2026 SQA Advanced Higher paper 1 review and quick solutions

Added: 2026-05-12

[00:00:04]Well, this was a straightforward paper.

[00:00:07]Quite a few easy questions. It was certainly a paper where you could harvest your marks to gather a bucketful of them in preparation for what paper two might have in store because they were fairly easy to achieve in most of these questions. So, what were they like?

[00:00:22]So, question one, just what you'd expect for differentiation, product rule, quotient rule. Only thing about this question was the resulting expressions were just that wee bit cumbersome.

[00:00:32]Enough to make you feel uncomfortable that maybe you've done something wrong.

[00:00:38]Well, there is the first one. So, it's Y so write dy by dx. So, it's just the usual. Differentiate one, leave that alone. Now, leave that alone and differentiate that. Can't remember? It's at the front. Don't forget it's a function of a function. So, there's that little two multiplying in there. Now, you just got to think, well, what can you do with a slot? Well, apart from just making that a six which isn't an awful lot. Now, it may well be that that's all they want. Quite often they do that, but a lot of it gathers up.

[00:01:02]They've got quite a lot in common.

[00:01:05]So, that just leaves a two for the first one and you can see all you've got for the second bit is an x tan x.

[00:01:10]Right, so there's that one. A wee bit A wee bit clumsy looking.

[00:01:15]Now, part B was the quotient rule.

[00:01:18]This time they're using f of x. So, make sure you write down f dashed x.

[00:01:23]Don't know if there's any marks for that. I'm really I keep forgetting to check that.

[00:01:27]Well, it's the usual pattern. Square the denominator so then they each take their turn. Now, the difference system is remember to do a subtract when it comes to the second part. Now, this one specifically said tidy it up. Well, you can see an e to the x in common so you could take that out first or you could multiply the brackets first of all.

[00:01:45]Like what I've done there.

[00:01:49]And then you've got a 10 x and you've got a three. You may as well put that linear expression first. So, 10 x plus three lots of e to the 5x over that denominator.

[00:01:59]And that's it simplified.

[00:02:03]Uh the Gaussian elimination. Quite a little innocuous one, a harmless wee beastie this time. There are no unknown constants for instance to juggle with.

[00:02:11]It's just a case of work this out and it all works out very [clears throat] neatly.

[00:02:16]So, if you know what to do, you transfer all of those coefficients into an augment There's the join. I can see the join. There's the augmented matrix.

[00:02:24]It's just very wrote this, isn't it? You just There's lots of micro arithmetic.

[00:02:29]Now, the plan is to get it into upper triangular form. You've got to get zeros in these two positions.

[00:02:36]So, the plan there would be subtract two lots of row one from row two. And the plan here would be subtract three lots of row one from row three. You just have to do these very carefully. So, I'm not going to go through the tedium of them.

[00:02:49]So, those two rows are fine. Now, you want another zero in the bottom there.

[00:02:53]So, it's going to be three of row three take away row two.

[00:02:57]And that should achieve what you want.

[00:02:59]Nice neat numbers here. Two is a good start and then a nice even number. There we go. Notice this Wait, that backwards negative two. Those were the z's. That's the z column.

[00:03:08]Step back up a line. That's negative three of the y's. But, I know what z is now.

[00:03:13]So, find y from that.

[00:03:17]I'm not going to go through it the trivial arithmetic. Back up to the top row to get x cuz now I know y and z. So, you just pop in them in and then finally you've got the answer for x and there you are. You probably do like that. I just like to finish off in this little line. So, there's the solution.

[00:03:33]Straightforward.

[00:03:36]Here we go, complex numbers. That first part is even if it just says express it in polar form, it's easy enough. I'm not for sure whether you can just draw a little quick sketch in an Argand diagram because it's crying out there. Look at that root three and one. So, you know the answers. Can you just state them from that or do [clears throat] you have to go through the rigmarole?

[00:03:53]Well, you're probably safer just going through the rigmarole, but apart from that it is a straightforward equation.

[00:03:59]Look, it's only power 3 as well.

[00:04:02]So, that are easy one.

[00:04:05]Right, well, on a diagram there is you've got a 1 2 root 3 triangle with 30° there pi upon 6. You could go in with that, but you're probably safer just spelling it all out.

[00:04:15]So, there it is just Pythagoras obviously for that part. And then for the argument you're just going to be using the tangent opposite over adjacent there.

[00:04:25]Then you can put it together, which you could have done straight away I suppose.

[00:04:29]>> [snorts] >> The old cos plus i sin form or polar form.

[00:04:34]Now, in part B they said, "What about z cubed?" Well, that just means you cube the two parts.

[00:04:41]You cube the three and you cube the bracket. And the bonus was if you're cubing the bracket you multiply the arguments by that. So, this will simplify right down to just pi upon two, so that's nice and easy.

[00:04:54]As long as you remember what they are you can always just draw a quick sketch.

[00:04:56]There we go. Cosine is at zero and the sine is at one. So, you end up with just a purely imaginary answer. So, now you just make that statement.

[00:05:06]It's purely imaginary. Maybe give a reason as well, but not sure. Because there's no real part.

[00:05:12]So, there's another four marks to stuff in your pocket.

[00:05:16]And here, the second order differential equation question. And look at that. Of all the horrors you could have like the ones where you've got a trigonometrical expression in x on the right hand side add the complementary function it throws another one at you on the left in the same angle.

[00:05:32]Here, this is a a veritable little cuddly toy in comparison.

[00:05:38]There you go, nothing on the right hand side, but you've got some initial values to get a particular solution. But you start off the usual way. What's the auxiliary equation? Just pick out those coefficients to form the quadratic and then you solve that and they've got nice neat answers, 1/2 and 1, which means that what would be the complementary function is just A lots of that one and B lots of E to the other one, but since there's nothing else there, that is your general solution.

[00:06:05]So, now you can just go and differentiate it and then use your initial conditions to find the A and B.

[00:06:11]So, differentiating it, easy enough. X is 0, that's good because it knocks all these E's out. That's just E to the 0.

[00:06:18]That's just A and B. So, A + B is 2.

[00:06:21]That's one equation.

[00:06:23]And the second one, again, X is 0.

[00:06:26]Y is -1 this time.

[00:06:28]So, what you've got is it's just going to be 1/2 of A + B is -1. It's the second one. Now, it's just trivial simultaneous equations.

[00:06:38]Subtract them to get rid of B. That leaves you 1/2 of A.

[00:06:42]1/2 of A is going to be 3, so A is 6.

[00:06:44]And then go back to number one because that was easy.

[00:06:48]So, 6 and B is going to be 2, so B is going to be -4. Now, go back up. Oh, wait. Wait. Wait. Underline all of these first of all so we don't get distracted.

[00:06:57]Now, you can go back up and fill that in. So, for your particular solution, Y is going to be What was A? 6. And what was B? -4 lots of that.

[00:07:08]And there we are.

[00:07:10]Ah, the 2 by 2 matrix question. A nice little I quite like this. A nice little twist here instead of asking us to find the inverse, it gives you the inverse.

[00:07:18]So, instead of just shuggling those elements there to form the adjugate, it gives you the adjugate and you've got to unshuggle them back into its original form. Plus, you've got to remember that the determinant of the product is the same as the product of the individual determinants.

[00:07:34]A nice wee question, though.

[00:07:37]Right. Well, first bit, what's the determinant of A? Well, you know the pattern. Product of the main diagonal minus the product of the other diagonals. A little bit of arithmetic there and you get that expression.

[00:07:47]Now, in part B, it gives you an expression for the determinant of the product of A and B.

[00:07:54]That one there.

[00:07:56]And well, you just need to remember that the product the determinant of product is equal to the product of the determinants. So, that if you want the determinant of B, then just take that determinant of A across and divide by it. And you can see it goes in exactly four times.

[00:08:12]As for the final part in part C, it tells you the inverse. You don't have to find this. It tells you the inverse of B. And there it is with these little suspicious fractions here. You can see the four is lurking in the background.

[00:08:23]Now, the way you get the inverse is you do one you divide the adjoint by the determinant. The adjoint's that shuffled up one.

[00:08:30]Well, that means that the adjoint would be cuz you know the determinant, if you multiply that inverse by it, that'll take you back to the adjoint. That was the shuffled up ones. So, just multiply them out. Now, you all you got to do is unshuffled that. So, B would be what was it did? You reverse the main diagonal, so just put that back. And you made the other diagonal into the negatives, well, just un-negative them.

[00:08:52]And there it is.

[00:08:57]And here, oh, it's actually the more demanding question, well, only demanding in compared to the relative ease of the previous ones. It's just a composite.

[00:09:05]They've knitted together substitution integration by substitution [clears throat] and a volume of revolution. So, it's just two standard techniques that have been put together.

[00:09:14]So, you should be fine as long as you don't go into panic mode at the illusion of complexity that might be there.

[00:09:22]So, differentiate to get the differential substitutions when there was a look to that. Now, you can just substitute it cuz that's just U. Now, that needs to get rearranged cuz X can be put in terms of U.

[00:09:34]And then dU and there you are. Multiply those brackets and you've got two harmless wee terms to integrate. Add one to the power, divide by the power. Same again. Don't forget plus C.

[00:09:45]Now, you have to go back to X's or so because that's the whole point of it. There's no limits there. You're not working anything out, so go back to X's.

[00:09:53]And that's the result.

[00:09:55]All right, popped up to top to get more room. So, volume pi r squared dx where y is the radius of the revolution. So, v equals pi times now from 0 to 1 of that squared. So, you've got when you square that you end up with what you had to begin with. Well, apart from that four, but that's just going to get popped out anyway. So, now just copy that down.

[00:10:17]So, you've done the work for that already in part A. But now you've got to work it out. But since you're in one, that's particularly easy. Now, when it's one, they all come to zeros, so that whole lot just goes. And when it's zero, unfortunately, they come to negative ones. So, just be careful with your signs here.

[00:10:33]So, working that out. So, that's still 4 pi.

[00:10:36]Take away a negative, so that's positive. Put that first and that'll be subtract that part. And then just go to do a fifth take away a sixth. They obviously they're going to 30, don't they? That'll be 60 could be five. So, you've got 30, but that'll cancel down because two goes into them, so you've got 2 pi upon 15 units cubed for your answer.

[00:10:55]Ah, and for number seven, what we've got here, we've got that old chestnut that makes a reappearance. I don't think I've seen it for a while. We've got that favorite of candidates of your factorizing a polynomial. We've got a quartic in this case, but one which has got some or maybe all imaginary roots.

[00:11:10]But it's a technique that you should know. So, it's just a standard question.

[00:11:16]All right, so for this quartic, if you've got one complex root, the other one must be its complex conjugate, one with the opposite sign in it. Because that's what happens when you solve a irreducible quadratic. You end up with plus or minus the discriminant. All right, now part B said now you've got to do the whole lot, because you can hardly see this." So, reconstruct that irreducible quadratic then. So, if those were the roots, those would be the factors. Now, just a simple pattern.

[00:11:41]Square the first, then you've got the two inner ones. Now, it's handy when you add conjugates, cuz the imaginary parts disappear. Now, when you multiply conjugates, you end up with the sum of the squares. And so, you just end up with this. So, there was that irreducible quadratic that you had to solve to get those two separate roots.

[00:12:00]Now, what was the other one? Well, if that's a factor, it's got to divide in exactly. So, now you're just going to have to go through this big algebraic division. But, you know the pattern.

[00:12:09]To make a zed four, it must be a zed squared. Multiply it out.

[00:12:14]Take it away to see if there's anything left.

[00:12:17]Ooh.

[00:12:18]Be careful here.

[00:12:20]And so, and then bring that down and do it again. To make that, I need a two zed. Now, multiply that out to see if it works. Or if there's still something I don't I still something left over. So, subtract it to see what's still left over. And bring the next one down. Now, you want that as your answer this time, or you've had it. So, hopefully this comes out exactly one 10. So, when it comes out exactly, you know that's all worked. Right. Now, you can rewrite that. I don't want to write it again. I'll give it a new one.

[00:12:46]So, that's going to be a There was that irreducible quadratic that you had. And here's the quotient. Which, in this case, it would be nice if it factorized, but just quickly check its discriminant.

[00:12:57]Its discriminant is four minus eight, or you've had it. So, it's another irreducible one. So, you're going to get another pair of imaginary roots. They'll be the complex conjugates of each other.

[00:13:07]So, square root of negative four is just going to be two I. Everything divides by two. So, you've got negative one plus or minus, so that's the reason for the conjugate pair I. So, what you've got in got negative one plus I and negative one minus I. And that's it done. Same with the whole paper. Woah, lovely marks.