Advanced Higher Maths 2026 Paper 1 Full Solutions

Hinzugefügt: 2026-05-09

[00:00:00]Qualification Scotland of Van Hamath 2026 paper one nice little paper this one hour it was on the 7th of May let's get into it straight away 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 UK.

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[00:00:47]Qualification Scot advanced high math 26 paper one question one we had differentiation we've got the product rule and the quotient rule so question a remember the product rule is v u dash plus u v dash so let me say that u is 3x 4 and v is se 2x then u dash I can see is 12 x cubed and v dash is well sec 2x at the start of the exam paper and that is a sec 2x tan 2x but then because of a chain rule I need times by two so the y by dx is equal to v u dash so 12x cubed se 2x and uv ddash So 3x 4 * 2 sec 2x tan 2x. So that is 12 x cubed sec 2x + 2 3s is 6 x 4 sec 2x tan 2x.

[00:02:11]It doesn't say simplifying your answer in this one. So I've got a feeling that we'll stop there. But let's just to be on the safe side, take out common factors. 6 x cubed se 2x is a common factor. 6 2's is 12. So that will just be 2 + x tan 2x.

[00:02:32]But to be honest, I think the previous one is fine. Part B, the quotient rule.

[00:02:37]quotient we remember is v u dash minus u v dash over v ^ 2. So in this case u is e 5x u dash is 5 e 5x v is 2x + 1. So v dash is just 2. So f-x subbing in we've got v * u dash. So 5 e 5x 2x + 1 - e 5x * 2 cuz it's u v dash.

[00:03:17]So 2 e 5x all over 2 x + 1 all squared. Now we have to simplify. So factorization is simplified not expanded. So the bottom is already factorized. I'll just leave it alone. The top I can take out a common factor of e 5x.

[00:03:38]So the first term is 5 * 2x + 1 minus 2.

[00:03:46]And then I can expand that bracket out on the top to see if it gives me anything.

[00:03:52]So that gives me e to the 5x 5 twos is 10 x 5 1's is five but you take away two so you got plus three all over 2x + 1 all^ 2 and I think we're done there.

[00:04:07]Qualifications Scotland advanc maths 2026 paper one question two gausian elimination solving these systems of equations. So I just set it up. 1 1 - 1 9 2 - 1 3 - 2 3 2 - 2 21. And the game with this is to get a zero in the lower triangle. So you can do two rows at once if you want. So I'm going to do row three minus 3 row one. And I'm going to go row two minus 2 row one. And that will give me the zeros in the right place. So we've got 1 1 - 1 9 row 2 - 2 row 1 2 - 2 is 0 -1 - 2 is -3.

[00:04:55]Then we've got 3 + 2 is 5.

[00:04:59]And then we've got -2 - 18. So - 20. And then we can do our bottom row. 3 - 3 is 0. 2 - 3 is -1.

[00:05:11]- 2 + 3 3 - 2 is 1 and finally 21 3 9 is 27. So it's 21 - 27 which is -6.

[00:05:24]Now we need to get a zero here. But we can only use these two rows. So we can just say I've got three row three then which will give me minus3 take away row two.

[00:05:37]So, we're still freezing these rows and we're just writing them out again.

[00:05:44]And then zero, just double check. -3 - -3 is - 3. Add three, which is nothing.

[00:05:51]3 1's is 3. 3 - 5. 3 - 5 is -2.

[00:05:58]And 3 * 6 is 18. So, I've got - 18 + 20 is 2. So you get a nice thing there. And you just take your time on these. It's really easy to just mess a number up and the whole thing's gone.

[00:06:11]Okay.

[00:06:13]So now we can just say when we're finished that that means the bottom one is remember it's xy z. So - 2 zed is 2 which means zed must be 2 / 2. Middle row - 3 y + 5 z = -20 - 3 y - 5 is -20. So - 3 y is - 20 add 5 and therefore y must be five. So we got a y we got a zed. Top row x + y - zed is 9. x + y - z = 9. Our x we don't know but we've got + five and then we've got minus one. So min -1 is + one that means that x must be three. 5 + 1 is 6 + 3 is 9. We've solved the system of equations.

[00:07:10]So we're done there. Just a note using gaus elimination to solve system of equations is actually find the point of intersection between three lines. So if you're ever doing vectors that's a way to do that. Scotland qualification Scotland 202 qualification Scotland 2026 advanced high maths paper one question three complex numbers were on a complex numbers divide by zed def qualification Scotland advanced high maths 2026 paper one question three a complex numbers defined by zed is root 3 plus i express zed in polar form so polar form is the cost and the sign you need the argument and the module modulus. So let's go for the modulus which is the size of r that's just by the square root of roo<unk>3 plus not i 2 it's just one i. It's just the part how many you've got roo<unk>3 2 is 3 + 1<unk> 4 is 2. So now we need the argument.

[00:08:13]So tan theta is imaginary over real because imagin goes up the way real goes along the way vertical horizontal one over root3 exact value triangle then usually leave it in radians but in case you think in degrees I'll put it in degrees for you 1 2 3 1 /<unk>3 is 30 so the inverse tan of 1 /<unk>3 three is not 30 but pi / 6. Now we do need to check what quadrant we're in. So just looking at your original one. If you just draw a quick argan diagram, you've got imaginary and real here.

[00:08:58]It goes along root three and up one.

[00:09:03]So you can see we're in the first quadrant. So that means that theta is pi / 6. If we're in another quadrant, we pick the one it matches. So now we write it in in polar form. So that means that you've got z equ= 2 cos<unk> / 6 + i sin p<unk> / 6 and we're done there for part a. Part B use theorem to show that z cubed is imaginary. Okay. So demo theorem for part B start of the exam paper. The vers the says that you can do two and put cubed here and then you just times the angles by the power which is three. This is quite an easy one this as well. 3 p<unk> / 6 I sin 3 p<unk> / 6. So I'd imagine that since it's purely imaginary cos 3<unk> / 6 is zero but we'll check 2 cubed is 8. That's cos of p<unk> / 2 i sin p<unk> / 2.

[00:10:02]In case you don't know, there's your cos graph there. It goes pi / 2 here which is zero and your sign graph pi / 2 is there which is 1. So that means that your z cubed = 8 cos i / 2 is 0 + i * 1.

[00:10:24]Well that means that that means that z cubic is equal to a i which is purely imaginary.

[00:10:34]So we're done there. PK Scotland advance higher maths 2026 paper one question four. Find the particular solution of the differential equation. It's a second order equal to zero as pretty much predicted. So auxiliary equation that's like a quadratic. You've got two squared m^ 2 - 3 m + y. So + 1 = z. And you solve that.

[00:11:02]Nice little quadratic to solve. We've got 2 m and m one and one minus and minus 2 m - 2 m - 3 m. Yep. So you get m = a half from the first one and m = 1 from the other one. So two there. So that means that you're So now we need to find y. Now usually you find it as a complimentary function and then a particular integral and combine them in some way. However, it's equal to zero.

[00:11:29]So we just say that y's general solution is a e to the half x plus b e to the x and then go to initial conditions. So when x= 0, y = a + b because e= 0 is 1 and a + b = 2. So one equation and then d y by dx is minus one when x is zero. So we need to find the y by dx. So dy by dx just differentiating the y general solution is a half a e x plus b e x and when x = z the y by the x is equal to -1.

[00:12:16]So subbing that in we get a half a + b = -1.

[00:12:24]So we got two equations to solve. A + B is 2 and a half A + B is minus 1. So let's write them together.

[00:12:32]Take away a half a minus A or A minus a half A. I'll just take away up the way is a half a 2 - - 1 is 3. So a is 6. Now we just sub back in. A + B was equal to 2. 6 + B = 2. So b must be min -4.

[00:12:55]So that means that our y is a e to the half x plus b e to the x.

[00:13:04]And we're done there. As higher maths 2026, paper one, question five. Matrix A is defined by this matrix. State an expression for the determinant of A in terms of X. The determinant of a just call it a or you can put bars around it is just that times that minus that times that. So let's just state 3x minus 5 * - 2 is + 10. That simple. Matrix A is multiply matrix B such that the determinant of A is 12 X + 40. State the determinant of B. So it's just testing.

[00:13:38]Do you know that the determinant of AB equals the determinant of A times the determinant of B? Hopefully you do know that determinant of A is 12 X + 40.

[00:13:51]That equals the determinant of A, which is 3X + 10 times the determinant of B. And therefore that B must equal some number * 3 is 12. Four 3es is 12. 4 10 is 40 of B must be four and we're done there.

[00:14:12]Okay, question C. The inverse of B is this. Find B. So question C, let's just take a note of that matrix cause inverse of an inverse is back to itself. So we just need to do the inverse of the inverse. So B is B inversed inversed. So how do you find inverse of a matrix?

[00:14:36]Well, it's one over the determinant times D minus B, C minus C, A. In other words, if that's A, B, C, and D, B, switch places, but B switch sides. So, let's do the determinant first. The determinant of our inverse is three halves minus and then we've got -1 * 5 quarters is 5 quarters minus * a minus is a plus. So that's 34 3 6 quarters - 5 quarters is a quarter. So that means that our in our B is 1 / a4 times 1 and 3 half switch places. So three halves and 1 and -1 and - 5 quarters become 1 and 5 quarters.

[00:15:37]But we need to tidy that up. 1 / quart is four. 3 over 2 1 5 over 4 1.

[00:15:47]Might as well put that back in to get rid of some fractions. 4 is 12. 12 divided by two six. Four ones is four.

[00:15:54]And then that gives you five. And that gives you four. And we're done there.

[00:15:58]Qualification Scotland advance high maths 2x people one. Question six. Use the substitution u= x - 1 to integrate that. So integration by substitution.

[00:16:08]Find the exact volume of a solid rotated around with a curve y = 2x x - 1^ 2. the x- axis 2 pi radians from 0 to 1. So if it's around the x axis it's p<unk> y^2 dx. So part a u is x - 1.

[00:16:27]Now this is a tricky one this one because if I differentiate that du by dx I'll show you du by dx is going to be well just one which means dx is du but I've still not got my x I've got u is x - one so what about this extra x where I can just rearrange that for x is u + one so now I can integrate because my integral becomes the integral of x is u + 1.

[00:17:04]x -1 is u and then du. Now that looks still like a thing that doesn't really do much, but you can just expand the brackets as an integration by parts. So that's the integration of u 5 plus u 4 du. That's easy cuz that's just a six U6 plus a fifth U5. Don't forget your C.

[00:17:31]And then you need to switch it back out.

[00:17:33]U is X - one. So that means it's a 6 X - 1 6 plus a 5th X -1 5 + C and we're done there. So part B says hence find the exact volume of solid form by rotating the curve with equation y = 2x - 1^2 about the x ais through 2 pi from 0 to one. Now doesn't look like this hence seems to be leading from anywhere. So just follow through and see what happens. The volume we'll just write what our volume is is the pi * the integral between 0 and 1 pi y^ 2 dx.

[00:18:14]So what is y^2?

[00:18:16]Well y^ 2 is the whole of 2<unk>x x x -1 squared squared. So we need to square all of that. So squaring all the terms separately. It's just a product. 2^ 2 is 4. <unk>x is x. x - 1 2 is x -1 to the power of 4.

[00:18:40]So now we're doing the integral between 0 and 1 and pi of 4x x - 1 4 dx.

[00:18:50]Well, does that look like anything we've just done? This integral start with xx - 1 4. So if I make it look like that by taking the four outside, it's integral between 0 and 1 of x x - 1 4 dx. So that bit is literally just the answer that we just got. So we can just sub that in instead of doing integration again. So that is 4<unk>i * a 6 x - 1 6 plus a 5th x - 1 5 all between 0 and 1.

[00:19:27]So subbing n that's 4<unk>i * 16 1 - 1 6 + 15 1 - 1 5 take away 16 0 - 1 6 plus a 5th 0 - 1 5 and just watch your signs on this. So taking that 4 pi still hanging out. If we look at our terms, 1 - 1 is 0 and that's zero.

[00:20:00]So then we've got a minus. Watch the minus times a - 1 to the^ 6 is a 1. So it's 1 * a 6.

[00:20:13]A - 1^ 5 is - 1 * a 5th - a 5th.

[00:20:21]So that is 4<unk>i - 5 - 6 over 30.

[00:20:34]Or do it any way you want. 5 - 6 is -1 over 30.

[00:20:42]So that's just - plus. So it's 4 pi over 30 or 2 pi over 15 units cubed and we're done there. Multification Scotland advance high maths 2026 paper one question seven more complex numbers.

[00:20:59]A complex number z = 2 + i is the root of a polomial and it's a quartic. State the second root. Well the second root is just complex conjugate. So zed is equal to 2 minus i is another root. So we're done there. Part B, find the remaining roots. Well, if you got two linear roots, you can form a quadratic factor. So our factors come from our roots. You need to start from there first. So we've got our factors are z - 2 + i and z - 2 - i. And we're going to just times them together.

[00:21:39]So expanding the brackets carefully we get z^ 2 minus z 2us i - z 2 + i and then - 2 + i - 2 - i is plus 2 + i 2 - Okay.

[00:22:09]Now, there's tricks to do this, but I just want to show the full thing. So, that gives you zed squared from the first term. And if you look at this, you're going to have - 2 zed - 2 zed's - 4 zed. But what about the i's? Well, you're going to have minus * minus is plus z i minus z i. They're canceling out. Similar at the end, you get a lot of cancization. So, we've got two. Remember my twos, my Z's have got a line through them. So, if it's not got a line through them, it's a two. So, I've got 2 * 2 is 4. But then I've got - 2 I + 2 I is cancelled. So, it's 4 - I 2. I - 1. So, - I + 1. 4 + 1 is 5. So, there's a quadratic factor. And now we can and now we can divide. So let's divide our thing using I always use log division. You can use there's other ways, but we've got z 4 - 2 zed cubed - z^ 2 + 2 zed + 10 all divided by z^ 2 - 4 z + 5. So zed squ.

[00:23:33]Then we times 3 by z^ 2 to get z 4 back - 4 z cubed.

[00:23:39]Z^ 2 * 5 is 5 z^ 2 taken away. Then - 2 + 4 is 2 z cubed - z^ 2 - 5 z^ 2 is - 6 z^ 2. I've still got + 2 zed. I've still got 10 zed^ 2 into 2 zed cubed is 2 zed * 3 by 2 zed then to see the remainder that's 2 zed cubed 2 4 is 8 zed^ 2 5 is 10 zed take away - 6 + 8 is 2 zed 2 - 10 is - 8 zed + 10 and then we can go again cuz Z 2 still goes in. Z^ 2 into 2 Z^ 2 is 2. So + 2 * 3 by 2 you get 2 Z^ 2. Two fours is 8 Z 2 is 10. If you don't get the same here, you've done something wrong above. We take away to get zero. Remainder is zero. So we've got a new factor. So zed^ 2 + 2 zed + 2 is a factor. And we're trying to factor find the remaining roots. Yep. So we got z^ 2 + 2 zed + 2 is a factor. So that equals zero.

[00:25:01]Well, you just need to try and solve that. It has to be the quadratic formula or completing the square. But most people do quadratic formula. So zed = -2 plus or minus the square root of 2 2's is 4 - 4 * 1 * 2 over 2 * 1.

[00:25:22]That's - 2 plus or minus the square root of two twos is four. Four twos is 8 all over two - 2 plus or - -4 over 2. But we can square root a minus number that's four. This becomes 2 I and - 1 plus or minus I is the final roots. Or to write that out better, minus one plus I or minus one minus I.

[00:25:53]One two and we ended up with the ones that we started with earlier on. 3 4 2 minus I 2 plus I and we're done there.

[00:26:02]It's been qu math today. We did qualification Scotland advanced higher maths 2026 paper one. Get on to paper two soon. Take care. Stay safe.