IGCSE Add Math | Feb 2026 | Paper 2 (0606/22/F/M/26 & 4037/22/F/M/26)

本站添加: 2026-05-12

[00:00:00]All right. Uh, question one. The function f is defined by this thing here. So 2x^2 - 6x + 4. Write f ofx in this form where a, b, and c are constants. In other words, it wants us to do completing the square. So how can we do that? So we've got f ofx is equal to 2x^2 - 6x + 4. So the first thing we want to do is to factoriize the two from this uh x^2 and the - 6x here. So we're going to write that. So we've got two bracket x^2 - 3x like that + 4 and then we can proceed to do uh completing the square with this thing here the x^2 - 3x. So if I do that what have I got to do? Well, I've got to take the coefficient of x, which is -3, and divide it by 2. And that will then give me x - 3 upon 2^ 2 - - 3 upon 2 all^ 2. Okay? So, in effect, what we've done is we have replaced uh this term here x^2 - 3x with this thing here. Right? Those two are equivalent.

[00:01:12]Well then if I simplify out um what I will get is 2 bracket um x - 3 upon 2 all^ 2 - 9 upon 4 + 4 and if I expand that I get 2 bracket x - 3 upon 2 all^ 2 - 9 upon 2 + 4 and plus 4 is 8 upon two isn't it right and so therefore if I was to write this I've got 2 x - 3 upon 2^ squared and - 9 upon 2 + 8 upon 2 is - 1 upon 2. Okay. So we've written it in the form that um has been asked. So basically a is 2, b is -3 upon 2 and c is 1 upon 2. Now um why do we write it in completed square form? Well because uh if if we've got a quadratic which is of the form let's write x I'll write it like this, right?

[00:02:11]a bracket x - b^ 2 + c. Then what that means is that the vertex is at the coordinate b comma c. That's the turning point of the curve. So in this particular case, right, we know that our turning point, our vertex is going to be at the point 3 upon 2 and a half, right? So you just got to pay attention to the minus sign here, right? So we've got x - 3 upon 2, which means that the x coordinate of the vertex is positive 3 upon 2. Then minus a half there. So the ycoordinate is at minus a half. Okay. And therefore, what what that means is that the graph, if I was to draw it out.

[00:02:52]Ah, let's do that a bit better, shall we? Come on. Right. So here we go.

[00:02:59]So this is a half one uh 3 upon two. Here we go. Here's minus a half. I guess minus half might be there.

[00:03:12]So the vertex is going to be at this point like that. And so our quadratic will will be something like that. And I'm I'm sorry, but I've I've got the x intercepts wrong, but it it doesn't matter. That's that's the basic idea.

[00:03:24]And this helps us because then part B of the question hence find the range of f.

[00:03:28]Well if if the minimum value right is uh where uh y is equal to a half then it means that the the range has got to be every single value of y which is abovegative a half right. So so the range is f is greater than or equal to half. Should just explain again what range means. Range means all of the y values all the output uh y values. So when we got a graph like this, right? Um what what uh y values can can our graph take? Well, uh it take a y-val of zero, y value of a half, y value of one, one and a half, two, it can even be like minus a quarter and minus a half. But if I go below minus a half, right, then I can't have that as a y value because negative a half is my lowest possible y-value. So what we know here is that the range the all the y values I can have is negative a half and everything above. Right? So f is greater than or equal to a half.

[00:04:26]Right? Um part c of this question this has become quite a popular question as a onem mark question in recent years. So the function f is defined by this. So notice it's the same equation as we had in parts a and b but they've called it g of x. Uh why? Because this function g has got a different domain. Right? This is x is less than or equal to k. Uh and given that g has got an inverse state the largest possible value of of k. So we need to understand when a function has got an inverse. Um and in order to do this we need to understand the concept of mapping. So let's try to think that one through now. Um mapping tells us inputs to outputs. So if what we've got is a one to one mapping, what that means is that one input corresponds to one and only one output, right? So the the classic example of that is you got a straight line, right? So a straight line is a onetoone mapping. Why is that? Well, because let's say I've got one input here, right? That corresponds to one and only one output. Okay? So that's one one. Um what about many to one?

[00:05:37]Many to one means that there are many inputs that correspond to uh one output.

[00:05:43]And the classic example of that which I always use is y is equal to x^2. Now if I was to draw the graph of y is equal to x^2.

[00:05:52]So like this. Sorry about that. I know it's a bit dodgy like that. Then let's imagine that we've got the value of two as an input value. That gives us output of four. But actually we could have got that output of four with an input of -2.

[00:06:06]So in other words there are two inputs right? There are many inputs that give us the one output. So this is a many to one mapping. And then lastly what do we got? We've got a one to many mapping and a one to many mapping right means that uh one input will give you many possible outputs. So the classic example of that is uh the square root of of x like plus or minus the square root or we could say that's x is equal to y^2 whichever way you want to say but in this case right let's say we had input is 9 then that would give us an output of positive three uh but it would also give us uh an equally valid output of neg -3. So here we've got one input which gives us many possible outputs.

[00:06:53]Now, now the key thing for you to understand is that uh a one to many mapping is not a function.

[00:07:02]Um and essentially the reason is that for something to be a function, we must guarantee that there's only one output.

[00:07:09]Right? So these guys are functions because it doesn't matter how many inputs we've got, we always end up with one unique output. Uh what we never want to have is we don't want to have a situation where there are many possible outputs to a function because that ends up confusing the system. Um so how does this relate to inverses?

[00:07:28]Well, essentially what an inverse does, right, is it is it switches inputs and outputs, right? So the classic diagram that I like to draw is this one, right?

[00:07:36]Let's say we got a function, we call it f. We stick in an input, let's call it, you know, input. We usually call it x and that gives us an output. We usually call it y. What does an inverse function do? Well, it it takes that output and should bring us back to the original input. So if I have got a function, I put in the number two and I get as an output the number five. The inverse function, right, should take that number five and bring me back to the number two. That's the big idea. So inverse functions switch inputs and outputs. So now let's think about it in terms of the mappings, right? If I've got a one one um function and I invert it, right? Then this is going to switch. And if I switch inputs and outputs, I'm going to get one one. Okay, that's fine. But in this case, right, when I've got a many to one uh function and I invert it, I'm going to flip these. So a many to one will become a one to many.

[00:08:33]And of course, if it becomes a one to many, then it means it's no longer a function because we're this guy here, right? And that guy is not a function.

[00:08:41]So essentially that the in order for a function to have an inverse it must be one one. That's the big idea. Right? So in order for a function to have an inverse it must be a one one. So now we can go back to the graph of our function. Right? We basically say right what we want is we want this guy to be one one. So we want to choose the domain in such a way that we guarantee that.

[00:09:03]Well um we're told that x can be anything um less than or equal to k. So let's go back.

[00:09:11]Basically, if I cut the graph in half and I take in this case the left hand side, right, because I want to be less than a value, right? So, if I get rid of everything here, right? That's all gone.

[00:09:24]Then what I've done is I've converted the function to be a one:one mapping, right? Because now there's only one input to one output. So the function g of x is this guy here on the left hand side and that's for the domain where x is less than or equal to 3 upon 2. So essentially the answer to the question state the largest possible value of k it will be 3 upon two. Okay I hope that's helpful. I know that's a very long-winded answer for just a one mark uh question but I think it's important you try to understand the rationale for what's going on.

[00:10:00]Okay, then uh part D. Uh the function h is defined by this guy here. E 3x - 1 for a domain x is less than or equal to zero. Given that gh exists, find gh of x. Now, this is a composite function, right? GH. Uh what what does that actually mean? Well, it means that you take the function h and you stick it into g, right? So you are doing g of h of x like that. So, we're going to take the function h and we're going to substitute it where we've got a g. Um, well, what is the function g again?

[00:10:32]Well, it's this this guy here, isn't it?

[00:10:34]Yeah. 2x^2 - 6x um + 4. Um, but actually, we've we've already simplified it. We simplified it earlier, right? To to be this guy here. So, I'm going to take that guy um and let's uh rewrite it. So I'm just going to go over here going to rewrite and say okay g of x was equal to 2 um bracket x - 3 upon 2 all^ 2 - 1 upon 2 and we're told that um x is less than or equal to 3 upon 2. Okay.

[00:11:07]All right then. So um g of h of x means that what I'm going to do is I'm going to take this guy here, this function h, and I'm just going to substitute it into here. So that's going to give me 2 bracket e 3x -1 - 3 upon 2 close bracket all squared minus one upon 2.

[00:11:33]So can can you see that? So um basically what have I done? I have replaced this x here with the function um h of x. So then if I simplify what am I going to get? I'm going to get two bracket e to the 3x -1 - 3 upon two right is going to be minus um 5 upon 2 uh all^ 2 - 1 upon 2. Now we end up with a little bit of a problem here, don't we? because we're kind of saying um look not really sure um uh we're not really sure exactly what it means by simplifying answer right is is this answer already simplified enough um I don't know I mean to me it is uh but if we're not sure then we we could do a little bit of expanding so I'll I'll just do it anyway so I've got e 3x^2 which is e ^ 6x I'm going to have e 3x * by -5.2, but it's going to be twice. So, it's uh -5 e 3x. And I'm going to have -5.2^ 2, which is + 25.4 minus a half. And then if I expand this, I get 2 e 6x - 10 e 3x um + 25.2 2 - 1 upon 2 and uh that will become 2 e^ 6x - 10 e 3x 25 upon 2 - 1 upon 2 is 24 upon 2. So that's just + 12 and uh I mean you could also factoriize by by two there as well that would also be fine but I think that's as simple as you can get. Okay, I hope that's helpful.

[00:13:17]Okay. Then uh the points A and B have got coordinates 1 3 and -71. The point P with coordinate 2 A lies on the perpendicular bis sector of AB. Uh find the value of A. Well, I think it's helpful just to understand what a perpendicular bis sector is. And in order to do that, I'm just going to draw a very quick diagram. I know this is a little bit of overkill, but I think it can be quite helpful. So, um we got this diagram here.

[00:13:44]So let let this be one and then like one 2 3 right so we've got one coordinate point there that's uh the point A so A is the coordinate 1 13 and we got the point B which is -71 so 1 2 3 4 5 6 7 so there you go that's -7 and uh point B is this guy so 71 now what is it meant by perpendic ular bis sector. Well, it's the line that is perpendicular to AB. So, it's got to be perpendicular to this this blue line here and it bicts it, which means uh it cuts it into two, right? So, by means to sect means to cut. So, we're going to cut this line in two in such a way that we're we're going perpendicular to the line. So, that means that we're going to go through the midpoint. So, we got the midpoint there. I'm going to go perpendicular to it like that. Like this. Okay.

[00:14:43]So that green line there is the perpendicular bis sector.

[00:14:48]So what would we need to find the equation of the perpendicular bis sector? Well firstly we want to find the gradient of the line AB.

[00:14:55]Okay. So how can we do that? Well y2 - y1 so 3 -1 over x2 - x1. So 1 - - 7.

[00:15:04]And if we do that we're going to get 2 over 8 uh which is equal to a quarter.

[00:15:11]and the uh gradient perpendicular to that will be the negative reciprocal. So it's -1 / a quarter which is -4. So that's going to be the gradient of the green line. And uh we also want to find the midpoint of AB. And how do we do that? Well, we add up the two x coordinates. So -7 + 1 is -6. We divide by two. So -7 + 1 / 2. And the ycoordinate is 1 + 3 over 2. Okay. Okay. Now, if we do that, -6 / 2 is -3 and 1 + 3 / 2 uh is 2. All right. So, the midpoint is -3,2.

[00:15:52]And lo and behold, of course, that looks right on the diagram. Now, of course, we we can't solve the problem with an accurate diagram, right? That's not allowed. But the diagram is there to help us to kind of validate that we're on the right track and help us to see what's going on. Okay. So then the equation of the line right the equation line is going to be y - y1 is equal to m x - x1 where x1 y1 is the coordinate um uh that the line goes through and uh m is the gradient. So we got y - 2 is equal to m which is -4 um x - 3 so x + 3 and of course if we want to we could rearrange this a little bit. y is equal to -4x uh this is going to be -12 and then we got to add two so that would be minus 10 I believe.

[00:16:42]Well what what are we asked to do? Uh we are asked to find the point p which has got coordinates um 2 comma a right.

[00:16:50]Okay. So basically we're asked to find um what a is. So when x is equal to 2, y is equal to a. So we can just substitute in uh a is equal to -4 * 2 - 10.

[00:17:07]Okay? And that will be uh8 - 10 which is -8. So in other words, if we extended this line on quite a lot further right down here, then when x is 2, then y should be uh -18.

[00:17:24]All right. Then um part B, uh the point Q is the reflection of P in the line A.

[00:17:29]Find the coordinates of of Q. Now um this question when it was first asked a number of years ago, I think it caught a lot of people out, but now it's been asked I think two or three times. So you should be familiar with it if you've done past year papers. Um, okay. I can't draw P on the diagram that I've got here because, you know, it'll be way way way down here. So, what I'm going to do is just use an example. Let's say I've got this other point. I'm going to call this guy whatever S. Okay? And, uh, I'm going to reflect S in the line AB. So, if I did that, right? Uh, the reflection of this point in this line is also going to be somewhere along the green line.

[00:18:05]Why is that? Well, because the green line is perpendicular to the line AB, isn't it? Right? So if I reflect any point on the green line in the blue line, then I'm going to get um uh uh also onto the green line. So I'm going to be somewhere over here. Now the best way to find this this point, which I guess I'll call t, okay, the reflection, um is not to use the distance formula.

[00:18:28]Instead, it's to try to think of it in terms basically of vectors. Um what what I'm going to say is um in order to go from S to the midpoint, we have to kind of go a horizontal distance to the left and we got to go a vertical distance up.

[00:18:42]And then to go from the midpoint to the reflected point T, we'd have to go the same horizontal distance across like that and then the same vertical distance up. Right? So in other words, right, this distance here is going to be same as that distance and that vertical distance is going to be same as that vertical distance. So um I know it's not to scale so uh please forgive me but I'm I'm going to now just put on and say okay this is point P and uh the point P is uh 28.

[00:19:12]So what's happened on this diagram?

[00:19:13]We've gone from an x coordinate of two to an x coordinate of -3. That means we've moved back by five right? So we've gone back by five to the left. What about the y coordinates? Well we've gone from8 to two which means we've gone up by 20. So plus 20. So, in order to get to to this point here, which is going to be Q, uh, we got to do the same thing, right? We got to go back by five and up by 20. So, if I'm at -3 and I go back by five, I'm going to be at8.

[00:19:42]And if my y coordinate is 2 and I go up by 20, I'm at 22. So, um, the coordinate of Q should be uh,8 um, 22.

[00:19:54]All right. Then, uh, question three. The diagram shows the graph of y = x + 2 * x -1 * x - 3. Right? Cuz it's got roots of -2, 1, and 3. By drawing a suitable line on the diagram, solve the inequality where this whole function is greater than 7. So basically what we got to do is we got to draw the line y is equal to 7. So I'm going to try to do that on this uh diagram here.

[00:20:19]I think that's roughly there. It's not not going to be too good. That is it.

[00:20:23]Let's just try to do it a bit lower.

[00:20:25]something like that. So that's supposed to be the line y is equal to 7. I apologize if it's a bit off. So um where is the function greater than that? Well, it's it's going to be uh this region here, right? And it's going to be this region here, right? So basically what we're going to try to find is the x value the x values from here to here.

[00:20:47]That's going to be one region. And then the x value from here onwards, which is going to be another region. Now unfortunately I can't really do it. As you can see, using a tablet and on the screen, it's going to be stupidly inaccurate. So, um I'm sorry, but I'm just going to cheat and use Desmos because that kind of gives me the exact values. So, I get uh -1.29 and 0.22. So, this is uh -1.29 and what's that one uh what's it say?

[00:21:17]Negative beg you pardon 0.22.

[00:21:22]Okay, so let's just undo that. So0.22.

[00:21:25]So those two x coordinates there. And the other x coordinate is 3.51.

[00:21:34]So 3.51.

[00:21:36]So um my guess is that answer will probably be something of the form, right? X is um between -1.3 and uh0.2 probably. That' be my guess. and uh x is greater than 3.5. So what what I'm saying is you know they obviously can't have such a high degree of accuracy.

[00:22:00]It'll be rounded up to some extent and there'll be some latitude right it'll be like negative 1.4 to negative 1.2 or something like that. But that's basically how you do it right you draw the line you identify the regions you read down on your graph right what gives you the the x values that you need and then you just state those x values as two separate inequalities.

[00:22:22]Okay. Then question eight, a uh six-digit number is to be formed from the digits 1 2 3 4 5 6 7 8 9. No digit may be used more than once. Find how many six-digit numbers can be formed uh in the case where there are no further restrictions. Okay. So firstly, are we using combinations or permutations?

[00:22:39]We're using permutations because permutations is when order matters when we are arranging items in a particular order. Combinations is when we are just choosing them. Uh so how do we know that? Well, because the number 321 for example is different from the number 1 2 3. Even though the digits are identical in each case, if I rearrange the digits, I end up with a different number. So the order of uh the digits matters.

[00:23:05]So basically what I'm doing is I am arranging six digits out of nine digits.

[00:23:10]So that's 9 p6. And so I got to type that into a calculator. So just give me a second. That's uh 60,480 according to my calculator. 60,480.

[00:23:24]Um okay. So then part two, the six-digit number includes the digit four. Well um let's approach this kind of two ways. So first way um we are going to choose the digits uh first before we arrange them.

[00:23:38]So I want to choose six digits and I'm going to going to arrange them. So how can I do that? Well, I've got to choose the digit four. So there's only one way that I can choose the digit four. That's kind of locked. One way of choosing it.

[00:23:49]After I've chosen that digit four, I've got to uh choose five more digits, right? And so how many ways can I choose those uh those five more digits? Well, there there are now eight digits to choose from because I can't have four.

[00:24:02]So that's going to be um 8 C5, right? So I've uh chosen the the digit four. I've chosen five remaining digits out of eight digits. So now what I want to do is I want to put those digits in order. So how can I put them in order?

[00:24:19]Well, how many ways can I arrange six items? Well, it's 6 P6 or 6 factorial.

[00:24:25]So in the end, I think what I should get is 8 C5 um times by 6 factorial. So what I'm going to do again, I'll type that into Desmos.

[00:24:34]So let me do that. So yeah, uh where is it?

[00:24:41]So 8 C5 um times by do beg you pardon times by 6 factorial.

[00:24:51]So 6 factorial which gives me 40,320.

[00:24:56]So 40,320.

[00:25:03]Okay. Well, um, what about another way of doing that question? Let's imagine we've got our six-digit number here. So, 1 2 3 4 5 6. Um, and we say, well, how many ways can we arrange the the four?

[00:25:16]Well, it could be placed in each of these spots, right? Right. So, you know, if we if we put the four here, we could put it in the next one, the next one, next one. Right. So, there are um six ways of arranging the four. And then what have I got to do? Well, I've I've got to take the remaining digits and put them in um these five remaining slots, right? So, I am arranging um five digits out of a possible eight digits. So, this guy here is going to be 8 p5. So, I think if I do 6 * 8 p5, I should get the same answer. So, let me just uh type that in. So, sorry, beg you pardon.

[00:25:56]Oh dear. uh 8 P5 and then I'm going to times that by six and I get 40,320 which I believe was the previous answer as well. Yes. So there are two ways of getting the same thing. Probably the second is the easier one to think of in this case but it's good to be able to do both.

[00:26:21]All right. Then part B. Um a class contains 10 girls and 12 boys. A group of six students who is to be selected from the class. There must be at least two girls and at least two boys. Find the number of different selections that can be made. Okay. Um well, this is going to be a combinations question because uh the order in which uh we have the students doesn't matter, right? So um if I chose Levi and Micah, that's the same thing as choosing Micah and Levi, right? If I switch the order, it's it's totally the same thing. Um so uh let's try to think through all the different options that we can have. Right? So we've got girls and we've got boys. So let's say we've got um two girls. Then how many boys would we need? We need four. Okay. What's another alternative?

[00:27:06]We could have three girls in which case we would need three boys. Or we could have four girls and then we'd have two boys. And then that's going to be it, right? because um we we we can't have uh fewer um boys than two because we've got to have at least uh two boys as well as two girls. So then the question is right how many ways can I choose two girls?

[00:27:26]Well uh there are 10 girls in total.

[00:27:30]So number of ways that I can choose two girls is 10 C2. Number ways I could choose three girls is 10 C3. Number ways I could choose four girls is is 10 C4.

[00:27:40]And uh then I got to say stuff. Well, how many ways could I choose four boys uh out of 12? Well, that would be uh 12 C4 and three boys would be uh 12 C3 and two boys would be 12 C2. And we got to understand, right? What what's happening in in each of these cases, right? So for each case, we want two girls and three uh four boys. Three girls and three boys, four girls and two boys. So whenever we've got an and we need to multiply because we want this and this together and uh we then have uh three separate cases right we've either got two girls and two boys or three girls and three boys or four girls and two boys and whenever we've got an or we add. So essentially what we've got is we've got um 10 C2 * 12C4 plus 10 C3 * 12C3 plus this times by that. So now what we need to do is just need to um enter all of these into our calculator. And you know I'm I'm going to pretend that we don't have a calculator. So this is 10 factorial over 2 factorial 8 factorial times by 12 factorial over 4 factorial 8 factorial.

[00:28:54]And the same thing over here, right? 10 factorial over 3 factorial 7 factorial times by 12 factorial over 3 factorial 9 factorial and uh 10 factorial over 4 factorial uh 6 factorial uh times by 12 factorial over 2 factorial 10 factorial. Right? So we use the formula for factorial that we're given and things will cancel off because uh 10 factorial is 10 * 9 * 8 factorial right so I can actually rewrite this guy as uh 10 * 9 / 2 because the 8 factorial cancels off same thing with 12 factorial right that's 12 * 11 * 10 * 9 and then the 8 factorial cancels off 4 factorial is 4 * 3 * 2 * on this guy is going to be 10 * 9 * 8 and the 7 factorial will cancel off 3 * 2 * 1 like that 12 factorial. So 12 * 11 * 10 the 9 factorial will cancel off over 3 * 2 * 1 and this guy is 10 * 9 * 8 * 7 over 4 * 3 * 2 * 1 and this guy is 12 * 11. The 10 factorial cancels off divided by two. So I should be able to do even more cancelling, right? Because uh four times by 3 is is 12. Um you know I can divide by by two that becomes five.

[00:30:26]Divide by two again that becomes 5.

[00:30:29]Let's have a look over here. So I can um say 3 * 2 * 3 * 2 * 2 is 12. So that cancels off and I can divide by 3 again.

[00:30:41]And that cancels off there.

[00:30:44]Uh 4 * 3 is 12. So that cancels with that. Uh 2 * 2 is four. So cancel cancel and that becomes two. So in the end what does this become? This becomes 5 * 9 * 11 * 5 * 9.

[00:31:02]And this is going to be uh 10 * 3 * 8 * 11 * 10.

[00:31:10]And this thing is 10 times by 9* 2* by 7 time by 11. And of course you know those things are are not easy to do in your head. But you know if you lost your calculator you you could do it um with pen and paper and kind of simplify it.

[00:31:26]Now at this stage because I've I've essentially demonstrated the trick of how to manipulate the um factorial notation. I am just going to be lazy and type into the calculator at this point.

[00:31:35]So 5 * 9 * 11 uh * 5 * 9 that's 22,275.

[00:31:44]So 2 275.

[00:31:48]Okay, let's do the next guy. So 10 * 3 * 8 * 11 * 10 is 26,400.

[00:32:02]And the last guy uh 10 * 9 * 2 * 7 * 11 is 13,860.

[00:32:13]So then we got to add up all of these.

[00:32:16]So 22,275 plus 26,400 plus 13,860.

[00:32:26]So I get that is 62,535.

[00:32:30]So, I think that's the final answer there. Now, of course, it is it's perfectly plausible, guys, that I've made a mistake somewhere here. Um, and so some of you say, you know, the the final answer is wrong when the mark scheme comes out. Look, I'm doing this in a bit of a rush to get it out. So, I'm really sorry if there's a mistake, but I'm also am trying to um I I basically want you guys to understand that in the absence of a calculator, you can make progress in in this, right? So you don't have to lose all your marks because your calculator goes bust. Uh you can still get the marks. So keep going. Uh but my apologies if it's not quite right.

[00:33:07]Right. Then uh question five, which is a radiance question. So the diagram shows the design for a badge. ABC is an equilateral triangle with side of X. Uh the arcs B, C, C, A, and A are parts of the circles with centers A, B, and C respectively. So, so in other words, right, you've you've got an arc which kind of looks like that. Yeah. So, that that guy there, uh, write down the size of the angle ABC in radians. Well, if this is an equilateral triangle, then um that angle there is going to be<unk> by 3, right? 60° by 3 in radians. And the area of the badge is um 2 *<unk> -<unk>3 cm 2. Find x. So, um, how do we get the area of the badge? Well, um, we got this sector here, right?

[00:33:53]Like that.

[00:33:55]So, so we kind kind of got this guy like that. But, um, we could also have this sector here. So, I'll do it in green like which is going to be that guy, that guy, and that.

[00:34:08]And then we could have the, uh, the last remaining sector, which I'll try to do in red, which would be like that guy, that guy, and then that guy. and and basically right all three of these sectors overlap and they're all identical. So if I added um the the green sector and the blue sector and the red sector all together, then what would I have? I would have the whole thing, but then I would have counted this central piece um three times, right? In other words, two more than I need, right? So that the kind of the triangle in the middle is has been counted three times, which is which is an extra two than we need. So how can we find the area of each sector? So the area of each sector is equal to a half r 2 theta which is the the formula right. So that's going to be a half x^ 2 * so half of x^ 2 * p<unk> upon 3 which is the same thing as um<unk> upon 6 x^ 2.

[00:35:10]That's that. And then um how can I find the the area of the of the triangle in the middle? So the area of the uh equilateral triangle is equal to a half b c sin a right. So in this particular case it's going to be a half of x * x * the s of pi upon 3 and sin pi upon 3 is roo<unk>3 upon 2. So what we end up with is <unk>3 upon 4 um times by x^ 2 and uh what did what did I say earlier? I said the the area of the badge rate what's it going to be equal to? Well it's it's going to be three lots of those sectors. So it's going to be 3 * um pi upon 6x^2. But then we've got to subtract off the um area of this triangle twice, right? um because we we've counted it more times than we need. So we've got to subtract uh two lots of um 3 u roo<unk>3 upon 4x^ 2. So let's try to simplify this now. Uh what do we get? we get<unk> upon 2x^2 -<unk> 3 upon 2x^ 2 which I believe we can um factoriize as being a half x^2 bracket um<unk> -<unk>3 well we're told in the question aren't we that uh the area is actually equal to uh 2 *<unk> -<unk>3 so if we just uh cancel off the pi minus roo<unk>3 that's gone and that's gone We've got 2 is equal to a half of x^2 which means that x^2 is equal to 4 which means that x is plus or minus 2 but of course um we can't have a negative length so x is equal to + 2 only. Okay. So uh that's the answer to that.

[00:37:05]Okay. Question six which is a trigonometric graph question and it's it's a tang graph which is a little bit more unusual. Okay. Okay, so the diagram shows part of the graph of y is equal to 1 + tan a half x where x is measured in degrees. We want to write down the period of 1 + tan a half x and write down the equation the asmtote a which is this asmtote here. Okay. Um so I want to just make a quick point right so let's say we've got a graph which is of the form y is equal to a tan uh bx + c right what what let me just write that a bit neater yeah plus c what does this b represent uh this represents the number of waves in 180° if we're using degrees or pi radians if we're using radians right so it's really important to get this understanding this definition, right?

[00:37:58]Okay. Um, this number does not represent the period. It represents the number of waves in 180°. And, uh, distinct from the s and cos graphs, right? Where the number represents the number of waves in 360°, uh, tan wave is unique. It's the number of waves in 180. So basically in this question, what we're told is that there is half a wave in 180°, right? So there is half a wave in 180°. Well, if half a wave is in 180°, right? Then what it means is that one wave is going to be 360° and and that's the period of the wave, right? So the period t is equal to 360°, right? So that that's the key thing. So what we actually learn here is that this one wave that guy has got a period of 360°, right? So the size of this guy is going to be 360° right from here to here. Well, if that's the case, right?

[00:38:57]If that's going to be 360°, then u what's going to be the location of these asmmptotes? Well, um this asmtote here will be plus 180 and this asmtote here will be minus 180, right?

[00:39:09]So this guy here is 180 and this guy here is minus 180. Okay? Okay. Well, then if we want to go to the next asmtote along, right, we're moving one wave across um uh to uh to the right.

[00:39:22]So, how big is one wave? Well, it's another 360. So, we got to do 180° plus um 360°.

[00:39:32]So, what's that going to be? Uh that's going to be 540. So, this guy here is going to be 540°.

[00:39:40]Okay. So um the essentially the the key thing for you to understand is um what does that number the half mean and it it what it means is the number of waves in 180. So there's half a wave in 180 which means one wave is 360 and once you know that right then you can piece together um all of this uh in information. Okay I hope that's helpful and clear.

[00:40:05]All right then. Um, part B of this question, another trigonometric graph.

[00:40:09]Uh, on the axis below, sketch the graph of y = 3 cos 2x + 1 for x between 180 and positive 180. Okay. Well, um, we need to understand if we've got y = a cos bx + c. What do each of those letters represent? Uh well, a represents the amplitude and the amplitude is the uh uh the distance from the center line up to the peak or the center line down to the trough. U what does this uh number B represent? That is the number of waves in 360°. Right? So again that is not the period. That is the absolute classic mistake. Right? So the period is not two. Instead, what we know in this question is that there are two waves in 360.

[00:41:00]And lastly, what does this uh number C represent? This represents the shift uh vertically in the y direction. Okay, so the best way to draw the graph, I think, is to start with C.

[00:41:12]So the whole graph has been shifted up by one, right? That's that's the kind of the the key thing. So the graph is shifted up by one. So what do we got here? Uh that's 1 2 3 4 five six, isn't it? Okay. So that's one there. So this is going to be the center line of our graph like that. Okay. Then secondly the amplitude um is three, right? So it means the graph is going to go up by three and it's going to go down by three from the center line at one. So if I start at one and I go up by three, then I go to four here. And if I go down by three, then I'm going to go down here to -2. Right? So I got to indicate and and write on right the the numbers. So it goes up to there. It's going to go down to there. Okay. Then um lastly, uh what else do we know? We we know that there are two waves in 360 degrees, right?

[00:42:04]Because this this letter, this number here is two. So two waves in 360. So two waves in 360. Well, if that's the case, um how big is one wave? Well, one wave is uh going to be 180, isn't it? So we're going to have one wave and 180°.

[00:42:22]So if we've got a cosine graph, right, the cosine graph always starts at the peak. And what do we know? We know that in 180°, in other words, from here to here, it's got to complete one full wave. Okay. Now, at this point, right, the if you want to draw a nice neat cosine graph or a sine graph, basically you just need to divide it up into quarters.

[00:42:42]In in the first quarter, right, the cosine graph is going to go from the peak to the center line. And then the next quarter it's going to go from the center line down to the bottom. The next quarter it's going to go back up to the center line. And then uh in the last quarter it's going to go up to the to the peak again. All right. So a lot of people struggle to draw neat uh graphs.

[00:43:03]And the reason is because they they haven't um divided up the wave properly into quarters. Right. Okay. I'm drawing it so badly. I'm so sorry about that.

[00:43:13]Okay. Let me just try to make a bit more neat.

[00:43:17]Okay. like that. There you go. So that's that's one wave there. So hopefully that's all right. And of course the same thing is going to happen on the other side. So down to the center, down to the bottom, back to the center, and then back to the peak. So let me just try to draw that one. Oh dear.

[00:43:32]Okay. Like that.

[00:43:34]Like that.

[00:43:38]Like that.

[00:43:41]And like that. Okay guys. So that's what your wave should look like. And please try to draw it as a nice smooth curve.

[00:43:48]Right? So it's got the curve. You can't do saw tooth. So like straight line, straight line, straight line that cannot. But a a reasonably smooth curve that looks, you know, roughly symmetrical and you should be able to gain all of the marks.

[00:44:01]All right. Then a ship is sailing with a speed of v km/h. The sailing cost per hour uh is given by this uh function here. uh find the speed that makes C a minimum and justify that this value of C is a minimum. Okay. So if we want to find the minimum of a function, we can differentiate and uh set uh our our differential equal to zero. Right?

[00:44:26]Because the the minimum is a stationary point. It's a place where the gradient is zero, right? So imagine we got a a curve that looks something like that. I don't know like that. At this minimum point, the gradient is perfectly flat.

[00:44:39]Um so you know dy dx is equal to zero.

[00:44:41]So basically that's what we're trying to find. We're trying to find where the gradient is zero. So we're going to differentiate this curve. So C is equal to V ^ 2 + 3,000 V ^ -1 + 100. Right? So I write it as indices form. And we're going to differentiate this with respect to V. So D c DV and that will give us 2 V minus 3,000 V to the -2. And if I differentiate uh 100 I get zero. And we can say that we've got a stationary point when um dc dv is equal to zero. So 0 is going to equal to 2v minus 3,000 over v ^ 2. And so if we just uh rearrange it a little bit, right? 2 v is equal to 3000 over v ^2. So let's multiply by V ^2 that will give us V cub / 2 that will be 1500.

[00:45:46]Okay, like that. And then um V will be equal to the cube root of 1500. My apologies, I don't have a calculator on me at the moment, but you're probably going to have to actually compute that.

[00:45:58]And uh what does the question actually ask? So find the find the speed that makes C a minimum and justify that it is a minimum. Okay, so how how can we justify that it's a minimum? Well, we can use the second derivative test. So, we differentiate a second time. So, d2 c dv^ 2. So, if I do that, I get 2 plus uh 6,000 uh v ^ -3, which is the same thing as 2 + 6,000 uh over v cubed. And what's the second derivative test? We'll basically write if the second derivative is positive at our stationary point, then we got a minimum. If it's negative at our stationary point, we've got a maximum.

[00:46:43]Um, and so what we got to do is we just got to take our value of v and we just got to substitute it in. So v is the cube root of 1500, right? So uh when v is equal to the cube root of 1500 then uh d2 c dv ^ 2 is going to equal to 2 + 6,000 over 1500 um which I believe is going to be four isn't it? So we've got 2 + 4 which is six. Now it doesn't actually matter what the number is. The key thing is it is um positive right? So d2 c dv ^ 2 is greater than zero. So therefore it is a minimum at that particular point. Okay.

[00:47:26]So that's that. Now I'll I'll just open up a calculator. I can use Desmos, can't I? So let's just open up Desmos calculator and let's find the cube root of 1500. So 1500 to the power of a3 gives me 11.4.

[00:47:46]Okay, so that's what Desol says 11.4. So that should be your answer.

[00:47:53]Okay, then part B of this question, um, hence find the minimum sailing cost for a journey of 150 km. Okay, we need to do a little bit of thinking here. So um, we we found the velocity that minimizes, right, which is the the cube root of 1500. And so probably what we want to do is we want to find um this cost per hour that that corresponds to and we'll do that by substituting in that value of v.

[00:48:20]So uh let me just write down here right.

[00:48:22]So c is equal to v ^ 2 + 3000 over v uh plus what was it plus 100.

[00:48:36]So when V is equal to the cube root of 1500, make sure I actually have got that right. Yeah, cube root of 1500.

[00:48:51]Then the cost per hour which is what we call C is going to be equal to well we substitute that value of V in right. So this is going to be 1,500 to the^ of a3 to the^ of 2. So it's to the^ of 2/3 um plus 3,00 times by um 1,500 to the^ of a3, right? Because the negative means one over and the third means we've got the cube root. Um plus 100.

[00:49:25]So so that's going to be the cost per hour. Um but the thing is we want to ask well how many hours we traveling, right?

[00:49:31]So, how many hours?

[00:49:34]How many hours?

[00:49:37]So, we can just use speed is equal to distance over time, right? Let's not complicate it too much. So, speed is equal to distance over time. Sorry about the terrible handwriting. So, um time is going to be equal to the distance divided by the speed.

[00:49:57]So our distance is 150 km.

[00:50:02]So 150 km divided by the speed which is the cube root of um500 like that. So that's that's going to give us the amount of time in hours that we've been traveling. And so of course if the cost per hour is uh is this guy here then we just got to multiply it up by um by this guy here. Right? So we we got to multiply it up. Um so how can I write that down? Well, let's let's just go across. So uh total cost is going to equal to this guy here. So 1,500 ^ 2/3 plus um 3,000 actually I can write 3,000 as 2 * 1500^ 1 * um500 to the^3 um plus 100 and I got to times all of that guy by uh by this. So 150 over the cube root of 1500. So so times by 150 times by 1500 to power negative a3 right because the negative means one divided by and the third is the is the cube root. Well I'm doing this kind of you know because I I I'm showing that it can sort of be done without a calculator. So I'm going to simplify a little bit. That's 1,500 ^ 2/3 plus 2 * well I've got 1500^ 1 * 1500^ of a3. So I add up those powers. That's going to give me um 1,500 to the^ of 2/3, right?

[00:51:45]Uh plus 100 uh times by 150 uh times by um 1500^ of minus a3.

[00:51:56]And uh of course what I've got here is I've got 1500^ 2/3 plus two lots of 1500^ 2/3. I can add them up, right? And get three lots, right? So this becomes three lots of 1500 ^ 2/3 + 100 times by 150 times by 1500 to power minus a third. Right? So I think I've probably simplified it as much as I can. I mean you know what I could do is I suppose I could multiply out and and what have you, but I'm I'm not going to bother. So, um, at this stage, I think what I'm going to do is I'm just going to pause and I'm going to type everything into my Desmos calculator and see what it spits out.

[00:52:37]All right, then. Um, if I do that on Desmos, I get 6,461.58.

[00:52:45]Uh, I think you should probably give it to three significant figures, I think.

[00:52:47]So, I'm going to write down 6,460.

[00:52:50]So, 6,460 uh to three significant figures. So I I don't know exactly what the mark scheme wants because the mark scheme is not out yet. Um but that's the way I would approach it. Now um just a final point guys right as I said I haven't got the mark scheme in front of me so it's possible that I've made a um arithmetic error and if if so then uh please forgive me um uh but I want to try to get this out before the May June exam.

[00:53:15]So the method is essentially correct but yeah just double check see that I haven't made any small slips here and there and apologies if that's the case.

[00:53:24]All right. Uh, question eight. A circle has got center of -20 and a radius of 5.

[00:53:30]A line has equation y is equal to 2x + k where k is a constant. The line does not meet the circle. Find the set of possible values of k. Okay. Well, I think it's important we write down the equation of the circle. So, if the center is um -20, then what we've got is we've got x - -2. So, x + 2^ 2 + y - 0 all squared. So, it's just just plus y^ 2, isn't it? um is equal to the radius squared. So that's equal to 5^ 2. And uh let's just pretend for the moment that this does uh intersect with the line y= 2x + k. And what we'd like to do is we'd like to find the points of intersection.

[00:54:06]How would we do that? Well, what we would do is we would substitute the line into the equation of the circle. So um let's got the circle there. We've got the line there. So what we're going to do is we are going to substitute um equation two into equation one. Now if we do that then uh what we going to get? Well I I'll expand the x + 2 right?

[00:54:29]So that's x^2 + um 4x + 4 um plus the 2x + k all 2 is equal to 5 2. So that's equal to 25. And I'll expand the other brackets. So I get x^2 + 4x + 4 + 4x^2.

[00:54:51]2x * by k is 2 kx, but I've got two of them. So it's 4 kx + k ^ 2 is equal to 25. So what I want to do is I want to group together all of the x squ terms. Group together all of the x terms and group together all of the constant terms. Right? So the x squ terms is that guy and that guy. The x terms is that guy and that guy. And the um the constant terms is that guy, that guy, and that guy. So let's uh get it.

[00:55:18]So x^2 + 4x^2 is 5 x^2. And how many lots of x we got? We've got four lots of x plus 4k lots of x. So I'm going to write that as um 4 + 4 k lots of x. And then uh what about the remaining constant terms? Well, we got um 4 + k^ 2. And I've got to subtract 25 from both sides. So that's - 255 is equal to 0.

[00:55:44]But of course it's more convenient, I think, to write that as k^ 2 - 21. So I'm just going to rewrite this guy as k^2 - 21.

[00:55:55]Okay. So let's write that k ^ 2 - 21 there.

[00:56:01]Right. My apologies. I had to pause recording just for a second. Um, so we end up with this quadratic. And uh what's this quadratic actually represent? Well, if we solved it, the solutions to this quadratic would tell us the intersection points uh between the circle and the and the straight line. Um, but we we're told that the line does not meet the circle, right?

[00:56:23]So, there are no intersection points.

[00:56:25]What does that mean? It means that this quadratic here should have no solutions, right? We want no solutions if the line and circle do not intersect, right? So if they don't intersect.

[00:56:42]Now when does a quadratic not have solutions? Um when the discriminant is is negative. Right? So we need to just recap this uh principle very quickly. So what's the discriminant? The discriminant is b ^2 minus 4 a c. And uh basically what this uh value tells us is the number of solutions to the quadratic. Right? So number piggy pun number of solutions.

[00:57:09]So if the discriminant is positive what it means is that we have got two distinct solutions.

[00:57:16]If the discriminant is zero it means we've got one repeated solution. So that would be the case where the line is tangent to the curve. But if the discriminant is negative then what it means is that there are no real solutions. Right? So no real solutions.

[00:57:30]And that means that the uh the line does not cut the curve. Right? So if the discriminant was positive, the line would cut the cut the uh circle in two distinct places. If the discriminant was zero, then it would cut it at one repeated place. Right? So it' be a tangent. And if it was negative, it doesn't cut. So we want to find when b ^2 - 4 a c is negative. That's what we've been asked to find. B ^ 2 - 4 a c to be negative.

[00:57:58]Okay.

[00:58:01]So then uh what's our b? Well, our b is 4 + 4k. But I think I'm going to try to be a little bit more intelligent. I'm going to write b as 4 * 1 + k like that.

[00:58:13]So that's that squared - 4 * a which is 5 * c which is k^ 2 - 1 is less than zero. So I want to try to solve this particular um uh inequality here. So how can I do that? Well, um 4 * 4 or 4 2 is 16. So I've got 16 * 1 + k all squared. Uh minus 20 lots of k^ 2 - 1 is less than zero. Okay. So um I think I can divide through by four, can't I?

[00:58:45]So if I divide through by four, that guy becomes four. I divide through by four, that guy becomes five. And I'm going to expand the brackets. So I've got 4 bracket 1 + 2 k + k^ 2 - 5 lots of k^ 2 - 1. Well, let's expand that. So it's - 5 k^ 2 um + 5.

[00:59:06]Okay, so let's uh expand this. So I've got four lots of k^ 2 plus 8 lots of k + 4 - 5 k^ 2 + 5 is less than zero. Oh, thank you pun. got that wrong way around. Less than zero.

[00:59:24]And um let's try to simplify this. So 4k^ 2 - 5k 2 is minus k^ 2. Uh I've just got 8k there. And 4 + 5 is 9, isn't it?

[00:59:37]Okay. So I think that's right. And um you know, I'm going to multiply everything through by -1, which will give me k^ 2 - 8 k - 9. And of course, if I multiply by negative, then I need to switch the sign of the inequality. So I end up with k^ 2 - 8 k - 9 is greater than 0. So I've got a quadratic inequality. How can I solve that? Well, let's factoriize. So I've got k - 9, I think, times by k + 1 is greater than zero. So now I'm just going to try to think about this. Um the best way to understand it is to draw a quick sketch. So I'm going to draw a sketch of my quadratic.

[01:00:15]This particular quadratic is going to have roots at negative - 1 and roots at positive 9 like that. And so the quadratic is is going to look like that.

[01:00:24]We Okay, sorry about the slightly bad drawing. Let's kind of do it like that.

[01:00:30]There you go. That's the quadratic. And we want this quadratic to be greater than zero. So when is it greater than zero? Well, uh this part here and this part here, right? Everywhere above the x axis or the k axis. So when do we um uh for what values of K is that? Well, either K is greater than 9 or K is less than -1. So we've we've got the condition basically K is either greater than 9 or K is less than -1. So I'm just going to go back and reread the question again. Make sure that I've under stood it correctly and I've answered. Yeah.

[01:01:04]Find the set of possible values of K. So there are two sets of possible values.

[01:01:09]Either k is greater than 9 or k is less than -1.

[01:01:16]All right. Uh question nine which is a trigonometric question. We start with an identity. Show that tan theta plus c theta plus 2 sec theta can be written as this guy here. Okay. So with an identity right what we're trying to do is we're not trying to solve. We're trying to show that this thing here um is the same as that thing. So I'm going to write we've got the left hand side is equal to tan theta plus co theta plus 2 * sec theta. Now what's tan theta the same as? It's the same as sin theta over cos theta.

[01:01:53]And what's cot the same as? Well cot is 1 / tan. So it's going to be co over s.

[01:01:59]Right? So in other words it's the it's the inverse of this guy here. Right? And so that's tan. Cot is 1 / tan. So basically, we're flipping this fraction.

[01:02:08]And then lastly, what's se? Sec is 1 / co. So we've got 2 * 1 / cos theta.

[01:02:17]So we we've got three fractions here and we want to combine them, which means we've got to give them the same denominator. So for this guy, I got to times top and bottom by sign. This guy top and bottom by cos and this guy top and bottom by sign. So if I do that, then what will I get? I will get sin^ 2 theta over sin theta cos theta.

[01:02:38]This one will be cos^ 2 theta over sin theta cos theta. Sorry about the handwriting. And this one will be uh 2 sin theta over sin theta cos theta.

[01:02:53]And of course then um if I combine these then what I end up with is sin^ 2 theta plus cos^ 2 theta plus 2 sin theta all divided through by sin theta cos theta and of course sin^ 2 + cos^ 2 right this guy here is equal to 1. So what I'm going to do is just going to replace it with one. So I get 1 + 2 sin theta over sin theta cos theta. So I end up with that. Okay. Now um just a a quick point.

[01:03:32]How did I approach this? Well, generally speaking, a a good idea is to replace like tan and cot and se uh with signs and with causes. That's usually a good approach to these type of questions. And in particular in this question, right, we know that what we want has just got signs and causes anyway, right? So we want to take tan, take cot, take second and completely replace it with signs and causes. Second thing is uh usually we or very often rather we have to kind of combine fractions with the same denominator. And the last point I would make um is that you you don't want to skip steps, right? So I would be careful about skipping from here to here, right?

[01:04:10]In other words, missing out this intermediate step. Um because the examiners are kind of quite sensitive to the idea that students are trying to hoodwink them, trick them, uh make it look as if they know how to do the identity, whereas actually they're they don't really. They're just kind of using the answer. Um so you want to be very clear about what you're doing. So write this out nice and clearly um so that it's very plain to the examiner that that you do understand the identities and you you know what you're doing.

[01:04:38]All right. Then um part B um is hence solve the equation tan 2x + c 2x + 2 se 2x is equal to zero. Well we we can um uh replace this guy here with the identity that we have just shown in part a right which is why it says hence. So we can replace this with 1 + 2 sin 2x over sin 2x cos 2x. Right? So we're going to write this as uh 1 + 2 sin 2x / um sin 2x cos 2x is equal to zero. And uh after that point right how are we going to solve this? Well, let's multiply um both sides of the equation by this denominator here, right? So, if I multiply the left hand side by sin 2x cos 2x in order to get rid of the fraction, then of course I've got to do the same thing on the right hand side. I've got to multiply that by sin 2x cos 2x. Well, what's the outcome of that? Well, on the left hand side, um it cancels off and I'm left with 1 + 2 sin 2x. And on the right hand side, what am I left with? Well, I'm left with zero, right? Because anything multiplied by 0 is zero. Well, then with a little bit of rearranging, I've got that sin 2x is going to be equal to a half. And at this point, right, it's just a simple uh trigonometric equation.

[01:06:05]Now, how do I do this? Well, um, we got to use the unit circle.

[01:06:10]So, I'm going to draw a unit circle like that. We Okay. And like this.

[01:06:17]Okay, good enough. Okay. So, here's my unit circle. So, right, unit means it's got a radius of one. So, 1 one.

[01:06:26]And if you've seen me do this on other videos, basically what what I try to highlight all the time is um imagine you got a point that's on the circle like that. And this point makes an angle of theta with the x-axis and you take that point and you drag it all the way around the circle. Well, as you drag the point around the circle, the x and the ycoordinate is always changing. And the x and the y coordinate is dependent upon this angle of theta, right? So when the angle is 90, the x coordinate is 0 and the y coordinate is 1. When the angle is 180, the x coordinate is ne1, the y coordinate is 0. Well, essentially, right, what is s and cosine? Cosine tells you the x coordinate on this circle for any angle. and sign tells you the ycoordinate on this circle for any angle. So what does that mean? It means that when we are solving the equation um sin 2x is equal to negative half. What we're really doing is we are finding um the um angles on the unit circle when the ycoordinate is equal to a half.

[01:07:28]Right? Because because sign means the ycoordinate. So we're trying to find where the ycoordinate is negative half which means we doing like this guy.

[01:07:36]Right. So, this is where the ycoordinate is equal to half. And that means that we've got um these two angles here.

[01:07:42]We've either got that guy or we got this guy. Well, if we um type the number into the calculator, right, that angle there will be -30. Now, are we are we using degrees or radians? We're using degrees.

[01:07:54]All right. So, so this angle here will be -30, which means that this angle here, right, will be 150, right? Why? Why is that?

[01:08:03]Well, because that angle there is is going to be the same size as that one.

[01:08:06]So, if the size of that is 30, the size of this is 30. This angle here is 180 and I go back 180 to 150. And the reason we've got a negative, right, is because positive angles go anticlockwise and negative angles go clockwise. So, that's that's kind of the reason. So, what we end up with is 2x, right? Because that's our angle is either -30° or it's 150. But there's a catch. The catch is if I take this point here, right, and I add 360°, then I land back on the same point. And if I take this point here and I add 360°, I land back on the same point. So, so actually what we're saying is the the angle could be -30°, but it could be -30° plus 360 or plus another 360 or plus another 360 or plus another 360. So I'm going to write that as -30° plus k lots of 360 where k is like any integer and 150 plus k lots of 360 as well. And what this means is that x is uh found by dividing everything by two. So 15° plus multiples of 180 or -75° plus multiples of uh of 180 180°. So what we want is we want x from 180 to positive 180. Okay. So um let's let's start with this guy. So we've got5. Well if I subtract 180 I'm going to be out of bounds. So uh -15 is a valid solution.

[01:09:40]If I add 180 to that, then uh what will I get? I'll get 165. And that's another valid solution. If I add 180 to that, well, that's going to be out of bounds.

[01:09:50]So, I don't care about that.

[01:09:52]So, then I move to -75. That's a valid solution. And if I add 180 to that, I'm going to get 105°.

[01:10:00]And that's a valid solution as well.

[01:10:03]Okay. Uh if I add another 180 to that, then obviously that guy is going to be out of bounds. Okay? So, I should end up with four solutions.

[01:10:10]-1575 uh negative uh 165 sorry about that I think the whiteboard failed and uh 105 and if you're not sure about that you can type the equation into desmos and you should be able to get solution by looking at the graph okay part C another trigonometric equation um solve the equation 3 sec x -1 -1 is equal to tan^ 2 x -1 we got this x -1 here which makes things a little bit awkward. So what we're going to do is we're going to use a substitution. Let u= x -1, which means that we're now trying to solve 3 sec u - 1 is equal to tan^ 2 u. And um we're going to need to use our identities in order to do this. So if you look at the front of your exam paper, you should have the following identities. Um cos^ 2 theta + sin^ 2 theta is equal to 1. And if we took that identity and we divided it all by cos^ 2, then we would get 1 + tan^ 2 theta is equal to sec^ 2 theta.

[01:11:19]And that's the identity that we're we're going to use here today. So um I'm going to replace the tan^ 2 with se^ 2 -1. So I'm going to take this equation 3 sec u - one and I'm going to replace tan^ 2 with se^ 2 uh u - one. So what I should end up with is quadratic isn't it?

[01:11:42]Right. So 0 is equal to sec^ 2 u uh minus 3 sec u. And if I factoriize that, I'm going to get se u bracket se u minus 3. So that's going to give us um two possible outcomes, right? Either sec u is equal to zero. So se u is equal to zero. Or alternatively um se u is equal to 3. Well, um what does se represent?

[01:12:16]SE represents 1 divided by cos. Right?

[01:12:18]If you remember the third letter 1 2 3 third letter is C. So it's 1 divided by cos. So 1 divided by cos uh u is equal to to zero. Well uh there's no case where where where that's true, right? So so we we kind of reject that.

[01:12:37]It's impossible for se to be equal to zero. Um, but we do have 1 upon cos u is equal to 3, which means if we rearrange it, we've got cos u is equal to a 3. So we're trying to solve cos u is equal to a 3. Well, um if you watched my part b to this, I said let's use the unit circle. So again, let's draw out the unit circle. Circle of radius 1. And um what does co represent? It represents the x coordinates on this unit circle.

[01:13:07]So if we're trying to find cos u is equal to a third what we're actually trying to do is we are trying to find where the x coordinate on this circle is equal to a3 right so this is where the x coordinate is equal to a third and so we're either interested in uh this point here which is that guy or we're interested in uh this point here which is that guy and you can use your calculator then to find that angle and of course um that angle there will be the same as this angle but only negative now I I at the moment do not have a calculator. So I'm just going to quickly shift to Desmos and use their calculator. Just give me one second.

[01:13:42]Okay, so Desmos tells me that cos inverse of a third is 1.23 to two decimal places, right? So we've got u is equal to 1.23.

[01:13:53]Um but it's also the case, right, that um we could be negative 1.23 cuz uh that would be this angle here, right? So this is positive 1.23 and this is negative 1.23. Why? because these points are reflections of each other.

[01:14:07]Um, but we've also got to consider, right, that we could be at this point, but then if we add 2 pi radians, in other words, if we go around the circle one time, then we end up back on the same point. And the same is true with this guy. If I add 2 pi radians, I end up back on the same point. So, the solutions are not just 1.23, but it's 1.23 plus multiples of 2 pi for for both instances. Okay, so we're almost there.

[01:14:32]Um, but the thing we need to remember is we don't want solutions for u. What we actually want is solutions for x, isn't it? And um, u is equal to x - one. So I'm just going to change this now. So um, u is x -1. So what's x going to be equal to? Well, I just add one to both of these, right? So it's 2.23 plus multiples of 2 pi or um, uh,0.23.

[01:15:01]2 3 plus multiples of 2 pi. So now let's try to finalize and get all of our answers. So uh where do we want it from?

[01:15:09]We want it from -1 uh to positive3. Okay. So let let let's start um so we will start with 2.23 right? So that is a possible solution 2.23.

[01:15:24]But if I subtract 2 pi right so that you know pi is roughly three. So 2 pi is going to be six. So if I subtract six from this, it's going to be like minus4 something. So that's outside of the outside the boundary that we want. And if I add 2 pi to that, it's going to be like eight. So again, that's outside the boundary that we want. So let's ignore that. Then the next one is0.23.

[01:15:47]Well, um, if I subtract six from that, that's going to be out of bounds. And if I add six to that, it's going to be like, yeah, 6 point something. Um, and again, is that out of bounds? Yeah, I think it's out of bounds. So my my assumption is that there are only two solutions 2.23 and negative 0.23.

[01:16:06]And uh I've actually plotted the curve on my Desmos graph. So I think I can probably see if it's right. Yeah, negative0.23 and uh 2.23 and those are the only solutions between 0 and five. Okay. So just two solutions there. Okay. All right. I hope that that was helpful.

[01:16:25]Okay. Then question 10. Um in this question time is in seconds and displacement is in meters. For t is greater than zero a particle moves on the positive x ais um at the time t the acceleration of the particle is 2t - 9.

[01:16:39]At time t is equal to 7 the velocity is 4. Find the times when the velocity of the particle is zero. Okay. Now in order to understand this uh basically we need to understand the relationship between displacement and uh velocity and acceleration. So let's write those down.

[01:16:53]So displacement we'll call s. Uh velocity is v and acceleration is a. Now what is velocity right? Velocity is the rate of change of your displacement right? It's how much your displacement is changing over time. So we can write that in calculus language as dst.

[01:17:13]And uh similarly with acceleration right acceleration is the rate of change of your velocity. So I can write that as dv dt. So what does this mean? It means that to go from displacement to velocity and from velocity to acceleration, what I need to do is I need to differentiate.

[01:17:29]And if that's the case, right, if I want to reverse the operation, then I'm going to need to integrate, right? So if I want to go from acceleration to velocity and velocity to displacement, I've got to reverse differentiation, which means I got to integrate. So in this particular question, right, what we're told is that acceleration is 2tus 9 and we want to find velocity. So velocity means we're going to integrate 2 t - 9 and if we do that we will get t ^ 2 - 9 t plus some constant c. Well we're told in the question right that when t is uh 7 the velocity is um is four. So let's write that down. When t is equal to 7 v is equal to 4. So let's uh substitute in 4 is equal to 49 - 9 * 7 which is uh 63 uh plus c.

[01:18:27]Oh my mental arithmetic's uh kind of rubbish today. Uh let's add 63 to both sides. So that's uh 67 is equal to 49 + c. And I've got to subtract 49 from both sides. So what's that going to be?

[01:18:41]It's 18, isn't it?

[01:18:47]Uh yeah 18 18. Okay. Um so v is equal to t ^2 - 9 t um + 18. Okie dokie. So we want to find when uh uh when the velocity of the particle is zero. So when is v equal to zero? So when v is equal to zero. So v is going to be equal to t ^ 2 - 9 t + 18. And does that factoriize? Um I think that's going to factoriize to be t - 3 times by t - 6. Right? Because we want two numbers that multiply together to give pos 18 but add up to give -9. So that means that uh t will either be equal to 3 or t will be equal to six. Right?

[01:19:38]Those will be the two values when the velocity is equal to zero.

[01:19:45]Okay. Then part B. Um at time t is 12, the displacement of the particle from the origin is 150. Find an expression for the displacement of the particle at time t. Find the distance traveled by the particle in the first 6 seconds.

[01:19:58]Okay. So um in order to find displacement, right, we need to integrate velocity. If we just look at the the diagram again, right? To go from velocity to displacement, we need to integrate. So velocity was t ^2 - 9 t + 18. So we're going to integrate uh t ^ 2 - 9 t + 18 dt. So if we do that, we're going to get a/3 of t cub uh - 9 upon 2 t ^2 + 18 t plus another constant which I'll call k. And we're told that when uh t is equal to 12, s is equal to 150. So when t is = 12, s is equal to 150. So let's sub substitute that in. So 150 is equal to a/3 * 12 cubed - 9 upon 2 * 12^ 2 um + 18 * 12 uh + k. Okay. Um well uh that doesn't look terribly fun, does it? But uh let's try to simplify. And you know we're going to try to simplify as if we don't have calculators. So we got a third of 12 * 12^ 2 uh - 9 upon 2 * 12 * 12 um + 18 * x 12 + k. Now why am I doing this? Because I'm going to try to simplify things if I can. Um so a third of 12 that's four and I can cancel out the two and make that into six, right? So that kind of makes my life a tiny bit better I guess. So I've got 4 * 12 which is 144 uh - 9 * 6 * 12 uh + 18 * 12 um + k is equal to 150 okay can I can I factoriize anything let's factoriize out 12 from this so 12 bracket um 4 * x 12 - 9 * 6 uh + 18 plus k and uh that may make things a tiny bit simpler. So that's 12 times by that's 48 minus uh 54 + 18 plus k.

[01:22:28]So that's going to be 12 * okay 48 - 54 that's min - 6 isn't it + 18 is going to be 12 uh plus k let me just quickly double check that uh 48 plus that that's going to be uh 66 uh yeah that's 12 okay so actually what we end up with here is 150 is equal to 144 um plus k so um k is going to be equal to So if we do that then the displacement right is so s is equal to a3 of t cubed uh what was it - 9 upon 2t ^ 2 uh + 18 t I think it was uh plus the constant k so plus six okay so that's the expression um for for for the displacement. So that's part one. Part two is find the distance traveled by the particle in the first 6 seconds. Now this is where we need to understand the the difference between um distance and displacement. So let me try to try to draw a diagram for you. So um here's like our uh our x-axis I guess.

[01:23:47]So here's our origin and let's kind of um imagine that this is like 1 2 3 4 whatever it is. Imagine the following, right? Imagine that I moved forward um three steps. One, two, three. And then I moved back two steps.

[01:24:05]One, two. My displacement, right, would be one because that's my um position relative to the origin. But my distance is going to be five, right? Because I move three steps and then I move two steps. So if I want to find the total distance, right? Um I need to be careful um in case the particle has changed directions. And when does a particle change directions? Well, when its velocity is zero. Let's think about that because what happens is if I'm moving forward, my velocity is positive. And then if I move backwards, my velocity is negative. So if I've gone from positive velocity to negative velocity, then I must have gone through where velocity is zero. Similarly in the other direction, right? If I have negative velocity, I'm going backwards and I change to be positive velocity, I'm going forwards.

[01:24:49]That means my velocity must be zero. So the the core concept for you to understand is there is a change of direction uh when the velocity is zero or or or the other way yeah change of veloc change of direction when the velocity is zero.

[01:25:03]So we know that the velocity is zero when t is three and t is six isn't it?

[01:25:06]So we change direction um at 3 seconds and at 6 seconds. Well we want to find the distance traveled in the first 6 seconds. So what do we want to do? um we want to find out firstly where we start right so when t is equal to 0 t is equal to 0 then the displacement is going to be equal to 0 + 0 + 0 + 6 okay so we're going to be at the position of 6 and then we want to find out when t is equal to 3 cuz that's when we change direction. So then the displacement will be a third of um 3 cubed - 9 upon 2 * 3^ 2 uh + 18 * 3 + 6. So again let's try to figure that out. So a third of t cubed is uh a third of 3 cubed is 3^ 2. So that's 9 uh - 9 upon 2 * 9. So that's um - 81 upon 2 uh + 18 * 3 uh 54.

[01:26:15]Yeah. Uh + 6. Okay. So let's try to simplify that a little bit. Uh so 54 + 6 is 60. 60 + 9 is 69. So we got 69 - 81 upon 2. So let's double this guy. That will be 138 upon 2 - 81 upon 2.

[01:26:38]So I think that's going to be 57 upon 2.

[01:26:43]Uh sorry, I just double check that with the calculator because my mind's getting a little bit tired at the moment. So uh 138 minus uh 81. Yeah, 57 upon 2. Okay, so uh that's where we are at 3 seconds.

[01:26:59]And then lastly, we want to find out where we are when uh t is equal to 6. So when t is equal to 6, displacement is going to be a third of uh 6 cubed uh what's it - 9 upon 2 * 6^ 2 uh + 18 * 6. Is that right?

[01:27:23]Yeah. Plus six.

[01:27:26]Okay. Okay. Well, again, I I like to kind of assume that we don't have a calculator. I just think it's such good practice. So, what I'm going to do is I'm going to factoriize out 6. So, we've got a third of 6^ 2, which is 36 - 9 upon 2 * 6 + 18 + 1. So, that's if I factoriize out six.

[01:27:45]Okay. So, this is 6 times by what's the third of 36? That's 12. Um, I can divide top and bottom, can't I? By two, right?

[01:27:55]So I end up with 9 * 3 is 27. So 12 - 27 uh + 18 um + 1 I think.

[01:28:04]So what's uh 12 + 18? Uh that's 30. Um 30 - 27 is uh 3 and 3 + 1 is 4. So I think it's 4 * 6 is 24.

[01:28:20]Okay. So we got the three key positions that that we want. And of course um 57 divided by two is going to be 27.5. So that guy's 27.5 sorry uh 28.5 to beg your pun.

[01:28:43]Um so uh what what what are you basically saying?

[01:28:48]We're saying that we start so go back go back to this diagram.

[01:28:54]draw it out again.

[01:28:57]So I'm not I'm not going to draw it very well to scale. Okay. So here's our displacement. This is the origin at uh time t is equal to 0. Our displacement is six. Okay. So let's say this is displacement is six. That's where we are when um time is equal to zero. That's our starting point. U when time is equal to 3, we're at 28.5. So um let's say 28.5 is somewhere over here. Right? So that's 28.5.

[01:29:22]That's where we are when uh t is equal to 6. Sorry, t is equal to 3. But at that point, we know that we change direction and we uh change direction and uh um at t is equal to 6 seconds, we're at 24.

[01:29:38]So let's say that uh 24 is over here, right? So 24 is over there and that's where we are when uh t is equal to 6. So let let's just think about what's actually happened.

[01:29:49]Uh we start here, we move all the way over here. So we're kind of moving forward. And then afterwards, what do we do? We kind of turn back and then we head head over here, right? So what's the total distance going to be? Well, it's going to be the distance from here to here or what's that distance? That distance is going to be 22.5, isn't it? Right?

[01:30:10]Because it's 28.5 minus 6. So the distance is going to be 22.5 plus whatever um this distance here is which is the difference between 24 and 28.5 which is 4.5. So that's going to be 4.5.

[01:30:27]So if I add those two together um that's what 23 + 4 I think that's 27 isn't it?

[01:30:34]Okay so the total distance I think should be uh 27.

[01:30:39]Now again um I don't have the masking in front of me so it's it is quite feasible that I've made an arithmetic error there. Uh definitely because you know I'm trying to do it mentally in my head without a calculator. Um but the the basic method is is correct right so what you need to do is you need to understand that the particle changes direction.

[01:30:59]Therefore the the distance traveled is going to be different from the displacement. I think one of the other things just to bear in mind is that um the particle does not start at the origin. And I think that's going to trap a lot of people. So you need to you need to know where the particle starts um when t is equal to zero. Okay.

[01:31:16]Okay. Then we are here in question 11.

[01:31:18]Uh find in descending powers of x the first three terms in the expansion of this. Uh give each term in its simplest form. Okay. Then uh so we've got x^2 plus 1 upon 2x to the^ of 9. So we want descending powers of x. So the powers of x got to go down which means we're going to start with this term here. So we're going to have x^2 to the^ of 9. Then the next term will be x^2 to the^ of 8. And the next term after that will be x^2 to the^ of 7. Right? So those powers go down 9 87 and then the one upon 2x those powers are going to increase. So 1 upon 2x to the^ of 0 1 upon 2x to the^ of 1 and 1 upon 2x to the^ of two. Okay, so those powers go up. And let's just check that the powers always add up to nine. So 9 and zero, 8 and 1, 7 and 2. So that's all good. And then uh next what we want to do is we want to put in the coefficients. So this guy will be 9 C 0. This one will be 9 C1. And this guy here will be 9 C2. So we've now written out the basic structure of the first three terms. Of course there are other terms. So now what we need to do is to simplify.

[01:32:38]So let's simplify each thing. So 9 C 0 is 1 cuz anything C 0 is one. Um X^ 2 to the^ of 9 to the^ of 9 is X ^ 18 because I multiply the powers. And 1 upon 2X ^ 0 is one because anything to the power of 0 is one. Next uh 9 C1 is 9 because uh 8 C1 is 8 and 7 C1 is 7 and so on.

[01:33:04]X^2 to the^ of 8 is X ^ 16 because I multiply the powers and 1 upon 2X ^ 1 is just 1 upon 2X.

[01:33:14]Next 9 C2. Well, we've got a calculator, don't we? But I think I'm going to try to do this uh without a calculator. So this is 9 factorial over 2 factorial 7 factorial. I'll return to that in a moment. X^2 ^ 7 is X ^ 14 and 1 upon 2X^2 is 1 upon 4 X^2 plus dot dot dot.

[01:33:38]Okay. So now let's multiply everything out. So we've got 1 * X^ 18 * 1. So that's X ^ of 18. Uh what's going to happen with the next term? Well, we've got 9 / two, isn't it? So right 9 / two.

[01:33:51]And we've got um x ^ 16 / x ^ of 1, right? It's x^ 1. So we subtract the powers and we'll get x to the power of 15. Now the next thing well um if we've got 9 factorial over 7 factorial, right?

[01:34:07]9 factorial is the same thing as 9 * 8 * 7 * 6. So it's can write that as 7 factorial there over 2 factorial which is 2 * 7 factorial there. So the key part is that the seven factorials are going to cancel each other out and we're left with 9 * 8 / 2.

[01:34:26]Um x^ 14 / x^ 2. What's going to happen is we're going to subtract the powers.

[01:34:32]14 - 2 is 12. So that's x ^ 12. And then we still got to divide by 4 here plus dot dot dot. Well, hopefully we can also notice right on the denominator we got 2 * 4 which is eight. And it's going to cancel out that eight there. Right? So 8 on the bottom, 8 on the top cancels out.

[01:34:50]So now if I simplify, I believe I've got x ^ of 18 plus 9 upon 2 x ^ of 15 + 9 x ^ of 12. And so those should be the first three terms in descending powers of x. And hopefully you've noticed that I didn't need to use a calculator in order to get all of that. So I think it's good practice for you there uh to try to uh simplify carefully uh each step.

[01:35:20]All right. Then um part B of this binomial expansion question. Uh we are told that the expansion of this thing here where n is a positive integer has a term that is independent of x. Show that n is divisible by three. Okay. Um so what is meant here by the expression independent of x? It means that there's no x in the term, which means we're just going to have a number. We're going to have a constant term. So let's think about that. Basically, every single term in the expansion is going to have the following form. It's going to be x^2 to some power times by 1 upon 2x to another power and then with a coefficient of n c something. Right? So so this is the this is the key thing, isn't it? Right? that um we just don't know what what these powers are.

[01:36:11]However, we know that the two powers here have got to add up to be n. So that's the first thing. And we also know that they've got to um uh be chosen in such a way that they cancel out the x's that are here. So what I would suggest is that we approach it by saying okay let's just call this power here r and that means that this power here has got to be n minus r.

[01:36:34]Now, if that's the case, right, then the the power of x for this term is going to be 2 r, right? So, therefore, we're going to have 2 r term there. And the power of x for this term is going to be x ^ of n minus r. And of course, if we want the x's to cancel out, then we want the power of this guy to be the same as the power of that guy. So, in other words, what we want is for 2 r to be equal to n minus r, right? because what I'm saying is this power here uh in order for it to cancel out with this power here these two powers have got to be the same and uh of course if that's the case right if I just rearrange then 3 r is going to have to equal to n well actually then if if that's the case right then we know that n must be divisible by three right because um n here is a multiple of uh of three because r is just an integer so in in a way that is kind of like A sufficient um sufficient answer.