation that you're very sure about first. Apply that transformation and see okay what happens after that.

[07:55:07]Okay. Otherwise you can often get wrong results.

[07:55:12]Let's have a look at this. It says the curve with this equation is translated by minus3. Find the equation of the translated curve giving an answer in the form y= ax plus bx plus c. Now this is just an application of a transformation.

[07:55:24]The transformation is already given.

[07:55:26]It's minus13. What that means is if you had the function f ofx that would become f of x + 1 + 3 inside the function you have + one because of this horizontal transformation. Outside the function you got plus three because of that vertical transformation. All you have to do is just replace x with x + one here and outside the whole thing have a + 3 and then simplify that. Completing square in this case is not necessary necessary because you had the transformations given and you start to apply them. But when you have to identify the transformation for example in the second part now you've got this equation and it's it's transformed to this equation describe fully this single transformation that has been applied to go from this step to this step. Now how do you think about this?

[07:56:15]Okay, let me actually come back to that that example in a moment. Let me do this first of all. I I'll go back to that because that that this single transformation that there's there's another another complication here. Let's do this one first. Before that, we've got this f ofx function as x - 2x + 5.

[07:56:28]This is another function g ofx = x^2 + 4x + 13. We have to figure out what transformation transforms the graph of y= f ofx to the graph of y= z of x. How do you go from this graph to this graph?

[07:56:40]How do you identify this trans transformation? You should first complete the squares. First complete the squares. Otherwise, it will become difficult to identify the transformations. It will not be clear.

[07:56:51]This is x - 2 x + 5. When you complete the square, it becomes x - 1 + 4. And then the final result is x2 + 4 x + 13.

[07:57:00]When you complete complete the square, it becomes this x + 2 + 9. And now you can look at this. What is changing in these functions? Outside the function plus 4 is going to plus 9. What do you have to do to go from four to 9? You have to add five. You have to do plus five. And that gets you to this final result.

[07:57:22]Two plus five outside. And then what else? Inside the function you have minus one. Right? Now that has to become + two. How do you go from minus1 to +2?

[07:57:32]How much do you have to add? You have to add three. You have to do plus three. So inside the function you say it's f ofx + 3. What transformation is that? You have to describe this transformation. + 5 means the vertical translation 05 + three means it's a horizontal translation minus 3 0. When you combine them it becomes minus 35. That's the transformation that you've got. It's a translation minus 35. It's very difficult to spot directly from this unless you complete the square first. So when you have quadratic function then you have to identify the transformations complete the square first and that will help you understand that better.

[07:58:10]Sometimes when they say it's a single transformation it's a single transformation. In that case it is sometimes possible to do it directly as well but you have to think a bit about that. For example, in this case, you've got this function x2 + 2x - 5 and another function x 4x2 + 4x - 5. The function on the left side is being transformed to the function on the right side.

[07:58:36]Now, what transformation is this? You you can think about this in your head as well.

[07:58:42]- 5 - 5 the constant is not changing that that's remaining the same. That means if anything was applying to the whole function, then this constant would also change.

[07:58:53]This is not changing. So this is not a vertical transformation right because in vertical transformation the whole function changes in this case minus 5 is remaining the same that's not changing so that means it has to be a horizontal trans transformation. Now what transformation x² is becoming 4x^2 so that looks like something is getting multiplied and 2x is becoming 4 * x something is multiplying so it's horizontal we figured that out and something is multiplying what h what transformation is it when we multiply something that's a stretch now what do you multiply x by if you think about it 2x is becoming 4x you're basically multiplying by two here x^2 if you replace that with 2x^2 that becomes 4x^2. So yes, that's what is happening. You have f ofx going to f of 2x. So that's a vertical stretch. That's a horizontal stretch scale factor 1 /2.

[07:59:47]Okay. Sometimes you can identify directly. But in very rare cases when when it's a single transformation, you you can think about it in your head as well. But when it's multi multiple transformations, you cannot do it directly. You have to complete the square for that necessarily.

[08:00:02]Okay. This is an alternative method for that. If you would do it using the building square method, this in this case it's a bit long but you can have a look at this. It gives the same result eventually. Okay.

[08:00:19]Let's move on.

[08:00:23]Now you have transformations in trigonometry as well. These are sometimes a bit tricky. Let's have a look at this example.

[08:00:32]So you've got this original curve. This is the graph of y= cos theta. Okay. And this is being transformed to the upper curve. This curve here they are saying find in terms of a cosine function the equation of the upper curve. What's the equation of this curve? How do you figure that out? Now what's the baseline of the original graph? It's the x-axis. What's the baseline of this graph? How do you identify that? The maximum point that you have here is five. The minimum point on on this graph is one. Where is the baseline going to be? The baseline is going to be in the middle. That's at the point three. So that means the baseline is going up. The baseline is going up by three units. What does that mean? It's a translation. It's a vertical translation by three units up. So you would say that becomes cos of theta + 3. That's one thing that's definitely happening. Okay.

[08:01:18]So baseline is moving. What about the amplitude? Is that changing? Let's think about that. The amplitude earlier was 1.

[08:01:25]Now the baseline becomes this at three.

[08:01:28]What about the amplitude now? That's two. So amplitude is becoming twice that means the vertical stretch scale factor 2. So we multiply by two outside it becomes 2 cos of theta + 3. What else?

[08:01:39]Is there anything changing horizontally?

[08:01:44]Is anything changing horizontally? Is the period changing? Is the graph shifting left? Right? The period is clearly changing. It's completing two cycles in 4 pi. In fact let's consider 2 pi the standard period in 2 pi. This is one cycle. This is one full graph.

[08:02:02]The period for the period for the transformed graph has doubled. It is it is it has become twice. It was 2 pi earlier. For the original graph, the period is 2 pi. For the transformed graph, the period becomes 4 pi. The period is becoming double. When period becomes double, that means it's a stretch scale factor two. If scale factor is two, you've got half theta here. This is going to be half theta.

[08:02:29]Okay. Yes. So some sometimes it's easy to figure out these two transformations the vertical transformations horizontal.

[08:02:38]Some people find it a bit difficult. One way to do that is that once you figured out these two transformations then you could say this is 2 cos of a theta + 3 and then input some value there. And for take some value of the graph for example at zero it's five. See if that works. If that gives you a value, zero will not give you a value because a will get eliminated. But take another value for example at two pi it's one. Input that and find the value of a from that.

[08:03:05]That's longer. You can do that. If nothing else works then yes you can do that. But it's it's much easier and much quicker to understand transformations.

[08:03:14]Well since the period is getting double that means it's a stretch scale factor too. It's a horizontal stretch scale factor two. If the scale factor is two inside here we have it divided by two like this. You can also understand this in terms of the period of the graph. The period is becoming so the original period is so this is the equation that we can use for this the new period of the graph the of new the period of the transformed graph it's always equal to the original period divided by that number that's multiplying. So let's say this number is P. You can just input values in this formula. Original period of a cause graph what's that in radians that's 2 pi the period of this transformed graph that's 4 pi find the value of v from that that turns out to be two and that's what you sorry that turns out to be 1 /2 and that's what we were looking for okay if you cannot think of it in your head directly you can use this equation and that will give you the value for v as well all right let's take another example in this if you have a tan function like this what about this now for time there's no amplitude as such what about the baseline. What did it tell you about the baseline of the time function?

[08:04:29]It says the diagram shows the graph of this function a tan of x - b + c. How do you figure out the values of a, b, and c from this?

[08:04:41]For a tan function, the baseline is when the direction of the graph is changing.

[08:04:45]So look at this graph. It's going in this direction and then it changes its direction like this. So where's the baseline? baseline is at this point here. That's where the direction is changing. If you look closely, that's the point where the direction is changing. So that's the baseline of this graph. Now this is the baseline that's one. So you say this number outside plus minus the graph is shifted upwards right by one unit. The original time graph it has its baseline at uh zero. So it looks like this, right? The original time graph its baseline is at this point that is moved one unit up. That's translation vertically translation one unit up. So you have this plus one outside the value of a C therefore is equal to one. Okay.

[08:05:28]What else is there any any stretch that that is happening? Look at the horizontal direction. What's the period of this graph? What's the period of this graph? Period of the original tan graph is pi radian.

[08:05:49]What about the period of this graph?

[08:05:51]This is completing one full cycle from -4 to 3 p<unk> over4. What's that distance? 3 p<unk>4 - -<unk> pi radian.

[08:06:01]Is the period changing? No, period is not changing. So, original period was pi. This is also pi. So, period is not changing. Okay, period is not changing.

[08:06:09]What else is changing? The graph is shifted. Original time graph, what does that look like? Original time graph looks like this. It starts from zero and it goes like this.

[08:06:20]Right? That's the original time graph.

[08:06:22]It looks like this.

[08:06:24]Now we have figured out that it's moving one unit up. So that would have moved somewhere here. But after moving one unit up, it'll look like this. But then this point, where is that going? That's going here. What's that? That's a translation. Pi or four units to the right.

[08:06:41]P4 units to the right. That's a translation. That means this number here, B, that has to be pi /4. It's it's a minus sign. minus means you're going right in in horizontal transformations.

[08:06:50]So that number is therefore p<unk> /4 because you're moving pi over 4 units to the right like this.

[08:06:57]Now what else? What about the number outside? They're saying there is a vertical stretch there as well. What vertical stretch? How do you figure that out?

[08:07:07]Now in this case one idea is because this is actually a bit hard to figure out. Once you have once you have the values for P and C, you input any point on this graph in this equation. For example, you have this function now Y = A into tan of X - 5 / 4 + 1. That's the function. Take any point in the graph, input that here the values of X and Y at that point. For example, you could write this point here. This is P<unk> / 4 and 1. That works. tan of 4. Sorry, this was like this into tan of x -<unk> /4 like this.

[08:07:48]If you input p<unk> /4 there, that's not going to that's not going to work because 4 minus p<unk> /4 that will become zero. It'll eliminate a as well.

[08:07:54]So you'll have to take another point.

[08:07:56]Take another point in the graph. For example, you can take this point. This is x=<unk> /2 and y = 3. Input this point here. Find the value of a from that. That's one idea that will work.

[08:08:07]Another another way to think about this is the original tang graph. Original tang graph it basically goes like this.

[08:08:14]It starts from zero and then it becomes 1 at4. So the value that it has at pi /4 is equal to 1. So the tan graph the original tan graph it moves one unit up.

[08:08:28]It moves one unit up as it goes p<unk> / 4 units to the right.

[08:08:34]It goes one unit up in in an interval of five or four units. Now what about this transformed graph? This graph is basically starting from this point, right?

[08:08:44]How much is it moving in the vertical direction in a distance of pi /4?

[08:08:49]This is 54 distance. In this 54 distance, the graph is going up by 2 units. So it was originally supposed to go up by one unit. In this case, it's going up by two units. So that means the value of a here it's a vertical stretch.

[08:09:05]It's a vertical stretch of scale factor two right. So the value of a there should be two. If you understand that that's good otherwise this is an easier way of doing this in this particular example that you input a point and find the value of a from that. Okay. All right let's move on just a couple more examples and then we end.

[08:09:28]Okay. Let's have a look at this. Now this is from a recent paper fairly complicated examples.

[08:09:34]You've got this diagram that shows the curve y= k cos of x - 5 / 6 where here's a positive constant and x is merged in radians. The curve crosses the x-axis at points at points a and b as a minimum and a and b is a minimum point. Find the coordinates of the points a and the point b. Now what's the function? It's a cause function. And they have shown you what transformation is happening in the horizontal direction already. It's x -<unk> / 6. What does that mean? X -<unk> / 6. That means the graph is going P<unk> / 6 units to the right.

[08:10:03]Now, what's the original cause graph?

[08:10:04]The original cause graph looks like this. It starts from 1, goes to 0 at pi / 2, and then goes on like this. That's what the original cause graph looks like. This is the graph of y = cos of x.

[08:10:17]Now, if you move it pi / 6 to the right, what happens?

[08:10:23]this point which was at zero earlier that moves to the right and it becomes p<unk> / 6 here. So the graph moves p<unk> / 6 units to the right like this.

[08:10:33]And then they're asking you to find this point a. How do you do that?

[08:10:39]How much is this distance? P<unk> /2.

[08:10:43]So that means this distance from that maximum point to this point this should be pi /2.

[08:10:51]And then from this point to the minimum point, this this is another p<unk> / 2.

[08:10:55]This is another p<unk> / 2. So that's<unk> / 6 plus 3 p<unk> / 2. And that gives you this point that's 5<unk> / 3. So the value of a therefore the coordinates of a they are 5 pi over 3 and 0. What about the minimum point b? You do the same thing here. You know the point a. Now let's go to the right. It's pi / 2 more here.

[08:11:19]Another pi / 2 here. another pi /2 here and that's the point B. So 3 p<unk> /2 more added to that that gives you 19 p<unk> / 6. The coordinates of b therefore what are they going to be? The x coordinate is 19 pi / 6. What about the ycoordinate? The graph uh the function that they have given it has k multiplied outside.

[08:11:40]So that means k is the amplitude of this graph. If k is the amplitude, this point is k units below the x-axis. It's k units below the x axis. So what's that point going to be? That's minus k. The ycoordinate therefore turns out to be minus k. Okay, let's have a look at this example.

[08:12:02]Similar to that, the diagram shows the curve with equation y= k sin of/x. The curve has a minimum point at a. Now it's a sign function. Think about the original sign function. What what does the original sign function look like? It starts from zero. It goes to 1 at pi / 2. It comes back to 0. Goes to minus 1 at 3 pi / 2 and so on. Now what's the function? It's k sin of half x. You know what transformation is happening in the horizontal direction? Half x. What does that mean? The period becomes double, right? It's a horizontal set scale factor two. What's the period of the original sign graph? For the original sign graph, the period is 2 pi. For this the period becomes 4 pi. It's a horizontal stretch scale factor 2. Okay.

[08:12:51]They want us to find the coordinates of this point a.

[08:12:55]Now the y-coordinate just like in the previous example is going to be minus k.

[08:12:59]Why? Because we've got k multiplying outside. So the amplitude is k. So this point here has to be minus k. What about the x coordinate? How do you figure that out? In the original graph, this minimum point comes at 3 pi /2, right? The minimum point comes at 3 p<unk> /2. In this original graph, when it is stretched by a scale scale factor of 2, that gets multiplied by 2.

[08:13:25]3 p<unk> /2 * 2, what does that give you? That gives you 3 pi and that's the x coordinate of that minimum point. So, it's 3 pi and minus min - k. That's the point that you get there.

[08:13:37]And then it says a sequence of transformations that is applied to the curve in the following order. Now this is the curve that you have. They're applying some transformations in that translation of two units in the negative y direction that's 0 - 2 reflection in the x-axis. Reflection in the x-axis that's a vertical reflection right the ycoordinate gets multiplied by y and multiplied by minus one. And then it says find equation of the new curve and determine the coordinates of the point on the new curve corresponding to a. So the corresponding to the point a the minimum point you have to find the new point. How do you do this? Now the original graph looks like this k sine of half x that's the original function. What are they doing?

[08:14:17]Translation of two units in the negative direction. What does that mean? You've got minus2 outside the function. And then that's the first thing that happens. And then they're saying it's a reflection in the x-axis. That's a vertical reflection. So you have minus multiplied with the whole thing.

[08:14:32]vertical reflection minus gets multiplied outside the function. So you get you multiply minus outside the whole function like this. Simplify that and that's the result that you get from that. And then what are the coordinates of the point a going to be? Originally it was 3 pi and minus k. Apply those same transformations on this as well.

[08:14:52]Translation of 2 units in the negative y direction.

[08:14:57]Minus k was the original point. You subtract two from that. It becomes minus k minus 2 initially after that translation. First of all k becomes minus k becomes minus k minus 2. All right. And then you have a reflection in the xaxis. It's a vertical reflection.

[08:15:15]You multiply that ycoordinate with minus1. So you multiply minus outside.

[08:15:22]That's the second thing that you do.

[08:15:24]This is the second thing that you do. So it becomes minus into minus k minus 2.

[08:15:28]And that simplifies to k + 2. There's no horizontal transformation. So the x coordinate remains the same. That's 3 pi. And the result that you get is 3 pi and k + 2. That's the final answer that you've got. These are most most of the difficult questions that you find in transformations in recent papers. We have these in the notes. Have a look at them in your um in detail and that should prepare you well for transformations as well. All right.

[08:15:55]I hope that was helpful. It again took a lot more time than we were expecting initially.

[08:16:01]But I hope that was helpful. If there any questions, you can let me know now.

[08:16:14]This Okay. So remember in horizontal transformation the rule is you do opposite of B mass. Whatever B mass is you say no I'm not going to do that. I'm just going to do the opposite of that.

[08:16:42]This is minus 2 into x + 4.

[08:16:45]This is horizontal because everything is happening inside the function. Right? So these are horizontal transformations.

[08:16:51]Now board math says you evaluate this bracket first. So you do plus4 first.

[08:16:54]Plus 4 + 4 means translation.

[08:16:57]But you have to do the opposite of that.

[08:17:00]So you don't do translation first. You do stretch stretch deflection first. So that's what you do first and then you do translation after that. So it's just opposite of hard mass, right?

[08:17:25]It's okay. Uh every p a paper would not have all difficult questions. You'll have like two three difficult questions, two three very straightforward questions and then six average questions. So if you get a difficult transformation question, you'll get some other easier questions.

[08:17:42]It's okay.

[08:17:45]It'll get balanced as long as you're prepared well. You can rely on the threshold as well. And of course das both.

[08:17:53]Yes, I'll be doing it for all components in uh that that's what the plan is. Uh for M1 and S1 at least I will be doing it P3 after the P1 experience. Uh I think that's going to take forever to end. I'm not too sure about that right now. I I do plan to do that. Uh but I still have to see how to schedule that. It'll be like 6 hours, 2 days. Uh let's let's see what we can do for P3. Can I state the updated quick notes before the work exchange? Well, uh you I I I thought you knew me from the class that you studied with me at ISL. I do everything at the last minute.

[08:18:41]So for this class I was preparing until 4:30. So I told you we will start at 4:15 but we started at 4:30. Why did we start 15 minutes late? Because I was still making these notes. I cannot do anything before time. That's not how I'm I'm built. So that's not going to be possible.

[08:19:02]they will only be available when uh the the the actual workshop happens because I will keep procrastinating. Leave it for the very end and then it'll be made at that time. But I'll try I'll try. I say that for everything but that's what mostly happens at the end. I'll do it at the very end.

[08:19:29]I will die. However, I I'll try if I can do that. I'll share them with you. But the older ones are available. But I'll be adding much more to that from the recent papers.

[08:19:42]But that will only happen at the end. I can't promise that I'll be able to do that earlier.

[08:19:48]-4 -4 uh I can't find that. Was it this?

[08:20:34]How much more does Amraadi? Is that fine?

[08:20:38]Can you read anything?

[08:20:48]Okay. Whatever it was, just remember this is the rule. This is the rule. Uh multiple vertical transformations, you follow B mass. multiple horizontal you do opposite aparts.

[08:21:02]Okay, that's the rule. So that's the rule that that will work. Now when you have vertical and horizontal both it does not matter whether you start from vertical or horizontal. You can start from any of them and that will be fine.

[08:21:15]But within vertical within horizontal you have to follow this order. Yes.

[08:21:23]>> Okay everyone let's begin then. P1 differentiation. This is the next thing that we're doing. So what's differentiation? First of all, in coordinate geometry, we learn how to find gradients of lines y2 - y1 / x2 - x1. For straight lines, we can find gradients like this. But what about curves? If you have a curve like this, how do you find gradient of a curve?

[08:21:44]It gradient of a curve changes at every single point. Right? So there's no one gradient at that exists at every single point. So it changes at every point. For finding gradients of curves, what we do is use differentiation. Now, what's the idea of differentiation? There's some background to it that that you also need to know for your exam. I'll show you some questions on this as well. The idea is if you want to find the gradient at any one particular point, for example, you want to find the gradient at the point A. This is.1 one. This point is right here. Now, what we do is we say let's take another point on the same curve. So, we want to find the gradient at the point A. We take another point as well on the curve. That's point B here.

[08:22:22]and we say let's find the gradient of the point of the line AB. That gradient turns out to be five. Now that's not a very good estimate for the gradient of the curve at that point because this direction and the direction of the tangent at the point A is not uh very similar. So it's not a very good estimate. However, if I take that point close closer to A than it than it currently is, for example, if it becomes C such that the coordinates of C turn out to be 3 and 9 and then I find the gradient of the line AC, this line that turns out to be four. Now, that's a better estimate than the earlier line which was AB. Four is a better estimate than before for the gradient of the curve at that point A. Now as we keep bringing this other point closer and closer to that point A, what happens is the estimate for the gradient of the curve at that point keeps on improving.

[08:23:16]It keeps on improving like this. And then from that pattern what we can say is if we were to reduce that difference a lot. For example, eventually we got a point that was like 1.01.

[08:23:28]The point A was x = 1 and here we got 1.01. 01. So that's point that's very close to the point A. At that particular point, the gradient of the line A that's turning out to be 2.01.

[08:23:43]Now looking at the pattern, you can see what value these numbers are approaching. It's it started from 5 4 3 2.1 2.01. If you take some more values, it'll be like 2.001 2.00001. And that's telling you that the gradient is approaching this value two. So you can say that the gradient of the curve that that particular point is supposed to be equal to that's the idea of differentiation that if we want to find the gradient at any one particular point we take another point and then keep bringing that point closer and closer to improve the estimate of the gradient.

[08:24:20]Now in general we could say something like this. If we have the point x we take another point that's x + h. What we want to do is we want to find the gradient of this line AB that turns out to be something using y2 - y1 / x2 - x1.

[08:24:36]And now if we bring this second point very close to this first point, how would that happen? We would want this value h to be as small as possible. So h is the distance between these two points. If this h becomes very small, this other point will become very will get very close to the point x and the gradient of the curve will be uh the gradient of this line will become a very good estimate for the gradient of the curve at that point. So we want to bring that second point very close to this first point and that will give us a a good estimate for the gradient of the curve. Now what do you need to be able to do about this?

[08:25:14]Let's look at this example. Let's say we've got this equation of a curve that's y = 2x - 3. We have two points.

[08:25:20]One is 2 and the other is 2 + h. Okay. First of all, it says find and simplify an expression for the gradient of the chord a in terms of h. We've got two points.

[08:25:33]One when x is equal to 2, the other when x is equal to 2 + h. We input those points and and that to input those values of x in this function. We get the corresponding values of y from that.

[08:25:43]Eventually when we when we do that we end up getting two points. One point turns out to be 2 and 5 and the other point turns out to be 2 + h and 2 h² + 8h + 5. Now we can find the gradient of that line using y2 - y1 over x2 - x1.

[08:25:57]This is y2 - y1. This is x2 - x1. And that simplifies to this value. This is 2 h + 8. And now they say explain how the gradient of the curve at the point a can be deduced from the answer two part a and state the value of this gradient.

[08:26:14]Now think about this. We took a point that was 2 and five and the other point that was 2 + h and 2 h2 + 8 h + 5. There were two points on the line like this.

[08:26:26]Two points on the curve like this. Two separate points a and b on the curve like this. Now how would we get the estimate for the gradient of the curve at that point? We would want the second point to be as close to the first point as possible because that is going to improve the estimate for the gradient.

[08:26:45]Now for the second point to be as close to the first point uh as possible, we would want the value for h to be approaching zero. So now in this gradient value, if the value of h becomes zero, what's the value of the gradient that you get from that? the value of gradient that you get is eight and they and you say that the that the gradient of the curve at that point would therefore be equal to 8. Let's take another example of this. So they have this function that they've given f ofx equals this and they say that these points lie on the curve. So you've got point a b c d e f points like this and you say and you see what they're saying first of all find the value for k. Now what's k is this point here. You can do that. That's uh you've got the xcoordinate of this point given and you have to find the corresponding ycoordinate. Just input that value of x in this function. You'll get that point.

[08:27:37]So once you have that point now they're saying this table is showing the gradients of the chords A B A C A D A F.

[08:27:43]Now what's a chord? This line that you're that you're drawing here in in the context of curves as well. We can call this a chord. Okay. So this is the chord that they're talking about and what they're doing is they're finding the gradients of all of these chords like this. We we have the point A here.

[08:27:59]They're finding the gradient of AB, A with B, A with C, A with D and so on.

[08:28:05]They're finding all of these gradients.

[08:28:06]Now, you see what's happening? The points are getting closer and closer to the original point A. So this point F, it has an X coordinate of three. E has an X coordinate of 2.1. This is 2.01. This is 2.001. This is 2.001. So that point is getting closer and closer to the point A. So when we find these gradients from here, you can see the pattern that you that it emerges from those gradients. So in in part B, they're just saying find the gradient of the line a a you can do that using a formula. But then they say deduce the value for fdash of two using the values in the table. Now what's fdash of two?

[08:28:46]That's the gradient of the curve at the point 2 because fdash represents the derivative. So we are looking for the derivative of the curve when x is equal to 2. Now that means the gradient of the curve when x is equal to 2. That's what we want to figure out. Now look at the pattern that that that emerges from these points 7.22 the next variant was 6.35 6.26 6.2551 6.2511 6.2501. And you can see these values seem to be approaching 6.25. 25.

[08:29:21]So that's what you say that the gradient of that curve at that particular point it should be equal to 6.25 because that's the value that that that it's approaching. Okay. So that's that's uh one thing that you need to know about differentiation. You'll find very few questions in this. You need to make sure that you practice these ones. Have a look at these again and make sure that you understand them well. For a detailed video on this, you can see the playlist for differentiation. You'll find the first lecture on differentiation that explains this concept in detail. Now the main part of differentiation is then this. You need to be able to differentiate functions using the power rule of differentiation. Now how does that work? Let's say we've got a function that looks like this f ofx.

[08:30:01]First of all, we need to know some notation. So f ofx is one function.

[08:30:05]What's the derivative of that? That's fdash of x. This is the notation that we have for differentiation. Second derivative that's fdash of x. That's one notation that you need to know. And if you have a function y the derivative of that is given by dy / dx. And if you have a fun and if you have dy by dx and you want and we differentiate that again we get d² by / dx² from that. So this is some notation that you need to know about if the original function is f ofx the derivative of that is denoted by fdash of x and the second derivative is denoted by fdash of x or frime of x f prime of x. If the original function is represented as y, we represent the derivative as dy / dx and the second derivative as d² y / dx^ 2. Now what does this derivative represent? It's the function of gradient for for this original function. It's the gradient function of the original function. It represents the gradient of the graph. Now dy by dx this basically represents that you're taking the derivative with respect to x. So d represents derivative dx represents with respect to x and you're finding the derivative of y with respect to x.

[08:31:14]Right? That's what dy dx represents. Now how do you do differentiation? We've got some rules for this that are derived from this background. Now how exactly that's out of scope for your syllabus.

[08:31:26]Now we have this these rules that are derived. Number one we've got this power rule this simple power rule and that works like this. But x^ n the derivative of that is given by nx raised to power n minus one. So what that means is whatever is the function if it's a power function what you do is you multiply by the power and you subtract one from the power. So whatever is the function you multiply by the power subtract one from the power. If it's x raised to power something for for example if it's x^ minus 5 that becomes - 5 raised to power - 5 - 1 and you get your result. Now if there's a constant multiplying outside for example 7 / 4 if a mult if a constant is multiplying or dividing that remains the same that does not change okay so 7 over 4 in this case that remains as it is and then what you do is you multiply by by the derivative you multiply by the power and subtract one from the power and then you just simplify that and you get the result.

[08:32:21]Have a look at this example. If you have something in the denominator like this, if you have x in the denominator to to be able to differentiate this, you first need to bring x x to the numerator and then you can apply the rule in that because that rule works when you have x raised to power n in the numerator. This isn't in the denominator. To be able to differentiate this, you first need to bring this with to the numerator. How would that happen? This is x^ 1 /2. When it uh when it comes to the numerator, it becomes x power minus 1 /2. Three does not come with come up with that. Three remains there because there's no power in three. We're not bringing three up that remains there. So now we've got this function and this constant outside -2 over 3. We apply the rule in that.

[08:32:59]What's the rule? We multiply by minus 1 /2 subtract one from the power and then simplify that to get the derivative.

[08:33:05]Right? And similarly for these other examples you can have a look at this.

[08:33:08]The rule is you multiply by the power, subtract one from the power and you get the derivative from that. Now if you have something like this 2 over 3 * x, what's the derivative of that? If you apply the rule in this, it becomes 2 over 3 as it is. But then the power is one. When you subtract one from the power, it becomes x^0. X^ 0 is 1. So X gets eliminated. If you have -4 * X, if if you apply the rule in this, it becomes X raised to power 1 - 1 that's X^ 0. Again the X gets eliminated. We can generalize this that whenever we have something some constant with x like this the derivative is just going to be the constant value that you have. You can also compare this with y = mx plus c remember. So m represents the gradient.

[08:33:56]Differentiation also gives you the gradient. So since the gradient is 2 over 3 for this line the derivative is also 2 over 3. So when you have a function of this form where you just have a constant multiplying by x, the derivative is this just this number that that's that that that you have outside.

[08:34:10]And what about constants? If you have a constant like this y= 3, what's the derivative of that? You can think of it like this. That's it's 3 into x raised to power 0. When you apply that rule in this now, it gets multiplied by 0. When you multiply by the power and everything becomes zero for that. So whenever you have a constant, the derivative of constant is always equal to zero. So these are two special cases that will help you do things quicker if you just remember them. If you have k * x some constant time x the derivative is just going to be the constant that's multiplying and that's it. And if you just have if you just have a constant y the derivative of that is going to be equal to zero. Derivative of of a constant is always equal to zero because all you can also think of it graphically it's a horizontal line that has a derivative of zero gradient of zero. So that's why it turns out to be zero. This is a straight line. Straight line has this gradient. So that's the derivative.

[08:35:03]So these two gradients you can write directly. For anything else you will apply the power rule and get the derivative from that. Now when you have multiple terms adding or subtracting like this. When you have plus minus between terms like this. All you have to do is just differentiate each term separately. So in this case you can just differentiate one term at a time.

[08:35:24]Differentiate this separately, this separately, this separately. and then just put the same signs with them that are there already plus or minus and you will get the result. Okay, you can have a look at this working.

[08:35:35]But then what if we have something more complicated? So sometimes you can have something like this 2x^2 + 5 raised to power 1 / 3. Now what about this? This is not x raised to power something. The previous rule was when you have x raised to power something the derivative of that was nx^ n minus one. What if you have something other than x inside some other expression in terms of x? This is a more general form for the power rule. If you have anything raised to power n. So this box represents that you can have any expression in terms of x in place of this. Okay? Any expression in terms of x. It could be anything. Now what's the derivative of this going to be? You multiply by the power. You subtract one from one from the power. And you multiply by the derivative of the base of the power. the derivative of that box. Whatever is that is in that box, you multiply by the derivative of that.

[08:36:26]Okay, so that's these are three steps that you follow for this. Multiply by the power, subtract one from the power and multiply by the derivative of the base whatever is inside the brackets inside the power. Now if you think about it for x^ n as well we were even if you apply this rule that will still work there because this is x raised to power n right x^ n you can consider this as the base it becomes n into x^ n minus one and then you multiply by the derivative of x that's the base and the derivative of x is one so when you multiply by one it doesn't change anything because of this we were not explicitly multiplying by one in the previous uh examples Because the derivative of x is one. It doesn't make any difference if you multiply by that by that or not. However, if x is replaced with anything else. For example, if x is replaced with something like this x + 5 whole - 4. Now the base that you've got here, the derivative of this is not equal to one. So we have to separately multiply by that derivative like this. So how would you differentiate x + 5 - 4? You would say multiply by the power subtract one from the power and multiply by the derivative of what you have in the base of the power and then simplify that you get you get your result from that. Take another example. If you have - 5 / 7 that's a constant that's multiplying that that will remain as it is. You multiply by the power that's -4 over 3. Subtract one from the power and you multiply by the derivative of the base of the power. Now when you subtract one from the power notice this part initially will remain as it is. Okay. So inside the power it's going to remain as it is. You are not going to differentiate this inside the power. That will remain as it is. XQ - 4X that that is the same thing here. But then the power only reduces. Then outside you multiply by the derivative of that base and that gives you the result.

[08:38:18]This is another example. If you have x in the denominator like this, first of all you have to bring it to the numerator and write it as power minus 1 /2. Simplify that and differentiate that and then simplify everything. You get you get the result from that.

[08:38:31]Now that's the only rule that you've got in differentiation. The power rule of differentiation. You need to be able to differentiate functions using this rule.

[08:38:39]Now next we've got some applications of differentiation. What is differentiation used for? As we said, differentiation is going to be used to find gradients of curves at different points. For example, this is a very simple case. You have this function. You want to find the gradient of this curve at this point when x is equal to 3. How do you do that? You first of all differentiate this function using using the power rule. Using the power rule, this is in the denominator. Right? Now you bring that to the numerator. It becomes power minus one. Right? Like this. You apply the rule. Multiply by the power.

[08:39:08]Subtract one from the power. Multiply by the derivative outside. And then this is the function of gradient. This is the gradient function for this curve. Okay.

[08:39:19]Now how do you find the gradient at this particular point? You input the value of x in this x= 3. And that will give you the derivative which is the gradient of the curve when x is equal to 3. So if you want to find gradient of a curve at any particular point all you have to do is find the derivative. Put that value of x in the derivative and that will give you the result. All right let's take another example. Let's say you've got u a curve like this. This this has this has equation y = xq + 2. And what we want to do is we want to find the equation of the tangent at a particular point. and the equation of a normal at a particular point. So this is another application of differentiation. Now you'll need to find equations of tangents and equations of normals. Let's have a look at this example. So we've got this curve xq + 2. And this curve has a point. This is point a here. And we want to find the gradient of the tangent at at this particular point. The tangent looks like this. It's the line that's in the same direction as the curve at that point. And it just touches the curve. And the equation of the normal at that point. What's the normal?

[08:40:24]Normal is a line that's perpendicular to both the curve and the tangent at that point. Now, how do you find this uh these equations? Now, these are both straight lines. In order to find equations of straight lines, we need two things. Number one, we need a point on that line. Number two, we need the gradient of that line. Do we have a point already? Yes, we know that the point A has coordinates 1, three. They could also just give the x coordinate and you may have to find the ycoordinate yourself first of all because if x is equal to one you can easily figure out what what the value of y is going to be by just inputting that value here. But now assuming that we have the point now we need the gradient of the tangent and the gradient of the gradient of the normal. Now how does that work? You will have the equation of the curve given.

[08:41:04]What you would do would be you you would differentiate that first. You'll find the derivative of that and in that derivative you'll put this value of x at which you want to find the equation of the tangent and and the equation of the normal and when you do that you will get the gradient of the curve at that particular point. Right? Derivative is going to give you the gradient of the gradient of the curve at that particular point. And now tangent has the same gradient as the gradient of the curve because its direction is the same as as the direction of the curve at that point. And what about the normal? normal is going to be perpendicular to the curve. So its gradient is going to be negative reciprocal of the gradient of the curve. So all you have to do now is once you've got those derivatives and you found the gradient of the curve at that particular point, you say what's the equation of the tangent going to be gradient of the tangent is three because that's the gradient of the curve at that point. So it'll be the same as that. And we have a point that's a we use y2 - y1= mx - x1 and find the gradient of that.

[08:42:01]Find the equation of that tangent like this. For the normal, we take negative reciprocal of that gradient because normal is perpendicular to the curve. So whatever was the gradient of the curve, the gradient of the normal is going to be negative reciprocal of that. So that's -13. We change the sign take reciprocal. So now we've got the gradient of the normal. Gradient of that point is uh the the coordinates of that point are also given. That's this. You can find the equation using y2 - y1 sorry y - y1= m into x - 6 and that that should give you the result. Now this is what you need this is what you need to remember here gradient of the tangent at a particular point is the same as the gradient of the curve gradient of the normal at a particular point that's the that's negative reciprocal of the gradient of the curve at that particular point.

[08:42:49]The next thing that you need to know about is how to find stationary points and their nature. Now what are stationary points? Let's have a look at this example. It's it's a quadratic curve. It's a U-shaped curve. Stationary point is just another name for the turning point. Turning point, vertex, stationary point. The this point has all of these names. How do you find this? So for quadratic functions, we have done multiple ways of finding the turning point before. For example, we know how to do that using minus b over 2 a. We can do that using completing square. But this is this is another method that we can use to find the turning point. And that's using differentiation. Now this is a general method that works for any kind of function. minus b over 2 a completing square they were specific for quadratic functions for any other kind of function differentiation is the general method that works for finding turning points. Now how do you do that at a turning point you can see if you were to draw a tangent that will always be a horizontal line horizontal lines have a gradient of gradient of zero. So when you have to find a stationary point we know that the gradient of the curve at that particular point is going to be zero. Now if the gradient is zero, derivative represents gradient. That means at any stage three point the derivative is always going to be equal to zero. So whenever you have to find stage three points what we do is we put the we find the derivative put that equal to zero and that will give you the give you that point that you're looking for. For example, in this case this is the curve. What you do is you find the derivative that turns out to be this.

[08:44:08]You put the derivative equal to zero for finding the stationary point and that gives you a value for x equals 2. put that value of x back in the function that will give you the corresponding value for y as well and this is the point that we were looking for. Now you do not always just whenever you find derivative always put that equal to zero. That will not work. You can't always you cannot always uh just uh put the derivative equal to zero for no reason. When you're finding stationary points you put that equal to zero only in that case not in general. Okay let's take another example. So this can also work for cubic functions now but for which completing square minus b over 2a would not work. So if you have a cubic function like this xq - 12x and so on and we want to find the stationary points of this function. How do you do that? You would find the derivative put that equal to zero you get two values of x from that x= 1 x= 7. You can find the corresponding values of y now by inputting those values back in the function and that will give you the corresponding value for y as well. that turns out to be uh the points turn out to be 125 and 7 minus 83. These are the two points that we were looking for. Now these are the stationary points. The next thing we want to figure out is whether these points are minimum points, maximum points. One is minimum, the other is maximum. How do you figure that out? Now this is what the curve is going to look like in this case. This is what a cubic function looks like. Uh this is one possible shape of a cubic function.

[08:45:30]In this case, this is one point that we're getting. This is 125 and the other point is 7 - 83. Now the first one that you see on the left side we call this a maximum point. So whenever you have a shape like this this point is going to be a maximum point. So this point we consider maximum point and when whenever you have a shape like this we consider this point as a minimum point. This is the minimum point here and this was a maximum point here.

[08:45:54]Now I showed you how the graph looks like. Now if the graph was not given and you had and you had to find the these find and you had to figure out whether these points are maximum or minimum yourself without the graph. How would you do that? This is where the second derivative comes in. So what you do is you have the derivative. If you differentiate it once more, that's called the second derivative of this function. So 3x - 24x + 21, that was the derivative. You differentiate it again and that turns out to be 6x - 24. And now if the second derivative is negative, that is going to mean that it's a maximum point. If the second derivative is positive, that that means it's going to be a minimum point. Now how do you do that? This is 6x - 24. It depends on the value of x whether the number is positive or negative. So if the derivative is in terms of x you have to input those points. So we found two stationary points there. One had an x value of one the other had had an x value of 7. So what we do is we input x= 1 in this ones and that gives us minus8 and then we input x= 7 in this and that and that gives us + 18. Now when the second derivative is negative that means that it's a maximum point. When the second derivative is positive, that means it's a minimum point.

[08:47:05]This is what you need to remember about the second derivative. If it's positive, it's a minimum point. If it's negative, it's a maximum point. Now, for stationary points, this is the summary.

[08:47:15]If you have a function and you want to find if you have a function and you and you want to find the stationary points, what you do for that is you put you find the derivative, put that equal to zero, and that gives you the x coordinates of the stationary points. put those x coordinates back in the function and that will give the corresponding uh ycoordinates for finding for figuring out whether that stationary point is maximum or minimum what you do is you find the second derivative put the values of x in that if the if the second derivative is positive that's a minimum point if it's negative that's a maximum point so that's what you need need to know about stage three points and whether they're maximum or minimum there's a terminology that you that you also need to be aware aware of sometimes they will not say whether They'll not ask for whether it's a maximum point or a minimum point. They will use this terminology find the nature of the stage point. So nature of the stage two point means if it's a maximum or a minimum, right? That's what nature uh represents.

[08:48:15]Okay. So sometimes they they will ask you to find the maximum or minimum value of the gradient of the function. Not of the original function but the gradient of the function. Now how would you do that? Let's say we've got a function that looks like this for which they have given you the derivative. So they give gave you that the derivative of a function dy by dx that is equal to x + 4x^2. Now we want to figure out the minimum value of the gradient of the curve and we have to clearly show whether it's a minimum value or a maximum value.

[08:48:48]Now how do you do that? This is dy / dx.

[08:48:50]Now normally to find the maximum point or the minimum point what we do is we put the derivative equal to zero. Right?

[08:48:57]But that's something that we do for finding maximum or minimum point of a function. We're not finding maximum minimum point of the original function as in we're not finding the minimum point or maximum point of y. We are finding the minimum point of the gradient. Now what's gradient? We can say dy / dx is the is the gradient function of the curve. So we treat this as a new function. We say this is the gradient function. This is the gradient of the curve. Right? So dy by dx which is x x + 4x^2. We say this is the gradient. This represents the gradient of the curve. And now what we're looking for is the maximum or minimum value of the gradient not the original function y. So this is the gradient function.

[08:49:39]What we do is we find the maximum or minimum value of this the normal way for any function. When you're finding the maximum minimum value what you do is you find the derivative for that equal to z.

[08:49:48]So this is the function. Now you find the derivative of this function and put that equal to zero and that gives you the value for x that is two.

[08:49:58]And that tells you at this particular value of x the gradient would be minimum. Okay. So you put that value of x back in this function and that gives you the corresponding value for gradient. When x is equal to 2, the value for gradient is three and that's the minimum gradient. Now how do we show if it's a minimum gradient or maximum gradient? We find the second derivative.

[08:50:17]we find the second derivative of this gradient function. The second derivative turns out to be this. You put that value of x in this that turns out to be positive. That means that gra gradient was a minimum value. Okay. So when they ask you for the maximum or minimum value of the gradient of a function what you do is you treat dy / dx as a new function. I would I would recommend write it as g of x. So that's the gradient function. And then find the derivative of this function. put that equal to zero to find that station three point and then you can differentiate it again once more to find whether this point is a maximum or a minimum. Okay, so that's that's an important concept that you need to know about. The next thing that we have is increasing and decreasing functions. Increasing and decreasing function that's another concept that we've got in differentiation. What what are increasing functions? Increasing functions are those functions that have a positive gradient. Whenever the gradient is positive, that's an increasing function. Decreasing function is when the gradient of a function is negative. For example, the curve on the left side here, you can see as we go from left to right, the curve is going in the upwards direction. So the gradient is positive. If gradient is positive, derivative is positive. This is an increasing function. On the right side, the gradient is negative. As you go from left to right, you're going going in the downward direction.

[08:51:35]Gradient is negative throughout. If the gradient is negative, that means dy or dx, that is supposed to be negative.

[08:51:41]These are examples of some increasing and decreasing functions. Le on the left side you've got increasing on the right side you've got decreasing.

[08:51:50]Now these functions are throughout increasing and these functions are throughout decreasing.

[08:52:00]But some functions can be increasing in some part and decreasing in another part. For example in this case on the left side the function is decreasing. On the right side the function is increasing.

[08:52:10]And this cubic function it's increasing here. It's decreasing then and then it's increasing. So some functions are throughout increasing or decreasing.

[08:52:19]Some functions are increasing in one one part and decreasing on another part.

[08:52:24]For example, let's say we've got a function that looks like this. Y = 1 6 into 2x - 3 whole cube - 4x. And let's focus on the last part of this.

[08:52:34]Find the set of values of x for which this expression is an increasing function. It's an increasing function.

[08:52:41]Now what do you do? What do we do for that? In the first part they made us find the derivative. So we have the derivative for this. The derivative of this turns out to be 2x - 3 square - 4.

[08:52:49]Now if it is an increasing function what that means is the derivative dy by dx that is supposed to be greater than zero. So what we do is if it's an increasing function we say let's find the find the derivative dy by dx that turns out to be this put that derivative greater than 0. So 2x - 3 square - 4 that should be greater than zero. You get a quadratic inequality from that.

[08:53:12]Solve that quadratic inequality.

[08:53:13]Simplify everything and you get the result. This is these are the values of x for which this function is increasing. Okay. So you find the derivative put that put that greater than zero and you find those values at which the function is increasing. This is what the function looks like. The original function was cubic.

[08:53:33]This is what it would actually end up looking like. It was increasing in this part decreasing here and then increasing again. You find that using calculation even without knowing the graph. You find the derivative put that greater than zero. Solve that inequality and that gives you these set of values for which the function was increasing. Now sometimes they will ask you to show that a function is increasing or decreasing.

[08:53:54]So sometimes they will tell you if it's increasing or decreasing. Sometimes they will ask you to show that it's increasing or decreasing. How do you do that? Now this is the function and they're saying show that this function is an increasing function. Now notice this word show. You cannot assume already that it's greater than zero. You would find the derivative the normal way. This is what the derivative turns out to be. 6 into 2x - 5^ 2 + 1. Now you're not going to say this is greater than zero and solve and find the values of x from that. You will not get any values from that anyway. But you would not do that. You would not put this greater than zero. You would argue that okay now this is the gradient that we have. How can we show that it is always positive? We have to show that this is positive. How do you do that? When you have this form you can say six is positive. This is a square that can never be negative. So the first term therefore this whole thing is positive or at least zero. and we have a plus one here. So 0 + 1. The minimum value of this expression therefore is greater than or equal to 1. That means the derivative is always greater than zero.

[08:54:54]Since it's greater than zero, that means it's an increasing function. And and that's the proof that this is supposed to be an increasing function. Similarly, here you've got this function and they're saying show that it's a decreasing function. How do you do that?

[08:55:06]You find the derivative. The derivative turns out to be this. And now you have to show that it's a decreasing function.

[08:55:11]How do you show that the derivative has to be negative? the derivative has to be negative. One way to do this that do that here could be you take minus common outside. Now this expression inside is always positive inside because we've got x^ 4 here. We've got x square here. Even powers can never be negative. All the numbers are positive inside and we have a negative outside. That shows that this overall number is always negative. So you say the derivative is negative. That is that is why it's a decreasing function throughout. So notice the difference. Sometimes they ask you to find the set of values of a function. In this case, for instance, they ask you to find the sale values of a function for which that function is increasing. In that case, you find the derivative. Put that greater than zero and make it an inequality. Solve that inequality to find those values of x. Sometimes they ask you to show that it's an increasing function or show that it's a decreasing function. In that case, you will not make an inequality. You'll find the derivative and show that it's always positive or always negative. And that will give you the result like this.

[08:56:09]Now in the question that you will see sometimes you will have menstruation as well and using menstruation you have to prove something in the first part and then use that to find the stationary points. Sometimes find find the nature of those stationary points. For example, this is one case in the first part they ask you to show this thing. Now as usual you'll have some information given in the question that you can use to make some sort of an equation and come to this result. For example, in this case, they're saying in the question that the area the surface area of this cardboard is 200. So what you do is you find the areas of all the surfaces of this cardboard just like we used to do in O levels. Find this find the areas of all the surfaces all the faces all the faces of of this of this uh uh container and then you put that equal to zero. Rearrange that you'll get this result at the end.

[08:56:58]And then similarly they ask you to find the volume. It's a prism. That's area of cross-section times the length or area of base into height. Whatever formula you want you want to use. And then based on this they will ask you to find something like this. Find the value of x for which v has a stationary value. What does that mean? Stationary value.

[08:57:15]Station value means it's either maximum or a minimum. How do you find stationary values? You find stationary values by putting derivative equal to zero. So for this you would say dv / dx in this function that should that should be equal to zero. You'll find the value of x from that and that's the result that you were looking for. Then they say determine whether it's a maximum value or a minimum value. How do you do that?

[08:57:33]You find the second derivative of that.

[08:57:35]Find the second derivative of that and put that and figure out whether that's positive or negative. If it's positive, that's a minimum point. If it's negative, that's a maximum point. Then you've got rate of change and chain rule. Now you need to understand this terminology. If you have dy over dx, what that means is it's the rate of change of y with respect to x, right?

[08:57:52]The rate at which y is changing with respect to x. Now mostly what happens is we do we talk about rate of change in terms of time. So you would have time in the denominator like this dy / dt for instance that implies it means rate of change of y the rate at which y is changing with respect to time.

[08:58:12]dx / dt this means rate of change of x the rate at which x is changing.

[08:58:19]So whenever they say something like this rate of change of anything rate of change of y that's dy / dt rate of change of x that's dx over dt even if they don't mention that it's with respect to time it's understood that if they don't don't mention with respect to what variable it is going to be with respect to time. So rate of change of y that's dy over dt rate of change of volume that's dv / dt rate of change of area that's da over dt and so on. Now how would you use that? Let's say you've got got a container that looks like this. Now this is a container and it is a it is in the shape of a cuboid and we are given that the water is being poured into the container at a rate of 2 m cube per second. We've got the rate of rate of a rate at which the water is being poured into the container. Now how do you measure water? Water is measured in terms of volume. So what do you understand from this statement that water is being poured into the container at a rate of 2 m cube per second. That means the rate of change of volume DV / DT that is equal to 2. DV over DT that is going to be equal to 2. That's what you understand from that statement. You can also understand that from the units here. It's 2 m cube. Meter cube is a unit for volume. So that's DV / DT. And then they're saying find the rate at which the height is increasing. The rate at which the height is changing. So we are given DV over DT. And we are supposed to find the rate at which height is changing. DH / DT. So we're looking to find dh / dt. Now for finding dh over dt, we make something that we call the chain rule like this. We say okay dh / dt that's what we're looking for. We make two fractions like this and we multiply them such that the numerator of the first fraction is supposed to be the same as the numerator here. And the denominator of the second fraction that's supposed to be the same as the denominator here. So these two have to match. And then in the third place here and the fourth place here we've got that whatever the third variable is. In this case, the third in this case, the third variable, the variable was volume. So, we say we've got DV there. And now, how exactly do we find the value of DH over DT here? We were given the value for DV over DT. That was two. We input that value of two here. And DH over DV, that's not something that we know. Now, for that, we need some equation that connects these two quantities volume and height. Now, what's the shape of this of this of this shape? That's a cuboid.

[09:00:36]What's the volume of a cuboid given by?

[09:00:38]It's given by length into width into height. We were given that the length is 8 meter and the width is 4 meter. So what's the volume going to be? 8 into 4 into into h base area times the height.

[09:00:50]That's the volume of the shape. So from that we get an equation like this. From this you can find the derivative dv / dh that turns out to be 32. Take the product of that and that will give you 1 / 32.

[09:01:02]You can input that value there and that gives you the result. Now sometimes it ends up being a bit more complicated. We have a look at some some examples for that. But quickly, how does chain rule work? How do you make equations like those? In case you get confused by that, whatever you have to find out. For example, sometimes you have to find dv over dt. Sometimes you have to find out dh over dt. Sometimes you have to find out da / dt. What you'll do is you'll write it on the left side. Whatever you have to find out on the right side, you'll have two fractions. Two fractions like this. You would say these two fractions are multiplying. The numerator of the first fraction is going to be the same as this. So that's DV. The denominator of the second fraction is going to be the same as this denominator. That's DT. And then in these places, the remaining places, you've got whatever the third variable is. So if it's VA and D, what's the third variable? That's A. So you would say that's DA. So it becomes DA and DA.

[09:01:52]Here, if you had to find DA over DD, you'll you'll make a fraction like this.

[09:01:56]You'll say it's a product of two fractions. The first numerator is going to be DA. The second denominator is going to be DT. And then in these places you've got whatever the third variable is that's DV and that becomes the chain rule. This is equal to this.

[09:02:09]Now this is a very simple example of this volume of a spherical balloon that's increasing at a rate of 50 cm cube per second. Volume is increasing.

[09:02:16]That's the rate of change of volume. So we say DV over DT is equal to 50. Now sometimes they might say it's decreasing. In that case if they had said the volume is decreasing at this rate we would say this is minus 50. In this case they're saying increasing. So we just write 50 here like this. And what we have to find out is find the rate of increase of the radius. Rate of increase of the radius. What's that?

[09:02:39]That's dr / dt. Rate of change of radius. How do you find that? You make a chain rule like this. You say d r / dt.

[09:02:45]That's supposed to be equal to dr the numerator of the first fraction, dt the denominator of the second fraction. And then in these places you've got whatever is the third variable. That's what the chain rule looks like. And now in case of dv dv over dt you put whatever you have there that's 50.

[09:03:03]However this is something that you do not know dr over dv. So you have to find that out. How do you do that? You need some relation between those two um variables that you've got volume and uh radius. Volume of of a surfer is given by this. You have this here. You find the derivative of that. Put that value of r. So they wanted the rate of increase of radius when the value for r was equal to 10. What you do is you input that value of r uh 10 inside inside this derivative that gives you some value which is 400 pi here that's dv over dr. Now you have dv over dr for this you want d r / dv. So what you do for that is you take reciprocal of this reciprocal of 400 pi that's 1 / 400 pi and that will give you the value for dr / d. So that's how this chain rule works. That's how you deal with this with these rate of change problem.

[09:03:48]Sometimes you end up getting something more complicated. For example, if you have something like this, water is being poured into a tank at a constant rate of 500 cm cube per second. The depth of water in the tank t seconds after filling starts is 8 cm.

[09:04:03]That's what we're given. And what we need need to find out is the rate at which h is increasing at the instant when h is equal to 10. Now what's the information that's given? It says water is being poured at a rate of 500 cm cube per second. So that's dv over dt. So that's one piece of information that we're given dv over dt and what we have to do is we have to figure out dh / dt. Now we can make a chain rule like this. DH / DT equals DH / DV into DV / DT. Now DV over DT is something that we know. DH / DV is not known. How do you figure that out?

[09:04:37]You've got the volume function given here. You find the derivative of that.

[09:04:40]You find the derivative of that. Put that value of H. They wanted the want wanted that rate when the value for H was equal to 10. You input that value of h here h equals 10 and that gives you the value for dv over dh and that's your final result. When you input that 4900 in this chain rule you get that rate of change of height.

[09:05:01]Sometimes they will want to find they will want you to find some particular value not the rate for example in this case they're saying at another instant the rate at which h is increasing is 075. So we were given the value for dv dt already in the first part. In fact in the in at the start of the question and that was DV over DT equals 500. And now we given that DH / DT that is equal to 0.075.

[09:05:26]These are two values that we're given.

[09:05:27]We are given both rates. We are given rate of change of volume and we're also given rate of change of height. We given both of these values and what we have to find out is DV over DH. We have to find out DV over DH.

[09:05:41]Okay. I'm sorry. We have to find the volume at a particular point. Not not DV over DH. You have to find the volume at a particular point. Now how do you do that?

[09:05:50]These are two things that we're given.

[09:05:52]DV over DT and DH over DT. In the first part, what we've done already is we had the volume function.

[09:05:59]We want to find the the value of V at this particular point when this is the rate of change of height. Now that will come from this volume function. Now for this volume function, you need the value for H. You need some value for you need you need the value of h at that particular instant when the rate is given to be 075. Now how do you find that particular value for h. What you can do is you can say if dv over dt is equal to this dh / dt is equal to this.

[09:06:25]You can use chain rule to find dv / dh.

[09:06:28]You already have an expression for dv dh from from the previous part when you differentiated the volume function that turned out to be this dv over dh. Its expression was given by this. What we can do is we can say using the chain rule we can find this find the specific value the the constant value of this uh this derivative dv over dh using chain rule and that turns out to be 500 into 1 / 0.075 075.

[09:06:56]This value if you input that in this derivative that gives us the value for h. Once we've got the that value of h we put that value in this function the volume function that gives us the corresponding volume value and that's the result that we wanted. So sometimes you're given both of these rates and what you're supposed to do is you're supposed to use both of those those rates in the using the chain rule and find the value of the derivative at at that particular point. Now in this case we found the value of that derivative and we had this expression for the derivative. We put that there and that gives us the value for h. When we input that value of h in the volume function that gives us the final result for the value for b. Now this is the more complicated type of question that you'll see in general rule. That's how you would approach that.

[09:07:43]Okay. So one last example.

[09:07:47]Sometimes you they say something like this. You have you've got these two curves. Look at the last part of this question. The the rest is integration.

[09:07:55]We're going to we're going to skip that for now.

[09:07:58]The tangents to these two curves, we've got two curves there. One is y = 9 - xq and the other is y = 8 /x. They're saying these two curves, they they have tangents that are parallel to each other. At some point, the two tangents of these two curves are parallel to each other. So there's some point x equal to c and that point is between 1 and 2. So that's somewhere here. Let's say that's the point c. At this point the gradient uh the tangents are parallel to each other. So you might have a gradient have a tangent that looks like this here for one curve and you might have a tangent that looks like this for the other curve. They are both parallel at this particular point.

[09:08:41]At this particular point, both of these tangents are parallel. Now if they're parallel, how do you find the value of that point at which they're parallel? If at any particular point gradient uh tangents of the two curves are parallel, what that means is both of those curves have the same gradient at that particular point. Right? If the tangents are parallel, that means at these at this particular point the gradients of the curves are going to be equal to each other. Now if the gradients are are equal, we find the gradients of curves using differentiation. So what we do is we say we've got this curve, we've got another curve, find the derivative of both of them and equate those derivatives because if the tangents are parallel, the the gradients of the curves are going to be the same at that point.

[09:09:26]Gradients of curves are found by differentiation. So we find the derivative of both for both of them.

[09:09:30]These are the derivatives and then we equate those derivatives and that gives us the value of x that we were looking for and that's what they're calling C.

[09:09:38]So C equals under root 2 theorem. Now sometimes they will use this uh statement in in in questions. They will say some line makes an angle of tan inverse of K with the positive X ais tan inverse of something with the positive x ais.

[09:09:55]What do you understand from state state statements like this? In this case they're saying normal. It could be any line. Whenever any line makes an angle of tan inverse of something with the positive xaxis what do you understand from that? In coordinate geometry, we've done a formula that's tan theta equals gradient. That gives us the angle of a line with the x-axis. If you make theta the subject here, that gives you tan inverse of the gradient. So gradient, the angle of any line with the x-axis, that's given by tan inverse of gradient. If they're saying the angle is tan inverse of K. What that means is this value that you would have after tan inverse that would always represent the gradient of of that line. So in this case if they want you to find the normal at a particular point you know from this statement that the gradient of that normal is supposed to be K. Okay. So this is something that you will see come up number of times in in cross paper questions.

[09:10:53]As I said, you would have some questions in which in the first part they will ask you to use some inservation. So you should be familiar with all of these formulas that you've done in O level.

[09:11:01]Some of them them are given in the formula sheet. Some are not given. You will need to make sure that you remember them. For cone, pyramid and sphere.

[09:11:08]You'll have the formulas given for cube, cuboid, cylinder, prism. The formulas are not going to be given. Now the volume formulas are important. For surface area, you can just find the areas of the faces and just then just add them up together. It's not necessary to remember these formulas as such. Uh just find the areas of the faces and then add them up together.

[09:11:30]Let's have a look at this one. One last example. I'm sorry I said previous one was S but this that was not the last one. Let's have a look at this. This is a curve that we've got. We need to find the set of values of X for which Y decreases as X increases. Now what does that statement mean? Y decreases as X decreases. Now think about it on an XY plane as X increases that means as you go to the right side Y decreases you're going downwards. So which direction is the curve going in? It's going downwards like this. Now it could be any shape. It could be like this. It could be like this. It doesn't matter. All that matters is as x increases y decreases.

[09:12:06]That means the gradient of the function is going to be negative. If the gradient is negative that means the derivative is negative. So it's a decreasing function.

[09:12:14]It's a decreasing function. That means the gradient is negative. What you would do in that case would be you'll put find the derivative put put that less than zero. Solve that inequality to find the values of x that they're looking for. So instead of saying it's a decreasing function, sometimes they can say something like this as well. That means the same thing as x increases y decreases. That means it's a decreasing function. So the gradient is supposed to be less than zero.

[09:12:38]If they say some line is a tangent to some curve and they ask you to find this value of k.

[09:12:45]If this function was a quadratic function and and we had that the line was tangent to the to the curve, we would say that b squareus 4 a c is equal to 0. But b square - 4 a c does not work for cubic functions. In this case, it's a cubic function. And in this cubic function, they're saying this line here is a is a tangent to the curve. Now, we can't use b square - 4 here. What do we do then? If this line is a tangent with the curve at a particular point, what that means is whatever is the gradient of the gradient of this tangent, can we find that? In this case, that's compare this with MX plus C. The gradient is 9.

[09:13:20]Gradient of the tangent to the curve at that particular point should be the same as the gradient of the curve at that point. Right? So, whatever is the gradient of the tangent at this point, that's nine in this case. The gradient of the curve at that point should be the same as well. Now this is what the curve looks like. We want to find the point at which its gradient is equal to 9. So what we do is we find the derivative because that's how you find gradients of curves. We find find the derivative of this function. This is what the derivative turns out to be. We put that equal to 9. That gives us the value for x. Once we have once we have the value for x, we put that back in the function.

[09:13:56]That gives us the corresponding value for y. And that's the point that we were looking for. Once we got that point, they're saying find the value of the constant k. Just input that point here.

[09:14:04]And that will give you the value of the constant k. So whenever it's a non-quadratic function and they say some line is a tangent of the curve, what you do is you find the gradient of that tangent, put the derivative of the curve equal to that number and and then solve that equation and that will give you the result. Okay? All right. So that's what you need to know about know about differentiation. That's all on this.

[09:14:28]We'll have a break now. If you have any questions, you can let me know.

[09:14:31]Otherwise, we're going to end this uh differentiation part here. When we come back from the break, we'll have a quick look at differentiation uh at integration as well. Any problems anyone?

[09:14:50]15 minutes. 10 15 minutes.

[09:15:06]Are the first part or second part? First part or second part.

[09:15:10]Okay. So the question says the line is a tangent to the curve. So whatever the curve is, let's say it looks like this.

[09:15:16]It's a cubic function though it's shape is going to be different. But let's suppose this is a random curve. At this particular point, this is what the tangent looks like.

[09:15:24]They're saying this tangent is y= 9x + k. Now what's the gradient of this tangent? That's 9. So the gradient of the curve at that point should also be equal to 9. So what we do is we say at this point dy by dx is supposed to be equal to 9.

[09:15:40]So we find the derivative and put that equal to 9 because whatever is the gradient of the tangent at that at that point the the derivative of the curve the gradient of the curve that is also supposed to be the same.

[09:15:53]Make sense?

[09:16:00]Okay. Any other questions on this quickly?

[09:16:06]So now was your question answer that's not stationary point. Why would it stationary point? It was if it was a stationary point its derivative would be zero.

[09:16:53]No other questions.

[09:16:56]Yes. Aan.

[09:17:03]>> Okay. We have to tell if uh it's increasing, decreasing or neither. I don't understand how to do those. Like you said about you told us about uh how to tell about increasing and decreasing. But how do you tell if it's like none of them?

[09:17:24]>> Okay, let me go back to one of those questions. Increasing and decreasing. Uh for example, this one when they ask you to find the set of values of x for which this function was an increasing function. Now this function we can say it's neither increasing nor decreasing because in some part it's increasing in some part it's decreasing. So if you have a function like this, the derivative turns out to be this 2x - 3 - 4. Now you think about whether this derivative is always positive or always negative. Whether it's always positive or always negative. Now we can't say that because in some part the derivative is negative and in other part it's positive. Right? So in this part it's negative between 1 /2 and 5 /2 and in the remaining parts it is positive.

[09:18:16]So in this case we would say it's neither increasing nor decreasing.

[09:18:23]So that will be uh that case. Does that make sense?

[09:18:28]All right. So next we've got integration.

[09:18:31]So first of all integration is just the opposite of differentiation. So we also sometimes call it refers differentiation. Just the opposite of what we do in differentiation. So let's say if if you got a function that like this y= 5x^2 what's the derivative of that that's 10 x if you want to go in the backwards direction that we have dy dx given and we want to find find y from that the process that we would have to use for that would be called integration right so for from 10x we can find that the original function was 5x^2 for that we do integration now how does this process work there are two steps that you do for functions that look like this x to power n You just reverse the process in differentiation. We used to multiply by the power and also subtract from the power. Subtract one from the power. Now we do just the opposite of that in the opposite order as well. We say add one to the power and divide by the new power. For example, if we have 10 x, what's the integration of that going to be? 10 remains as it is. That's a constant. Just like in in differentiation, if you have a constant multiplying that remains as it is, that that does not change. We add 1 to the power and we divide by the new power. So x + x^ 1 + 1 / 1 + 1 that gives us the integration.

[09:19:43]This is what the rule looks like. In fact before that we've got this notation as well. In differentiation the notation used to look like this. Derivative with respect to x of x cub for instance that is 3x^2. That's the differentiation notation that you need need to know about. In integration we have this notation. This symbol that represents integration and this dx that represents that you're integrating with respect to x. the variable in terms of which you're integrating that is x. So that's the notation that we use for integration. So if you have something like this for instance x raised to power n and then the sign dx that means integrate x raised to power n with respect to x.

[09:20:19]What's the integration going to be? We add 1 to the poweride by the new power and we get the result of that. Now whenever we integrate we always have this plus c. That's the constant of integration. I'll tell you I'll tell you the reason for that briefly later on.

[09:20:31]But we always have this plus c whenever we integrate. So let's say we've got 5 x^ - 3. How do we integrate this?

[09:20:39]Five is a constant that remains as it is. We add 1 to the power divide by the new power. And that gives us the result - 5 over 3. That's a constant that remains as it is. Add 1 to the power. So 1 / 4 + 1 divide by the new that new power 54. And that gives you the result after simplification. And you have a plus c with that. So that's how integration works. If you've got x in the denominator just like in differentiation we have to we have to bring x to the numerator first of all before we can apply that rule. So this is x^ 1 over3 in the numerator in the denominator when it comes to the numerator it will become x^ - 1 over 3 and then we apply that rule we say 7 is a constant that's outside we add 1 to the power of x divide by that new power and that gives us the result. Similarly 5 / 2x cq x cq comes to the numerator becomes x^ minus 3 we add 1 to the power divide by the new power we get the result. If you have a constant what's the integration of a constant we just have an x multiplied with that like this. Why is that? Because 7 can be thought of as x^0 7 into x^0 and then we can say add one to the power divide by the new power and that gives us x like this. So whenever we have a constant just remember the integration of that is going to be just that constant multiplied with x like this. All right. So if you have k integration of that is k * x.

[09:22:00]If you have multiple terms just like in differentiation we can just integrate each term separately. If there's a plus minus between them. So if you have plus minus like this you integrate one term at a time and that gives you the result.

[09:22:14]This is often quite helpful in integration when you have complicated integrals. Just remember if you have a constant that's multiplying or dividing by the whole thing since that is not going to change after integration as no integration rule applies in that constant that's multiplying you can just write down that constant outside the integration sign as well to simplify things sometimes. So this k can be written outside and that will not change anything. For example, if you have something like this 2x - 4 whole^ 7 and you have to integrate this to simplify things, we could write the seven outside so that it becomes 1 / 7 into 2x - 4 and then 1 / 7 outside remains as it is.

[09:22:50]Just apply the integration rule on what is inside and you get the result from that.

[09:22:57]Now this is why we have a c there.

[09:23:00]Should I tell you about this quickly?

[09:23:02]Okay. So if you have a function like this, this function, this function, all of them have the same derivative of 16x.

[09:23:08]Now if you were to integrate 16x, what's the integration of 16x? That's 8x squ that only gives us this. What if there was a constant here? If it if it was + 5, if if it was + 3 here, their derivative were also the same. How do you get those those other functions? Why are we only getting 8x^2 when we integrate this? Because when we differentiate something, this constant basically gets eliminated, right? we don't have any constant in the derivative that we have. So when we integrate we consider this possibility that there could be a constant in that original function as well. So we always have a plus C. This is called constant of integration. We always have a plus C. Whenever we integrate we always put plus C with that unless when when we have limits in that case even if you ignore this uh you can get away with this because it gets cancelled there.

[09:23:59]But otherwise apart from that you always put c there otherwise you can get end up getting wrong answers. You need to make sure that you remember that. Now we talked about how to integrate x raised to power n. What if you have had some other expression in ter in place of x inside the power and you have to integrate that expression. For example, if you have 2x - 3^ 4, how do you integrate this? Now we call this the general power rule of integration. It's it's the generalization of that previous rule that was just a specific case of this 2x - 3^ 4. How do you integrate this? This is the rule that we use for this. The these are the steps that we follow to integrate this. We add one to the power just like before. We divide by the new power just like before. But then we have an additional step and that is we divide by the derivative of the base.

[09:24:46]So in differentiation we used to multiply by the derivative. In integration to reverse that we divide by the derivative of the base of the power.

[09:24:54]So for example if you have something like this mx + c all raised to power n what's the integration of that? We add one to the power we divide by the new power and we also divide by the derivative of the base. What's the derivative of the base? If it's mx plus c derivative would be m. We also divide by m in the denominator and that will give you the integration of this. Now for this rule we have a condition and that condition is that the part that is inside the function the inside the power this part has to be something linear it has to be linear it cannot be nonlinear in differentiation we did not use did not have any such condition in integration we have this condition the part that is inside it has to be linear if you have a nonlinear expression inside the power that does not work you cannot integrate that using this rule okay this rule don't would not work if the part inside is not linear. These are some examples of that. You can have a look at them. If you had this 7 over 3 into 4x - 1, again the first thing you have to do is you have to bring this to the to the to the numerator. So it it comes to the numerator and becomes 4x - 1 raised to power - 1 /2 - 7 over 3 was just a constant. Simplify things. You can write that outside like this. And then you apply the the power rule on this. You add one to the power divide by the new power and also divide by the derivative of that base. and simplify that to get that integral. If you have 7 / something like this - 3x + 1^ 2. Now when you bring this to the numerator first of all it becomes this power minus 2. So if you had to integrate this you want to integrate this thing.

[09:26:31]What how would the integration of this work look like? You would say first of all bring this to the to the numerator.

[09:26:37]The power becomes -2. 7 remains as it is. You add 1 to the^ - 2 + 1 that becomes - 1. Divide by the new power and also divide by the derivative of the base that is -3. Now what about this?

[09:26:49]Now 3x 2 - 1^ 2. How do you integrate this? If you apply the power rule in this like this that would be wrong. This is not acceptable because the power rule does not work here. You're dividing by x here. That is problematic. This will not work. Okay. The power rule does not work in this case. The inner part of the function it has to be linear. If it's not linear, we cannot apply this. So how exactly do we integrate this function?

[09:27:13]Like if it comes up, how would we integrate this 3x - 1^ 2? We would first have to expand this. This is very important. Now when you have a nonlinear function inside generally with a power two because for higher powers it'll become very complicated to expand that. But with power two what you'll do is you will say let's first of all expand this using a minus b whole square or a plus b whole square. And then after you you've expanded this then you just integrate each term at one term at a time. This separately this separately and this separately it can't be done directly because the inner part is nonlinear. If the inner part was linear we could apply the power rule directly. In this case we cannot do that. We have to expand it first. Now the these were the rules that you need to know about integrating functions. Now what are the applications of integration? Number one you need you should be able to find equation of a curve from its derivative. Sometimes what happens is they give you the derivative of a function for example like this and they want you to find the equation of the curve from that. Now if you want to go from dy by dx to y. y is the original function. Right? How do you do that? You do that by integration. In order to go from y to dy dx you use to differentiate. In order to go the other way around now we have to integrate. So you have dy dx equal to this. You integrate that function using the power rule. Do not forget the plus c. Whenever you integrate you always have a plus c.

[09:28:29]Now when you have a plus c here how do you find the value of that c? You're given a point that lies on that curve.

[09:28:34]That's the point 38. You input that point here. Input the value of X as three. Input the value of Y as eight.

[09:28:40]Find the value of C from that. And then that's the that's the so that's the equation of the curve that you end up getting from that when you input that value of C. So that's one thing that you have to do. Then you need to be able to integrate using limits. So sometimes what happens is they have these numbers written on top and at the bottom. These are called the limits of this integration. The the number at the at the bottom that's called the lower limit. the number on top that's called the upper limit. Now how do you evaluate this? You integrate like you normally do. So 2x + 3 the integration of that that turns out to be x2 + 3x + c. Now what you do is you input the upper limit once input x= 5 in this ones and then you input x= 3 in in this ones. That's the lower limit. Subtract from the upper limit case the lower limit case and you get the result from that. Now notice c is getting cancelled here.

[09:29:29]So you can get away with not writing C here at all. So when you have limits you can ignore C as well but otherwise you will always write this. And that's how you evaluate these limits. You integrate the function. Input the input the upper limits subtract from that the case and you have the lower limit and that gives you the result. Now what exactly does this represent? This number that you get from this it actually represents the area under the graph of this function.

[09:29:52]So this this 2x + 3 that's a straight line. That's a straight line that looks like this. It's 2x + 3. That's the straight line. The limits that we input there they were from 3 to five. Now the these these are the points three and five on this on this graph that value that we got there 22 that's basically the area of the region below that line with the x-axis the shaded part this number represents the area under this graph. Now this is the trapezium you can find that area using the area of trapezium formula as well and that also turns out to be 22. However, integration is very helpful when instead of straight lines, we've got some curves and we want to find areas like that. And that's brings us to this how to find area between a curve and x-axis or y-axis.

[09:30:39]We've got these formulas for that that we can use to find areas between a curve and the x-axis and the y-axis. Area between the curve and x ais, the formula for that is integration of y with respect to x. And we have these limits here. Limits are supposed to be values of x in this case. And the other formula is area between the curve and y-axis.

[09:30:55]That's the formula x with respect to y.

[09:30:58]Integration of x x with respect with respect to y with limits of y there. In this case, this is y equals something.

[09:31:04]And this is x= something. Now let's talk about these a bit more detail. Let's say we've got a function like this. y= x + 2. And we want to figure out what this area is. Okay. And we're given that this is the area between x= 1 and x= 3.

[09:31:19]How do you find this area? This is this is an area between graphs and x-axis.

[09:31:23]Now the formula for that is integration of y with respect to x and we have these limits x= a to x= b. These are the values of x from which the graph uh between which we have to find the area.

[09:31:36]Now when you have to use this for when you use this formula you have to make sure that y is the subject of the function that we're given. So in this case it already is the subject otherwise you have to first make it the subject so that you can input y equal to that thing inside this equation then inside this formula then y in place of y you write x² + 2 and the limits that you input there when you're finding a graph and x-axis the limits are supposed to be values of x these are the limits from 1 to 3 and all you have to do now is integrate this input those limits and you'll get get a result that will give you the area between graph and x-axis now this this formula and the volume formulas They are not given in the formula sheet. You need to memorize these. They're not given in the formula sheets. So, so when you have to find area between graph and x-axis, the formula is integration of y with respect to x. When you have to find area between graph and yaxis, the formula has x in it. Right? So, it's the other way around. With x-axis, the formula has y. With y-axis, the formula has x.

[09:32:35]Okay? Now, this is another example of this. You can have a look this.

[09:32:39]Sometimes what happens is this. The area turns out to be below the graph like this below the x-axis like this. Now how do you find this area? The formula that you have that is basically for the area between the graph and the x-axis. Now in this case since this area is below the x-axis you will find when you apply the formula the end result that you would get from that formula that's going to be negative that's - 36. The reason for that is this area is below the x-axis. If it was below above the x-axis that would turn out to be positive. If it's below the x-axis that turns out to be negative.

[09:33:10]Now since we we're just looking for area we would just make it positive and we would say the area is actually equal to 36 even if it's a negative value. That's only representing that the area is below the x-axis. Now sometimes what can happen is you have an area like this.

[09:33:23]Some part of the graph some part of the shaded region is above the x-axis and other part is below the x-axis. How do you find this total area? Now since the part that is above the x-axis that's going to be that's going to turn out to be positive from the formula and this area is going to turn out to be negative. we cannot do uh calculate it all together. We would have to split this into two parts. So what we do in this case is we say let's integrate the function. Let's integrate the function from 0 to 2 separately and that gives us some value that gives us 20 over 3 here and then integrate the function from 2 to 6 separately and that gives us another value that's minus 128 over 3.

[09:34:00]Now the area above the x-axis this a1 that is 20 over3 and the area below the x-axis that's 128 over3. This negative sign is only representing that the area is below the x-axis. What's so what's the total area going to be? You will make both of these numbers positive. So this is already positive. You make this positive as well. And then you add them up together. The total area turns out to be 148 over 3. That's the total area that that you have from this. So when some part of the area is above the x-axis, some is below the x-axis, you have to break it down into two parts and then integrate.

[09:34:33]This is also an important point here.

[09:34:34]How do you integrate function like this?

[09:34:36]X into x -2 into x - 6. Now it's a product. We've got multiple expressions multi. We've got multiple expressions multiplying with each other. We cannot integrate it separately. We can't do something like this that it's x square over two and then x - 2 square over 2 and x - 6 over2. We can't use the power rule on each of them individually. It's a product. We could do that if there was a plus between them. We could integrate each of them separately. But this is a product. This whole thing is one. We can only integrate it if we first simplify this. So we simpl simplify it first like this and then we integrate one term at a time and that's how that's how I've done it in this example on the left side. All right. So that's what happens when you have some area above the x axis some area below the x axis. You break it down into two parts like this and then you integrate.

[09:35:22]You could have area between a graph and y-axis as well. That's the formula that that we have for this integration of x with respect to y. Now in this case x needs to be the subject. Limits are supposed to be values of y. Now what does it look like graphically? It looks like this. We could have a function that that that is this and we're looking for this area between the graph and y-axis.

[09:35:41]Now, how do you use the formula for this? If the function is given like this y = xq + 5, the formula has x in it. The e the x variable should be the subject.

[09:35:52]So, first of all, you have to rearrange this and make x the subject. How do you do that? You've got y= x cub + 5 here.

[09:35:58]You make x the subject. Take five to the other side. Take cube root on both sides. And that gives you x equals this.

[09:36:03]And now this is something that you can input in the formula. You get y - 5^ 1 / 3 with respect to y. That is what you integrate here.

[09:36:13]Okay. So now only the variable name has changed here.

[09:36:19]It's y in place of x. So you just integrate integrated like you normally would and then input the limits to get that area that you're looking for. Now in case in case of y-axis area on the right side that will turn out to be positive. Area on the left side that will turn out to be negative. from the formula again area cannot be negative you will just make that value positive and that will be your final result if they just want you want an integral by the way so sometimes what they do is they ask you the value for an integral so they do not refer to area they will just write find the uh find this integral x + 2^ 2 with respect to x and maybe uh let me actually use another function that could actually be negative as well let's say they ask you to find the area of this integral x cub minus something with respect to x and limits are let's say from minus1 to zero. Now in this case if you were to find this area find this integral it might end up having a negative value you will keep it negative because they're not referring to area if they just ask you to integrate this evaluate this this result if it turns out to be negative you just leave it as negative. If they specifically mention that they want the area from a function, in that case you will evaluate it like you normally do. If the value turns out to be negative, you will just make that positive and say the area is positive.

[09:37:40]All right? So that's what you do. You don't nec necessarily need to put put a modus sign there. That's not necessary.

[09:37:46]If it turns out to be negative, just make it positive. If they're referring to area, if they're not referring to area, you keep it negative. That remains as it is. you just leave uh you just write the negative number as the final answer.

[09:37:59]Okay. Now a lot of times what happens is it's not just area between a graph and x-axis or area between a graph and y-axis. It's area between two functions, two graphs, two different graphs. So if you have something like this, you've got these two curves. We say one of these is the upper curve. So in the shaded part, look at whatever is the upper curve. So in this case, this green one, that's the upper curve. and the blue one that's the lower curve. So what we do is we find the area between the upper curve and the x-axis. We subtract from that area between the lower curve and the x-axis.

[09:38:32]So we find all of this area area between the upper curve and the x-axis and subtract from that all of this area between the lower curve and the x-axis and that will give us the area between the shaded reg area of the shaded region. So you find one area subtract from that the other area. For the upper curve, you find the area first and then subtract from that the area for the lower curve and that will give us the result. Now, one way to do this is like this that you do it separately for each function. You find it for the upper graph separately, you find it for the lower graph separately and then you find the difference and that gives give that gives you the result. That works. You can do it separately. So that uh saves you from careless errors a lot of times that you do these two calculations separately. You say let's do the calculation for the upper curve separately and let's do the calculation for lower curve se separately. And then at the end what you do is you find the difference and you say that's the result. For example, if this is a1, this is a2. You say the final area therefore is going to be a1 minus a2. That's one idea. Now to make it more efficient sometimes because in recent papers sometimes papers do do tend to be long as well. So you need to be efficient with your time as well.

[09:39:37]If the limits if the limits are the same if the limits are the same for both functions. So in this case for example both the upper curve and the lower curve they have the same limits. If the limits are the same you can do it directly like this as well. So you can say one function is 9 - x cub and the other function is 8 / x. If the limits are the same, you first subtract the functions like this 9 - x cub - 8 / x and then you integrate it. Input the limits directly.

[09:40:03]If you understand that and kind of comfortable with this uh you can do do this as well. Otherwise if you're u confused about it, just do it the normal way. If you're used to doing it this way, that that's okay. It just take maybe one more minute and that's that's okay. that doesn't make much difference.

[09:40:18]But if the limits are the same, it is sometimes more efficient to subtract the functions first like we have done on in this case and then input the limits directly and that gives us the result. Okay. Now, what about volume of revolution? Let's say we've got a line like this y=/x and we've got this shaded area, the yellow shaded area. If you were to rotate this area about the x-axis, in this case, if it's a triangle, when it gets rotated 360° about the x-axis, the shape that we end up getting from that is a cone. Now, for cone, we have a direct formula that's 1 over 3 p<unk> r square h. Now, instead of a straight line, what if you had a curve like this, a curve like this. And the area between the curve and the x-axis or the y-axis, what if that was rotated about the axis?

[09:41:08]In that case we got a shape. It's a 3D shape but that's not a standard shape for which we have any direct formulas.

[09:41:12]So for cases like cases like these but what we do is we use integration to find these volumes as well. Now what are the what are those formulas that we've got?

[09:41:19]So this for volume of revolution about x-axis we've got this formula pi into integration of y square with respect to x. But for volume of revolution about yaxis we've got this formula pi into integration of x square with respect to y. Again when you're doing it about about xaxis the formula has y^2. When you're doing it about y-axis, the formula has x squ. So the opposite of the axis, that's what you have in the formula.

[09:41:43]Let's start with this one first. Volume of revolution about x-axis.

[09:41:48]This is the formula that we've got. V= pi into integration of y^2 with respect to x. Right? And then then you input limits of x and this.

[09:41:58]Now for using this formula, the first thing you have to do is you have to make sure that y² is the subject in the given equation. If it's not the subject, you have to make it the subject. First of all, make y square the subject. Then you input that in this integral. And the limits that you have, they're supposed to be values of x.

[09:42:14]And what does the formula give you? The formula gives you the volume when the area between the graph and the x-axis.

[09:42:23]Area between the graph and the x-axis is rotated 360° about the x axis. So the if you're rotating about x-axis, the area should be between the curve and the x-axis as well. Okay? And then you'll get that correct volume. That's that's what the formula is for. The it's the volume when the area between the graph and x axis that is rotated 360° about the x axis.

[09:42:46]Now how does it work? If you had a function like this y = x² + 5 and you had to find the volume of revolution from 0 to 3. In this case, you would first of all find y square. From that y square turns out to be this x square + 5 square. We input that in the formula. It becomes x2 + 5 square. Now how do we integrate this? It's a nonlinear function inside this inside the power.

[09:43:08]We cannot apply the power rule directly.

[09:43:09]So we expand it first. We expand it to this x^ 4 + 10 x + 2 25 using a + b whole and then we use integration on each term separately. Input the limits and at the end remember to multiply pi there and that gives the result. A lot of times in this these cases they ask you for exact answers. Always write them in terms of pi. In that case if they want an exact answer do not use decimals uh in your working. What about volume of revolution about y-axis?

[09:43:36]If it is about yaxis then the formula has pi into integration of x² with respect to y. So in this case the subject that you have that you need to have in the function is supposed to be x² and the limits are values of y and that this formula gives you the volume when the area between the graph and y-axis when the area between between the graph and y-axis that is rotated about y-axis. Now this is important as you'll see in a couple of examples later on.

[09:44:09]How do you use this formula? Now find this this volume of revolution for example. This is with the y-axis. How how would you do this? You would say this is the function y = x2 + 5. Let's make x² the subject first of all. So it becomes x² = y - 5. Now you input that in the formula. It becomes y - 5. What are the limits? Limits are supposed to be values of y. They're going from 5 to 14. These are the limits that you input there. Integrate it and input those limits. You will get that volume. Now what if you have to find volume of revolution of the area between two graphs. So you have this area between two graphs. How do you find this volume of revolution? Now it works very similar to how we used to do this in area. We find the volume of revolution of the area under the upper curve. Subtract from that the volume of revolution of the area under the lower curve and that will give the result. Find one volume, find the other volume, find their difference and that gives you the result. So just like we used to do in area in area, we do the same thing here as well.

[09:45:10]For example, if you have something like this, we want to find the volume obtained when this shaded region is rotated 360° about the x-axis. Now, how do you rotate this about the x-axis? If you think about this curve, if you apply the formula directly on the curve, that would not work because when you're rotating about x-axis, the formula, if you just apply that that in the curve, that will give you the volume when this green part is rotated about the x-axis, right? that will give you this volume and this green part is shaded about x-axis but the shaded part is not the area between the graph and x-axis. So how do we think about this? We think of it as two graphs. We say okay we've got this one line. It has an it has the equation y = 2, right? That's just a constant. It's a horizontal line. So y = 2. That's that's a horizontal line. And the other graph is the other graph is this curve. So we've got this graph that's a straight line horizontal line and the other graph is y = 3x + 1. So what we do is we find the volume of evolution under the upper graph. So that's y = 2. We integrate that separately. We apply the formula on that and then we integrate the lower part separately. Apply the formula on that as in first of all you have to make y^2. So this becomes y^2= 4. This becomes y^2= 3x + 1. apply integ apply the volume of evolution formula on both of them subtract this one from the other and that will give you the volume. Okay?

[09:46:40]So you can't apply the formula directly on directly only on the curve because that will give you the volume of revolution for this part for this area.

[09:46:48]That's not what we're looking for.

[09:46:51]So again this is what you have to do volume of revolution of the of the upper graph that's y equals 2. do that separately. Do this separately and find their difference and that will give the result.

[09:47:05]Okay. Then another concept that you have here is of improper integrals. Sometimes you get integrals in which infinity is one of the limits. Infinity is one of the limits or this is not something that you'll find a lot in past papers but it is something that you need to know as well and that is when you get zero in the denominator after inputting one of the limits. So improper integrals is when one of these two two things happens either infinity is in the limit or when you input any limit you get zero in the denominator. If you have one of these cases we call that an improper.

[09:47:41]Now let's take an example. Let's say we have a curve that looks like this. Y= 1 /x^2. Now this is what the shape of the graph is. We want to find the area under this graph from two onward to the right side. Now this looks like it's an infinite area because it's going on forever like this on the right side.

[09:47:57]There's no limit to it on the right side. What what are the limits going to be for this? If you want to find all of this area, we apply the formula integration of y with respect to x.

[09:48:06]Limits from x= a to x= b. Now what are the limits going to be in this? Y is equal to 1 /x². We can say that's x^ -2.

[09:48:13]That's the function that we've got. What are the limits? It starts from 2 and goes up to infinity. The limits are from 2 to infinity like this. Now how do you evaluate this? You just do it the normal way. You say integration of x^ minus 2.

[09:48:24]What's that going to be? You add 1 to the power. You divide by the new power.

[09:48:27]And the integration turns out to be minus 1 /x from this. Now input those limits. The upper limit is infinity.

[09:48:33]Just input it like this. And that's okay. At this level, this is fine. That will work. You just input infinity here.

[09:48:40]Minus min - 1 /2. And then you evaluate this 1 / infinity. One over an extremely large number. What's that going to be?

[09:48:47]You say this approaches zero. and uh then this this is + 1 /2. So you evaluate that that turns out to be 1 /2.

[09:48:55]Now another case could be that when you integrate you end up getting 0 in the denominator. For example in this case the lower limit is zero the upper limit is two.

[09:49:06]When you integrate this you end up getting - 1 /2 - - 1 / 0. And that's something that's undefined. You can't have zero in the denominator. But in this case we consider this as an extremely large number 1 / 0. That's basically telling us that so in this case for example when you look at the limits here it's going from 0 to two.

[09:49:25]Now it's not becoming exactly equal to zero because the graph is becoming an asmtote here. So this value is basically approaching zero. It's becoming very small. So when you have zero in the denominator you think of it like this then it's a very small value in the denominator and that basically gives you infinity. In case of that you write infinity and the overall area in this case turns out to be infinity. So the take away from that is when you have infinity in the denominator when you input the limits when you have constant over infinity we approximate that as zero. When you have constant over zero we approximate that as infinity. Now the first case you'll find a lot in cross papers. The second one not so much. So this is more important the first one. When you have infinity in the denominator just remember you would get you would uh replace that part of the fraction with zero. Then let's simplify that.

[09:50:14]Okay, there's a design going on here.

[09:50:15]We'll have a break and we'll have a look at a couple of questions quickly and then that's it.

[09:51:18]Okay. So this is what happens when we uh have infinity as one of the limits or we get zero in the denominator. When we have infinity as the limit and it turns out to be in the numerator then the result would be infinite. But that's again something that you would not see in past papers until now because they will normally ask you for something that has a finite value. Something that results in infinity at the end. I haven't seen any questions in past papers at least that in which the result is infinity. But that's what what would happen if infinity is in the numerator then that means that the overall thing is going to be infinity. Now sometimes you end up having areas like this in which you have to add two parts of uh the area. For example, if you have something like this, this shape, how do you find this area? How do you think about this? Is it area under the curve minus the area of triangle or something else? How do you think about this? It's not going to be done as one of one of those examples in which we say find the area under the under one curve and subtract from that area under the under the other curve. This is actually two different parts um that you have to do separately. For example, this is one triangle. So you find the area of this triangle separately and then you say find the area under the curve from this point A to the origin separately. So sometimes you have to break your area into two parts like this. Even in integration questions like you do in menation you do in se do it in separate parts like this. You say okay this is a triangle find the area of this part separately using area of triangle and the left part this is the area under the curve find the area of that area under that curve separately and then just add those two areas together in this case you don't subtract one area from another but you find two areas separately and then add their values so this is how it would be done you find the area of the triangle separately so in the first part we have found uh the coordinates of B and C using that we find the area of the triangle that turns out to be one in this case and then for the curve part we find the x intercept for that first of all inputting y equal to 0 and then we use these limits find the area under the curve from that point to zero and that turns out to be after this integration minus 1 / 6 since it's a negative area I calculated it with using the y-axis formula if you calculate using calculated using x-axis formula that would turn out to be positive. Using the y-axis formula, it turns out it turns out to be negative. Either way, the area is 1 / 6. The numerical value is 1 / 6.

[09:53:50]You just add these value this value to one and that gives it the total area that's 7 / 6 like this.

[09:53:58]So that's something that you need to be aware of as well that sometimes you can break your area into two parts like this, evaluate them separately and then add them up together. Now this is another question another type of question that you will see in past papers that you have the second derivative given and using that second derivative they somehow want to somehow uh sometimes want you to find out the original curve or something else sometimes. So in this case you've got the second derivative given and they're gi giving you that the minimum point of this function is 3 minus 10. Now that's the minimum point. Now if it's the minimum point that means it's the stationary point. At stationary points we know that the derivative is equal to zero. So what we know about this point is at this point the value for x is equal to 3. The value for y is equal to - 10 and at this point the derivative dy by dx that is equal to 0. x= 3 y = - 10 and dy by dx= 0. That's what we know about this point. So how do you find the coordinates of the maximum point from this? How does that work? You have the second derivative. First of all you integrate it once. That gives you the first derivative. When you integrate it, you get a plus c there. When you get a plus c, how do you find the value of the c? Now for this value of c you need some value for x and dy / dx at a particular point. Now in this case for example you know when x is equal to 3 dy by dx is equal to zero. So you input that value here. Input x= 3 and dy dx dx equal to 0. That gives you c= minus 6. So now you've got the value for the derivative.

[09:55:27]Now what were you looking for? You were looking for the coordinates of the maximum point. How do you find the coordinates of of maximum point? by dy dx equal to 0. So now you've got dy / dx= dy by dx. The expression for that you put that equal to zero. When you put that equal to 0, you get two values of x. One is x= -2. The other is x= 3. x= 3 was already given. That was the minimum point. We already found that before.

[09:55:50]So x= x= -2 is the other stationary point and that's what we were looking for. But now the problem is they wanted us to find the coordinates. That means we also need the y-coordinate. How do you get the ycoordinate? For that we will need the original function y. How do you get to that? Now we've got dy / dx. So we integrate it again. Integrate dy by dx again. And that gives us this function.

[09:56:14]But then you have another c. So for that you use the point that was given. The point that was given was 3 - 10. You input that point here. 3 - 10. Find the value of this c. Now once you've got the c, now you've got that function. You know the value of x x of the stationary point. Put that value of x in this function. That will give the corresponding y value. and you finally get that maximum point that you were looking for. So sometimes you have the second derivative given and you have to go back to the original curve. For that what you do is you integrate it twice.

[09:56:40]So so you integrate twice like this. You say you have the second derivative first. You integrate it and that will give you dy / dx. You integrate it again and that will give you y. Now in each step you are going to get a plus c. In each step you'll get a plus c. For finding the value of c you'll have to use this information that's given. If it's a stationary point, you know at this point the derivative should also be zero. So you can use this point now to find the values of both of those C's both in the derivative and the original curve and that will give you that final result that we were looking for. Let's have a look at this one last example.

[09:57:13]Again we have to find this shaded region. Now how do you think about this?

[09:57:17]Sometimes it's it's a bit confusing uh from the shape that's given. So how do you find this shaded part of this? So the way to think about this would be you could say it's area under the curve from the point A to the point X = 2 there. This is the curve that we've got area under the curve that's all of this region. If you want the shaded part from that we would have to subtract the area of this triangle.

[09:57:50]So how do you find the area in the curve? You just integrate it. input the limits from 1 to two and that gives you area under the curve that turns out to be 0.2 and then to find that shaded part you find the area of the triangle as well. How do you do that? For that you have to use some information from the previous part to because you'll have the point B there. When you have once you have the point B you know this point is one you have the height somehow this is this is also one in fact height is one and the base can be found using the point B you have the base you find the area of that triangle and then subtract those two and that will give you the in uh the area of that shaded part as well.

[09:58:24]So you have to think about which area you have to subtract from which area or sometimes which area you have to add to which area some this is going to be important in these problems. Now for area just one last point for area mostly what you can do is for finding the same area you can use the x-axis formula and also the y-axis formula for example this part on the left side here you could use area between graph and x-axis here with these limits and you could also use area between graph and y-axis formula with these limits both of them will work. Both of them will work. In area, you can use any of the two formulas in some cases and they will work. But when they talk about volume, if they're saying find the volume of revolution about x-axis, then it has to be the x-axis formula. You cannot use the other other formula because the shape totally changes when you rotate about the other axis. When they say rotation about y-axis, it has to be the volume of evolution about y formula. It cannot be the other formula. You have to use the Y formula. In area you can sometimes use the other formula as well and still get the correct answer but with as in think of that that area as area between X-axis area between graph and X-axis or Y axis and use the limits appropriately and you will get the result from both of both of them. But for volume that does not work.

[09:59:48]If the question says find the volume of of revolution about x-axis it has to be the x-axis formula. If they say find the volume of revolution about y-axis it has to be the y-axis formula. you cannot use the other formula in that case. That's all about integration. I hope that was helpful. We'll end this here. If you have any questions, you can let me know.

[10:00:07]And that finally completes paper one. Uh took a lot of time, but we got done with it finally. Let me know if you guys have any problems now, any questions, any feedback. Uh because it's been 10 12 intense hours of paper one.

[10:00:28]Let me know if you have any purposes.

[10:00:45]Okay. Can you get a different angle other than 360? I haven't seen any in the past papers until now. But if you get, you can figure that out just by ratios. So this formula is for angry for for when the angle is 360. If they say the angle is 180, it's rotated 180°, you just half that volume that you found there. Whatever the angle is angle is, you just find the find the ratio for that. For example, if it's 270, you just multiply that by 3 over 4. If it's 90°, you divide that by four and so on. Just find the ratio and you will get that volume. I haven't seen any question like that in past papers until now. uh but if it happens you just find ratio okay just use the same this this method find the volume for 360° at the end find the ratio for whatever angle that's that they have given okay anything else anyone Any other questions?

[10:02:06]No.

[10:02:13]You're welcome. Good to know that. Happy to hear that.

[10:02:19]For the next workshops, those of you who would be attending, just any uh suggestion about the speed at which event we went and any other uh suggestions you might have for for the next workshop, how we could improve something M1, S1, P3.

[10:02:40]Those of you who will be attending any of them, if you have any feedback as in in terms of suggestions for for what we could what else we could do or how we could improve it, do it quicker than this, do it slower than this or remove some of the parts. If some some of things were very straightforward, could we just ignore them or to save time? Or any other suggestions that you might have, you can let me know.

[10:03:12]Topical for a particular topic, functions, function transformations, quadratics, recent questions uh last couple of years, equation of circle questions.

[10:03:32]These are some topics that you can attempt this time. Equation of symmetry, function transformations, some recent quadratics questions, functions questions in general, they should be helpful.

[10:03:57]I think the pace was fine for the people who are well prepared relatively.

[10:04:02]But then I felt that uh some people probably went black and blank at some point because of the difficulty level of some of things some of the things that we discussed and at a quick pace maybe they could not understand much but there isn't much that can be done in a revision session to help with that actually because if it's a difficult question uh then you have to spend time on that and unfortunately in a revision class if we do do that then it ends up taking like double the time.

[10:04:45]Yeah, P3 will be like 7 8 hours 2 days.

[10:04:50]It it's going to be like that. Although uh uh if you think about it in paper three there isn't much variation in questions.

[10:04:58]So in this case you know since the concepts are not very difficult they try to change a lot in in the questions in P3 if you've done enough practice on integration if you've done enough practice on defensiation there's only a few different types of questions that they asking them there isn't much variation there. So if you if you understand the concepts very well in paper 3, it becomes kind of repetitive at some point.

[10:05:31]P3 is difficult. P3 is difficult. uh although uh if you understand the concepts well once as in you make that effort of doing a lot of practice questions then eventually you can still score 65 plus 70 plus in that as well to find increasing function we have to use x values what do you mean by Um, okay. So, the video that I shared in the group, uh, that was in in another group. So, there's a playlist for the calculator function that I have on my channel. I can share that with you here.

[10:06:26]Let me So what I'll do is I'll after the class maybe I I'll share that playlist if I don't give me a reminder about that. I tend to forget things sometimes.

[10:06:36]Uh but I'll do that. I'll share the playlist link with uh view and the group that has like four or five videos and some useful functions. Some are not going to be relevant to you because I think some of them are for P3 as well like complex numbers but some of them are going to be useful for paper one as well. For example, the table functions one, the sketching trigonometry graphs and things like that.

[10:07:01]Yeah, complex numbers is one in which in recent papers they're trying to change a lot. That's true. Apart from that, it's pretty repetitive.

[10:07:11]Calculator functions, I'll do that. I'll share that with you.

[10:07:15]For S1, for S1 it will be uh 10th of May inshallah, Saturday, the Saturday before your exam. So 3rd is a Saturday and then 10th. Yeah. So for S1 most likely Saturday, 10th May because on Friday some people have an exam already and uh I think it's it's in PM as well.

[10:07:41]So we might not be able to do that on Friday.

[10:07:45]9th of May. What's the exam on 9th May?

[10:07:47]I think I was looking at the sheet and I I think there was some big exam on 9th maybe chemistry or biology one of those.

[10:07:56]So it'll be on S uh it'll be on 10th most likely 10th May for S1 and again next week uh before paper 3 there is an exam on Friday but since we need do the two days for P3 uh that will have to be and your P3 exam is on Monday we can't do it on Sunday for that reason because that will be too late so it'll be Friday Saturday for paper 3 we can't do before Friday because Friday is a major exam from what I remember from the that I saw. So P3 will be 16th. No, P3 will be 16th 17. Yeah, 16th and 17th will be P3.

[10:08:35]16th and 17th and S1 will be 10th inshallah.

[10:08:45]P3 is difficult but when you are well prepared on that once then it becomes easy. S1 is like uh you can say it's it's sometimes unexpected the permutations and combination questions. it's unpredictable uh what question you'll end up getting in the exam on permutation combination because sometimes it makes you think a lot in the exam and it's you you cannot figure that out unless you've done like 80 100 questions in that P3 is not like that if you've done sufficient practice there isn't much that they change in the exam or they can change in the exam in S1 and permutation combination and probability they can sometimes end of giving some totally you know unusual scenarios that you need to think about a lot. So that requires you to think logically in the exams. That's what a lot of people find difficult and that time pressure it becomes difficult for that reason.

[10:09:50]P4 is mechanics. Mechanics will be uh this coming Saturday in so that's 3rd May. Third May will be the ember workshop in yeah permutation combinations is a difficult topic. It is a difficult topic. Although the standard questions on once you do enough practice they are fine but sometimes you you end up getting something that's totally different and that becomes difficult.

[10:10:27]Yes, you have P1 this week. You have it on Friday. You have three days left. So, yeah.

[10:10:37]All right. So, that's it then. I will see you again in another class and insh all the best for your exams.

[10:10:45]What I'll do is I I'm thinking of making uh a group in which I can just add everyone of my students because uh I'm getting so many questions recently that it's becoming humanly impossible to respond to all of them. Uh so I will what I try to do is I'll try to connect everyone so that you know you guys can help each other out as well and that might help.

[10:11:15]I'll of course keep responding to your questions but it's just that a lot of times uh students get impatient and I uh cannot respond uh immediately. So what I'll do is I'll be adding you guys to another group in which you can help each other out as well.

[10:11:36]U there are actually some people there are some people um okay by the way I would I would want to get this feedback as well.

[10:11:52]So, how many of you would prefer having these workshops workshops in Uru or the normal way we speak generally it's bilingual like Okay. Actually there I think we're just as well.

[10:12:45]So it might be difficult uh to switch to UDU.

[10:12:51]Unfortunately that'll be difficult although you should all you should all uh know your own language as well. So some people when I go to these elite schools I I'm very disappointed sometimes.

[10:13:16]That's not good.

[10:13:19]You totally lose connection to your culture to your home to your home country.

[10:13:36]Even if you're more comfortable with one language, Yeah, unfortunately it'll have to be that because of some international students because some audience from outside as well. So when they get registered that totally switch to okay all right so all the best for your exams everyone and I'll see you in the next session inshallah if you have any questions you can keep messaging in the groups try to help each other out I'll try to respond to your question as well and I wish you all the best in Thank you everyone. Allah says inshah inshah I have all um I have my duas for all of you in I will be making duas for everyone as well. So you you all of you as well make sure that you prepare well and make lots of duas not just for exams but for you know the whole situation that we all Muslims are in right now.

[10:15:06]Hopefully everything stays fine.

[10:15:09]I'll see you again in another session in Allah.