2026 SQA Higher Mathematics paper 2 review

Hinzugefügt: 2026-05-10

[00:00:05]Well, I thought this seemed a fair exam.

[00:00:06]It was just full of standard surd equations, quite a good coverage. One or two little tricky points, perhaps, but you'd expect that in an higher exam.

[00:00:15]Right, so let's have a look.

[00:00:18]Oh, here's an innocuous wee question to start off with. What is it? Roots of a quadratic equation. You're just going to form an equation with the discriminant and then solve it appropriately. That should be an easy three marks.

[00:00:32]Right, equal roots, well, the discriminant should be zero. Pick it out. B squared, four, one times and there's C, C is K plus three. So, that's going to be a quadratic. Multiply that out. Not a lot to it.

[00:00:45]Now, factorize it.

[00:00:48]It's fairly straightforward. It's just a case of what what's that 12? Is it three or four or two or six? It's two or six cuz they've got a difference of four.

[00:00:54]So, it gives you your two answers. K is -2 or K is six.

[00:00:59]So, that was easy.

[00:01:02]Standard question, scalar product leading to what's the angle between two vectors. And it's all at the front. It tells you how to work out the scalar product and it shows you the formula you'll use to work out the angle. So, again, this should be a straightforward question.

[00:01:18]Right, so there it is. There's the components of them. You multiply the corresponding components together, the X's, the Y's, and the Z's. And then it's just a wee bit of arithmetic. You get the multiplications and you add them up and that's that part done.

[00:01:33]Then in part B, that's where you have to find the angle between the two vectors.

[00:01:37]Well, you've got that formula. So, you just write it down. Okay, it's not exactly like that in the formula sheet.

[00:01:41]That's just it rearranged. But you'll need those two magnitudes. Well, that's just Pythagoras in three dimensions.

[00:01:48]Square and add the components.

[00:01:51]And you don't really need to tidy it up.

[00:01:53]I know it comes to seven, but you're going to be using your calculator. Same with that one. Square and add the components. That'll give you the magnitude of that vector.

[00:02:03]Again, that one, 45, that could be simplified. There's a nine in there, it's 3 root 5, but just leave it because you'll be using a calculator. Now, pop them in. You've got five from the first part and you've got the root 49 and the root 45. So, get rid of that cos and it'll be over and it's just the one square root now.

[00:02:21]Oops, that was a 45. Put it in your calculator.

[00:02:25]Out it comes and you write down the answer. Come on now.

[00:02:31]83. etc. Round off to one decimal place.

[00:02:34]83.9 degrees. There you go.

[00:02:38]So, inverse of a fairly trivial little function. There's not a lot going on there. Divide by five and add on six.

[00:02:43]You could just do it in your head. Take off the six and multiply the whole thing by five. But you'd have to state you'd have to state it out. There's three marks at risk. You might get nothing if you just put the answer down. So, state it out. One way or another, I'm going to use the fact that a function acting on its inverse takes you back to X.

[00:03:02]So, in this case, let G act on its inverse.

[00:03:07]So, it'll do what it does. It'll divide it by five, it'll take its inverse, divide it by five, and add on six. And since it's acting on its inverse, that should take you back to X again. So, now it's just a case of getting rid of the bits and pieces.

[00:03:21]Get the six over and subtract it. Get that five across and multiply it. And as you suspected all the way along, that's the answer.

[00:03:31]This is a straightforward little question. Get the equation of a circle.

[00:03:34]You just need to know its center, it tells you that, and its radius. So, that's the whole point. Work out its radius and you can pop it into that little equation that you know.

[00:03:44]Right, well, you didn't you did a diagram. You just need to know the center and its radius. The radius joins the center to any point you know, and then you can use Pythagoras in that little triangle. So, difference in the x's, square it, difference in the y's, square them. Add that to the Pythagoras.

[00:04:01]And you'll get r squared. No, just leave it as r squared cuz when you pop this into that equation, x minus x coordinate, y minus the y coordinate, it's r squared that you want. So, just leave that 29, and that's it, finished.

[00:04:20]So, the reconciliation, that's another standard equation. You know how to use that to work out the following term.

[00:04:25]Again, you also know why a limit exists, when it exists, if you're multiplying by a proper fraction. Part C, for part C, there's nothing clever about that cuz there's no formula overall for any particular term. So, there you just have to plod through them, and that's perfectly valid.

[00:04:43]So, u one, just multiply u naught, which was the preceding term, by what it says, and add the five. Wee bit of arithmetic.

[00:04:51]There you go.

[00:04:52]Part B, why is it not got a limit?

[00:04:54]Because that number that you're multiplying by is too big.

[00:04:58]It's bigger than one.

[00:05:01]Or you could have said, well, that 1.125 doesn't lie between the negative one and the one. It's not a proper fraction.

[00:05:10]Now, in part C, where you've got to work out how many terms it takes, just use the answer button on your calculator, and then create that reconciliation with answer instead of u n. Then you just keep pressing the buttons, and they'll all pop out as required.

[00:05:27]So, if you set that up, start with two, put in the reconciliation with answer, keep pressing the button, and they'll all come out.

[00:05:36]I know in England they all come out with that as a fraction. And then you have to change them to a decimal, but I do it after anyway. So, you just keep doing that until you get to a number bigger than 30. Nothing very clever about this, I'm afraid. So, there there is. It happened at the fifth time. So, it takes five terms to get more than 30.

[00:05:56]So, the smallest number, value of n, is five.

[00:06:04]So, to transform the graph, you can expect to get that anyway, but there's not a lot going on with this one. This is a particularly easy one. All you do is you make it negative and you and you add one.

[00:06:15]But, just don't forget to mark in all the points.

[00:06:18]So, there's the graph there then of f of x. Now, there's no change to x or to sign, so nothing happens across the way.

[00:06:25]Now, f stands for the old y coordinates and they're getting made negative. So, what's below goes above and what's above goes below. So, in other words, it just flips over, flips over about the x-axis, and then it's got a plus one. So, up it pops one and that's it done.

[00:06:40]Now, get some get some of these points in.

[00:06:46]You were given four points, so put four points in it. It says make that negative and then add one. Keep the x the same.

[00:06:53]Make the y negative and add one.

[00:06:59]And that's more or less it. Let's finish that off there.

[00:07:03]All right. Here's a straightforward wee question about finding an area in a diagram. Usually, it's area between two curves, but this is the area between two curves. That'll save you having to think about all these negatives. The area is enclosed by the x-axis above y = 0 and that parabola below. So, that'd be the way to set it out. That takes care of all the signs. You'll get a positive answer that way.

[00:07:28]So, here it is then. The area is bounded by the x-axis above and the curve below.

[00:07:34]So, when you write down the area, the integral from where you start to where you finish of the what was above, it was y = 0. What was below, was that quadratic there.

[00:07:44]Now that will give you the proper answer with the correct sign.

[00:07:48]Zero makes no difference. Take that negative out. So answer's going to be the negative of it.

[00:07:54]Standard integration. Add one to the power, divide by the power.

[00:07:58]And then work it out. Now you could flip those side those intervals cuz of that negative, but I'll just leave it the way it is.

[00:08:04]And now it's just plugged.

[00:08:07]Just replace all those X's with a two.

[00:08:10]And then from it subtract replace all those X's with a negative one.

[00:08:16]And we have done that. We've got the whole thing be negative. I could have reversed that subtraction as well.

[00:08:22]But I just left it.

[00:08:24]You'd have been better off just reversing the limits to begin with.

[00:08:28]Pop it all into a calculator.

[00:08:30]And there you are. Nine upon two, nice positive answer straight away without you trying to mess about making excuses for yourself at the end.

[00:08:42]Right. The lines in a triangle creation, nobody expected that earlier. This is a particularly easy one. They'll only ask you for one line, a perpendicular bisector, and you're getting four marks for it. That's particularly generous.

[00:08:55]Part B though, for two marks, when it's working out those angles between lines, that's when it could be a little bit tricky for some.

[00:09:04]So it's a perpendicular bisector. That means it cuts it in half and it's at right angles to it. Now you need an equation. So you need a point on it and its gradient. Now we get that point first of all because that's just halfway between A and B.

[00:09:17]Half the coordinate add them and half them. Get the average of the coordinates. A little bit of arithmetic.

[00:09:22]And that's you've got the point on the line, the only point you know. Now the gradient. Get the gradient of AB because you can compare it to that difference in the Y's over difference in the X's.

[00:09:32]And that cancels down to a nice negative two, which means that that line must be the negative of the reciprocal. So, that's a half.

[00:09:40]Now, pop it into the y - b = mx - a. y - the y coordinate, gradient, x - the x coordinate. Take the two close and multiply.

[00:09:49]Don't forget that's a plus.

[00:09:52]And then, make sure the numbers are combined. And that'll do.

[00:09:57]Now, in part b, it's asking for an angle. It says, "What's the obtuse angle between the line l1 and the line ac?"

[00:10:05]Well, it's going to be the gradient and the tangent of the angle you're going to be using. So, it gives you the equation of ac.

[00:10:12]But, you'll need to rearrange that into y = mx + c so you can pick out the gradient. So, there the gradient's -1. A gradient of 1 is 45°.

[00:10:22]So, that's 45° down that ac is going.

[00:10:26]What about the line l? How's that heading off?

[00:10:29]Well, you know it's gradient. It's gradient is a half. And you know that the gradient's the tangent of the angle.

[00:10:34]So, that means it'll be the inverse of the gradient, which is inverse tan of a half. Calculator time.

[00:10:41]So, pressing the buttons will give you the angle. That's the angle up from the x-axis. In other words, up from the horizontal.

[00:10:48]It doesn't need to be the x-axis.

[00:10:50]So, the clue to doing this part is don't think of the x-axis. Just think of the horizontal through it.

[00:10:56]One of them is going down at 45. The other one's going up at 26. And if you want that obtuse one, it's going to be the whole thing adds up to 180. So, take them away.

[00:11:06]Gives 108.4 for the final answer.

[00:11:12]So, this one, determine the range of values for which it's strictly decreasing.

[00:11:17]Again, that's just a standard question, isn't it? You just get the derivative and find out when it's negative.

[00:11:23]So, there's nothing to complain about in this one.

[00:11:28]Well, get the derivative then. Same routine. Multiply by the power, take one off the power, and if it's a constant, it doesn't change. So, it'll decrease if that derivative is ever negative. So, find out when it's negative, you form this inequation.

[00:11:42]Now, there's a common factor there, 12x.

[00:11:44]You take that out.

[00:11:46]Looking for Those are the answers for the equation, not the inequation. The x is 0 and 2.

[00:11:53]So, to find out what the range is, sketch the graph of it. You know that it cuts at 0 and 2, but you're not including including 0 and 2 is negative between them. That's the justification for saying that x must be less than 2 and greater than 0.

[00:12:10]So, the double angle equation. Well, of the two, the cosine one, that's the worst one because you can end up with a quadratic, so there's there's more working to do.

[00:12:18]But, apart from that, it's still again a standard question.

[00:12:24]So, the first part there is change that cos 2x, the double angle, into cosine to match at the end. So, just get the formula, which you should remember or you get from the front.

[00:12:36]Right, so that means you've got 6 cos squared, then the 10 cos x, and then the number part. So, there's a quadratic. Now, take out the two because that'll make it easier.

[00:12:47]Left with this quadratic here then to factorize. So, you can forget the two because that's not zero. So, it must be 3 1 for the first part, and it must be 1 2. 3 2s are 6. So, those are two solutions. Either the cosine is equal to positive a third or negative two. Now, it can't be negative two because it could only go up to 1 or down to negative 1. So, no solution for that part. So, down to just two possible solutions now. Get the inverse cos, pop it into your calculator, and the answer that it'll give you is what? It's going to be 70.5 etc. degrees. That's only one of them, of course.

[00:13:22]So, the other one, you can either think of the graph or use let all sin tan cos table. It's going to be 360 minus it.

[00:13:30]So, it's either the 70.5 or 360 minus that.

[00:13:34]So, those are two solutions. 70.5 or taking that away from that, 289.5.

[00:13:41]And notice there's no degree signs there anymore.

[00:13:44]There, that wasn't too bad then.

[00:13:48]Oh, the logarithmic graph question.

[00:13:50]That's probably one that quite a few don't want to see, but it is a standard question. There's two ways of going about it. Either get the equation of that line the way it is and try and remove the logs to get back to Y equals or start from Y equals and introduce logs and compare. I'm doing it that way.

[00:14:06]I'm going to start with the Y equals AB to the power X and then convert it and compare.

[00:14:12]So, either the two ways, you can either start with the graph and go to the Y or start with the Y and go to the graph.

[00:14:17]Starting with the Y, right. It's in log form, so take log base 4 of both sides.

[00:14:22]Now, you start to start to use the rules. So, if it's a product, that will split into the sum of the logarithms and then that X can pop out. I'll just put that at the front there.

[00:14:31]So, it looks a bit more like the form of that line cuz that looks like Y equals MX plus C.

[00:14:37]Where log 4 B would be the gradient and C is the log A. So, that M the gradient of that line is difference in the ups over difference in the alongs is two.

[00:14:49]So, that log B, which is multiplying X, must be the gradient. So, that must be two. So, that means B is going to be Take the four across, four to the power of two, which is 16. C is where it cuts the Y axis. It cuts it at three. So, that log base 4 of A, that's meant to be an O, it looks like an A. That little log base 4 of A is three. So, again, take the four across, it'll be four to the power of two three, which is 64. So, there's the two solutions.

[00:15:14]Now, you don't need to write it out in the original form, they didn't ask for it. But, if you were to, then you would just write 64 * 16 to the power X.

[00:15:22]Ah, here's that type of equation which can get confused with, you know, find the stationary points and determine their natures because the nature of the stationary points don't match in this.

[00:15:31]This is a max- maximum and minimum value in an interval creation. The only significant thing here is that there is a stationary point there to contained with the two limits of the interval.

[00:15:43]But it is again, it's a It's a standard technique.

[00:15:48]Now, it's just asking for the X coordinates of stationary points. You're still looking for stationary points to differentiate it. Usual technique, multiply by the power, take one off the power. And since it's a stationary point you're looking for, that derivative should equal zero. So, that means you've got that quadratic equation to factorize.

[00:16:04]So, you factorize that. Must be two and two to give a six and a two.

[00:16:08]Which gives you two solutions. X is 2/3 or X is -2.

[00:16:12]And that's the first part done.

[00:16:14]No finding the nature of the stationary points or anything like that. Now, the second part says if X is only allowed to be from -1 to 1, what's the biggest answer and what's the smallest answer you can get? Well, the two main contenders are the two limits, -1. Put it through that.

[00:16:33]And that comes to six. The other major contender is the one because usually the maximum and minimum lie at the outsides of the interval.

[00:16:42]Unless there's a stationary point inside and there is one. There's a 2/3. That's a possible contender. But now you're going to have to put 2/3 through all.

[00:16:53]Now, that's a bit of a pain but you're using a calculator for this, but just use the answer button.

[00:16:58]Just do 2/3 equals and then use answer cubed, answer squared and so on. And then when you press the buttons, you'll get -13/27.

[00:17:07]So, that did turn out to be one. So, the maximum was six when X was -1 and the minimum was at -13/27 at X = 2/3.

[00:17:18]Now, they didn't actually ask for that part. It was just those values that wanted.

[00:17:22]Uh the standard wave equation question then where So, you've got the first part where you have to turn it into the form of a sine or a cosine with an amplitude and phase, but there's always a part B.

[00:17:32]Here the part B is actually not too bad because they've already drawn a graph.

[00:17:36]So, it's not like one where you have to draw the graph, especially in radians.

[00:17:40]Just watch out for that little point P at the end though. That's the one that we sort of capture it.

[00:17:47]All right. So, just expand that then the way we usually do, but you should remember the formula. If you don't, copy it from the front. The degree signs are up there.

[00:17:55]Now, I like to pop the coefficients together separate from the variable part of the term, but you don't need to do that. It just makes it easier to identify. So, that K cosine must be the two for the sine x term, and that K sine A must be the seven, and they're both negative, so that just keeps it as a negative seven.

[00:18:13]Now, if you square them using Pythagoras, you'll get the size of K, that's root 53, and if you divide them, you'll get a tangent. The tangent should be 7 up on 2, so the inverse will give you the answer to that. Pop it into your calculator. 24.1. It's just 24.1, it's not degrees.

[00:18:32]And to confirm that, everything was positive, so there's only one place that happens, so that is the correct place for the angle.

[00:18:39]I'll just write out again there, root 53 and then 74.

[00:18:44]All right. Let's not We're not going to talk to get through there. There's that original curve from that equation you had to start with and then y = 3, you're looking for that point P of intersection. So, make the equations equal to each other, and of course, that's what you just changed. And that's much easier to solve cuz there's only one mention of x. So, go ahead and do the bits and pieces. Divide, do the inverse sine. Just look at that 74 though, but you'll have to work this out first of all on your calculator. So, So, put this into your calculator and it will give you one of the answers, but you know the other one is going to be 180 minus it.

[00:19:17]Those are the two answers for the sign.

[00:19:19]So, it's 24.3 and 155.7.

[00:19:22]So, now take that to 74 across and add it to them both.

[00:19:26]So, now the two solutions are the final solutions are these two, but neither of them are the point P because P is negative. The ones you've got are the ones over here. You've got one there, and that 229 is further on, but you can get to the point P by going back a wavelength from the further on one. So, take 360 from the 229.8, and you'll get the negative 130.2, which is the X coordinate of P.

[00:19:59]The growth and decay equation questions, you got one of them as well as that logarithmic graph question. Sometimes there you have one or the other, but here you've had them both. But again, they are standard questions. So, in this one, it's fairly straightforward as long as you're quite happy with the fact that logarithm and exponential are inverses of each other.

[00:20:18]So, at the start, T is zero. So, pop T into that equation for the temperature.

[00:20:23]So, that just means it's E to the zero.

[00:20:25]That's quite handy. E to the zero is just one. So, you've got is 825° C.

[00:20:32]So, the first part, using the equation forward is easy enough. Now, the second part, you have to use it in reverse. It says, "How long will it take for the temperature to fall to 150?"

[00:20:42]Well, just put it into the equation. 150 will be whatever it said, 800 E to the etc. But there's got the unknown T in it.

[00:20:50]Now, you're solving it for T.

[00:20:52]So, bring that 25 across. It drops that.

[00:20:55]Now, get rid of the 800. Take that across and divide.

[00:21:00]Now, get rid of the E. So, that will be logarithm on the other side. Logarithm of that 125 over 800. Now, divide by that negative 0.124 and it's all ready to put into your calculator.

[00:21:15]So, when you put it into your calculator, out comes the answer 14.97 etc. So, 15.0 seconds to three significant figures. And that's it. Not too bad.

[00:21:29]And for the last question, you've got one of those trig identities, which isn't really that bad. I know probably a few would like it, but really if you knew your identities, there's only a couple of them. sin² + cos² is 1 and tan is sin over cos. Apart from the formulas you should know or you can find at the front, it's just a case of looking at what you've got and figuring out which of those can be used to change it.

[00:21:50]So, it is a standard question again.

[00:21:55]So, if that's to look like the one over there, multiply out the brackets to see what terms you've got lying about that you can rearrange into a well-known phrase or saying.

[00:22:04]So, multiplying them out into these four terms, you can see that the sin² and the cos² can go together cuz you know sin² + cos² makes 1.

[00:22:13]And that part, when you put them together, that make 10 of sin x cos x, that should ring a bell. Take that three out so you've got specifically the sin² cos². Take a five out so you're left with another pattern, two sin cos because you recognize them both. That's 1 and that's the sine double angle, sine 2 theta. So, they are 3 + 5 sin 2 theta for the result.

[00:22:38]And it's done.