Higher Maths 2026 Paper 1 Full Solutions

Added: 2026-05-12

[00:00:00]Mythical maths. It's high math paper 1226 full solutions. You'd sat this paper on the first of the 7th of May.

[00:00:06]Probably today you sat this paper Thursday for me 9 to 10:15 a.m. I've spent ages going through the solutions trying to make sure I get 100% right. Hopefully I do well. Let's go. This video is sponsored by Leki and they are offering viewers of this channel 30% off of any of their books. All you have to do is use the code cleon maths and the link to their books is down below. Now let's look at a couple of their books to highlight. We've got National Five Maths and Higher Maths. And these are complete revision and practice. Two books in one.

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[00:00:52]Qualification Scotland higher maths 2026 paper one question one express 2x2 + 20x + 3 in the form px + q ^2 + r so this is called completing the square so we take 2x^2 + 2x a common factor so 2 is a common factor of that so that's x^2 + 10 x and we still got plus three just hanging out so that's two and then I complete the square x + 5 half the term squared take away 35 but we still got + 3 on the end. So that becomes 2 * x + 5^ 2ar 2 25s is 50 + 3 2 x + 5^ 2 then - 50 add 3 is - 47 and we're done there. Propagation Scotland higher math 2026 paper 1 question 2 integrate 15 x^ 2/3 + 7 x^2.

[00:01:49]Well get it ready. integrate first. So I've got luckily nothing to do with this one. But then 7 x^2 - 2 then dx add 1 to the power. So 2/3 3 4 5/3ide by 5/3. I've got an x and then + 7 x -1. So divide by minus one. And then don't forget the c. Now dividing by a fraction is the same as tsing by upside down. times it by the reciprocal. So that's 15 x^ 5/3 - 7 x to the -1 then plus c.

[00:02:32]Tidy it up slightly.

[00:02:35]5 into 15 goes 3 * 3 is 9. So you've got 9 x^ 53 - 7 x^ -1 + c. And this say just fine for four marks. I think you'll be fine there. So we could stop there, but just in case you wanted to or you went the extra step, you could put it into 9 cube root of x 5 - 7 /x + c. But I think you'll be fine there. So we're done there.

[00:03:07]Qualifications Scotland Higher Maths 2026 paper 1 question 3. A circle has equation x^2 + y^2 + 2x - 4 y - 20.826 A26 lies on the circle.

[00:03:21]Find the equation of the tangent at A.

[00:03:23]So to find the equation of the tangent, we need our center. So to find our center, it's - G minus F. So 2 G is 2 in front of X and 2 F is -4.

[00:03:39]So G is 1 and F is minus2. Change the sides.

[00:03:45]-1 and 2. So we've got min -1 and two here and we've got point a which is 2 six. So now we can just do the gradient of that in perpendicular gradients. So that means we're going to say that we've got the gradient of AC is equal to 6 - 2 over 2 - -1. 6 - 2 is 4 over 2 - - 1 is 3. And then we can say the gradient of the tangent is equal to minus 3/4 because they're perpendicular.

[00:04:24]So m1 m2 = -1.

[00:04:29]Now we move on to using our point which is 26.

[00:04:33]So y - b = m x - a. So I usually take the four up to get 4 y - 6 - 3 x - 2. That's 4 y -4 = - 3x - * a minus is a plus.

[00:04:57]So you can move everything to the same side. So if I move the 3x over, I'll have 3x + 4 y. And if I move the 6 takeway 6 is -30, that equals zero. or equivalent to that. So you could also have x and y on the same side and then the 30 on the other side or you could divide through by four. Whatever way you've left it is probably fine. So we're done there. Scotland higher math 2026 paper one question four diagram shows two right angles with P and Q.

[00:05:26]Find sq and then it's going to be addition formula. So quite standard stuff but we need to use Pythagoras on the first one. Let me just flatten that out. You've got three and four and the P's up here. 3^ 2 + 4 2 is 9 + 16 25. We get 25 is 5. So we get a five nicely up here. And then we've got the other triangle which is drawn like this with the Q up here.

[00:05:56]But the hypoties of the first one is equal to the height. So that one is five here, 13 there.

[00:06:04]And I need to do Pythagoras again. 13^ 2 - 5^ 2 16 9 - 25 is 144 <unk>144 is 12. So we get 12. And now I'll answer the question. I just did that in case I need it all later. Find the exact value of sin q for a sin q is equal to opposite over hypotuse 12 over 13. Nice and easy there. So there's part b just to remind us. So we need sin p we need sin Q. We need them all really. So sin P we can just use the first triangle.

[00:06:44]Opposite over hypotenuse is 4 fths.

[00:06:49]Sin Q we already know is 12 over 13. And then we'll just do the same for coses.

[00:06:55]Then cos p cos q.

[00:07:01]So our cos p adjacent over hypotes is 3 fifths.

[00:07:07]and our cos Q adjacent of hypoties is 513. And now we do an expansion. So for part one, sin p + q sin p cos q start the exam paper cos p sin q. Just taking your time. Sin P is 4 fths.

[00:07:30]Cos Q is 513 plus cos P is 35s.

[00:07:36]Sin Q is 123.

[00:07:40]Now since I probably got a combinator anyway, I won't bother to actually cancel out first. So I'll just say that's 20 over 5 10 are 50 65 plus 3 10 is 30 36 over 65 36 + 20 is 4656 over 65 and then we just need to check can you simplify that. So that's that one done. So we now move to part B which is part two is cos p + q that's cos p cos q minus sin p sin q.

[00:08:23]So subbing in again both the causes 3 fifths and 513 minus both the signs 4 fths and 1213 that's 15 / 65us 48 over 65.

[00:08:51]So taking away 15 25 35 45 63 - 33 over 65 three doesn't go into them. So I think we're done. That's minus that's the answer part B a little bit of curve tan P plus Q.

[00:09:11]Well if you remember for higher mass can is S over cost. So I can just do sin p + q over cos p plus q.

[00:09:24]But the sign is 5665s.

[00:09:32]Divide means times by the reciprocal.

[00:09:34]I'm dividing by a fraction. So I just flip it upside down. So it's - 65 over 33. So all I've done is I've divided by a fraction. So times by the reciprocal.

[00:09:44]The 65s cancel to leave - 56 over 33.

[00:09:54]Check it simplify, but I'm pretty sure that is as well. So with that there qualification Scotland Ham 26 paper one question five. This one's on functions.

[00:10:04]We've got f of x is 2x2 + 5 and g of x is x + 3. Find g f of g of x. So part one for f of g of x you just sub g into f. So f is 2x^2 + 5. So I can write 2 g of x^2 + 5.

[00:10:22]That is 2 x + 3 all^ 2 + 5. Now I get a feeling that you'll get your marks there and you'll be done. But for those of you that expanded the brackets, you've got two that will be x^2 + x + 3. x + 3 makes 6 x and 3's are 9 + 5. So 2 x^2 + 12 x + 18 + 5 2x^2 + 12 x 18 + 5 is 23. But to be fair, I think you're getting a mark right there. Okay. G of f of x and g subn f ofx. So g is just x + 3. So f be that becomes f of x + 3. But f of x is 2x^2 + 5. So it's just 2x^2 + 5 + 3, which is 2x^2 + 8, which again, I think you'll get a mark there, but you could take two out as a common factor if you really wanted to to get that. And we're done there.

[00:11:29]State the range of g of f ofx. So for part B, the range of G of F of X. Now that's an interesting question. I think that'll be tricky for most people. You've got domain and range. So domain is what X can go in, which obviously is anything for this. Any negative, any positive.

[00:11:49]The range is what can come out. Now if you think about this, if X was negative, then when I square, I get a positive number. So like the smallest it can be is one or smallest can be 0.1 or whatever but a positive number plus 4 * 2. So the smallest that the bracket can be x^2 + 4 is smallest when x = 0. So that would be 0 2 + 4 but then I'm tsing it by 2 on the outside. 2 4 is 8. So the smallest that g of f of x is is is 8. It's bigger than eight then all the time. So the range is g of f ofx is always greater than or equal to 8. And we're done there. Now this is sort of a technical way to write that but I don't think you'll need to. It would be to say something like let y= g of x of x.

[00:12:49]Then you've got a set. The range is just a set of numbers such that y is a member of the real numbers and y is gal to a but again I don't think you'll need that. I think you're getting the mark there. So we're done there. Scotland high maths 2026 paper one question six.

[00:13:08]A bcde e f g h is a cuboid. A is u. BC is v. G is w. So in vectors m is the midpoint of hg. The point N divides EA in the ratio of 1:2. Just keep all that in mind. Express M to N in terms of UVW.

[00:13:29]So what we asking here? Where's M? We want to go from here to there. So a little bit tricky this one, but we can trace it out. So since we know AB is U, then any vector parallel to that is you.

[00:13:45]So this whole G H is U. So I need to go along there. But what is G H? M is the midpoint of HG. So halfway along U. So that's minus a half U. I then need to go along here, which is the same as going along there. So minus V.

[00:14:07]Just tracing it out. And then you go up here, which is the same as going up here some way. Now, how far?

[00:14:17]Well, the ratio that N divides E to A is 1 to two. So, you've got one part and two parts. So, I need to go up 1/3 of the way cuz the total it would be three.

[00:14:30]So, 1/3 of W. So our final answer is mn is equal to minus a half u minus v plus a third w.

[00:14:47]And we can't really simplify it anymore.

[00:14:49]You could make it look neater by putting the w first I suppose but I don't think there's any real need to.

[00:14:58]So we're done there. of Scotland higher math 2026 paper one question seven determine the gradient of the tangent to the curve with equation 2x + 4<unk>x at a point where x is 9 so it's the gradient of a tangent sometimes it's a gradient sometimes it's the whole equation of the tangent but this time just the gradient gradient of a tangent is a power word and that means the derivative gradient equals the derivative so gradient equals d y by dx so I need to get it ready to differentiate first so y getting it ready to differentiate is 2x + 4 <unk>x is x to the half. So now I can differentiate it.

[00:15:34]2x just becomes 2 and then times by a half. 4 * a half is 2 x and take away one for the power. But I'm going to have to sub 9 with a calculator. So I need to get that back to a normal form. So it's 2 over x to the half cuz it's minus. But x to the half we know is a root.

[00:15:56]So then we can just sub in 9. So when x is 9, since the gradient is dy by dx, it's 2 + 2 over<unk> 9.

[00:16:06]So that is 2 + 2/3.

[00:16:11]I suppose you could just write 2 and 2/3, but I think more commonly you would want it as a fraction. Two is six, seven, eight, eight.

[00:16:21]And we're done there.

[00:16:23]Qualification Scotland qualification Scotland high math 2026 paper 1 question 8. Solve log 2x plus log 2x + 2 = 3. So a logarithmic equation with algebra. So we're probably going to get a quadratic.

[00:16:36]Let's look. We keep the base the same.

[00:16:39]And add means times we logs. So it's just x x + 2 = 3. Now the way you get rid of that is the base to the power of 3 equals this. And the log disappears.

[00:16:51]So x x + 2 is 2 cubed which x x + 2 then must be 8. Now since I've got x bracket, you can't just say x= 8. It has to be a quadratic. It's going to give you an x squ. So you do have to expand the brackets and then move the eight to the other side so that it's a proper quadratic and then solve it by double bracketing it up again. X and X eight and one. No, it's not going to be eight and one. It's going to be four and two. Four twos is eight. And I want to make plus two. So plus 4 - 2. So x = 2 or x would normally equal -4. However, x is greater than zero. It actually tells you in the question. It doesn't even have to tell you that. You can't have a log of a negative number. So we reject that one because x is greater than 0. And we just say x is equal to 2 is our answer. So we'll just score it out and we're done there. Qualifications Scotland. Hi, I'm M 26 paper one question nine. Show that the points D E and F are colinear. Point G is such that F divides DG in a ratio of 44. Find the coordinates of G. So quite standard colinear or a lot of trick on part B. Coronarity with vectors means you need to show that we're parallel and you've got a common point.

[00:18:11]So we do D to E which is E minus D which is 127 - - - - - - - - - - - - - - - - - - - - - -1 61 oops - 1 61. So our first one 1 - - 1 is 2. 2 - 6 is -4. 7 - 1 is 6.

[00:18:33]If you ever get a multiple it's probably easier to just take out a common factor.

[00:18:38]So 2 is a common factor of that. So that's 2 * 1 - 2 and 3. Now we do E to F.

[00:18:51]F minus E minus.

[00:18:56]So my F is 210.

[00:19:01]My E was 127.

[00:19:05]2 - 1 is 1. 0 - 2 is -2. and 10 - 7 is 3. Notice I've got 1 - 2 3 in two places. We can clearly see that two of 1 - 2 3 is one is ef two of this is one of them. So we can say that two ef equals d And that's it. Therefore, E to F and D to E are parallel.

[00:19:57]E is a common point.

[00:20:03]This is a full statement and therefore points D, E, and F are collinear.

[00:20:16]And we're done there. Okay. Part B. The point G is such that F divides DG in the ratio of 3 to 4. Find the con of G. Now, you're better off visualizing this because it is not a standard one. We've got D to G, point F in the ratio 3 to 4.

[00:20:33]Let's say there.

[00:20:36]So you could use the section formula for this or you could use So you could use the section form for this or you could do what I normally do which is to say that DTF over F to G must equal 3 over 4. And therefore 3 FG equals 4 DF and then just rearrange to get what I want. And I'm trying to get the coordinates of G this time not F. So 3 G - F = 4 F - D 3 G - 3F = 4 F - 4 D we want G I think was it or E um G so 3 G 4 F + 3 F is 7 F - 4 D so G finally is 7 F - 4 D over 3 subn And then then we've got that equals seven lots of f.

[00:21:39]If remember way back up here is 20110 minus four lots of d.

[00:21:50]D way back up here is - 161 all over three.

[00:22:00]So a bit of work to do. Times the first one by seven. You get 14 070 minus - 4 4 6 is 24 4 1's is 4 all divided by 3. Work out my top. 14 - - 4 is 18. 0 - 24 is -4.

[00:22:22]70 - 4 is 66.

[00:22:26]And we're dividing all of them by three or other words doing a third of them.

[00:22:30]So that means that we've got 3 and 18 is six. 38 is 24.

[00:22:39]3 11s is 3 into 6 is 2 22. Now that's not a point. Remember that is a vector.

[00:22:45]That vector is called G. So to answer the question, the point G is equal to 6 - 8 22.

[00:22:53]And we're done there. Question 10.

[00:22:56]Qualification Scotland. Hi, I'm Math 26.

[00:22:58]Paper one. Question 10. Integrate between p<unk> / 6 and p<unk> / 3 6 sin 3x -<unk> / 2. Start exam paper tells you how to integrate sin ax.

[00:23:08]It says that equals minus cos ax / a.

[00:23:21]Now it does not mention this bit here but that's okay. That bit just goes along for the ride because it's just a chain rule. So you just need to remember divide by three. So our integral becomes 6 - cos 3x -<unk> / 2 / 3 between<unk> / 6<unk> over 3. That's our step. 6 / 3 is 2 cos 3x -<unk> / 2. Then between p<unk> / 6 p<unk> / 3.

[00:23:57]So then we sub in. So that's - 2 for the first one cos 3<unk> over 3 -<unk> / 2 take away - 2 cos 3<unk> / 6 -<unk> / 2.

[00:24:19]And now we need to tidy that up and solve it. So that's - 2 cos. Now pi over 3 / 3 is just pi. So it's p<unk> min - p<unk> / 2. That's 1 minus a half, which is a half. So it's p<unk> / 2 half of pi. So that's our first one. - 2 cos / 2 take away - 2 cos 3<unk>i / 6 is p<unk> / 2 take away p<unk> / 2 is zero because that's pi / 2.

[00:24:52]Now we need to do exact values. So cos graph goes 1 and minus one 90 180 270 360. So at 0 it's 1 and at 90 it's 0. So that's -2 * 0 take away -2 * 1 - 2 * 0 is 0 - - 2. And we're done there. Public case of Scotland higher math six paper one question 11. Show that x plus2 is a factor. Explain why x+2 is the only linear factor. Give a reason. And find the coordinates of the points where the curves with equations y = 2xqub + 20x^2 + 27 x + 9 and 6x^2 - 9x - 23 inter which is a little bit of curve ball. So let's go with part a. To show that's a factor. We can do some division. So we've got 1 x cub 7 x^2 18 x and 16 sub in min - 2 cuz we sub in a root. So 1 comes down 1 * - 2 is -2 7 - 2 is 5 5 2's is 10. So that's - 10. 18 - 10 is 8.

[00:26:06]8 * 2 is -6. So we get a remainder of zero. Remainder equals zero we write.

[00:26:14]Therefore, x + 2 is a factor.

[00:26:19]Part two.

[00:26:22]Why is it the only factor or the only linear factor? So, let's look at it factorized. So, factorized, we now know that we've got x factorized, we've got x + 2 and then it says 58 x^2 + 5 x + 8. And usually we then try and factoriize that.

[00:26:42]But in this case, since x + 2 is only linear factor, it won't be factorizable.

[00:26:47]So how do we show? We check b ^ 2 - 4 a c. Because if you get a negative when you do that, it can't be factorized because there's no real roots. So that's 5^ 2 - 4 * 1 * 8 25 - 32. That's - 7. So no real roots.

[00:27:09]Therefore, x + 2 is the only linear factor.

[00:27:18]That's what I would do. Part B. Again, find the points of the coordinates where they intersect.

[00:27:25]Now, points of intersection means symmet equations. We both got a y. So we can just write 2x cub + 20 x^2 + 27 x + 9 must equal 6 x^2 - 9 x - 23.

[00:27:40]We always need to make it equal to zero when we got powers. So I'll leave the 2x cubed 20 x^ 2 20 - 6 is 14.

[00:27:51]27 - 9 becomes plus. So + 9 27 becomes 36 and then 9 + 23 cuz it's minus. So 23 becomes 32 and we're going to say that equals zero.

[00:28:07]Now when you try to factoriize something, the first thing you do is look for a common factor before you even get anywhere. So two is a common factor of all that.

[00:28:15]7 x^2 18x and 16.

[00:28:22]Now we can just check. Is there anything in part eight that looks like that? I presume there is. Oh look, xqub + 7 x2 + 18 x + 16. It's the same thing. So we can just do part we can already factoriize it because I've done it. We factoriize it way up here. x + 2 x^2 + 5 x + 8. So we just go ahead and do that.

[00:28:42]x + 2 x^2 + 5 x + 8.

[00:28:47]We already know then that that means that each bracket equals zero. And we already know that this one on the right has no real roots because we just did it. No real roots.

[00:28:59]So we can't do anything with that.

[00:29:02]And that means that we ask the other one is just x= -2. So we've actually only got one point where they intersect at x= -2. Now we just sub that in to y. So when x is -2, y = just pick an equation 6x^2 - 9x - 23.

[00:29:22]6 * - 2^ 2 - 9 - 2 - 23 6 * 4 9 * 2 is 18 - 23 24 + 18 - 23 then 24 - 23 is 1 so it's 19 so the point of intersection POI is - 219 and we're done Qualifications Scotland higher math 2026 paper one question 12 an exponential function is defined as f ofx is drawn for on a graph like so the inverse function exists good in the diagram you ask about sketch a graph of inverse function so that's part A and then part B is something else so we'll do part A first and get to what part B is asking us so for part A we've got a graph Inverse functions can always be drawn because you reflect in the line y = x. So you're reflecting in this line here.

[00:30:29]The points just switch places. So we've got loads of points here. So we just switch those places out. We've also got an asmtope at minus4. That just becomes a vertical because it's switched out. So minus4 becomes minus4 on here.

[00:30:46]And there are points 25 becomes 52. So along five up two. And since it is an exponential, we know we're looking for a log.

[00:30:55]0 - 3 becomes -3 0.

[00:31:00]Let's call that -3 0. And 1 - 1 becomes - 1 1 - one 1. So a log graph goes a bit like that. I'll just note what that point was again minus one one because this we flipped them and there you go. Never touching that.

[00:31:24]There's our graph. So we're done there.

[00:31:26]Okay. Part B of this it says remember functions of the form log a x plus b deter value of a and b. So log a x + b.

[00:31:35]For part b remember the graph of a normal log function goes like this where the vasintop is at zero. So that's the graph of y= log x with some sort of base, right? We'll get to base in a minute, but it means that I can immediately say since it's y= log base a is it b or a log base ax + b. This is just a graph transformation that z's moved back to minus4.

[00:32:05]So therefore, since it's a shift to the left, B is positive left by four.

[00:32:13]So B equ= 4. And then we need to get A.

[00:32:21]So we just pick a point on the graph. A point would maybe be -1 1 - 3 0 52. So I think we'll pick 52.

[00:32:35]52 is an x and a y. So we've got y = log base a of x + b, which is x + 4.

[00:32:46]So log base a of x is 5 + 4 = y, which is 2. So that gives me a squared = 5 + 4 is 9. So I can square root. So a must be free because you don't get negative bases. And we're done there.

[00:33:07]State the domain of the inverse function.

[00:33:12]So the domain of any function is what x can be.

[00:33:18]So let's have a look at our inverse function. The inverse function we've just worked out is log base a which is three of x + 4. I think it was four.

[00:33:29]Yeah.

[00:33:32]Now, so for the domain, we've got log base 3 x + 4, but you can't have a negative or zero log. So that means that x + 4 must be strictly greater than zero. So x is greater than minus4.

[00:33:45]And we're done there. You could write that again as your x is greater than minus4 or x is a member of r x is greater than minus4, but most people are going to write it like that. So that'll be fine. Let's be clear math. Today we did qualifications Scotland higher math 226 paper one. Hopefully you found that useful. Be back soon.