Higher Maths 2026 | Essential Skills Paper 1

Added: 2026-05-06

[00:00:00]Hi, I'm Alister, thanks for Scales.

[00:00:01]Paper one for Mr. Lenny versus a excellent little paper, so give it a go, download it at clanmath.com and see how you get on. Remember, it's a week to go if you're watching this today. Last minute live streams are on the sixth for me, 6:00 p.m. That's Wednesday. Join now £4.99. It is going to be so epic. Last year was epic. All my predictions, everything's going to be in there. Let's go. Okay, question one.

[00:00:27]The point B 2 7 lies on the circle.

[00:00:31]Find the equation of the tangent at B.

[00:00:33]So, if you want to find the equation of a tangent, you need a gradient and a point, and the gradient is perpendicular. So, I'm going to need the center. Now, start of exam paper, the center is -g -f, where g, well, 2g is the number in front of x, so g is -3, and 2f is the number in front of y, so f is 2. So, my center, so for the center we just change the signs signs cuz it's -g -f, so it's 3 -2.

[00:01:02]And now we can do our gradient.

[00:01:04]y2 - y1 7 - -2 x2 - x1 2 -3 That's 9 over -1 -9.

[00:01:15]So, that means straight away the gradient of a tangent is equal to 1/9 because of perpendicular.

[00:01:28]m1 m2 = -1 is what you need to say. So, change the sign, flip it upside down. If it's a whole number, it becomes a fraction.

[00:01:36]We've now got our gradient and we've got our point. Point is 2 7.

[00:01:41]So, that is y - b m x - a 9y -79 is x - 2. So, we've got 9y - 63 is x - 2.

[00:01:59]You can move everything to the same side if you want.

[00:02:02]So, left or right, doesn't really matter. If you move it to the left, you get 9y - x. Add 2, so -61 = 0.

[00:02:12]Or you could have 9y - x = 61, or you could just change the sign on all of them to have a positive x, a -9y, and a positive 61. And we're done there. This video is sponsored by Leckie, and we are offering viewers of our channel 30% off of any of our books. All you have to do is use the code clanmaths and the link to their books is down below. Now, let's go cut a couple of their books to highlight. We've got National 5 Maths and Higher Maths, and these are complete revision and practice. Two books in one.

[00:02:38]We cover the whole course. Here's a random page, for instance, working simultaneous equations algebraically.

[00:02:43]You can see how good that book is and how clear it looks. But also, at the end, practice exam papers that look exactly like the SQA exam papers that are published. Just scan code clanmaths, click the link to order now. Question two.

[00:02:57]A line passes through 0 2, makes an angle of pi over 6 with the x-axis as shown.

[00:03:02]Determine the equation of the line.

[00:03:03]Gradient equals tan theta, you should all remember. But it's a trap because theta has to be in the positive direction of x-axis. That's negative, the other side. So, I need to actually find the other side by doing pi - pi over 6.

[00:03:19]That's 6 pi - 1 pi 5 pi over 6. So, that's the little trap there. And now we can do gradient equals tan theta.

[00:03:29]So, our gradient is tan 5 pi over 6.

[00:03:34]Now, if you can't think in terms of pi's and things, you can just change that back to angles.

[00:03:40]180 / 6 is 30, which means we're talking about 150 on this side. So, it's going to be related to 30, so if I do a little diagram 60 30 1 2 root 3.

[00:03:55]So, the tan of 30 is equal to 1 over root 3, but because it's in the quadrant of S, it's negative. So, our gradient straight away is -1 over root 3.

[00:04:09]Excellent. Find the equation of a line.

[00:04:12]Well, it goes through 0 2.

[00:04:14]Now, you can use y - b as mx - a if you want, but since it goes through 0 2, 2 is c. So, I can just use y = mx + c. -1 over root 3 x + 2. Now, if you're looking at the exact answers to these, it won't look like that, so this is to show you what you can change it to. If you times through by root 3, you get root 3y = -x + 2 root 3.

[00:04:38]So, you can have root 3y + x - 2 root 3 = 0. But I would have done it like this if it was me.

[00:04:48]Question B. Establish coordinates of point A. Well, A is on the x-axis, so that's when y is 0.

[00:04:54]So, for part B, when y = 0, we'll just sub in A this new equation or the top one. I'll just use my one actually. -1 over root 3x + 2 = 0.

[00:05:08]So, that means -1 over root 3x = -2.

[00:05:13]Times them through by root 3 then, and minus actually, then x is going to be -2 times - root 3.

[00:05:22]In other words, x is 2 root 3.

[00:05:25]So, the coordinates are 2 root 3 and 0.

[00:05:30]And we're done there. Question three. A sequence is generated by the recurrence relationship un + 1 = mun + c. The first three values are 25, 15, and 11. Find m and c. So, it's simultaneous equations.

[00:05:41]The second term, 15, is given by the first term times m + c.

[00:05:48]And then the third term, 11, is given by the second term m + c.

[00:05:54]So, then we can take away to get rid of the c's. 15 - 11 is 4 = 10m. That implies that m is 4/10, of course, dividing by 10, which is 2/5.

[00:06:07]Now, once we've got 2/5, we can just sub in. So, I'll just pick 15.

[00:06:12]25 * 2/5 + c 15 = Well, 25 / 5 is 5, * 2 is 10 + c, and therefore c must be 5.

[00:06:26]And there we are. Find the limit of the sequence. So, for part B, well, we know that un + 1 now = 2/5 un + 5.

[00:06:39]Even though it tells us to calculate the limit, so it must exist, we need to say why it exists. So, limit exists as it's a linear recurrence relationship, linear linear recurrence.

[00:06:56]And you probably won't lose a mark if you miss that statement, but you'll definitely lose a mark if you don't say as -1 is less than 2/5 is less than 1.

[00:07:06]And now we calculate our limit b over 1 - a.

[00:07:11]I've written + c, of course. I mean + 5.

[00:07:14]So, 5 over 1 - 2/5.

[00:07:19]5 over 3/5.

[00:07:22]So, that's the same as 5 * 5/3.

[00:07:27]5 * 5 is 25. 25 over 3 is simplified, so we're done there.

[00:07:33]Question four. The points P, Q, R are collinear. Find the value of k. Now, if they're collinear, that means that line segments are parallel. So, PQ is parallel to QR. And if they're parallel, it means that one's a multiple of the other. So, let's [snorts] do PQ.

[00:07:52]Q - P k -2 and 1 take away 1 2 -1 k -1 on the top -2 -2 is -4, and 1 - -1 is 2.

[00:08:07]So, there's our first one.

[00:08:10]And QR through our arrows in that, you can always lose a mark if you don't. R - Q R is 7 -10 5 and our Q is k -2 and 1.

[00:08:27]So, that gives me 7 - k -10 + 2 is -8 and 5 - 1 is 4. Now, if you look at these top and bottom, you should be able to see quite easily that that is double to get that, and that is double to get that.

[00:08:46]So, we can say that QR = 2PQ. Now, if you couldn't see it, just do one divided by the other. -8 / -4 is 2. 4 / 2 is 2.

[00:08:56]Or it could be a fraction, of course.

[00:08:58]Since it's double, then we can say that QR's part, which is 7 - k on the top, = 2 lots of k -1.

[00:09:08]Just solve that equation now. 7 - k = 2k -2.

[00:09:15]So, I'll move the k over here, and I'll move the 2. 7 + 2 is 9. So, k is 3.

[00:09:22]And we're done there.

[00:09:23]Part B. State the ratio that Q divides PR.

[00:09:27]Now, if you've already got it down, QR = 2PQ.

[00:09:34]And you want to know the ratio that Q divides. Well, we want to look at PQ over QR, where Q is stuck in the middle of the first one and starting the next one.

[00:09:42]So, if I take the two, if I keep the PQ there and take the QR there, then I'm going to have one on this side and divide by two.

[00:09:52]Since PQ over QR is 1/2, the ratio is 1:2.

[00:09:57]And we're done there. A function is defined as f of x equals x cubed plus 1 between 0 and 4. Find the inverse function and state the domain of the inverse function. Nice little question that. For part A, the most common way that people do it, so I'll do it that way, is to just say that y = x cubed + 1 and change the subject to x. There's a little trap here, but be careful. I'll do y - 1 = x cubed.

[00:10:20]So, it's the cube root of y - 1 = x. If you write y equals now, you'll lose a mark.

[00:10:27]If you write the inverse function as cube root of y - 1, you'll lose a mark.

[00:10:30]You need to just jump in and go, "Therefore, the inverse function of x is the cube root of x - 1." Okay, part B, domain and range.

[00:10:41]The domain of a function is the numbers that can go in. The range is the numbers that it gives you back, right?

[00:10:50]Now, you can't really see what numbers can go in in the domain of the inverse function, except a little trick.

[00:10:58]The domain of of original function is the range of the inverse function. And the range of the original function is the domain, they switch places. So, let's work out the range of f of x.

[00:11:10]Cuz we know the domain is 0 to 4. So, we sub in 0 and 4 to see where it goes between.

[00:11:17]So, f of 0 equals 0 cubed plus 1, which is 1.

[00:11:23]f of 4 equals 4 cubed plus 1 is 65.

[00:11:32]And therefore, the domain of the inverse function is just 1. And was it less than or equal to?

[00:11:43]Yeah. Is less than or equal to x is less than or equal to 65. That is what can now go in to the inverse function. And we're done there. Question 6, the diagram shows the graph of y equals f of x. There's a minimum turning point at 1 - 4 and a point of inflection at 2 or -2.

[00:12:00]Sketch the graph of 2 - f of x + 1 and then sketch the graph of y equals f dash x, the derivative. So, we'll do it bit by bit. Let's look at the first part.

[00:12:09]We've got quite a few things happening.

[00:12:12]x + 1, let's see what that does. That means left by 1.

[00:12:19]Then the next bit, minus means flip upside down, so reflection x axis.

[00:12:28]And then finally, the two in front means up by two, cuz that's a positive number.

[00:12:33]So, we can do that on each of the coordinates. So, let's start with -2 0.

[00:12:38]If we go go left by 1, left means -3 0.

[00:12:44]If we reflect that over the x axis, 0 is the one that's changing, but that's still -3 0.

[00:12:50]But then we add two, so -3 2.

[00:12:55]Can do the same with 1 - 4.

[00:12:57]1 - 4, left by 1, so that's the x part, so that's 0 - 4.

[00:13:05]Reflecting the x axis, so the -4 becomes positive 4.

[00:13:11]And then up by two, so it becomes six.

[00:13:15]So, we've now got our two points, so you can then just draw a graph. So, I'll draw an x y axis.

[00:13:22]I'll put my points on first.

[00:13:25]So, I've got -3 2, maybe there. Just label it -3 2.

[00:13:30]And we've got 0 6, well, maybe up here somewhere. Then we just need to examine what happens to this. Well, if I flip it upside down, instead of going near, it's going to go like that and then turn. So, that's what I'm going to draw, cuz it's just upside down.

[00:13:44]So, that's my point of inflection.

[00:13:47]That's my turning point, and then I'm going down somewhere. So, something like that.

[00:13:52]And we're done there. Part B wants the graph of the derivative. The graph of the derivative, the easiest way to think of this is I'll draw an x y axis again.

[00:14:00]But think of this as y equals f dash x.

[00:14:03]That's actually the values of the derivative at that point. So, if you think at a turning point, f dash x is 0 because it's horizontal. So, I can just put that on the x axis. And then I've got 1 -4, so along 1 is 0.

[00:14:19]So, I'll put that there. And then you're just examining when is the derivative negative and positive. So, it's negative when it's sloping down. It's sloping down there and sloping down there. So, before -2, we need to go below the axis and below the axis because we're sloping down the way. So, we need to be staying down there. So, I'll just put a little dot point there just to remind me for when I get there. Little point. And I'm still at a little point until I get to 1. And once I get to 1, it goes up the way.

[00:14:52]Goes up up up in a way.

[00:14:55]So, then we go up means above x axis.

[00:15:04]So, I need to be above the x axis after 1. So, I now need to draw some sort of picture that would allow that to happen.

[00:15:11]There's a couple of options.

[00:15:13]You could have that never touching and then sloping up, then down, and turning at some point and up.

[00:15:22]And then it just keeps going.

[00:15:23]Or a probably more common one, I think, would be that that's a turning point.

[00:15:31]But it's down. So, it tries to rise up, it goes down, it goes back up, and it just keeps going.

[00:15:39]There you are, turning somewhere. Don't know where it turns, so that's good enough. And we're done there. Question 8, express x cubed + 2x + 3 in fully factorized form. So, we need to find a factor first. To find a factor, find a factor.

[00:15:54]Question 9, find factorize this. We need to find a factor cuz it's a cubic. So, we need to try numbers. Now, don't try any numbers. It starts with x cubed and it ends with three, so it's some some factor of three.

[00:16:07]So, if I try 1, I'll just sub on in, I'll get 1 cubed + 2 * 1 + 3. And I just prefer to do it quickly rather than a full synthetic division. That's 1 + 2 is 3 + 3 is 6. That's not a factor.

[00:16:19]f of -1, I'll now try. -1 cubed, two -1s, two -1s + 3. Well, that's -1 -2 is -3 + 3 is 0. So, we now know that x -1 is a factor. x -1 is a root, sorry.

[00:16:39]So, we can then just do our synthetic division. And here's the trap. It's 1x cubed, 0x squared, 2x, and 3.

[00:16:46]And we're putting in -1.

[00:16:49]Just checking it works again. 1 * -1 is -1. Plus 0 is -1. Times 1 again is 1.

[00:16:56]Plus 2 is 3, then -3, then 0. We need a statement, remainder equals 0.

[00:17:05]Therefore, x + 1 is a factor.

[00:17:11]So, now we've got x + 1 x squared - x + 3.

[00:17:18]Usually, you can factorize the second one, but there was no numbers that times together to make three that add together to make one. So, it's done. But I would like to see why it's done. And here's a way to check.

[00:17:28]Let's do b squared - 4ac just on this bit.

[00:17:33]Then that's -1 squared -4 * 1 * 3.

[00:17:39]That's 1 -12, which is negative.

[00:17:44]Uh, -11, in other words.

[00:17:46]So, it's got no real roots.

[00:17:49]And that means that since it's got no real roots, that's fully factorized.

[00:17:58]And we're done there. Question 8, given g of x is cos 3x + sin 3x, find g dash pi over 2. So, differentiate with cos.

[00:18:06]So, start of the exam paper, cos goes to minus sin.

[00:18:11]Cos ax is minus a sin ax. So, that's minus a.

[00:18:18]I'm writing a. That's -3 sin 3x. That's because of the chain rule.

[00:18:23]You differentiate cos, it gets minus sin, and you times by the 3x differentiated, which is three.

[00:18:29]Similarly, sin goes to cos.

[00:18:31]So, that's cos 3x times by three.

[00:18:35]And then I'm just subbing in pi over two, which is 90. So, that's -3 sin 3 pi over 2 plus 3 cos 3 pi over 2.

[00:18:47]And you can check which what they are by just checking a normal sin and cos graph. A sin graph goes like that.

[00:18:54]And it goes 90 180 270 360. So, there's my 3 pi over 2 here.

[00:19:01]It's minus. Usually, minus one if it's a normal sin graph.

[00:19:05]So, then that's -3 * -1.

[00:19:09]And then for the cos graph, 90 180 270 is here. 3 pi over 2 is 270.

[00:19:20]It's 0. So, it's plus 3 * 0. So, our final answer is equal to three cuz -3 * -1 is three.

[00:19:29]And we're done there. Question 9, 2 log a 3 plus log a 4 minus log 3 log a 2 in the form log a b over c. So, use the logs.

[00:19:40]Two can come up to be a power.

[00:19:44]Then log a 4.

[00:19:45]And then three can come up to be a power.

[00:19:48]That's an a, sorry. 2 cubed.

[00:19:50][clears throat] Now, add means times. Take away means divide.

[00:19:55]So, log just in the same order. 3 squared is 9 * 4.

[00:20:01]And then minus 2 cubed is 8.

[00:20:05]So, that's log 9 * 4 is 36 over 8.

[00:20:12]4 goes into both of them. 4 * 9 is 36. 4 * 2 is 8. 9 over 2. And we're done there. Question 10. Express root 3 cos x minus sin x in the form k cos x + a. So, this is a wave function. k cos x cos a minus k sin x sin a.

[00:20:38]So, the bit next to cos x on this side is k cos a. So, I can see that k cos a is the bit next to cos x on this side, root 3.

[00:20:47]And then k sin a, well, minus is minus 1. So, they're both minus, so it's just 1. So, k sin a equals 1.

[00:20:58]To get our k, it's just Pythagoras. The square root of root 3 squared plus 1 squared.

[00:21:05]That's root 4, which is 2.

[00:21:08]To get our a, tan of the angle sin over cos.

[00:21:14]And then we can do our little diagram.

[00:21:17]60 30 will be an angle. It will probably be in radians, but I'll do it in radians in a minute. 1 2 root 3.

[00:21:25]So, the tan of 30 is 1 over root 3.

[00:21:29]So, the inverse tan of 1 over root 3 is 30.

[00:21:35]30 is pi over 6.

[00:21:38]And we need to confirm that it is that one. So, cast diagram. Sin was positive.

[00:21:43]Cos was positive. So, a is pi over 6.

[00:21:47]So, to write it in the correct form, you've got 2 cos x + pi over 6 as our final answer.

[00:22:00]Now, part b says, hence or otherwise find the minimum value of this. Well, that's the same as the original question plus 1. So, for part b, I've already got y equals 2 cos x + pi over 6 for the first bit, and then it says add in 1.

[00:22:19]So, the minimum of that Well, normally it's going to be 2 and minus 2. So, minus 2, but then I've got to add 1. So, that's minus 1.

[00:22:31]For part 2, the value where it occurs, well, it's a shift to the left >> [snorts] >> by pi over 6 because of the x + pi over 6.

[00:22:43]So, usually with a normal cos graph, the minimum happens here.

[00:22:49]The minimum there is at pi.

[00:22:55]So, it's pi minus pi over 6 because it's going left, which is 5 pi over 6.

[00:23:05]And we're done there. Okay, question 11.

[00:23:07]Some vectors.

[00:23:08]And the diagram shows a triangular prism, and we've to find AC in terms of t and u. So, AC I'm just starting at A and going to there, so I can go backwards, so minus t add u or u minus t. Nice and easy. Part b.

[00:23:27]M is the midpoint of AC.

[00:23:30]And it divide N divides BE in the ratio 3 to 1. So, we've got 3 to 1 here.

[00:23:36]And then I've got a half here just to be on the safe side. I have to go MN. So, that's a half of what I just found out because that's AC. So, a half So, a half of u minus t, the whole thing.

[00:23:53]Once I do that, I go along here, so minus u.

[00:23:57]And once I do that, I need to go 3/4 of the way along here. Well, along here is the same as v. So, it's 3/4 because it's 3 1, which is 4. 3/4 of v.

[00:24:10]Now, we're just trying to simplify that in some way. A half u minus a half t minus u plus 3/4 v.

[00:24:21]A half u minus another u, well, a half minus 2 halves is minus a half. So, minus a half u minus a half t plus 3/4 v.

[00:24:32]Or you can rewrite that so it looks neater, but that's fine. We're done there. Question 12. The point Px lies on the curve 6x squared minus 2x cubed.

[00:24:40]Find the value of x so that the gradient of the tangent is 6 at P. So, if it's the gradient of the tangent, m equals dy by dx.

[00:24:51]12x.

[00:24:52]2 * 6 is 12. 3 * 2 is 6, so minus 6x squared, and that now equals 6 because it says it equals 6, not that x is 6.

[00:25:01]Then we try and solve that. So, 12x minus 6x squared minus 6 is 0.

[00:25:09]Take out a common factor, but let me just rewrite that first. Minus 6x squared plus 12x minus 6.

[00:25:15]Now, most people wouldn't want a minus 6, so I'll take that out as a common factor or change all the signs. And I get x squared minus 2x plus 1.

[00:25:28]Now, you should get when you factorize this the same answer. If you don't, you'll make a mistake because it's a tangent.

[00:25:34]I've got 1 and 1, minus minus.

[00:25:37]Good. So, x equals 1.

[00:25:41]Find the value of x where the gradient tangent is 6. x equals 1. Find the equation of the tangent at P. Okay, so for part b, I know you know that x is 1.

[00:25:48]So, I can sub that in to get my y. 6 * 1 squared and minus 2 * 1 cubed.

[00:25:58]That's 6 minus 2 is 4. So, my point is 1 4, and my gradient I was told, remember, is 6.

[00:26:08]y minus b is m x minus a.

[00:26:14]y minus 4 6x minus 6 or y equals 6x.

[00:26:21]Minus 6 add 4 is minus 2.

[00:26:23]And we're done there. Okay, last question. Given tan x is root 5 over 2, find cos 2x and cos x. Now, it's between 0 and pi over 4. So, we're starting with 2x instead of x, which is fine. Opposite over adjacent, root 5 over 2. And then you can do Pythagoras on that. So, root 5 squared is 5. 2 * 2 is 4. Square root, that's square root 9. So, we get So, straight away, cos 2x is equal to opposite adjacent over hypotenuse, 2/3.

[00:27:01]Now, don't think you can just half that to get cos x. That'd be a mistake.

[00:27:05]We need to use the trig identity. So, I'll use 2 cos squared x minus 1.

[00:27:11]You could use any other one and just find that you've not got anything but you try to get cos x, so it has to be that one.

[00:27:19]So, cos 2x is 2/3.

[00:27:21]Equals 2 cos squared x minus 1.

[00:27:24]So, I need to add 1. Well, adding 1 to 2/3 is adding 3/3. 3 + 2 is 5/3.

[00:27:30]5/3 is 2 cos squared x.

[00:27:34]And then dividing by 2, 5/3 divided by 2 is 5/6.

[00:27:42]And that means that cos x is the square root of 5 over 6 or root 5 over root 6.

[00:27:50]Now, you might be asking, "What about the negative version of this?" Well, no, because it's between 0 and pi over 4 where x is, which means the inverse cos of root 5 root 6 has to be in the first quadrant. So, there isn't any other answer. So, we're done there. And there we are. Essential Skills Higher Mathematics Paper 1 for 2026 is done.

[00:28:11]Thanks to Mr. Graham again for that. And remember, not long to go. Last minute live streams. 6:00 for me, 6:00 p.m. for Higher Maths. Get on it now. Members only, £4.99, and I'll see you soon.