Advanced Higher Maths 2026 | Last Minute Paper 1

Ajouté : 2026-05-09

[00:00:00]math students. This is from Arcade Maths. This is 2026 mock papers. Going to give them a go today and see how we go and I'll give you a bit of last minute boost. I'll try and get paper two done if I can. But remember, last minute live streams are on tomorrow, weddednesday, 7:30 p.m. for advanced higher maths. Last year it was epic.

[00:00:18]Went through every single topic. I think it's going to come up. All the little tips and tricks, all my predictions, and just that last minute boost just to get you through before that exam. You don't want to miss it. So get on it. £4.99.

[00:00:30]Just click the link and join now. So question one for Arcade Maths 2026 advanced higher maths paper. 5x + 7 over x + 5 x - 4 partial fractions. So we're going to say that that's a / x + 5 + b / x - 4.

[00:00:48]And then that's a x - 4 + b x + 5 = 5 x + 7.

[00:00:56]sub in four to get rid of that a. So that's 9 b equals 5 fours is 27. And therefore b is 27 over 9 which is three. And then we'll sub in min - 5 to get rid of that b - 9 a then equ= - 255 + 7 - 25 add 7 is -5 16 17 18.

[00:01:24]So a is two. So therefore, our final answer is 2x + 5 + 3x - 4.

[00:01:33]And we're done there. Clever Math is sponsored by Leki, the educational publisher for Scotland. They offer viewers of this channel a massive 30% discount. Just use the discount code clever maths at leiscotland.co.uk.

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[00:02:11]Question two. Prove that n plus2 factorial over n - 1 factorial - n + 1 factorial of n - 2 factorial is 3 n + 1.

[00:02:21]Little bit a tricky one this one, but we'll just start by expanding. N + 2 factorial is n + 2 n + 1 n then * n -1 factorial all over n minus one factorial take away n + 1 factorial is n + one n minus one nus2 factorial. just stop there so I can get rid of that on the bottom as well. So we get a cancellation here, a cancellation there. That just leaves n + 2 n + 1 * n minus n n + I'll just write it in the order it was given actually so you don't mess this up and don't see it.

[00:03:11]Now I could expand all these and then try and get it back, but I've already got common factors. I've got a common factor of n and I've got a common factor of n + one for both of these. So I'll just do that. And then the first term gives me n + 2.

[00:03:27]And the second term I've got n + one. So take away what's your brackets here? n minus one. So you've got that. And then we can just simplify that down to n + one.

[00:03:39]n + 2 - n -1 is + one. Finally, that's n's cancel 3 n + one and we're done there. Question three. Given that y= a where a is the area of this rectangle, I presume show that the y by the x equals that. So our area is 2 tan x 3 co x 4 cosec x.

[00:04:11]So now I'll just expand that bracket just to make it neater. Two threes is six. 6 tan x cot x two fours is 8 tan x cosec x. Now you might be tempted at this moment to think well I'll just do the product rule twice. But cot remember 1 / tan. So that's just 6 tan x over tan x which is just six. So watch out for that. Then 8 tan x cosec x.

[00:04:42]So now we can do the product rule. Let U= 8 tan X and V equ= cosec X.

[00:04:52]Then U dash from the start of the exam paper I believe is se² x. So a sec² x and v dash. So v is cosec. So v dash from the start of exam paper is minus cosec x co x. So dy by dx is v u dash. So that's 8 se 2 x cos sec x plus u and v dash minus cose x c x taking out 8 cose x as a common factor I get 6^ 2 x and cot and tan cancel each other out so I've just got minus one left as required 8 cosec x^ 2 x - 1. Question four, prove by contradiction that roo<unk>3 is irrational. So if we're going to prove by contradiction, we'll just assume that root3 is irrational. Assume roo<unk>3 is rational. Now what is the conditions for something to be rational? It has to be a fraction in its simplest form. So how do you write that? That implies that roo<unk>3 equals a over b where a over b with no common factor. There's other ways to write that but I'll just write that.

[00:06:28]So roo<unk>3 is a over b. So we're going to start from there. If root 3 is a over b then we can say that 3 is a square b 2.

[00:06:40]So that would imply then that 3 b^2 equ= a^2 3 b²= a^2 that implies that a² is divisible by 3.

[00:06:56]Now here's an interesting fact that you should probably know. If a square is divisible by three that implies that a is divisible by three.

[00:07:07]If a square number is divisible by a prime number then the standard number is also. So therefore we can write a is equal to 3k. Now we do the same trick by going up from the start and saying that 3 is a square over b squared. So remember 3 b ^ 2 = a 2. So 3 b 2 = 3 lots of k^ 2 say. So 3 b^2 = 9 k^ 2. So b ^2 = 3 k^ 2. So b ^2 is a multiple of 3 and therefore b is a multiple of three and that's a contradiction but we'll just show it out. So b is some number 3 * k l and that means that roo<unk>3 is a over b. <unk>3 is 3 k over 3 l so we share a common factor that's a contradiction.

[00:08:25]Therefore, roo<unk>3 must be irrational.

[00:08:30]And we're done there. Question five.

[00:08:32]Integrate 2 over<unk> 4 - x^ 2. So that equ= 2 over 4 - x^2 to the half. So again, I think this will be start exam paper of exam paper. of 1 a^ 2 - x^2 is the inverse s of x a + c. Don't be tempted to use a chain rule here because it isn't. So we can go and say that the integral of this is equal to 2 * the inverse s of x over a which will be two because 2's is 4 plus c and we're done there. That's it.

[00:09:14]Question six complex numbers zed ^ 4 - 6 zq + 15 z^ 2 - 18 z + q is equal to zero. Given that zed equals too many as a root state another root well another root is the complex conjugate 2 + i just change the sign. So part b hence find the values of q. So we've got two roots factors. So z minus the first root is 2 - i times that by to get a quadratic root z - 2 + i.

[00:09:51]Now that's z^ 2 - 2 - z * 2 + i - z again * 2 - i. Then - is + 2 - i 2 + i. So that gives me z^ 2.

[00:10:15]The i's will cancel in the middle because it's minus z i plus z i. So you're just going to have -2 - 2 is -4 zed. And then you've got two twos is four. The i's cancel again. So minus i^ 2 which is + one. So + 5. So that's a quadratic factor.

[00:10:34]So then we can just divide by our quadratic factor. So long division z 4 - 6 z cubed.

[00:10:47]Then we've got + 15 z^ 2 - 18 zed.

[00:10:54]And then we've got + q z^ 2 - 4 z + 5. So z^ 2 z 4 is z^ 2. So we times by z^ 2 to get z 4 - 4 z cubed + 5 z^ 2 take away.

[00:11:14]So -6 - -4 + 4 - 6 add 4 is - 2 zed cubed 15 - 5 is 10 - 18 zed + q.

[00:11:30]So zed^ 2 - 2 zed cubed is - 2 zed.

[00:11:35]So we get - 2 z cubed times and through by - 2 z + 8 z^ 2 and - 10 zed. So take away again - 2 - - 2 is 0 10 - 8 is 2 z^ 2 - 18 + 10 is - 8 zed + q divide again zed^ 2 into 2 zed^ 2 is + 2. So you get 2 zed squ is 8 and two is 10 take away finally 2 - 2 is nothing - 8 + 8 is nothing we get q - 10. There's our remainder but that must be zero. So therefore q - 10 is zero. So q is 10 and we're done there. Hence find the other roots.

[00:12:31]So we'll go for part C z^ 2 - 8 z + 10 because we got 2 zed.

[00:12:57]So we're divided through. So z^ 2 - 2 z + 2 is a factor. So to get the roots we make it equal to zero.

[00:13:16]Just use the quadratic formula. Zed is - 2 plus or -<unk> a - 2^ 2 - 4 * 1 * 2 all over 2 that's 2 plus or minus 2 2's is 4 4 2's is 8 so that is 4 - 8 is - 4 over 2 so that gives me 2 plus or minus the square root of 4 is 2 but it's a minus 2 I / two cancel out the twos 1 plus or minus I and we're done there. Question seven, matrices A and B are singular. Find P.

[00:13:58]Singular means no inverse.

[00:14:02]Inverse doesn't exist when the determinant is zero. So for part A, I can do the determinant of A is equal to 1 * P - 4 * 2, which equals 0. 1 * p is p min - 8. So p is 8 or b. State the matrix 3 half a b.

[00:14:25]I want three halves a b two 4 p 221.

[00:14:41]So that's three halves.

[00:14:45]1 * 4 is 4. 2 * 2. So remember with these ones, you go along and down. Along and down. That's the first two. Along and down. Along and down. That's the next two. So I do 1 * 4 is four. Two twos is four. That makes eight. 1 * 2 is 2. 2 * 1 is 2. 2 + 2 is four. 4 is 16. 8 2 is 16 makes 32.

[00:15:09]4 is 8. 8 1's is eight. So that's 16.

[00:15:14]We want to do three halves each of a numbers if you can. So a half of eight is four, 3 is 12.

[00:15:21]Two freeze is 6.

[00:15:24]32 is 16. 32 48.

[00:15:29]And 16 is 8, freeze is 24.

[00:15:33]So if I've done that right, I've got 12 6 48 24. And we're done there. Question C. C is a matrix defined as 213051.

[00:15:46]Find CA.

[00:15:49]CA 213051.

[00:15:59]A is 1248.

[00:16:08]So two two 1's is two 1 fours is four 2 + 4 is six and then along and down two twos is four 1 eight is eight that's 12 then middle 3 1's is 3 plus nothing then middle again 3 twos is six plus nothing and then finally five 1's is five 1 fours is four that makes nine 5's is 10 18 is eight so that's 18 you could say that or you Say since a third is a factor of all of them, you could say two 4 one 2 3 six. So you could say that if you wanted to and finally question eight, f ofx is this. Determine if it is even odd or neither and then state the asintope of f ofx. So we're going to do a test. We can do the f of minus x test. f - x we put that in. So we get - x cubed squared + 1 take away - x 5 + - x cubed all over x^2 + 3 - x^ 2 + 3. So we need to check what that gives us.

[00:17:39]-* a minus time a minus is a minus. So that's - x cub take away x - x^2 is x^2 take away - x 5 - x cubed all over x^2 + 3.

[00:18:03]Now, you could simplify that at the start if you wanted to, but I'm just going to go for what I've got. Is that minus f ofx?

[00:18:13]Well, does that equal minus f ofx? Is what I'm going to check just by looking.

[00:18:20]If I put a minus in front of this, I get - x cub - x. Say x^2 + 1. Yeah.

[00:18:29]then take away x to 5x cubed. I could have that inside.

[00:18:36]Yep. And then all over. So that is minus f of x because minus f. Yeah. That's it.

[00:18:42]So therefore it is odd cuz f of min - x = - f ofx. A quicker one other way you could have done it is simplify the fraction first and then try and do it to see if it's an easier thing to look at.

[00:18:53]But there you go.

[00:18:55]Question B. State the asmtope of f ofx.

[00:18:58]So, I'm going to expand this back and see if I can simplify this. So, I've got x 5 + x cubed plus another x cubed + x - x 5 - x cubed over x^2 + 3.

[00:19:15]X 5 cancel and x cub cancels. Given x cub + x over x^2 + 3 x x^2 + 1 over x^2 + 3 and with the state the asintope of f ofx well we've now simplified it as much as we possibly can but we could divide the top by the bottom just to check it's not just a state here so it's going to be plus or minus root3 but I just want to do this check so if If I divide x cub + x by x ^2 + 3 x2 goes into x cub x * x * x 2 is x cubed x * 3 is 3x take away you get x - 3x is - 2x.

[00:20:06]So f ofx is x - 2x over x^2 + 3. So we've got a slant asintope at y = x because f ofx= x and then usually you can't normally it can't be zero but if you look at this x^2 + 3 then if x2 + 3 = 0 x^2 would be - 3 well you can't square a negative number. So that is not going to give you an asmtope. So the only asintope you've got is actually y= x for this one. So watch out for that and we're done there. Find the first derivative of this. So product rule in the first bit but is implicit differentiation. So implicit differentiation. Let's do the first one.

[00:20:50]We've got y ^2 * 2x v u dash + u v dash.

[00:20:56]And then the next term 2y is 2 y by dx.

[00:21:00]And then the right hand side log x is 1 /x - 6 x is - 6 and 3 y is 3 e 3 y d y by dx. Move everything with dy by dx to the left. x^2 2x^2 y d y by dx + 2 d y by dx - 3 e 3 y d y by dx equ= 1 /x - 6. And then you've got 2x y^ 2. So - 2x y^ 2. And now just divide by everything next to the y by dx. The y by dx is 1 /x - 6 - 2x y^ 2 all over 2x^2 y + 2 - 3 e 3 y and we're done there. It's been math today. We did arcade maths 2026 advance higher math people 1. See you soon. Remember last minute live streams sec for me half past 7 for advanced higher maths. Join. No, it's on the4