CAPE Pure Maths Unit 1 2026 Paper 2 Solution - Question 5

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[00:00:19]on the screen.

[00:00:22]Inside you can see the pure math unit one 2026 paper two.

[00:00:29]I'm just completed today.

[00:00:32]Seems like a very very easy paper.

[00:00:35]I've not yet gotten much feedback from the students in my program.

[00:00:40]Um, but one person is asking to look at module three.

[00:00:46]I said it's a easy paper, but I just want to confirm work in module three.

[00:00:52]So, I'm just going to start working on this paper at question five.

[00:01:00]I don't think I'll be able to do all of it all at once.

[00:01:03]But I'll just do those two questions in module three.

[00:01:08]5 and 6 to begin with.

[00:01:11]Let me be able to I'll look at the other two modules, module two, module four.

[00:01:19]So, let me see what module three says.

[00:01:27]Okay, here's module three. All right, so this question remind me reminds me of a question on a paper three a period ago.

[00:01:38]Something that we would have done in class, all right?

[00:01:41]So, clearly that's a limits question. It says the diagram below shows the graph of this function right here. All right, so that's the graph of a function f.

[00:01:52]And we can see it's a function that is It's continuous everywhere, all right? So, it's continuous through here.

[00:01:59]No, actually it's not continuous here.

[00:02:01]It's not continuous here.

[00:02:03]It's not continuous here, but this point is here.

[00:02:07]Uh it's also not continuous here.

[00:02:11]All right? So, you have two points of discontinuity.

[00:02:16]Here at x = 5 or at x = -5. It's obvious from the graph.

[00:02:22]All right? Um this one is called a jump discontinuity, and this one is a removable discontinuity.

[00:02:30]All right? But anyway, let's answer the question. It says to determine the limit. Remember, as x tends to 5, x tends to -5. This means from the right of -5. So, -5 is right here.

[00:02:44]And you're analyzing what is happening, what the function is approaching around here at x = 5.

[00:02:52]The thing though is that this plus means that you're approaching from a value of greater than -5.

[00:03:00]And so, you're like you're approaching -5 along the graph from this end.

[00:03:07]This way.

[00:03:08]From a value greater than -5.

[00:03:11]All right? And so, the function is approaching here, which has a y value of 5.

[00:03:18]So, basically the answer for this is just 5.

[00:03:25]So, if it was the limit as x approaches the same -5, but from the left of f of x, then that answer would be that you would now be on here.

[00:03:41]You'd be now approaching the -5 this way.

[00:03:45]So, you you'd be on this leg of the graph.

[00:03:48]And you'd be approaching a y value of what? That would be 8.

[00:03:52]So that would have been 8.

[00:03:54]If they wanted the other 8, it would have been an 8.

[00:03:59]And so I'm just leaving that on the screen.

[00:04:02]So maybe we can appreciate the telecobalt.

[00:04:06]So it's a Y value that this function is approaching as X approaches the value that is given here.

[00:04:14]And that's what a limit is.

[00:04:15]On here, it says the limit as X approaches 5. No, as X approaches positive 5. Now this positive 5 means from the left and also from the right. That's what this means, from both sides.

[00:04:28]This means from the both sides of 5.

[00:04:30]So this is where X equals 5.

[00:04:33]So we're talking about the function around this point.

[00:04:36]All right? And from both sides, as you come along the graph from both sides, the graph is actually approaching a value of -5.

[00:04:46]Yes, it is clear that f of 5 is not -5.

[00:04:51]f of 5 is this value right here. It's a shaded dot. That is f of 5.

[00:04:57]So this point represents f of 5, which in that case is -3.

[00:05:05]-3.

[00:05:07]All right? But the limit is the value of Y that the function is approaching as X approaches 5 from both sides, and that is -5.

[00:05:17]So the answer here is -5.

[00:05:21]This is -5.

[00:05:23]We can make a note as well that f of 5 is actually -3.

[00:05:32]Just again for information.

[00:05:35]So we see the difference.

[00:05:37]All right? That's -3 right here.

[00:05:40]All right? That's a Y value at this point where X equals 5. That's 5 That point is five negative three.

[00:05:50]All right. So, that is that.

[00:05:54]Go on with the assignment.

[00:05:56]It says, "State the value of K such that the limit of f of x is negative two as x approaches K." In other words, we want to know when the y value is when the y value is negative on the curve, what is the value of x? What is What is the value of x at that point?

[00:06:20]It's going to sit on the graph, right?

[00:06:22]That's what that means. This is when All right, this is where y approaches two.

[00:06:27]All right, where the function is approaching two.

[00:06:30]Okay, this is basically where y approaches where y equals two. In a sense, that That's what it is. That's what you're basically going to look for.

[00:06:38]Where y is equal to two, what the x value is at that stage, what the function is approaching.

[00:06:43]So, is when when y approaches two, put that arrow on it. When y approaches negative two at two negative two.

[00:06:54]All right, so when I look, let's see, at negative two, it's right here.

[00:07:03]It's negative two.

[00:07:05]And this is the point on the graph that they're talking about.

[00:07:09]It's clear that the x value right here is two.

[00:07:13]So, it is as x approaches two, that the y value is approaching negative two.

[00:07:21]So, this is the key. That will come to key.

[00:07:24]K is equal to two.

[00:07:26]So, as x approaches negative two, as x approaches two, the function approaches negative two.

[00:07:32]So, the key is clearly equal to two.

[00:07:35]All right, okay. I'll see that on the graph there.

[00:07:38]So, the key is equal to All right, so that is it.

[00:07:44]Um here is the question. Easy easy easy.

[00:07:46]All right. So, this question gives us this piecewise function.

[00:07:50]f of x is this g of x is this when x is less than one.

[00:07:55]Which is equal to 3x when x is greater than or equal to one.

[00:07:59]Question one says us to find g of one.

[00:08:01]So, g of one is on this one.

[00:08:05]So, g of one becomes 3 * 1.

[00:08:13]Which is three. Simple as that.

[00:08:15]Next part now says the limit of x approaches one. That's from both sides.

[00:08:21]So, what we can do what is g? What we can do is work both from the left and from the right. So, from the left, the limit as x approaches one from the left of g of x becomes the limit as x approaches one of And And from the left of g of g of x means that you're now on this function.

[00:08:46]Cuz that's on the left and number that is less than less than one.

[00:08:49]When x is less than one, you're on this part of the function.

[00:08:52]And if you're approaching one from the left, it means that you're coming from the a value that is less than one. So, you'll be on the 2x cuz one part of the function.

[00:09:03]And that is just simply equal to three.

[00:09:07]All right? Cuz if you put one here, 2 * 1 is 2 + 1 is 3. And then from the right, the limit as x approaches one from the right of g of x means that we are now on the other side of the function. It becomes the limit as x approaches one of 3x.

[00:09:32]Cuz we are now approaching from about greater than one. So, we're are inside of the function.

[00:09:39]On the three x side of the function. So, three times one is three.

[00:09:44]So, since the limit from the left and from the right is the same way is the same value, it means that this which means the limit from both sides is three is equal to that value. So, the limit as x approaches one of g of x is equal to three. That's the answer.

[00:10:01]Answer is three.

[00:10:02]From the left is three, from the right is three.

[00:10:05]All right?

[00:10:06]I don't know from no here.

[00:10:08]It says, "Hence state whether g of whether g is continuous at x equals one.

[00:10:13]Give a reason." Now, what are the conditions for continuity? Let's write it down.

[00:10:18]For g for g to be continuous at x equals one, what do we must have?

[00:10:38]The limit as x approaches one of g of x must be equal to g of one.

[00:10:48]And both must be defined.

[00:11:00]Otherwise, you have to know both of them.

[00:11:08]Now, what do we have?

[00:11:12]What do we have?

[00:11:21]What do we have on g?

[00:11:23]We have a one.

[00:11:24]We have the limit as x approaches one of g of x is equal to g of 1 is equal to 3.

[00:11:37]So, this condition for continuity is satisfied.

[00:11:41]And so, we can draw the conclusion that yes, therefore yes, g is continuous at the point where x equals 1.

[00:11:58]That is that.

[00:12:00]So, basically this is the reason that they're looking for.

[00:12:04]If I read the question, they're looking for that.

[00:12:06]All right, that's the condition for continuity at the point.

[00:12:09]Chain rule says given that y is equal to that and x is equal to that, we want to determine dy by dt.

[00:12:19]All right, so this is a chain rule question.

[00:12:21]But it does not have to be a chain rule because we can take this and substitute for the x in there and then calculate dy by dt and give the answer in terms of t. But the problem is they did not tell us what they wanted in terms of.

[00:12:34]It's just that well, y is in terms of x and x is in terms of t. So, we can use a chain rule method to do it. But you could also substitute. It's just that um we can substitute afterwards as well.

[00:12:45]So, we can leave it in terms of x and or t, I guess.

[00:12:49]But we're given both functions. So, it's a chain rule question. So, we're going to say um we're going to write what dy by dt is first by differentiating this one.

[00:12:58]So, dy by dx becomes as a composite function, so you treat it as if it is x to the 7th first. So, it becomes 7 * 2 x cubed plus 4 raise to the power of 7 to 6. Then we multiply by the differential of what's inside here, which is 6 x squared. 6 x squared.

[00:13:20]Then we can simplify. This 6 * 7 is 42.

[00:13:25]That is 42 x squared into 2 x cubed plus 4 to the power of 6. That's the dy by dx right now.

[00:13:38]Then we can calculate the dx by dt from this one.

[00:13:45]The dx by dt is simply -2. Differentiate this with respect to t, you're going to get -2.

[00:13:53]We're going to connect these two rates of change in order to get this dy by dx.

[00:13:57]So, dy by dx is in a different color for this one. So, the dy by dx that they want, dy by dt, sorry.

[00:14:05]Is equal to the product of two rates of change.

[00:14:09]So, we don't have a formula connecting y and t directly. So, we're going to connect them using these two rates of change. So, dy goes on top of the first fraction here and dt goes on the bottom of the second fraction here.

[00:14:24]Second derivative.

[00:14:27]And the third variable is x was here and there.

[00:14:31]So, we end up multiplying this by this to get the final answer.

[00:14:37]So, it's simple becomes equal to negative This one is a negative two.

[00:14:45]And that one is negative two.

[00:14:48]Uh and that's it. So, we get -2 * this.

[00:14:52]So, it is 42 x squared into 2 x cubed plus 4 to the 6.

[00:15:03]That's the dy by dx times the dx by dt is negative.

[00:15:08]So, it's -84 x squared into 2 x cubed plus 4 to the 6. But, they Or they tell us if it must be written in terms of x and t.

[00:15:26]I'm going to leave it in terms of x.

[00:15:29]If you like, you can make a substitution for the t expressions here for the x. So, we can say or dy/dt.

[00:15:40]Or we can say dy/dt is under the fact that the t is equal to let's write over here.

[00:15:47]I want to write the t is equal to 1 minus 2 2 x.

[00:15:52]Sorry, the x is 1 minus x is 1 minus 2t.

[00:15:56]The x is 1 minus 2t.

[00:15:59]So, we can make those substitutions here here.

[00:16:03]So, we can say or in terms of the t, dy/dx is minus 8t over and the x is 1 minus 2t. The square on that.

[00:16:14]Into 2 times 1 minus 2t cubed right here.

[00:16:20]Minus 2t cubed.

[00:16:22]Plus 4.

[00:16:25]And that's all we have to raise to the power 6.

[00:16:28]So, we can do that either one by parts.

[00:16:32]All right, what else?

[00:16:34]Question part two says we are to find the first derivative using first principles of sin x. Now, this is what we are going to do directly in the class. This is directly on the syllabus. So, I find that in case that I come that is directly stated on your syllabus, you should not have a problem with it.

[00:16:56]So, if you differentiate sin x from first principles, all right?

[00:17:00]All right, so now I know that f prime of x it comes from first principles becomes a limit as h tends to zero of 1 over h into f of x plus h f of x.

[00:17:20]So, that's what we are going to be doing.

[00:17:22]To get rid of this h, we're going to have to use the um the limit.

[00:17:29]All right, that says um rule that says the limit as h approaches zero of sin of h over h is equal to one. All right? That's going to be used somewhere around the way.

[00:17:48]That's how we want to eliminate this h.

[00:17:50]Remember the whole the whole intent of this process is to get rid of this h.

[00:17:55]To eliminate it from the expression, then you are going to apply h equals zero everywhere else.

[00:18:01]All right? So, we're just going to setting this up and simplifying.

[00:18:04]This then becomes the limit as h tends to zero of one over h into Now, f of x + h. Remember, I want to put x + h here.

[00:18:18]For the sign the signs that become the sign of x + h minus and then f of x becomes the sine of x, right? Sine of x.

[00:18:29]And this now become the difference of two signs, which you should know.

[00:18:33]Difference of two signs I think will be on the formula sheet as well.

[00:18:37]Well, difference of two signs is cosine.

[00:18:41]Well, as a cos of alpha so kind of sine of alpha difference, right?

[00:18:45]So, cosine.

[00:18:46]So, it becomes the limit as h tends to zero of one over h into Remember what I did over that I'm using here.

[00:18:59]All right, this is where I stopped here what I'm what I'm going to be using now.

[00:19:04]The difference of two signs.

[00:19:06]All right? More things you're having here the better. I'm going to use um sine of A minus sine of B.

[00:19:17]Difference of two sines is twice the cos of half the sum, sum of the A and the B.

[00:19:24]So, A plus B over two times the sine of half the difference. I should have solved the sum first.

[00:19:34]Times the sine of half the difference.

[00:19:37]That's A minus B over two.

[00:19:41]That is what I'm using.

[00:19:43]Yeah, this slide.

[00:19:46]Uh-huh.

[00:19:47]So, that this is the A and this is the B.

[00:19:51]All right. So, it becomes twice the cos of half the sum. So, we're adding these two things now.

[00:19:58]So, it becomes, um, 2x plus H over two times the sine of half the difference.

[00:20:07]Of half the difference, it becomes X plus H minus X over H.

[00:20:15]So, we're going to get H over two, half the difference.

[00:20:19]All right. Half the difference.

[00:20:23]And what happens now is that we actually need half of H under this.

[00:20:30]So, that this becomes, we'll get the limit of that. We'll get we get, um, we get one, all right? Because this is half H. We need half H under here.

[00:20:40]So, we'll get that half H under there.

[00:20:42]Well, let me move this first.

[00:20:45]All right. So, let's set it up now. So, this becomes equal to the limit as H tend to zero and I'm going to put back two cos of 2x plus H over two right here.

[00:21:03]Put that back.

[00:21:04]And when we kind of just meant on this now, so this H I'm going to bring it under here first.

[00:21:09]So I'm going to bring that H under here first. I'm going to put by the sign.

[00:21:13]Let's write that as half each.

[00:21:16]All right? Let's write that each of what I do as half each.

[00:21:21]Oh, but this H this H ain't coming under here.

[00:21:28]Cuz it's a half each I need. So what I'm going to do, I'm going to multiply the half under here.

[00:21:33]I'm going to multiply by the half up here.

[00:21:36]All right? So that I know how what I want.

[00:21:40]All right? So for all intents and purposes, that is part of it.

[00:21:44]It's going to become equal to one when I apply that limit.

[00:21:53]Oops.

[00:21:54]That part is going to become equal to one.

[00:21:56]Because this when I apply the limit now, this part is going to become one.

[00:22:01]And here this H right here is going to become zero.

[00:22:04]All right? So this thing becomes equal to. I know the answer is supposed to be cos x, right? Because you know that when you differentiate sin x you're getting cos x. So this final result is supposed to be cos x.

[00:22:15]So we're going to get two cos of width plus zero. This is going to become zero.

[00:22:25]over two times this half times this becomes one.

[00:22:35]when we apply the limit.

[00:22:37]And then it's obvious now that this becomes 2x over 2, which is x. So this becomes an x.

[00:22:44]This two cancels this half.

[00:22:46]Final result is what it is supposed to be, cos of x.

[00:22:50]That is the answer.

[00:22:52]Nice and easy.

[00:22:54]All right.

[00:22:56]Then we have part B. Wow, [clears throat] part B.

[00:22:59]So different question. So so us now to integrate between these limits. Making a substitution U equals X minus 4. So, we're changing a variable. Now, let's go. If U is equal to X minus 4, we have to differentiate first to the change of variable.

[00:23:15]Differentiate U on this side not with respect to anything becomes dU.

[00:23:20]Then, differentiate this side.

[00:23:22]Differentiate X becomes 1 dX. So, it becomes just dX.

[00:23:26]So, we can be we need to replace this this dX with dU when we change the variable.

[00:23:31]We also need to change these limits.

[00:23:32]These are X limits. So, here now, when X equals 2, the lower limit down here, the U becomes 2 minus 4, which is negative 2. So, this limit is going to change to negative 2.

[00:23:46]And when X equals uh upper limit is 4, the U becomes 4 minus 4, which is 0.

[00:23:54]So, this limit when we change the variable is going to become 4.

[00:23:58]All right. And then, we just need to make the substitution also from this.

[00:24:03]If U is equal to If U is equal to X minus 4, then if we make X the subject, then we transpose the 4, we can say also that X is going to be U plus 4. So, that substitution might be needed right here.

[00:24:18]U is equal to X plus 4. Well, now, let's call Let's call this integral I so we can reference that. So, I'm going to say let I equal this, all right?

[00:24:29]So, I want to say let the integral be I. I equal that thing.

[00:24:34]So, I'm going to call it I.

[00:24:36]So, the integral I becomes when I apply the limit change, you know. Apply the the variable change. So, it becomes the integral. The lower limit, which was 2 uh X value of 2 is now a negative 2 U value.

[00:24:49]Upper limit was the X value of 4 is now a 0 U value.

[00:24:55]This X becomes U plus four.

[00:25:01]And this in the bracket becomes a U. So, we're going to be getting U to the fifth power.

[00:25:08]And then this DX becomes a DU right here. DX becomes a DU.

[00:25:15]And all we need to do now is just expand this out and separate and differentiate term by term. So, I becomes equal to the integral from negative two to zero.

[00:25:26]And we [snorts] expand this, we get U to the to the sixth power plus five, right?

[00:25:31]U to the sixth power plus four times U to the fifth power integrating with respect to U. And then becomes easy sailing after this. I becomes U to the seventh power over seven. We're going to integrate this one.

[00:25:48]Plus four times U to the sixth power over six between the zero and negative two.

[00:25:58]So, I becomes equal to write the zero first, everything becomes zero.

[00:26:04]Minus then we're going to apply the negative two here and there.

[00:26:08]So, we're going to get negative two to the power seven negative two to the power seven over seven plus Well, this is actually two thirds, right? Four over six is two thirds. So, we can put two times negative two to the sixth power over three.

[00:26:32]And then we can get a calculator and get a final answer there, right? Let's just put that in a calculator, I think.

[00:26:38]I get the final answer there.

[00:26:41]Let's just do that to the calculator.

[00:26:44]So, it's just got the negative of this.

[00:26:48]We can put negative bracket. Let's write the same thing there.

[00:26:51]fraction.

[00:26:53]That's -2 to the 7th power. So, it becomes it becomes a negative.

[00:26:58]If if if we have a negative number to an odd number power, the result is negative.

[00:27:03]So, it's the same as -2 to the power 7.

[00:27:10]Dividing that by 7.

[00:27:13]Then we're adding we're adding another fraction.

[00:27:19]That becomes an an even power is an even power.

[00:27:24]So, the -2 the -2 um to the power 6 just like 2 to the power 6 and then multiply by another 2 here. That becomes 2 to the power 7 again. 2 to the power 7.

[00:27:38]2 to the power 7 again.

[00:27:41]Interesting.

[00:27:42]2 to the power 7. What is that? We're dividing by 3.

[00:27:47]And I'm sure that's I think it's a fraction.

[00:27:52]That's it.

[00:27:54]That becomes negative.

[00:28:00]This one right here.

[00:28:03]Let's make sure I'm writing it correctly. So, this rule called to put the negative on the outer bracket. It becomes -2 to the power 7. This one over 7, which is fine. Plus this 2 to the power 6 as a negative is the same as 2 to the power 6 times another 2 which makes it 2 to the power 7.

[00:28:23]That's over 3.

[00:28:25]And so, this is the answer.

[00:28:26]Um -512 over 21 the answer here.

[00:28:32]Or or I equals -24 8.

[00:28:42]That's the one.

[00:28:44]This is our answer.

[00:28:46]All right. So, that is number five. I'm going to do number six in a different video.

[00:28:51]I want to stop this recording and do number six in a different video. All right, so let me So, I have to stop the recording.

[00:29:17]So, let's stop it.