2025 Pure Maths Unit 1 Paper 1

Ajouté : 2026-06-03

[00:00:00]All right. So 2025 paper one. All right.

[00:00:05]So contraositive. If JMO is not rich, he's not a minor.

[00:00:09]F of X + 2.

[00:00:13]Um logically equivalent. So this is the repeat not P and not Q. F G inverse of F of 3 is a half. This can be simplified to 7<unk> 5 using summation. This will be the r + 1 and 3 to the r. Yeah, because it's alternating. Good. Which of the following factors of this? This is a repeat question. I know one and two are factors and not no I know two and three are factors. I'm not so sure about one.

[00:00:46]If I remember correctly, the remainder we got when the function was divided by x -1 was I want to say -4.

[00:00:58]So let me check it back and see right x + 1 x will be minus one. So I'll get 4 take away 8 um - 1^ 2 is going to be 1. So I have the minus2 then I have + 6 - 4 = -4. So part one is not a solution. So it has to be 2 and 3 only.

[00:01:22]Right? So this here the modulus function is always positive.

[00:01:28]This is bjective.

[00:01:31]The range of this. So if the range is that the domain right so the domain would have been this here.

[00:01:39]This would be log 16. It was 16, right? 2 4 2 and multiply by two. Yes, it was 16.

[00:01:53]Um, two is the only one that will give us non non- negative.

[00:02:02]The third root is 1.5 which is 3 / 2 a half ln of x - k repeats coming in 2025.

[00:02:14]So in this one here it was it was x is less than a over three and it was a on that side and a over three on this side.

[00:02:29]Let me go back and see.

[00:02:32]This is a common repeat question.

[00:02:35]Let me check back 2019. I know it's one of I know for sure it here. Yeah. Less than a over 3. Let me make sure it's modulus of 2x - a is greater than mod x.

[00:02:51]2x - a is greater than mod x. Yep. A.

[00:02:55]The value of theta that satisfies this is pi on 3.

[00:02:59]The angle between these vectors check it back and see 2 - 4 + 2 pi on 2 minimum value is minus<unk> 89.

[00:03:13]All right. So number 19. Let's go through the proof for this one.

[00:03:18]Sin^ 2 x - cos^ 2 x. It have three identities. For that I just going to put 1 - 2 sin^ 2 x and see where we go. tan^ 2 x - 1 is going to be sin^ 2 x over cos^ 2 x - 1 which is sin^ 2 - cos^² cos^ 2. We can leave this up as that one um sin^ 2 x - cos^ 2 x. So this is equal to sin^ 2 x - cos² x * cos^² x over sin^ 2 xus cos 2 x bam and we get cos square x for number 19.

[00:04:09]Number 20 is going to be b 21. This one was a tangent. So it touches the curve at the 0 24. So again you could watch the answers and I watch the question and know what the answer is. You could kind of get the answer from the question from the answers. All right. Um for this one here we did do this question just now. Let me do it over. um x1 y1 x2 y2 y2 - y1 / x2 - x1 4 fths is going to be my gradient of the radius the gradient of the tangent will be sorry 4 - 5 gradient of the tangent is 5 / 4 so we're going to get y - y1 is equal to 5 over 4 into x - 5. y is = 5 over 4x + 25 on 4 + 4. Multiply everything by 4. 5 y + 25 + 16 we get 41 day. Um 4 y is 5x sorry.

[00:05:31]So if 5x is positive and I move the 4 y across I get 5x - 4 y + 41. So the answer is d.

[00:05:43]Our curve is defined by the parametric equations that and that the cartisian.

[00:05:47]So from this t is = y - 2. So x = 3 + 2 into y - 2. We have the x first. Then we have the y. So 2 y bringing across becomes - 2 y and this becomes 3 take away 3 take 4 is minus one x - 2 y is -1 d is the answer um this is a repeat number 24 we get -10 this one here would be is a repeat but just double checking For the x coefficient, I'm going to get -2 + 8, which is 6. For the y coefficient, I'll get - 6 + 8, which is 2. And for the constant, we get -5.

[00:06:40]Divide everything by 2. 3x + y, sorry, - 30 - 15. 3x + y - 15. The answer is C.

[00:06:53]The length of the vector PQ is Uhhuh. So you see how they change and notice that they change the position of 5 <unk>2. So I don't know if the size of this vector is going to be 50 roo<unk> 50 sorry. So let me get the vector pq. So pq is o p which is -28 + -1 + 4 + 4. So we will get we will still get back the the um 50 9 25 and 16 we getting back the 50.

[00:07:34]So it's going to be 5 <unk>2 a unit vector parallel. So we go check back and see which one is going to be unit vector. The only one that's a unit vector is d 1/5. When I square that again 1/5 I'm sorry 1 over roo<unk> 5.

[00:07:52]when I square I get 1/5 2 over<unk> 5 when I square that I get in two um four fifths add them together I'm going to get one right so the angle between the vectors all right so this is a repeat question it's going to be<unk> 19 over<unk> 42<unk> 46 cos inverse of that so -9 /<unk> 42 *<unk> 46.

[00:08:24]Okay, that not looking right. So, calculator playing up. 42 x 46 is 1932. So, it's -9 / the roo<unk> of 1932 and it's going to be cos inverse of this to give me 115.61.

[00:08:45]Now I want to check something here, right? Cosine of the answer is that I minus one and cos inverse of that is this. So we we got cosine to be negative is in quadrant 1 and so 180 take away this is 115.61 good. So so confirming the answer is indeed 115 61. All right. So this one here is another repeat. So the answer for this is D because this vector here is the vector A B but in the reverse. So it's the vector BA.

[00:09:22]A curve C is given by the parametric equations X and T. The equation is um I know it was - 56.

[00:09:32]I think it was a. So let me confirm it is indeed a. Right. So x - 2 over 3.

[00:09:43]That's going to be t. So it's t ^2 + x - 2 / 3 x^2 - 4x + 4 / 9 + x - 2 / 3 - 6. So multiply everything by 9. So when I multiply by 9, this going to be 3x - 6. So - 4x + 3x is going to give me a min - x. So a is the answer.

[00:10:17]Um the differential of this we're going to have x cos x + 3 sin x. So the answer is c. Which of the following real functions has a point of discontinuity?

[00:10:30]It going to be d because I can solve the denominator. x will be minus one on the denominator.

[00:10:38]33 is - 20 34. Let's double check and see what the differential is going to be. Right? I know it's going to be 108 sin 6x but I don't know if it's plus or minus. When I differentiate sign once I'm going to get cosine so the minus persists. When I differentiate cosine I'm going to get negative sign. So the minus changes to plus. D is the answer.

[00:11:03]35 Lita looking like the play here. 2x 32 are 6 um plus or minus a quarter.

[00:11:18]This one was 48 root y 4 2's are 8 and then 8 are 24 by 2 is 48. Yep. This one, this was minus1.

[00:11:34]Let me check it back. I think it was, yeah, it was indeed minus one. But let me check it back. Right. Y is equal to 2x + 1 to the power a half.

[00:11:43]Differentiate to get the first derivative a half * 2 2x + 1 to the - a/ differentiate again. This becomes 1.

[00:11:55]I'll get - a/2 into 2 into 2x + 1 - 3 / 2. This becomes -1. So the only one that has minus one on the numerator is d.

[00:12:07]y prime is cos x - sin 2x.

[00:12:11]I mean I know it's a repeat but it wouldn't hurt us to go and rework it out. When I integrate cosine I get s.

[00:12:17]When I integrate s I get negative cosine. So that changes to cosine 2x.

[00:12:22]And I had to divide by the angle differential of the angle which is two.

[00:12:26]So it's positives all around with a half on the cosine 2x. Positive all around with a half on the cosine 2x.

[00:12:36]All right. So um similar scenario here.

[00:12:38]This is a repeat question. Nothing will hurt if we do it over. So we're going to get cossine of 3x^ 2 - 1 to the^ of 3 3 into cossine of 3x^2 - 1^ 2 multiplied by the differential of the angle which is going to be um 6x - 6x sin of 3x^2 - 1 multiplying the 6 and the 8 the3 - 18x are going to get cosine^ squar of the angle and I'm going to get s of the angle. So - 18 cosine^ 2 d.

[00:13:22]So again look at number 41. They change the way option d looks right. It's still correct but it just changed the way it looks. The gradient of the normal to the curve 6 x - 2 6 into 1 - 2 which is 4.

[00:13:39]That's the gradient of the tangent. The gradient of the normal is minus a quarter.

[00:13:45]Um this would have been 8 pi for to be the answer.

[00:13:50]The volume of the solid generated when the region enclosed by this um the y ais. So we have the integral between 4 and 0 of pi y d y cuz I x squ. Yep. So pi y^ 2 on 2 which is going to be 4^ 2 on 2 which is 16 on 2 which is 8 pi. So I doing a lot of working mentally here right?

[00:14:23]Okay, looking like newish question territory. I want to say is minus 21 v. No, there's a repeat question by du dx minus u by dv dx.

[00:14:39]I'm not going to work out v ^ 2 3 - 4x + 4x - 24 - 21 and lastly this will be two- fifths and that is it for 2025.

[00:14:57]So again if it is you see a repeat question and you don't know the working for it watch the videos from the first one 2018 come up the road. Right.