When solving exponential equations where x appears in the exponent with plus/minus signs between terms (e.g., 4×2^(x+2) - 5×2^(2-x) = 3), use substitution by letting u = 2^x, rewrite the equation using indices rules, solve the resulting quadratic equation, and then solve for x using logarithms. The solution requires careful attention to the domain (u must be positive) and rounding to the specified significant figures.
安装我们的扩展，即时搜索任意视频内容
A-LEVEL MATH | MARCH 2026 PAPER 32 | COMPLETE SOLUTION | P3 Past Paper Sessions 2026本站添加:
Okay, let's begin. March 2026 paper 32.
This is what we are doing today in Shara. We will do this full paper. Uh since the marking scheme is not available, I would highly recommend that you do the calculations alongside me.
Let me know if there's any credit error at any point. I will also try that I do the verification sats as well uh with each question. Uh wherever we can verify quickly, we will do that I'm sure. So you know how to do this in the exam as well.
All right. March 26 paper 3 2. This is a level maths. Let's begin. So we need to solve this equation. Question number one. 4 into 2^ x + 2 - 5 into 2^ 2 - x that is equal to 3 giving our answer correct to three significant figures.
Now how do we solve this equation? Can we take log in both sides? Can I take log on both sides? Is that going to work?
Taking log on both sides. Is that going to be a good idea?
Yes. No. What do you think?
Generally in such questions since x is in the power x is in the power we could uh we generally we take log on both sides so that x comes down right. But in this case the problem is in this case the problem is we have this minus sign in the middle. So taking log on both sides that is not going to help because we will end up getting minus inside log and there's no no way to simplify that thing. So what do we do in such problems? What do we do? What do we do in such problems where we have plus minus sign between the terms and x is in the power. Can anyone remind me of the of the method that we use in such problems? We have to use substitution, right? We have to use substitution. We will do it by using substitution. Now, how what substitution do we use? We want some number raised to power x that we will substitute with another letter.
Now, what power what number is x coming in in in the power of? It's two, right?
2 raised to power x. That is what we can use for substitution. For that we will need to separate these terms first of all. Now how can we do that? 2 ^ x + 2 I can write that as 2^x into 2^2. We can break it like this.
Right? Using indices rules 2^ x + 2 is the same thing as 2^ x into 2^2. And then similarly here as well we have 5 into 2^ 2 - x. What I can do is I can write this as 5 into 2 ^ 2 / 2^x on the other side I've got 3 as it is. So first of all using indices rules you write it like this. Now what we can do is 2^x is appearing here. 2^ x is appearing here.
What we can do is we can use a substitution. We can say let some other letter let u= 2 ^x and then this equation becomes 4 into 2 ^ 2 that's 4 * 4 which is 16 16 * u - 5 into 4 which is 20 - 20 / u that is equal to 3. This is the equation that you now need to solve.
Okay let's solve this equation now.
Multiply the whole equation by u and that gives you 16 u ^2 - 20 = 3 u on the other side. Let's bring 3 u to this side and that gives us a quadratic equation.
16 u ^2 - 3 u - 20 = 0. Just put this whole thing on your calculator. Uh after inputting in the quadratic formula - into -3 + -<unk> of -3 2 - 4 into 16 into -20 whole divided by 2 into 16. Do show this step even though in paper 3 sometimes they get away you you do get away with not showing this particular step as well but it's a good idea to make it this make this habit. It doesn't take a lot of time just writing this one step.
Write down this one step. Then you can write down the answers after that. What are the values of u that you're getting after this point? You can do the calculation directly from your calculator and that's okay. Use the equation solving function on your calculator and that will be fine.
Okay.
Now what are the values of U? U is 1.2157.
This is one possible value of U that you're getting. And the other one.0282.
These are two possible values of u that we're getting.
Okay. Y the let's now put that substitution back. 2^x is equal to 1.2157 or 2^x is equal to - 1.0282.
Now this is not possible. If you take log on both sides you will get an error from this. So you say this this is not possible. The other equation if you take log on both sides you can say ln of 2 ^x that is equal to ln of 1.2157 x comes down there and it becomes x ln2 equals ln of 1.2157 divide ln2 on the other side and that will give the value for x you can see what that is what's the value of x that we get can multiple people please do those calculations so we are not making any careless mistakes 1.2157 2157 divided by ln of 2 and that is giving me 0.281787.
Now in the question they're saying write down this number to three significant figures. This will definitely make you lose a mark if you don't round off in this case otherwise they often do not penalize but still the instruction is you should write three significant figures in general as well. So you should follow that 282. That's the value of x that we're getting and that's our final answer.
Is my screen freezed for everyone.
It is lagging.
Screen is lagging a lot.
It might be the air server that I'm using maybe. Let me check that.
Okay. Can you see me making all of these random lines here or not?
Okay. Did it only happen for a few seconds or has it been happening since the beginning?
Few seconds. Okay. All right. So, I think there must be some temporary issue there. Hopefully, it's it stays fine after this.
Let me check the connection on the other device as well. Is it fine?
take this is the answer that we're getting. No, it's not necessary to show more steps than this. Uh he might have said if you want an exact value, if you want an exact value, in that case you need to show more steps. But if you don't want an exact value, this is enough. This is sufficient. Okay, then you can write down the answers after that. Now, uh if you want to verify the answers, you can just put that back in the equation and see what you get on the left side. 4 into 2 to power x + 2 whatever was the answer. Just put that here. Answer + 2 - 5 into 2 power 2 - the answer. And see what that gives you 2.999. And that that is almost the same as three. And that confirms that our calculation is correct. Okay, let's move on. That was question number one. Question number two, let's look at this. Now, let's have a look at this diagram in vectors uh is not tested a lot in recent papers. That's what you're saying. U that does not tell you anything. Um because it is part of your syllabus and it is actually tested a few times at least in the recent papers. So, you can't skip anything. Please don't skip any topic thinking or any particular concept thinking that it will not be tested. That's not a good idea.
That's a very bad idea. You have to make sure that you know all the concepts in our syllabus. Okay. Question number two.
Let's look at this now.
On the ar diagrams, case the locus of the points representing Z. Now think about it. modulus of Z= 9 and argument of Z equ=<unk> /2 uh between P<unk> /2 and pi. Now, how can we do this?
We've done this before. We've done this before in other recent papers as well. What does this mean? Modulus of Z= 9. What is this locus going to be? It's a circle.
Modulus of Z can be thought of as Z minus 0, right?
It's a circle with center zero and radius 9. Okay, so that's one thing.
So there's a going on here. I have a few seconds break and then we'll see.
Can you guys see my screen again?
Yes.
Now I can see it on my other device as well. Let me see if it is still lagging.
So if you're sending uh me any message, make sure that you send it to no the host account. Send it to the host account or just send it to everyone.
That will be safer. If you're sending a direct message, a private message, send it to the host account. Okay? The one that has the video on otherwise I'll not be able to see your message. Okay? CH.
So we were given the complex number Z satisfies this equation. Modulus of Z is equal to 9. That means we have a circle with center at origin and radius 9. Now the circle should be drawn with a compass. It should pass through these points 9 - 9 9 - 9.
Now they're saying another condition is the argument of Z is between pi / 2 and pi. So what we can do for that is we can make a line like this from the origin 90° pi /2 radians. That's that that's going to look like this pi / 2 radians.
This will be one line, right? And then another line which is pi radian line.
What is that going to look like? that line will be like this. Now you can be anywhere between these two. What that means is this is the region that satisfies both of those inequalities.
Now this is one way of doing this. But in the recent marking schemes, in the recent marketing schemes, the way they actually show this is rather than shading that part of the curve, what they do is they just show the part that is valid. So they don't have anything else showing there. and they just do something like this.
So that's what you should do ideally.
This is the region that the this is the curve uh this is the arc that satisfies both of those conditions. Its modulus is 9. Its distance from origin is 9 units. And the angle is between pi / 2 and pi. This angle is pi / 2 and this angle is pi. So you'll make a line like make a curve like this.
So yes, you should erase the rest of the circle.
This this will be locus of the points representing Z.
Okay, in this case since we don't have an area as such, it's just all the points on this arc.
So it's a good idea to just not shade any region. Instead of that, just show this arc and erase everything else.
Another saying on the same diagram sketch the locus of Z conjugate + 3. What is this going to be now? Z conjugate + 3.
Z conjugate + 3.
What is Z conjugate in when when we find conjugate of something what happens is the imaginary part flips. And we have seen questions like this in recent papers as well where they where they ask you for the geometric effect of find of making a of uh taking a conjugate of a complex number. What is that geometric effect?
Reflection in the line x-axis. Why does that happen? Because the imaginary part flips. So if you were just thinking about z conjugate that would look like this. That will be an arc like this.
This would be z conjugate. It will be reflection of that in the x-axis. Okay.
However, we also have a plus three there. We also have a plus three there.
What do we do about that? Plus three means in the real part, we are adding three. What happens because of that? Uh when we add three to the real part, the horizontal axis represents the real part. What will happen would be this graph is going to move three units to the right side. Okay, the real part is going to increase by three units. So what happens is this curve is going to shift to the right by 3 units. So now it'll be starting from -6 here.
Erase the rest of it. It'll be starting from -6 here. and going up to this point which is 3 and - 9. So this is what it looks like. Now you can draw this roughly like this and you say this is the locus of Z conjugate + 3.
Make sense to everyone?
Isn't it a change in the imaginary part?
Do we have iota with it? This is just three, right? Three is a real number.
Three is a real number.
Okay, no problems with this. Let's move on. So, plus three, that's the real part. We going to add three to the real part, right? So, that means it moves to the right. It moves to the right. And yes, you should be making this arc using a compass. Use a compass for making this arc. Put your compass needle at three here. Open it 9 units and make an arc like this.
Okay.
Question number three.
Another saying find the exact value of this thing. Integral of 3 sin x into sin 2x with limits from<unk> / 4 to p<unk> / 3 give an answer in this form. How do you integrate this thing? sin x into sin 2x. These are two trigonometric functions multiplying.
Is integration by parts going to help?
Is integration by parts going to help in this? Generally no. Because remember trigonometric function it doesn't get simpler. Derivative of sin x is cos x.
Integration of sin x is minus cos x. So it is just going to keep changing from one function to another.
Loop method might work in this. You you you'll have to try but uh sometimes it works sometimes it doesn't.
What else can we do with such problems?
When we have two trigonometric functions multiplying that method is used quite a lot. The extended power rule that we discussed. If we have anything raised to power n and you have the derivative of the expression inside multiplying outside then the integration of this what is that given by this thing as it is. we add one to the power and divide by the new power.
Okay, this is what we may have to use. Now let's simplify this. First of all, let's simplify this. First of all, we have sin x and sin of 2x two different angles. We could simplify it like this. 3 sin x into 2 sin x into cos x, right? sin 2x can be written as 2 sin x cos x.
Now let's simplify this further. It becomes 6 sin² x which we can write as sin x^ 2 into cos x.
And now you see it is exactly that form where you have some expression inside brackets and there's a power on top of that. You have this expression inside.
There's a power on top of that. And we have the derivative of that multiplying outside. It's exactly this form. We can just use this directly. We can use this directly.
And that will give us six constant as it is. Integration of this thing. What is that going to be? Just sin x raised to power 3 / 3. That's your integration.
Input limits. Now limits are from p<unk> / 4 to p<unk> / 3.
Now we just have to input those limits.
Okay, input the limits 2 into sin x whole cube. That means 2 into sin<unk> / 3 whole cube - 2 into sin<unk> / 4 whole cube. Now since we want an exact answer, do not write down the answer directly.
You have to show working for this. You need to show working for this. Do not write down the answer directly.
Okay.
Sin pi / 3. Put this on your calculator.
Your calculator will give give you some number for this. I think it is under root 3 over2. Please confirm this as well.
You can just find it from your calculator and that's okay. Sin<unk> / 4 that is under <unk>2 /2 and then we have whole cube on top of that as well. So let's simplify this further. What does this become? Under roo<unk>3 whole cube.
What can you write that as under roo<unk>3 whole cube under roo<unk>3 whole cube? We could say that is under roo<unk>3 squared into under roo<unk>3. Right? And that becomes 3 under roo<unk>3.
Okay. Under roo<unk>3 whole cube that can be written as 3 under roo<unk>3. So in the numerator we have 3 under roo<unk>3. In the denominator 2 cube is 8. And similarly here as well under <unk>2 whole cube. We can write that as 2 under <unk>2. In the denominator we've got 8. Simplify this.
This gives you four. This also gives you this gives you two. And what are we left with? 3 over 4 under <unk>3 - 1 /2 under <unk>2. They wanted you wanted you to give your answer like this under root p under roo<unk>3 + q under roo<unk>2. That's exactly the form that we're getting here. P and q were supposed to be rational numbers. That means some fractions. And that's our final answer.
Yes, you're right. You could use substitution as well.
You could use substitution as well. The substitution you could use could be you say u equ= sinx. Use that substitution.
Right? Like we used to do for odd powers of sin x cos x. You could do that as well and that will also work.
Is that okay?
IBP integration by parts. Okay, I had to think about it what that means.
Integration parts won't be simplifying the angle because trying to do this when you have two trigonometric functions.
Okay, so remember Iate why do we use to decide between U and V dash?
We say whatever is on the left side that should be u. Whatever is on the right side that should be vdash because we want u to be something that simplifies after differentiation.
We want u to be simplified after differentiation.
Now I is inverse trigonometric that means tan inverse. So tan inverse when you differentiate that when you find u dash of that that becomes something simpler 1 / 1 + x^ 2. So that in the next step of integration by parts you will have a simpler integral. Similarly when you have log u = ln x that becomes 1 /x. When you have algebraic u = x cq for instance or x² for instance it becomes simpler after differentiation. It gives you 2 * x.
When you have t or e however t and e when these two functions come together then you have a problem because they don't become simpler after differentiation. So integration with parts we are going to get stuck at some point if you're using that. So if you have both trigonometric functions if you have both trigonometric functions that's going to be a problem because they will not become any simpler. If we say u is equal to sin x in this case it was uh sin x let's say and vdash was sin 2x. Now what is u dash going to be? U dash is cos x and v that is - cos 2x / 2. Now the next integral is again going to have a product and you're again stuck. So when you have two trigonometric functions multiplying integration by parts generally is not going to be helpful. The loop method works sometimes but this is more efficient. You can just you use this.
The loop method is going to be helpful when you have one trigonometric and one exponential. So for example, if you had something like this e to power x into sinx and you wanted to integrate this.
Now in integrals like this you can use that loop method that we discussed.
Say this is i equal to this and then you simplify this using integration by parts. At some point you will end up getting the same thing again and then you can write that as I and then you can evaluate whatever I turns out to be equal to okay let's move on.
So when you have two trigonometric functions when you have two trigonometric functions you will often have to do this. Okay. And this is happening recently in the last 2 three years. You will see at least two three questions like this where you have trigonometric functions multiplying like this. This is how you would do it. You can verify your answer quickly at the end as well if you want if you have time. Of course we are already late but let's just do it anyway. So if you can if you input this integral in this 3 into sin x into sin of 2x integration with limits from<unk> / 4 to<unk> / 3 just input those limits what do you get from this 5919 keep that in mind or just write it down somewhere 5919 3 over4 into under roo<unk>3 the answer that you're getting I'm just inputting that here minus under<unk>2 2 / two and that's coming out to be the same number and that confirms that your answer is correct. For a four mark question, it'll probably be too much effort. You can do it at the end of the paper once you're done with the rest of the questions.
Let's look at this. Now, now we've got binomial the coefficient of x cube in this expansion 1 - axis^2 over 5 that is 1. Coefficient of x^2 is 1.
x cub sorry that is one find the value of a how do we do this let's expand it let's expand and see what we get from this 1 - ax raised ^ 2 / 5 now input in the formula what's the formula 1 + x raised to power n that is equal to 1 + n x + n into n -1 / 2 factorial it's given in the formula sheet x^2 + n into n -1 into n -2 / 3 factorial x cq. Okay, that's the formula that's given in the formula sheet.
Apply the formula. See what you get from this 1 + Now what term are we really looking for? The coefficient of x cub, right? Do we need the rest of the terms in the next part? coefficient of x^ 4 in that expansion will be need that x into x uh x^ 4.
Okay. So since in the first part we are just looking for coefficient of x cq what we can say is where are we going to get x cube? x cq is going to get going to come in this term. So we can ignore the rest of the terms. We'll see in the next part if we need any of them. But for now, we just focus on this term, right? Ignore the rest of the terms.
N into N - 1. That means 2 over 5 into 2 5 - 1 into 2 5 - 2 / 3 factorial into - ax. Do not forget the negative sign with that. - ax whole cube. Right?
That's the term that we are concerned about. Just focus on this term. Ignore the rest of them.
Okay.
Now let's simplify. Put this whole thing on your calculator.
2 over 5 into 2 over 5 - 1 into 2 over 5 - 2.
Okay.
/ 6 and that will give you some number 8 / 25.
Uh and then you have this negative sign from inside as well. So it'll be -8 over 25 -8 over 125 a cub x cub. Now they told us the coefficient of x cub is equal to 1.
That means this thing should equal one.
So we just equate it to one and that will give us the value for a. Let's see what that turns out to be.
-8 over 125 a cub = 1 a cub = -25 / 8. Take cube root of that and that will give you - 5 / 2 or 2.5 - 2.5.
Right? Cube root of 125 is 5. Cube root of 8 is 2. That's 5 /2 which is 2.5. And that's our final answer. Okay, that's what we were looking for. A is 55 /2 or -2.5. That's the first part.
So now they're saying it's very similar to P1 binomial. Now the next part they're saying find the coefficient of x raised to power 4 in this expansion. Now first of all in problems like these we have to figure out which terms are we looking for? Which terms are we looking for?
This is x, right?
What will we need to multiply x by to get x^ 4 x cube, right? That's one term that we need. And now this is interesting because I've not seen this in any paper before.
What do we multiply one by to get x^ 4? x^ 4. Generally they do not require you to expand any more than this. But in this case they want you to expand up to x^ 4. Now so from that previous expansion you also need the x^ 4 term. These are two terms that you need from that expansion. Now the x cub term we already have right? x cub term we already have.
Let's just find x^ 4 term and see what that turns out to be. So now we know the value for a right the value for a is 2.5 -2.5 or 5 / 2 we can input that there. So this expression would become 1 - a x that means it is going to be 1 + 5 / 2 * x whole raised to power 5 / 2 2 over 5.
That's the expansion that we had.
Let's do this expansion. We know the coefficient of x cube is 1. So we have 1 x cub. What about x^4 term from this?
Now this is not given in the formula sheet.
This is not given in the formula sheet.
Normally they don't ask you to expand up to x^ 4. Right? Only up to x cq.
Now x^ 4 term is not given in the formula sheet. So you have to look at the pattern. What what's the next p next term going to be after this? Looking at the pattern, next term is going to be n into n -1 into n -2 into n -3 / 4 factorial into x^ 4. We have to input values in that now. Okay, let's input values in that. Is my screen fine now?
By the way, is it still lagging for anyone?
No. object.
Let's input values in this n which is 2 over 5 into n -1 2 over 5 -1 into n - 2 2 5 - 2 into n - 3 that's 2 5 - 3 whole / 4 factorial into x^ 4. That means 5 / 2x^ 4 5 / 2x whole raised to power 4. Okay.
Now we can simplify this. Put this whole thing on a calculator apart from x. Put the whole thing on a calculator. See what you get. I did some part of it before. Okay. It's not going back anymore. It is 2 over 5 and 2 over 5 - 1 and 2 over 5 - 2. Now we add 2 over 5 - 3 to this and then in the denominator we have 4 factorial which is 24 4 factorial multiplied by 5 / 2 whole raised to power 4. Please confirm this calculation as well. I'm getting -3 8 from this -3 / 8 x^ 4.
That's this term that we're getting. Now these were the two terms that we needed from that expansion. We know the rest of the terms. Now we can do do the multiplication. 2x + 1 into x cub -3 / 8 x^ 4. How will we get x^ 4? x^ 4 will be when we multiply 2x with x cub and 1 with -3 / 8 x^ 4. These two multiplications are going to give us x^ 4. Let's see what we get from this. 2x into x cub that is 2x^ 4. 1 into -3 / 8 that is -3 / 8 x^ 4. Put this on your calculator now. And that should give you 3 over 8 x^ 4. They asked you for the coefficient. So you can write the coefficient of x raised to power 4 that is 3 over 8. And that is what we were looking for.
-3 over 8 is the final answer from this.
Okay. Is this clear from 38? Is that what you're getting?
Right.
minus 2.875.
You're getting something different. Has somebody else tried this as well?
Somebody else done this calculation.
Okay. All right. Take Let's move on.
for an A and A as it really depends on whether you're taking M1 or S1 depends on that because M1 was hard generally generally what happens is uh from what I I have seen one of P1 and P3 is difficult and the other is easier and one of M1 and S1 is difficult and the other is easier. So you might get a relatively easier S1 but there's no show it to you of course but that's what I think is the pattern state the set of values of x for which this expansion is valid now when is this expansion going to be valid 1 + 5 / 2x raised to power 2 5 that's the expansion that we have done all we have to do is whatever is the second term whatever is the second term you say modulus of that that has to be less than one so you Okay 5 / 2x it's modulus that needs to be less than one. Now take these numbers to the other side and that gives you modulus of x that should be less than 2 over 5. That's the condition for that expansion to be valid.
Okay.
Does that mean you might get an easier P3? P1 was fairly easy as far as I've heard. So if B1 was easy then B3 may be hard.
Yeah. So my expectation would be that B3 will be a bit difficult. S1 might be easier.
Yeah.
Let's look at the next one. Question number five.
Now it's given that Z is equal to 3 + lambda iota / lambda plus 2 iota where lambda is a real constant. Find the value of lambda for which argument of z is equal to this.
Okay. So we find the argument of this complex number z and we put that equal to pi /4 and that will give us the value for lambda. How do you find the argument of this number?
We would have to simplify this first, right? We have to simplify this. Let's simplify this first and then we will try to find the argument. We can say 3 + lambda iota / lambda + 2 iota. We can multiply and divide by the conjugate of the denominator. So multiply and divide by lambda minus 2 iota.
Now the numerator what's that going to be? 3 + lambda iota into lambda - 2 iota. In the denominator it's a + b into a minus b. So that becomes lambda^ 2 - 2 iota squ.
Now 3 into lambda that's 3 lambda 3 into - 2 iota that is - 6 iota plus lambda 2 iota - 2 lambda iota 2.
Okay, that's the numerator. What about the denominator? In the denominator, we've got lambda^ 2 - 4 iota 2. Let's simplify this. In the numerator we have 3 lambda - 6 iota plus lambda^ 2 i iota minus 2 lambda iota squar that is going to become + 2 lambda. In the denominator we have lambda^ 2 + 4.
Now write the real part separate imaginary part separate 3 lambda + 2 lambda that will become five. So we can say 5 over lambda 2 + 4 that's the real part. And then the imaginary part, the terms that have iota in them, we could say they are - 6 + lambda squar / lambda^ 2 + 4 * iota.
This is what z simplifies to. If we if we separate the real part and the imaginary part, is that okay? Any issues with this?
Five lambda five lambda in the numerator.
Now that we have simplified this, they have told us the argument of Z is p<unk> /4. Now which quadrant is this in? It is in the first quadrant. Keep that in mind. It's in the first quadrant such that this angle is p<unk> / 4. Now with angle pi /4, real parts and imaginary parts, they're equal to each other, right? But you can just input in the formula as well. What's the argument going to be? It's in the first quadrant.
So whatever is the basic angle, that's the argument. So we can say argument of Z that must equal tan inverse of imaginary part divided by the real part.
Okay, that will be a bit too complicated fraction. Let me just write it down anyway.
- 6 + lambda 2 / lambda 2 + 4 whole divided by 5 lambda / lambda square + 4.
Okay, so what I've done is I've divided the imaginary part with the real part.
Now this must equal the argument that they gave us that's 4. So this should equal<unk> /4. Now what happens is lambda square + 4 lambda square + 4 they get cancelled. tan inverse goes to the other side becomes tan p<unk> / 4. On the left side we're just left with - 6 + lambda 2 over 5 lambda. The right side it becomes tan / 4 tan 4. Put this on your calculator that will be 1. So we have 1 = - 6 + lambda^ 2 over 5 lambda.
And again we get a quadratic equation.
lambda square - 6 that is equal to 5 lambda lambda^ 2 - 5 lambda - 6 is equal to 0. Now how do we solve this quadratic?
There was another part to this as well.
Where did that disappear?
Okay. Lambda square - 5 lambda - x.
We have to factoriize this lambda squared. We can say there are two factors. lambda 6 and 1 right so - 6 and + 1 lambda - 6 and lambda + 1 these will be the two factors that we get from this and lambda is either equal to 6 or lambda is equal to ne1 these are two values of lambda that we're getting now the question said find the value of lambda that's a singular so that means there has to be only one answer we're getting multiple answers that means one of them is not going to be possible.
One of them is not going to be possible.
Right?
Now, how do we ignore one of them?
We know the angle is supposed to be acute. So, we are in the first quadrant.
We in the first quadrant. In the first quadrant, that means the real part should be positive and the imaginary part should be positive. So we can just check if any of them is giving us maybe a negative real part or a negative imaginary part. Let's check for that.
This is lambda squar. This is lambda squar.
If we put minus one. However, what happens is the imaginary part here becomes negative. The real part also becomes negative with negative 1. Right? Both parts become negative. That means it can no longer be in the first quadrant. Then it'll go in the third quadrant. If the real part is negative and the imaginary part is also negative.
So for that reason we would have to ignore lambda equal to -1. We'll say this is not possible and the only possible value of lambda therefore is lambda equal to x. Whenever you get multiple answers always go back to the question statement and see if all of them are going to be possible or not.
I guess in this case negative 1 was not possible.
Any issues with this? Any problems anyone lambda is wavelength second part that is missing from here.
Let me copy that.
When lambda has the value found in part 5 a find the exact value of modulus of z making your method clear. Now once we have this value for lambda, this was the Z value that we were getting right after simplification. We could just use that input lambda= 6 in this and then find modulus from this input lambda= 6. What do you get? 5 into 6 / 6 + 4 + - 6 + 6 2 / 6 + 4 iota. What does this simplify to?
Can anyone tell me 5 * 6 is 30? 30 over 40 that will be 3 over 4. Right? Please confirm this. Uh this is 30 over 40 as well. 34 - 3 over4 iota + 34 iota right yeah that looks okay what about the modulus of zena now do not write down the answer directly show the step 34 2 + 3 4 2 and then take square root of that and that will give it the modulus what's the value for modulus that you're getting what's the modulus value.
What's the modus value? Has anyone done this?
75^ 2 + 75 2 square root of this and that is do they want an exact answer? Yes. Since they want an exact answer, uh we actually should show further working as well. We'll say this is 9 / 16 + 9 / 16 and that becomes 18 / 16 and then we could say 18 can be written as 3 under <unk>2 / 4. Yeah, that's what we were getting right 3002 over 4 and that's the final answer. Whenever they ask you for exact values, make sure to do it step by step. Do not skip any working in that case. Okay.
All right, let's move on to the next.
Move on to the next one. Person number six, though.
Question number six. Let's try this now.
Okay, the polomial 2x^ 4 + ax + so on that is denoted by p of x. It's given that this is a factor of p of x. Find the values of a and b. Now, how do we do this? This also looks like some complicated long division. Now, it's not a linear factor. That means we cannot use remainder theorem, right? That will not work. remainder is not going to work. So what do we do then? We have to use long division. Use long division.
Find the remainder. If the remainder turns out have to be zero.
In fact you will put the remainder equal to zero and that will give the values of a and b. Let's do this. Start with the division process. We have 2x raised ^ 4 + ax cub + 4 x^2 + b x - 3.
Divide this by x^2 + x + 1. What do we multiply x² by to get 2x^ 4? 2x 2x 2 into this whole thing. What is that going to be? 2x raised ^ 4 + 2xqub + 2x^2 change the signs it becomes - - gets cancelled. Now in the next step, make sure that you're writing it exactly like this. Coefficient of x cq, you need to write it together like this. So a -2 whole * x cq. This is how you must be writing this coefficient. Do not write down the terms separately. Ax - 2x cq that will make you do errors. Okay?
Write this whole thing as a single thing like this.
Okay?
What about this? uh 4 - 2 that will be 2x^ 2 and then + bx - 3.
Okay.
What next? Now what do we multiply x² by to get a - 2 * x cub?
a - 2 * x. Right? Multiply this with the whole thing.
What do we get? a minus 2 whole * x cube. Now you have to be a bit careful.
Uh it's very prone to careless errors this working. So slow down it a little if you need to. Now a - 2 whole * x into x. What's that going to be? A - 2 * x^ 2. Okay. Again write it inside brackets like this.
Next a - 2 * x into 1. That will be a - 2 whole * x.
Okay. Now change the signs again. This becomes - -.
This whole term gets cancelled.
Now 2 - a + 2. What is that going to be?
2 - a + 2. That's 4 min - a. Right? So these terms when when we add them up together they will give us 4 - a whole * x^ 2. Okay. So 2 - a - 2 that's becomes 2 - a + 2 which is 4 - a. So 4 - a whole * x^ 2. Okay.
Next b - a + 2.
We will have to write it like write it as it is b - a + 2 whole * x and then minus the outside.
Now again what do we multiply x² by to get 4 - a whole * x^ 2?
We multiply by 4 - a right 4 - a into this whole thing. Now what does that give us? 4 - a whole * x^ 2 + 4 - a into this that will be 4 - a into x and then 4 - a into 1 that is just 4 - a.
Okay. Again it's important to write down this number in brackets like this so that when you put negative sign you remember that this negative is going to be with the whole thing.
Okay. Now we change the signs again.
This becomes minus. This becomes minus.
This becomes minus. This gets cancelled.
Now let's do the subtraction. B - A + 2.
It's a good idea to write it down so you don't do any careless errors. B - A + 2 - 4 + A - 4 + A. What does that simplify to? A and A get. You're just left with B minus 2. So that's what you write down there b - 2 whole * x and then this last term what is this going to be again you can write it down this on the side - 3 -4 + a what does that become a - 7. So we've got a - 7 as the constant term. This is what we're getting. Now they said it's a factor. This thing is a factor. If it's a factor then that means the remainder must be equal to zero.
This is the remainder right? We cannot continue with the division anymore. The remainder must be zero no matter what the value for x is. What that means is the coefficient of x must be zero. So 0 in this case we would say is 0x + 0. The x term must be zero and the constant must be zero. So you put both of them equal to zero individually. put b - 2 equal to 0 and you put a - 7=0 and that will give you b - 2 = 0. The value for b turns out to be 2 and a - 7 = 0 and the value of a turns out to be 7.
These are the values of a and b that we get if this is a factor.
Any issues with this?
Ma, did you actually do this yourself or not? Or are you just watching me do it?
You can do yourself. Of course you can.
But why are you not doing it right now?
effort.
Okay.
Okay. Those of you on YouTube, you what you can do is you can maybe pause the video for a minute or two and then attempt the paper and then watch in 1.5x the solution. That might be a good idea as well if you have not attempted the paper attempted the paper beforehand.
Okay. Hence show that x + 3 is a factor of p of x. Now that's straightforward.
Now that you know the values of a and b you can say the polomial that you had is 2x ^ 4 + ax cq that means + 7 x cq + 4x^2 + 2x - 3. Now that will also be a confirmation of your previous answer.
If it does turn out to be a factor that means that the previous part was correct. So now how do you show this is a factor? Just put x equal to -3. Right?
Use remainder theorem. Equate it to zero. The value of x that you get input that input that in the polomial. 2 into - 3^ 4 + 7 into -3 whole power 3 + 4 into - 3². You have to show this working. You have to show this working where you're inputting the value x= -3.
A lot of very good students as well what they end up doing in the exam is they would do something like this.
Put x equal to -3 and then they say it turns out to be zero. That is not acceptable. They deduct marks for this.
They deduct marks for this. Make sure that you don't do it like this. You have to input values. You have to input values otherwise you'll not get full marks.
So show this step and then you can see if that indeed is equal to zero or not.
Can somebody check this? What do we get?
2 into 81 - 7 into 27 + 4 into 9 - 6 - 3 and that is turning out to be zero. And that confirms that x + 3 is a factor.
Okay.
If the quadratic factor is something we can solve, can we find x values of factor? Put those in the polomial.
Yes. So this only works in case it's a factor. It only works in case it is a factor. Not if they give you some remainder. Then it will not work. If the quadratic expression that is that is given that's a factor and it can be factorized. For example, if they tell you x² + 7 x + 12, this is a factor of some polomial.
You could say this can be factorized to x + 3 into x + 4. Which means x + 3 is a factor and x + 4 is also a factor. So in that case you could use remainder theorem as well. Input x= to -3 in this.
Input x= to -4 in this. you'll get two simultaneous equations. You could solve those two simultaneous equations. So yes, that is possible. Okay. But in this case, we could not factoriize this.
So we won't do that. We have to use long division.
Okay.
This should be the next one. That was question number six, part B. Let me just confirm if there's no other part to this.
This was six, right?
Yeah. Okay. Let's look at the next next one. Question number seven. Now they're saying by sketching a suitable pair of graphs, show that this equation l x equal to coseant/x has exactly one root in this interval.
Okay, this is the question that I recommend that you should leave for the end of the paper. Do not do this in the middle of the paper. It's too much more x takes too much time. Don't do that.
Don't do this question in the middle of the paper.
Now, let's mark X and Y axis like this.
You'll have a blank page for this, blank space for this.
You'll have to make your own lines.
Now, they're saying sketch the graph from 0 to pi.
So negative part we can ignore. We just have to go from 0 to pi.
Okay. 0 to pi. Let's break it down in the middle. That will be p<unk> / 2.
Break it down again in the middle. That will be p<unk> / 4. P<unk> / 4 * 3 that will be 3 p<unk> / 4. This is 3 p<unk> / 4. You can break it down further to these points if you want. Okay, that's now we can input this in the table function on our calculator.
Let's input this in the table function.
Make sure calculator is in radian mode.
Table f ofx= coseant/ x. Coseant means 1 / sin/x let's input this on our calculator l and x we will do separately then you can also do it together with say let me just do it together l and x let's just put that inside as well then we'll think about the asmtote for that as well separately but for now let's just input it as it is 1 / sin of/x this is 0.5 * x okay that's the other function We input these two functions start at zero end at pi what about the step we generally take whatever the interval is we divide that by 8 right so p<unk> / 8 that's what we can do p<unk> / 8 now these are the values that we get f ofx is the log function let's ignore that for now I'll draw the sin half x function because for log we don't need so many values just a couple of values are enough let's look at the coseant graph You have values going from undefined at zero. In trigonometric functions, an undefined value means you have an asmtote. So you have an asmtote at zero for coseant. So there's an asmtote at zero. You can make dotted lines like this for that.
Then it's 5 something.
Let me take a scale like this. Maybe 2 4 6 8 -2 5 point something.
Let's plot that somewhere around here.
This first value. Then the next value is 2.6. 2.6 maybe around here. Now exact points are not important. What's important is the shape that you get from this. So just by plotting a few points you'll understand the shape. It's going up all the way up to one at the end, right? The last value is one.
So, it's stopping at one here.
So, what is the graph going to look like then? It's going to go like this. No, not like this.
Something like this. Right? So, you draw a shape like this. You don't have don't have to pass through all the points.
You'll remove the points from there.
and just write this. The important thing in this graph is the shape of the graph. It should be clear that there's an asmtote at the y-axis and the lowest point on the right side that is at one. This is an important point and the asmtote is important. These are two things that are important. The points that we plot they are just to understand the shape of the graph.
Okay.
How will we do it without the calculator? We will not do it without the calculator. Why would you do it without calculator?
How will you find values of coseant without calculator? You have to memorize values for that. I don't recommend that.
You have your calculator. Just do it on that reciprocal trick function. You should sketch them using a calculator. Okay?
There's no need to memorize values.
A now what about the log function? What about the log function? Graph of L and X. That's a standard graph. You could actually remember the graph shape. You should remember the graph shape. Where is the important point for this graph?
Important point is when this thing is one, right? So x= 1, that gives us y =0 for the log function. That's one point.
x1 and y 0. Where is that going to be?
Pi /4 is 0 something, right? P<unk> / 4 is divided by 4 is 0.785.
So one will be slightly to the right of it. At one, we've got zero. So the log graph is going to pass the x-axis somewhere around here.
Where is the asmtote going to be? ASM tote is when this thing is equal to one uh equal to zero sorry this is the asmtote.
Now that's the same as for the coseant graph as well. Now what about the rest of the graph? Now how is it going to look like for that? We had that table function.
Why did I get rid of that?
Shoot. Okay, let's just put that again.
ln x start zero and pi we had that table already right so we could use that for a couple of other points so at pi I'm getting 1.14 at pi I have 1.14 that will be around here maybe it's going to be slightly above the coseant number right because this was one at co for cosecant for log it is 1.14 so it's going to be here we have another point here on the x-axis and you can see the other points they're slightly reducing from this point to that point and the asmtote is is at x=0 the graph is going to look something like this it's going to come here like this and here it's going to suddenly shape change its shape and become almost parallel to y-axis like this. This is the graph of the log function. You will label both of them. You'll say this is log of x and the other one is coseant of x. Take Makes sense, right?
Any issues? Any problems? Anyone?
Okay. Verify the calculation of this root lies between 2.6 and 2.9.
Now, how do we do that? We take all terms to one side. ln x - coseant of/ x we can write that as - 1 / sin of/ x take all terms to one side like this first of all that equals zero now you could say this thing is f ofx we have fx equal to0 let's input those x values in this input 2.6 six and then input 2.9 and see what values you end up getting from both of them.
Has anyone done this?
Has anyone done this?
No.
Can multiple people please do the calculation so we can verify quickly as well. We need to speed up a little. We have S1 today as well.
ln of 2.6 minus 1 / sin of x /2 which is 2.6 /2.
Okay. You can see what that is. This is turning out to be a negative number.
-0.0823.
Then you input 2.9 and see what that gives you.
2.9 gives you 0.0574.
One value is negative, the other value is positive.
So you write down the statement now. F of 2.6 6 and f of 2.9 they have different signs or the signs are changing. That's the key word. Signs are different or the signs are changing.
You need to make sure that you write that.
That is why the root lies between 2.6 and 2.9. Okay, that's what we had to prove in this part. That's a standard iteration point. Now they're saying use this iterative formula to determine root correct to three decimal places. Give the result of each iteration to five decimal places. Now what starting point do we use? We have shown that the root lies between these two values. So use the midpoint 2.75.
That will be your starting point. X1= 2.75.
Now put this on your calculator.
2.75 that's the starting point. E raised to power coseant. Now what is cosecant going to be? Let me write that down on the side.
On your calculator it'll do something like this. E ra to power 1 / sin of half in place of x n you put answer. That's what you would input on your calculator. So, e to power 1 / sin of half times the answer.
All right. So, x2 x2 is turning out to be 2.77175.
We need five decimal places for each iteration. X3 2.76 584 X4 2.76741 X5 2.76 699 X6 2.7 76 7 1 0. It's fairly stable. Now it's rounding out to 2.767, right? You can stop here. The difference between these values is only 11 at the end, right? So it's not going to change much after this point. You're allowed to stop if you want. If you want to write a couple of more iterations just to be safe, you can do that as well. So if I write one more, uh the next one is 2.767.
But this is sufficient. Now write down the answer. Now to three decimal places x = 2.767 and that's our final answer.
Okay that was straightforward. Let's move on to the next question. This was question number six seven from this paper. Okay.
The variables x and y they satisfy this differential equation. Variables x and y that satisfy this differential equation.
It's given that y is equal to 0 when x is equal to 0. solve this differential equation to obtain an equation in X and Y.
Okay, so first of all we separate the variables.
dy is in the numerator on the left side.
So y terms on the left side and that will become y / y + 5 d y x terms on the right side and that will become x e ra to power 3x when when it goes to the denominator we can say its power becomes negative right if you bring it to the numerator after that x e rais^ minus 3x dx now variables are separated we can integrate both sides how do we integrate the left hand side how do we integrate this thing y / y + 5 ln rule does not work.
It's not tan inverse form. Partial fraction does not work. Splitting the numerator, there's only one term in the numerator. What do we do?
How do we integrate this fraction?
You need to remember all the rules that we've got for fractions, right? Which rule is going to work on this?
It's an improper fraction. It's an improper fraction. So, we do long division. We do long division.
y / y + 5. What do we multiply y by to get y? 1. So, 1 into y + 5 that gives us y + 5. Change the signs. It becomes -.
This gets cancelled and we have minus 5 as the remainder.
Quotient plus remainder over divisor. So the integral on the left side quotient is 1. So integration of 1 plus integration of 1 with respect to y plus remainder over divisor remainder is - 5 - 5 over the divisor which is y + 5 integration of this separately. Now what's the integration of one?
Integration of one is just y integration of this thing. We can write minus 5 outside 1 / y + 5 integration of that is going to be ln of y + 5 because the derivative of y + 5 is 1. So we need one in the numerator. We can write -5 outside and that gives us - 5 ln of y + 5.
That's the integration of the left hand side. What about the right hand side?
That's a product. To integrate the right side, we have to use the product rule of integration. Integration by parts. Let's do integration by parts. And that use iate to decide between u and vdash.
X is algebraic. This is exponential. So you would decide that u is equal to x therefore and vdash is equal to e rais power - 3x u dash is equal to 1 v is equal to e power - 3x / - 3 right we are just using iilate for this I a t e on the left side we've got you on the right we've got vdash the two functions that we've got are algebraic and exponential so exponential must be vdash and algebraic should be u. Now input values in the formula. What's the formula? The formula is uv minus integration of v dash.
Let me write that down here. Integration of uv dash that is uv minus integration of v dash. The simple values in that.
Now on the right side for this u as it is u as it is means x as it is into v which is e rais power - 3x / - 3 plus integration of v u dash which means e^ - 3x / - 3 into u dash which is 1.
Simplify this as well. We have - x / 3 into e^ - 3x. Write this -3 outside. So it will become - 1 / 3 outside. Then integration of this how do you integrate e power - 3x e power - 3x / minus 3. So this another three is another negative -3 is coming in the denominator from this integration.
Okay. Now integration is done at this point. You have a plus c at the same point. Okay. Okay. So once you're done with all once you're done with the complete integration right in that step you should add a plus c. Okay. Now let's simplify this further and then we can input the values of x and y that we know x is equal to 0 y is equal to 0.
But before that I'll simplify this a little. y - 5 into ln of y + 5. On the right side we have - x / 3 into e^ - 3x + 1 / 3 into e^ - 3x + c. Input those values you put x =0 and y =0. To find the value for c 0 - 5 ln >> 1 9 that's 1 over 9. Yes. Thank you.
This is 1 over 9.
Okay.
So good that we spotted that there otherwise endus 5 into ln of 0 + 5 that is equal to 0 because x is multiplying plus 1 / 9 into e^ 0 + c. Let's simplify and we get we get - 5 l 5 and then 1 over 9 from the other side when that comes to the left side we have - 1 over 9. So c = -1 / 9. Now we input this back there and we get that equation. What's that going to be? y - 5 ln of y + 5 that is equal to - x / 3 e^ - 3x + 1 / 9 e^ - 3x + c. The value for C is -5 ln 5 - 1 over 9. They wanted a relation between X and Y. Some equation between X and Y. They don't want us to make any specific variable the subject. Are we allowed to just leave it like this? Can we just leave it like this and say this is the final answer or no? What do you think? Is it okay to leave it as it is?
This was considerably harder than other exams. So it was not marked.
I haven't heard anything like that.
Can we leave it like this?
The question does not ask us to make any variable the subject. Right? So is it okay to leave it like this or should we still do something further after this?
We have multiple log terms.
We cannot leave the fin leave multiple log terms in the final answer. We need to convert them to a single log term.
That is necessary. They often have one mark for this in the marking scheme at the end. So make sure that you do that.
Convert it to one single log term. Bring it to the left side and you can say this becomes y + 5 ln 5 - 5 ln of y + 5 on the other side you've got whatever there was take five common here and then you have 5 into ln of apply division law on this 5 / y + 5 on the other side you - x / 3 e^ - 3x + 1 / 9 e^ - 3x - 1 over 9 and that will be your final answer.
That's an equation in terms of x and y and we have simplified the logs as well.
We've converted them to a single log like this.
Okay, that was this differential equation question. Not that difficult. It was eight marks.
Okay, let's move on to the next one.
Now, question number nine.
Question number nine.
There was no partial fractions until this point, right? There was no partial fractions.
Okay, let's do this. Now they're saying let I be this. Use the substitution X= under root 3 tan U to show that it can be written like this in this one. Now how do we do this? They have told us exactly what to do.
I is equal to integration of X cub / 3 + X 2. We have to integrated with respect to X. Limits are from 1 to 3. The substitution that is given is that x is equal to under<unk>3 tan u. Find the derivative of this dx / du.
What's the derivative of this going to be? Under root3 remains as it is.
Integration derivative of tan is secant squ. So secant square u. You have to get rid of this dx from here as well. So make du the subject. And that will give you du = dx / under <unk>3 see square x. Now do the substitution.
See what that gives you.
X cube. Take cube of this number and that will give you under <unk>3 into tan u whole cube in the denominator. You have 3 + x². So that will be under root 3 tan u 2 and then du can be replaced with this thing dx.
There's something wrong. We had to replace dx, right? We have to make dx the subject.
If we make dx the subject that actually gives us under <unk>3 sec² u du. That is what we would input there multiplied by under <unk>3 into secant square u du. This is what we end up getting. Now everything is converted to you. At this point we can change the limits as well. Now what are the limits going to be? clearly write what the upper limit is and what the lower limit is because sometimes they can get switched. Up uh upper limit is 3. So x is equal to 3 and x = 1. Now 3 = under <unk>3 tan u tan u = under roo<unk>3. Put this on your calculator.
tan inverse of under root 3. What's that? I think it's pi over 3, right?
Please confirm this.
1 equ= under <unk>3 tan u tan u is equal to 1 / under<unk>3 that's the same thing as under roo<unk>3 over 3 and this I think is<unk> / 6 please confirm both of these one is p<unk> / 3 and the other is p<unk> / 6 right these are the values of u that we're getting now the upper limit is still the bigger number lower limit is still the smaller number so we can just input them here now the smaller number is 5 / 6 and the A bigger number is pi / 3. These are the limits that we're getting. Okay, thank you. Now we just have to integrate it. Now input limits.
No, we don't have to do that yet. They just want us to show this thing first of all. Let's do that. We have to simplify it and show that it becomes this. Now you see the limits are already given here in what we have to show. So in this working you need to make sure that you clearly are showing these steps because the answer is already given the marks are going to be for this working you have to show that working okay now let's simplify this a little under root 3 whole cube and this under root 3 that will become under root3 power 4 that's the same thing as 3 squar which is going to be 9 in the numerator you'll get 9 from this under root 3 cub and under root3. Okay. And then we have tan cube u into secant square u. This is what we have in the numerator.
What about the denominator? The denominator is 3 + 3 tan square u.
Okay. Now let's simplify this further.
In the denominator what I can do is I can take 3 common that leaves me with 1 + tan square u seeant square can also be written as 1 + tan square u.
So these two get cancelled this whole thing and this whole thing they get cancel 9 / 3 that is 3. What are we left with?
3 tan cq and that's what we wanted.
That's the first part. 3 tan cub u with respect to u limits are from<unk> / 6 to<unk> / 3. Take that's the final result. That's what we wanted in the first part.
Take it. Now this is done.
Okay. Next. Now they're saying hence or otherwise find the exact value of i.
Give your answer in this form where pq and r are rational numbers. Now that's a standard part from our standard methods. tan cube of u. How do we integrate that? Does anyone remember integration of tan cube?
We've done this separately. How to integrate odd powers of sin cos tan?
Right?
How do we do this integration of tan cube?
Does anyone remember? We can break it down into two parts. You can break it down into into two parts. You can say this is three. We can write three outside as well. tan² u into tan u.
Okay. And then we used to use a substitution. Right? Now you can do it using substitution as well. The substitution that we used to use was u equ= tan. Now in this case the variable is also u. So you'll have to use some other other letter. I don't know let's say y equals tanu. Have they used y anywhere? No. So you can say y equals tanu. Use the substitution and then do it using that. That is one idea. But now that we've done the extended power rule as well.
We can actually make it much quicker using that.
How exactly will that work? tan square we could write as tan square we could write as secant square minus 1 right because 1 + tan square theta we know this is identical to see square theta so tan square theta that can be written as sec square theta - 1 let's write it like that see square theta - 1* tan u so supposed to be u not theta So in place of tan squ u what we can do is we can write seeant square u minus 1.
Okay. Now let's simplify this simplify multiplyant square and tan u and then we have 1 into tan u let's integrate that separately.
Now this tan u how do we integrate tan u? Does anyone remember this integration of tan? How do we used to do that? Using log rule, right? Ellen rule. Tan can be written as sin / cos. So we can write this as sin u over cos u.
And then this integral that is going to become the derivative of cos is minus sin u. Right? So we want minus sin u in the numerator. So what we can do for that is we can have a plus outside and put minus inside.
So this integral will become ln of cos of u. So the second fraction the second term that is easy to integrate. What about the first one? Is there a way to integrate this? Substitution is one method we've done that. Can we use the extended power rule on this? Can we do extended power rule? Can we use extended power rule on this? Yes or no?
Yes or no? Extended power.
Yes.
G. How do we integrate this? Seeant square U into tan U. What can we do for this?
Seeant square U into tan U. Extended power rule. How do we use that? How exactly do we use that? Yes. So we'll have to do it the other way around.
We'll have to rearrange it. Write tan u and then because if we do it like this, if we say we have secant u squar and then tan u outside. Now the derivative of seeant is not just tan. It could also be done that way. V. Uh if you want to use seeant as the box in that extended power rule, you could say seeant u that is inside power one and then outside you have secant u tan u derivative of seeant is seeant into tan.
So you could use that as well.
The other idea could be that you rearrange and you write it like this tan u into secant square u. And now you could say you have tan u power 1 derivative of tan u is multiplying outside which is seeant squ. So the integration is just going to be tan square u / 2.
Okay. So the integration part is done.
Now once you're done with the integration input limits limits are from p<unk> / 6 to p<unk> / 3.
Okay. P<unk> / 6 to P<unk> / 3.
Input those limits and that will give you the result that three was multiplying with both of them, right?
Three was multiplying with both of them.
Yes. So, we have to be careful here.
Three was multiplying with the whole thing.
I could write that outside the limits.
Then just multiply with the whole thing at the end. Actually, let's input limits. / 3. When we input p<unk> / 3, that gives us tan p<unk> / 3^ 2 / 2 + ln of cos<unk> / 3.
Subtract from this the case when you input / 6 and that will give you tan / 6^ 2 / 2 - ln of cos<unk> / 6. Now do they want an exact answer at the end?
Yes, exact answer in this form. Okay, let's simplify this further. tan of p<unk> / 3. Put this on a calculator.
tan pi / 3 uh tan p<unk> / 3 is under root3 I think right? Please confirm. I think it's under roo<unk>3 squar. We did that in an earlier question as well. / 2 + ln of cos<unk> / 3 that's cos 60 cos 60 is 1 / 2 - tan / 6 that is under<unk> 3 / 3² / 2 - ln of cos / 6 cos / 6 is under roo<unk>3 /2. You'll just do this on your calculator. Evaluate each one of them on your calculator and then you simplify this.
Three is outside as it is inside.
This becomes 3 /2.
What do we want at the end?
PQ and R are rational. Okay.
3 /2 + ln of 1 /2. Put this on your calculator. The calculator will simplify this for you. Uh it's going to be 1 / 6, right? Under roo<unk>3 over 3. That's 1 / under roo<unk>3 1 over 3 / 2. So it will be 1 over 6. Please confirm this as well. I think it's 1 / 6 minus ln of under<unk> 3 /2. Okay, that's what we're getting.
Now let's move on.
What's the next step going to be now?
Simplify these constants.
Simplify the constants. 3 /2 - 1 / 6.
What's that?
32 - 1 / 6 9 - 1 8 8 / 6 which is 4 over 3 4 over 3 okay plus these are log terms apply division law on them ln of 1 /2 / under<unk>3 /2 2 and 2 gets cancelled and we're just left with 3 into 4 / 3 which is 4 + 3 into ln of 1 / under <unk>3.
Now could we leave this as our final answer? Could we leave this as our final answer? They told us your answer should be in this form. P P P P P P P P P P P P P P P P P P P P P plus Q LNR R P + Q LNR and P Q and R are supposed to be rational. Rational numbers are it could be inteious, it could be fractions, but you can't have square roots, right? Square roots are not rational. We know that from O levels IGCSE as well.
This is not rational.
Okay. So we can't leave it like this. 1 / under<unk>3. How do we write that? We could write this as 4 + 3 ln of 1 / 3 power 1 /2. Right? Under root 3 could be written as 3 power 1 /2. And we could put that as a whole power like that. Now this 1 /2 will go outside multiply by 3 and that gives us 4 + 3 /2 into ln of 1 / 3 and that will be your final answer.
Okay.
Take 3 to the power uh no we bring 1 /2 outside right that will be this form.
They don't have anything as a power on top of r here.
Okay, this is one way of doing this. Another possible result that you could get from this could be 4 - 3 /2 ln3 that will be equivalent. If I take 3 to the numerator, it becomes 3^ -1 and then I can put that negative outside.
So this will also be the same right? So any of them will be correct. Now this was a long integral nine marks. It's a good idea to check that.
It's pretty quick, right? X cube. Input that there.
You will generally do this at the end of the exam. 3 + x².
So try to finish early so you have time for this verification. So in case there's any careless error then you can catch that. We're getting 2.352. Right now the answer that we are getting here.
Let's check that. 4 + 3 /2 into ln of 1 / 3 and that's the same thing that confirms that this is correct.
Okay. Okay. That was question number nine part B. Let's look at the next one now. Question number 10.
Question 10. The variables X and Y satisfy this equation. Haven't we had done differential equations already? No, this is differentiation. Okay, not differential equation.
Okay, the variables x and y they satisfy this equation where is a constant. Show that dy by dx is equal to this.
Now should we take 1 / 2 to the other side as the power? Okay. No. So the derivative has 2 y in it, right? The derivative has has 2 y in it as well. So we have both x and y in the derivative.
When when does that happen? When we do implicit differentiation, right? So we should do implicit differentiation rather than taking square root on the other side. That will also introduce plus minus. We will do implicit differentiation. So y^2= k * k is multiplying with the whole thing, right?
It's with the whole fraction. You can see it's not just with x. So k into x - 2 /x + 2.
Differentiate both sides with respect to x. What's the derivative of the left side going to be? 2 into y into dy / dx on the other side.
That's weird. Uh dy by dx they have 2y in the numerator here.
Okay. on the right side we would have to use quotient rule in this case right what I can do is I can simplify this before differentiating that will be a good idea bring x +2 to this side so we avoid using quotient rule that complicates things right if I bring that to the left side x + 2 into y^2 I can do that right that will become y^ 2 into x + 2 on the right side so I have k into xus 2 so before differentiating you can actually simplify this a little okay that will become easier to solve Now you differentiate both sides. What's the derivative of the left side going to be? It's a product. It's a product. So we use product rule on this. We can say this is u and this is v. U vdash plus vu dash. U as it is that means y^2 as it is into derivative of this derivative of this is 1 plus v u dash that means x + 2 as it is into derivative of y^ 2 that is going to be 2y into dy / dx. Okay. Now what about the derivative of this? K is just a constant.
K is a constant. So K remains as it is.
Derivative of X is 1. Derivative of minus 2 is zero. We just have K into one on the right side. Now we need to make dy by DX the subset. Now now notice in the final result that we want there is no K. In the final result that we want there's no K in that. Now if there's no k in that what we can do is from this equation we can make k the subject and that gives us y^ 2 into x + 2 / x - 2 and replace k with that thing. Replace k with that thing now. So we input this thing in place of k here.
Okay. Now what do we get from this? y^2 + x + 2 into 2 y into dy / dx. On the right side we get y^ 2 into x + 2 / x - 2.
Okay. So we get rid of k in the beginning.
We get rid of k in the beginning. Okay.
Now we can make dy by dx the subject.
Take y^2 to the other side and that will give you x + 2 into 2 y into dy / dx. On the other side you have y^ 2 into x + 2 over x - 2 - y^ 2ide by x + 2 or you could first simplify this a little x + 2 if I take y square common that leaves me with x + 2 / x - 2 - 1 take lcm y^ 2 into x + 2 - x + 2 / x - 2. And now from the other side we have x + 2 into 2 y into dy by dx.
Okay.
Divide it on the other side. Divide this on the other side and you'll have 1 / 2 y from this side and also x + 2. x + 2 and 2 y. And now you simplify everything and that will give you the result. Y^ 2 gets cancelled with this y in the numerator. X and minus x get cancelled. And at this point we're getting 4 y divided by x - 2 into x + 2. And there's a two outside as well. Now 2 and 4 get cancelled to give you 2. So you have 2 y in the numerator. In the denominator x - 2 into x + 2 that's x² - 4. And that is the derivative that we were looking for.
That's our final answer. That was fairly tricky. U thinking about that you have to get rid of k like that. But that's how you do it.
Make sense? Right?
Any issues with this?
Here goes.
If we used quotient rule, the y square would not have been there. If we use quotient rule, y square is coming from the left side, right?
A yes, if we use quotient rule, y square would not be there in the first place.
Yeah, that's also right. So using quotient rule would make it maybe slightly more efficient as well. So we would not have to take this LCM in that case. If we used quotient rule, we would not have to take this LCM ourselves. We could save another step in that case as well. Yes, that's correct.
Uh yeah, I suppose the threshold for this is going to be actually quite low because it is a lengthy paper as well and also some parts were quite difficult.
Let's look at the next one. Let's look at the next one and see what we do here.
the S1 class has become really late.
Given that K is equal to 5, find the angle between the tangents to the curve when X is equal to 3. Give an answer in the form a tan inverse of b / c where a, b and c are inteious. Okay, now that we're given that the value for k is five, we need to find the angle between the tangents where when x is equal to 3. How do you find angle between any two lines?
But remember the formula that you do in paper one uh m1 - m2 divided 1 + 1 + m1 m2. That's the formula that you can use here. angle between two lines that is given by tan theta= m1 - m_sub_2 / 1 + m1 * m2.
Now what you need for that is the gradient of the two tangents. Gradient of the two tangents. Now how do you find the gradient of the two tangents?
We have the derivative given.
We want to find the gradients of the tangents when x is equal to 3.
We need the y values because the derivative is in terms of y, right? It has y in it as well. So we need to find the y values from that. How would you do that? That's straightforward. Now y^2 is equal to these are easy marks. If you remember this formula, the value for k is 5. So we have 5 into x - 2 / x + 2.
Okay.
Input x= 3 in this. Input x = 3. Now when we do that we get y^ 2 = 5 into 3 - 2 / 3 + 2 5 and 5 get cancelled and we're just left with 1. So y ^2 = 1 and that gives us y = 1 or y = -1. We can find the derivative from both of them.
Now the derivative dy / dx that was equal to 2y / x² - 4 that was the derivative. Input those values.
Let's find gradient of 1 tangent. 2 into y that means 2 into 1 / x² - 1 that will be 1. No that will be 3^ 2 - 4 2 over 3 square - 4 that is 2 over 5. And then we input this in the other gradient m_sub_2 that is going to be 2 into -1 / 3 ^ 2 - 4.
What is this going to be? -2 over 5. So the gradients are 2 5 and -2 over 5. Now input them in the formula. tan theta is equal to m1 - m_sub_2 / 1 + m1 * m2.
Now what does this turn out to be? tan theta = m1 which is 2 over 5 - - 2 over 5 / 1 + 2 over 5 into -2 over 5 you'll have to simplify this see what you get from this 4 over 5 in the numerator in the denominator what's the denominator what is the denominator Enter 1 - 4 25 which is 1 21 over 25. 21 / 25.
Put this on a calculator. Now what do you get?
20 over 21.
I'm not going to trust myself on the calculation. Now can you let me know what is it? What this is?
M1 was not leaked.
M1 was not leaked.
20 21 right? Tan theta is 20 / 21. Uh now what about the value of theta? That's tan inverse of 20 over 21. That's what we were looking for. That's our final answer.
Ka any issues with this?
No. Let's look at the last one now.
Question number 11.
We haven't seen vectors until now. So most likely yeah. So that's vectors.
Okay. Question number 11. Let's do this.
Now points A and B have position vectors these.
Show that the perpendicular distance from A to the line through O and B is this. Now that's a standard question.
Perpendicular distance. How do you find that? You first of all have to find the equation of the line. Line through A and B. This line that passes through O and B.
Uh okayish Aisha uh I I'll do that and sure I'll do that line through O and B O is the origin right? So we can say OB can be taken as the direction vector of this line.
We can say this is the direction vector.
Let's write down the equation of the line OB. R= A + lambda D in place of A.
I could take the origin, right? Origin is just 0 0 0. That's a point plus lambda into the direction vector. This will be the direction vector OB. That will be 4 2 4. So the equation of the line OB that turns out to be 4 lambda 2 lambda and 4 lambda. That's the equation. Right? The point is origin. So it just simplifies to this. That's the equation of the line. Now distance from the point A. So we can make a rough sketch. Now this is a line.
We know the equation of the line.
There's a point A somewhere. We know the coordinates of that point A and the vector for that point A as well. We have a perpendicular coming from that point A to this line. Let's call this point X.
Let's call this point X.
What are the co what are what's the position vector of X going to be? X is a point on the line. The line has this equation. Every single point on the line is going to be of the same format.
Right? So we can say o x vector is going to be 4 lambda 2 lambda 4 lambda. Right?
x is a point on the line. This equation represents every single point on the line. So the position vector of x should also be of the same form. Okay. Now we say ox is uh ax is supposed to be perpendicular to the direction of the line. We can find the vector ax. Now what is ax going to be? a x vector is o x vector minus o a vector. What does that turn out to be? 4 lambda 2 lambda 4 lambda minus o a is 2 2 -1 2 - 1 this simplifies to 4 lambda - 2 2 lambda - 2 and 4 lambda + 1. This is the vector ax. And now we use that idea that ax vector is supposed to be perpendicular to the line. So ax vector it's dotproduct with the direction vector of the line that must be zero.
So ax vector dot o vector this must equal z.
So we say 4 lambda - 2 2 lambda - 2 4 lambda + 1 dot OB what's OB 424 this must be equal to zero and that will give you something 4 into 4 that is 16 16 lambda - 8 + 4 lambda - 4 + 16 lambda + 4 = 0.
Okay, this simplifies to what? 36 lambda - 8 = 0. Can somebody please confirm this? I not sure if I'm doing the correct calculations. 8 over 36. Is that what you get?
2 over 9. Can somebody please confirm this quickly? Is this calculation correct?
I'm too tired to check this now.
Yes. Okay. Lambda is 2 over 9. Lambda is 2 over 9. Now all we have to do is find this perpendicular distance ax vector.
Input this value of lambda back in that ax vector. What does that turn out to be? Now 4 lambda minus 2. That means 4 into 2 / 9 - 2 2 into 2 over 9 - 2 and 4 into 2 over 9 + 1. Simplify this. What do we get from this? 8 / 9 - 2 that's - 10 / 9 4 / 9 - 2 -4 / 9 8 / 9 + 1 17 / Now let's find the magnitude. That will be the perpendicular distance, right?
Square<unk> of 10 / 9 2 + 14 / 9 2 + 17 / 9 2.
He can that will give you something. Now 100 over since you want to show that it turns out to be that it's important that you show the steps. Uh you can add this up first inside the square root. 100 + 14^ 2 + 17^ 2 you get 585 / 1 uh divided by 81.
Okay. Now the denominator is square root is just 9 square<unk> of 585 that's 3 under<unk> 65 and that's what you wanted to prove.
This is the final result that we have.
We can simplify this further 1 / 3 under<unk> 65 and that's the final answer.
Okay, make sense? That was the first part.
That was a standard part uh perpendicular distance. Now we have another seven mark questions.
Okay, let's do this. Now the point C has position vector this. Uh given that OC is perpendicular to AB and angle, this is equal to this. Find the values of P and Q. And this is also not that difficult. It may just be slightly lengthy but not difficult as well. OC vector that is equal to 3 PQ.
What about AB vector? Let's find the AB vector.
Yeah. So you can't end up getting stuck at any question in the middle, right? So for example, we spent quite some time on one question, right? And the differentiation one. So you would probably in the exam skip that part, do the rest of the parts and hopefully you'll get these seven marks at least even if you don't get those marks.
AB vector uh did we find that earlier?
No, we did not do that. OB minus that will be 2 0 and 5 20 AB vector from the first part I can say that's 205. I'll just confirm that again. 4 - 2 that's 2. 2 - 2 is 0 and 5 20 5 that's the vector AB the dot product is supposed to zero supposed to be equal to 0 that gives us 6 + 0 + 5 q that is equal to 0 the value of q is turning out to be - 6 over Okay, that's the value of Q that we're getting. So, we're getting that value directly from this OB minus OA. Uh the J component is indeed zero, right?
Okay, that's interesting. So, Q value, we got that directly. The angles are supposed to be equal. Now angle at the point O in both of them which vectors do we use? O A O C O O C O B these are the vectors that we'll be using for that.
Let's make an equation for that as well.
We can say if the angles are equal and then their cos are also equal to each other. Right? Cause of A O C that should equal cos of C O B.
Now that means O A dot OC divided by O A magnitude and OC magnitude that should equal OC vector dot OB vector divided by OC magnitude into OB magnitude.
Input values in this.
What are the vectors? Let me write them down here. O A vector what was that? O A from the first part was 22 - 1 and O B from the first part was 424.
Okay.
Let's input them here. the dot product OB what's that ob minus Q that's the dot product for OA and OC divided by magnitude of O A magnitude of OA let's figure that That will be square roo<unk> of 2 square + 2 square + 1 right which is square root of 9 and that's 3 magnitude of OB that will be square root of 32 + 4 that's going to turn out to be six.
Let's input that here. Magnitude of OA is 3.
OC is there on both sides right? We could just cancel that out as well. OC and OC the magnitudes they get cancelled. So we don't have to show that work end on the right side. OC dot OB OC dot OB 3 into 4 that is 12 + 2 * P + 4 * Q in the denominator we have magnitude of OB which is six that's what we get now we can cancel this to get 2 here cross multiply that gives us 12 + 4 P - 2 Q that is equal to 12 + 2 P + 4 Q now the value of Q we have already figured out we can input that here make P the subject 2 P = 6 Q P = 3 * Q is that correct 12 and 12 is getting cancelled on both sides right and then 4 P - 2 Q that is 2 P 2 P = 6 Q P = 3 Q and that will give you the value for P as well 3 into - 6 over 5 and the value of P turns out to be -8 / 5 that is what we were looking for. Q is -6 over 5 and P is -8 over 5. That was pretty straightforward. Easy seven more unless there's a careless error somewhere.
And but apart from that, it was fairly straightforward, right? The last part.
So you see this is happening again and again. Now the last three papers that we did in all of them, the last question was fairly straightforward. The last question in all of them was fairly straightforward. So you can't afford to get stuck in the middle. That's extremely important. So there were a couple of questions which were quite tricky in the middle. You would not do them in the middle. Make sure that you complete the questions first that you know about. And these are easy 12 marks at the end. A lot of people would not have reached this point.
So if you get to this point and make sure that you get these marks, the rest threshold will take will take care of that. I don't think it's threshold is going to be very high. It'll be around between 50 and 53 maybe somewhere around that region for an A. So that's it. I'll see you in the next session then inshallah. This has been a long long class. We started late as well. Okay.
Allah
相关推荐
Olympiad Mathematics | Indian | Can You Solve This One?
PhilCoolMath
650 views•2026-06-03
Escaping the Fog
LogicLemurGaming
760 views•2026-06-03
A Brutal Radical Expression Made Easy! The Shortcut Changes Everything.
tamoshop
112 views•2026-06-02
V : jee main /advance class 11 mathematics : Binomial Theorem class-1 ( 29 may 2026 )
dcamclassesiitjeemainsadva9953
125 views•2026-05-29
Is This Pentomino Tileable?
3cycle
241 views•2026-05-30
This Sudoku Has Many Lines!!
CrackingTheCryptic
2K views•2026-05-29
Olympiad Mathematics | Indian Can You Solve This One?
PhilCoolMath
268 views•2026-06-02
Olympiad Mathematics | Indian | Can You Solve This?
PhilCoolMath
669 views•2026-06-02
