Advanced Higher Maths 2026 Paper 2 Full Solutions

Added: 2026-05-10

[00:00:00]Mr. CL here for car maths. It's advanced higher maths 2026 paper two full solutions. This is going to be big and epic. Let's go. Okay. Scotland high maths 2026 paper 2 question one.

[00:00:12]Differentiate f ofx is free inverse s of 7x. So if it's an inverse sign the start of the exam paper gives us the formula.

[00:00:21]So there it is there. Inverse sign differentiates to 1 over the<unk> of 1 - x^ 2. And you just need to remember the chain rule. So f-x I've still got a three and then I've got the inverse sign becomes square<unk> 1 - all of 7 x squared but because of the chain rule you need to differentiate the 7 x so you need to times by 7 3 * 7 is 21 of course square<unk> of 1 - 7 7 is 49 x^2 a nice little warm up and we're done there Scotland advance hammer Maths 2026 paper 2 question two write down the binomial expansion of x^2 - 5x 4 and simplify your answer. So far so good. So we have got binomial theorem from the start of the exam paper is a + b to the n is sigma r n of n choose r a n minus r b r n choose r of course is this. So we can start by saying that I've got x^2 - 5x 4. And this bit here can be worked out using Pascal's triangle. Pascal's triangle side 1 1 2 1 1 33 1 just found by adding up the numbers above and I go up to the four because it's to the power of four, right? So now we can start a^2 - 5 x 4 is equal to n choose r. So 4 choose 0.

[00:02:01]And I'll write it out fully so you can see it all. And then it's just the first term x to the n minus r. So it's x^2 to the 4 and - 5x 0 which will always just disappear from the first term. And then we just start again. Four choose one. Four choose two. And what happens is that goes down a power by one and that goes up a power by one. So 4 choose 1 x^2 cubed - 5 / x to the 1 4 choose 2 x^2 - 5 / x^2 4 choose 3. Now x^2 goes down a power again - 5 / x goes up a power and finally 4 choose 4 x^2 goes down a power again. So that's 0 - 5 over x to the^ 4. So we put our choice in 42012 but we can just use Pascal's triangle to actually write them out. And quite often you can just skip this bit and go straight into it. But I just want a full solution for you. So we've got one x^2 4 is x8 - 5x 0 is 1 + 4 x^2 cub is x6 and then - 5 /x and then the next term 6 x^2 is x 4 - 5^ 2 is 25 over x^2 next term is four again x^2 to the 1 and then - 5 cubed is -125 over x cubed and finally at the end back to one x^2 0 is 1 - 5 4 is 65 all over x 4 and now we just need to simplify that so 1 x to 8 4 5 is 20. So it's - 20. X6 /x is x5.

[00:04:22]6 * 25 is 150.

[00:04:26]And then that will give me x4 x^2 which is x^2. 4 * 125 is 500. So - 500. And we've got x^2 over x cubed. So I could put it over x or x to the minus one. But over x is neater. plus 16625s all over x to the 4 and we're done there. Qualifications Scotland advance higher math 2026 paper 3 question three Mcllon expansions find and simplify Mcllen expansion up to x cubed of e 3x log 1 + x and then hence find and simplify up to and including term of x cubed of e x 3x log 1 / 1 + x interesting right mic examinion start exam paper so for each one you need to basically just keep differentiating the sum and zero eventually it's a nice part 2 over 2 xq cube over 3 factorial and so on. So for part one, I want e to the x e to the 3x. Now with some of these, some teachers just tell you to memorize them and you can memorize some of them or you can actually just do it. Okay, f dash x is 3 e 3x.

[00:05:43]F dash x is 9 e 3x.

[00:05:48]The third derivative is 27 e 3x and f of 0 is 1. f-0 is 3. f d-0 is 9.

[00:06:04]Third derivative of 0 is 27. So that third x cubed. So we now just sub in f of x is f of 0 plus f-0 x plus the second derivative of 1 is 9 x^2 / 2 2 factorial is 2 + 27 x cubed over 3 factorial 3 2's is 6. Simplify a little bit.

[00:06:35]1 + 3 x + 9 x^2 / 2 + 3 into 27 is 9 over two. And we're done there.

[00:06:51]Part two, do it for log 1 + x. Okay. So for part two, f ofx is log 1 + x. So f of 0 is log one. The log of 1 is zero. f - x is 1 / 1 + x.

[00:07:12]f-0 is 1.

[00:07:17]Second derivative.

[00:07:21]So now I need to get that ready to differentiate 1 + x to the minus1. So that means it's minus 1 + x^2 because it's -1 1 + x - 2. Second derivative at zero then is -1 cos - 1 / 1^ 2. Now the third derivative well that's -1 1 + x -2 so that's 2 * -1 is 2 and that'll be 1 + x - 3 or 1 + x cubed on the bottom and then at zero we get 2 over 1 cubed which is 2. So now we just sub in again. f ofx equals f of0. So f of 0 for this one was zero.

[00:08:19]Then f-0 x. So + 1 x.

[00:08:25]Second derivative -1 x^2 / 2 factorial. And the third derivative + 2x cub over 3 factorial.

[00:08:36]Tidy up.

[00:08:37]x - x^2 / 2 + 2 over 3 factorial is 2 / 6 which is 1 over 3 and we're done there. Part B. Hence find this formula expansion up to x cub 3x log 1 + x. A little bit of manipulation required because it's we had a log 1 + x but we can use the rules of logs. 1 / 1 + x is the same as 1 + x to the minus1.

[00:09:06]So that equals minus e 3x log 1 + x.

[00:09:11]Hopefully you did that. And that means that we can just then times our two series together. Times together. And not forgetting our minus. So just put a minus here. We have got our original one. 1 + 3x + 9 / 2x^2 plus 9 / 2x cubed.

[00:09:36]And that gets times by the answer to this one which is x - x^2 / 2 plus x cubed over 3.

[00:09:47]And we just ignore any term it goes above x cubed. So keeping that minus in check, we times by one to start with. So 1 * x - x^2 / 2 + x cubed over 3. So that's just tsing through by 1. And I take a little check for myself. Now times in through by the 3x. So 3x * x is 3x^ 2. 3x * - x^ 2 / 2 is - 3x cubed / 2. And then it should be obvious that if I times by this this one I go above. So I stop.

[00:10:26]Now the next term 9 / 2x^2 * x is 9 / 2x cubed.

[00:10:32]And then again if I times by anything else I'll go above. So I've done that one. So I'll just check it off. And similarly if I times this one by any of these ones I'll go above x cubed. So nothing times. So I've done them all.

[00:10:47]Okay. So now I write out my answer simplifying it in best I can. I'll leave the minus to the end. I've got x. I've got - x^2 / 2 + 9. There's no other x2 + 3 x^2. So 3 - a half. 6 - 1 is 5 halves.

[00:11:08]You can use a calculator anytime you want. Of course, the other terms we've got is x cubes. We've got one, two, three of them. So I've got 9 / 2 - 3 over 2 is 6 / 2. 6 / 2 is 3. 3 + 1/3 3 is 9/3 + 1/3 is 10/3. So + 10 over 3 x cubed.

[00:11:33]There's a minus. So I'll take a minus outside. So finally - x - 5 / 2 x^2 is 10 / 3x cubed. And we're done there.

[00:11:44]Qualification Scotland advance high maths 2026 paper two question four had the uklidian algorithm using uklidian alum algorithm to find the greatest common divisor of 1428 and 567. So let's do part eight. We start off by saying well we've got 1428 equals so many multiples of 567. So how many? Well calculator is going to help.

[00:12:10]1428 / 567 is 2 something. So it's 2 * 567 and we need to get our remainder.

[00:12:19]So we do two 567s.

[00:12:23]1428 - 2 * 567s is 294. So plus 294.

[00:12:31]We take our 567 and we divide that by the 294 the remainder. So that's just going to be 1* 294. I can see that already. Again, just in case I make a mistake, 567 minus 294 is 273.

[00:12:51]Again, we just start again. We take our 294. We divide by 273. So, 273s plus 73 83 93 is 21.

[00:13:05]Let's keep going. 273 equals I don't know how many 21s go in. Can't be bothered. 273 / 21 13. Exactly. How good is that? 13 * 21 + 0. There we are. We got zero. So the greatest common divisor is 21. So the GCD is equal to 21 or I guess D equals 21 considering that that's what we call it. Anyway, now we go to the next bit. Part B says, find integers A and B such that 1428 B + 567B equals D.

[00:13:43]So we just sub in backwards going up the chain. All right. So starting at the second line, we rearrange that to get 21 = 294 - 1 * 273.

[00:13:59]And then we can go up a line because we can say sub in 273. So that equals 294 - 1 * 567 -1 * 294.

[00:14:16]Tidy that up into multiples of 294 and 567. I've got 1 - 1 * -1 is pos1. So 1 2 * 294s in -1 * 567s.

[00:14:31]Go up the chain again. 7 294.

[00:14:35]So that gives me 2 * 1428 - 2 * 567 - 1 * 567 still hanging out. So that's 2 * 1428.

[00:14:58]2 * -2 is -4 minus another one is -5 * 567.

[00:15:05]So 21 equals that and we have to write find integers a and b. 14 28 a + 567.

[00:15:13]So a is = 2 and b is = -5 and we're done there. Qualification Scotland math 2026 paper 2 question five. Given that y = x 4x, use logarithmic differentiation to find the y by the x. Well, this looks not too bad, this one. We just take the log of both sides.

[00:15:37]Much easier than the one I did in my live stream.

[00:15:40]And then we differentiate both sides implicitly.

[00:15:44]So when you differentiate a y, you always get dy by dx back. So log y goes to 1 / y dy by dx.

[00:15:52]And I can take this 4x. Remember 4x can come down to log x. So I've now got the product rule. So v u dash 4 log x + u v dash 4x / 4. So 1 / y dy by dx is equal to 4 log x + 4.

[00:16:17]4 log x + 4. I can times by y now. So dy by dx is equal to y 4 log x + 4 which is equal to well y with x 4x 4 log x + 4 and 4's a common factor 4 x 4x log x + 1 would be fine as well. Some people may even have put the four up here but I think we're fine. So we're done there. qualification Scotland advance maths 2026 paper 2 question six an arithmetic sequence has terms u3= 6 and u 111 10 for the sequence common difference first term is sum of the first 109 terms so remember for an arithmetic sequence it starts off with an a then it goes a plus a difference then a + 2d all the way up to a + n minus 1d in other words if you had the third term you've got 2d and If you've got the 11th term, you've got 10 b. So u3 is 6. So u3 is a + 2d which is 6 and u1 which is a + 10 d is given by 10.

[00:17:37]Some simple simultaneous equations. Then if we take u 11 - u3 10 - 2 is 8 d. 10 - 6 is 4. So d is a half.

[00:17:50]And we're done there. Next bit, calculate the first term. Calculate the first term. So I just sub back in u3. A + 2d 2 * a half is six. And that will just give me my a back. A + 1 is six and therefore a is five. Nice. Question three. The sum of the first 195 terms.

[00:18:12]Start the exam paper. So we want up to 109 terms. Is it 109? Yeah. 109 terms.

[00:18:19]Start exam paper has sn= a half n 2 a plus n minus 1 b. So s 109 then = a half * 109 2 * a + 108 cuz that's take away one times common difference of a half and we just work that out. 109 / 2 10 + 54 109 / 2 * 64.

[00:18:56]So that equals I'll just get a calculator at this point. 109 / 2 * 64 I get 3 488.

[00:19:04]So we're done there for that bit. Part B, the terms V3 and V4 form a geometric sequence. How interesting. So for this sequence, find a common the first and an expression for the first n terms. Okay.

[00:19:18]So what's a geometric sequence? It's a sequence where there's a common ratio.

[00:19:23]So we got a a r squ a r cubed and so on.

[00:19:28]And that means that the power is one less than n. So for part one to get the common ratio I can say that v3 is equal to a r^ 2 which equ= 18 and v4 is a r cubed then which is 27.

[00:19:45]Dividing them together a r cub over a r 2 is 27 / 18 and therefore r must be 3 over two because 9 goes into both of them. Part two find the first term.

[00:20:00]Well, we can just sub in. We'll just pick a r squar. A r squ was 18. So that implies that since a * 3 / 2^ 2ar is 18.

[00:20:12]Then that is a * 9 / 4 is 18, a * 9 over 4 is 18. So 18 / 9 over 4, which is same as times by 4 9. Or you can use a calculator, but you'll get a = 18 / 9 is 2 * 4 is 8. So we've got our a. And then part three, an expression in terms of n for the first n terms. So I want the first n terms. So I go to start the formula sheet. I grab the sum of the first n terms and then I'll just sub in.

[00:20:41]So there's the start of the formula sheet there. So I'm just going to sub a and r into that. Subing in sn is equal to 8 1us and it was 3 / 2 to the n all over 1 - 3 over2. We just need to try and tidy that up a little bit.

[00:20:58]That's 8 1 - 3 / 2 I'm just going to I'll keep it as 3 over two I think to the n over 1 - 3 is - a half so that is -6 1 - 3 / 2 to the n and if you put a minus inside you don't have to but I will you get 16 - 1 plus 3 over 2 I'm just going to write 1.5 to the end since I'm going to use a calculator anyway in a minute or in other words 16 1.5 to the n minus one and we're done there. Okay, question C says find algebraically the least value that the sum of the geometric sequence exceeds the sum of the first 109 terms of the arithmetic sequence. So in other words, 16 1.5 to the n minus one is greater than the answer we got up here for our sum which was can't even find it 3488.

[00:22:01]Doing a bit of maths we get 1.5 nus1 or 3 over 2 n minus one using a calculator 3488 / 16 is 218.

[00:22:14]So 1.5 to the n is greater than should be greater than 219.

[00:22:19]Log both sides.

[00:22:24]Take the n to the front and that will get us our n. n is greater than log 219 for log 1.5.

[00:22:35]So we get a calculator to work that out.

[00:22:37]So that is n is greater than 13.291.

[00:22:45]But n is a whole number. Remember n is a number of terms. So the least value of n is 14. So the least value of n is 14.

[00:23:00]To answer the question, what's the least value of n? So it's greater than because at 13 it will not be bigger than it. So it's 14. And we're done there. Okay.

[00:23:08]Qualifications Scotland advance high mass 2026 paper two. Question seven. And a curve is defined parametrically by x = 3 log 2 t + 1 and y = t - t ^2. Find the express for the y by dx and simplify your answer. Find the coordinates of a stationary point on the curve. So for the y by dx, you need the y by dt because it's parametric.

[00:23:30]So the y by dt is 1. 2 * a half is 1. So it's minus t. dx by dt log goes to one over so it's 3 over 2t + 1 but don't break the chain you need to times by 2 t + 1 differentiated which is two so that's actually 6 / 2 t + 1 don't try and cancel the 6 and the two because there's not a was a one in the way okay so we've got these two answers so the y by dx is the y by dt t * the t by dx or some people think it is the y by the t / dx by the t. The y by the t is 1 - t and dx by dt is is 6 over 2t + 1. So if I flip back to 2t + 1 / 6.

[00:24:28]Let's make that a little bit nicer then.

[00:24:34]2 t + 1 1 - t / 6. You know, I probably wouldn't expand the brackets because factorized is more simple than unfactorized. It's fully factorized. But if you did, you would have got 2 t - 2 t ^ 2 uh + 2 t minus t is + t + 1 / 6. But I don't see why you would do that. Part B, find the coordinates of the stationary points on the curve. Now find the constitution of the curve. That means we just need to sub our t and x and y.

[00:25:07]There's a slight trick to this, but I'll show you it down here.

[00:25:17]We've already got our t's stationary points, but it says t is greater than zero. So the only valid t is when t equals 1. So we sub one in. We don't sub minus a half in. So x = 3 log 2 1's is 2 + 1 is 3. y = 1 - a half * 1^ 2 that's 1 - a half is a half. So your coordinate of a station point is 3 log 3 and a half. I suppose if you were using a calculator you could work that out.

[00:25:54]3 log 3 is 2.295. 295.

[00:26:00]So about what did I say two? 3 log 3 is about 3.3.

[00:26:05]So you could have 330 and a half but I think exact values are better than non-exact values. So with there for case of Scotland the vans math 2026 paper 2 question 8. The volume in cubic meters of water held in a reservoir is given by 18 2 + roo<unk> h 6 - 1152 where h me m is the depth of water.

[00:26:32]Water is pumped out at a constant rate of 0.6 m cubed per second. Find the rate of change of a depth of water in the reservoir when the depth is 9. So this is related rates of change, right? And a rate of change remember is a derivative.

[00:26:47]So since we're given that V is equal to this and what have we to find? Let's translate the question. First of all, calculate the rate of change of depth of water in the reservoir. The rate of change is always by DT.

[00:27:04]The depth of water is H. So we have to find DH by DT. And what do we know? Well, what we know is water is pumped out of the reservoir at a rate of 0.6 m cubed, which is a volume per second. So, in other words, the V by D T is straight away 0.6. Can we find another derivative? Well, yes. We've got a V and a H. So, we can find the V by DH.

[00:27:37]So, let's go ahead and find the V by DH.

[00:27:40]So got 18 times a bracket minus a number. So take the bracket 6 * 2 6 * 18 6 10 is 60 68 is 48. So that's 108 2 + root h to the five.

[00:28:05]So I've took the derivative down took away one power and then times by the in the brackets differentiated well root h remember h to the half. So the two differentiated is nothing h to the half differentiated is a half h to the minus a half if minus a number the number goes away. So we'll just tidy that up. A half of that is 54. So it's 54 2 + roo<unk> h to the^ of 5. I suppose I could put the h on the bottom as a root as well because h to the minus a half is h to the half on the bottom which is root h. So I've got dv by the h. So now I've to find the h by dt.

[00:28:49]So in terms of t, I've got dv by dt.

[00:28:53]I don't want a dv. I want a dh. So if I take my DV by DH and flip it upside down or divide by it, I get DH by DV.

[00:29:02]Essentially, it looks like these cancel.

[00:29:03]We don't actually, but it works like that. V by the T is 0.6.

[00:29:09]So that's 0.6 times this upside down because we're flipped upside down. So root h over 54 2 + root h to the 5. Now we have to do that when the depth of the water is 9.

[00:29:28]So when h = 9, do h by dt is equal to 0.6<unk> 9 / 54 2 +<unk> 9 to 5 0.6 6 * 3 over 54 2 + 3 54 * 5 to the 5. Now I'm just getting to the point of I can't do what to do anyway.

[00:30:08]Sums in my head. So 6 * 3 is 1.8 8 all over 54 * 5 ^ 5 is 168750.

[00:30:27]Since you got a calculator, just do it.

[00:30:30]168750 0.00001666.

[00:30:37]Anyway, you leave that as a fraction 1 over 93750 or 0.00 1 0666 which in scientific notation would be like 1.07 07 * 10 ^ of 1 2 3 4 5 - 5 our units dh by dt m/s and we're done there. Qualification Scotland advance maths 2026 paper 2 question 9 express 3442 base 5 in base 9. So steps to converting bases are convert to base 10 first then go from 10 to 9. So we've got 3 42 and base 5 3 * 5 cubed 4 * 5^ 2ar 4 * 5 1 I suppose plus 2 * 5 to then nothing just like if it was base 10 it would be like 3 * 1,000 3 * 100 3 * 10 and 3 * nothing 3 * 1 but it's five powers instead 10 powers all right So we then work that out.

[00:32:10]That's three 125s.

[00:32:13]425s 4 20 4 * 5 is 20 I suppose plus two 3 * 125 is 375 4 * 25 is 100 20 and 2. So in base 10 375 475 495 497 497 base 10. Step one done.

[00:32:42]To convert a number from base 10 into base 9, you just keep dividing by 9. So to go to base 9, we do 497 / 9.

[00:33:00]which is 55 and the remainder on that is we do remainders now. First time I've done a remainder since primary school 55 remainder 2.

[00:33:18]Now we take the 55 and divide by 9. 9 fives is 45 9 sixes is 54 remainder 1.

[00:33:26]So then we take the six and divide by 9 which you can't of course. So you've got zero and still remainder six. That is it in base 9. So you're done. So the answer is 49.

[00:33:39]Oh not actually write it out. I'll just write out 612 base 9 and we're done there.

[00:33:46]Qualifications Scotland math 2026 paper two question 10 and a curve is defined by x^2 6 e 6 x 6 y + x^2 + y 5 is 50 find the y by dx just implicit differentiation there then explain why the derivative is never zero we'll see when we get there I suppose part a I've got the product rule first so it's v u dash 2x e 6 y plus u vdash which is x^2 time 6 e 6 y. Every time you differentiate a y, you get the y by the x back. So, times the y by the x plus x^2 is 2x plus y 5 is 5 y 4 d y by dx and that equals nothing cuz 50 differentiates to zero. I'll take out I'll tidy up and take out all the stuff with the y by the x in it. So that's given me 2x e 6 y + 6 x^2 e 6 y d y by dx + 2x + 5 y 4 d y by dx equals nothing.

[00:35:05]Taking out the y by dx stuff as a common factor, I get 6 x^2 e 6 y + 5 y 4.

[00:35:15]And then I've still got the other stuff, but I can move that to the other side.

[00:35:19]So I've got minus 2x e 6 y minus another 2x because I've just moved these ones over to the other side and take a factor of them. So now dy by dx is the right hand side divided by the left hand side - 2x e 6 y - 2x over 6 x^2 e 6 y + 5 y 4.

[00:35:55]Now we'll just try and tidy it up a little bit. Always look for common factors. So on the top - 2x or positive 2x the common factor I'll take out minus 2x so I've not got two minuses in the brackets so I'll have e to the 6 y and then + one but you could take positive 2x if you want in the bottom there's nothing common between any of these terms so there's nothing I can do with it.

[00:36:25]So I'm fairly confident that's the y by dx. Part B, given that x is greater than zero, why is the derivative never zero?

[00:36:33]Part never zero. Yeah, never zero. X gater than zero. We just need to identify which bit of this fraction makes sense. Well, it's a fraction for one. So the bottom bit of the fraction never produces zero for it anyway cuz it's just a fraction. The only thing that can be zero is the top bit for the whole thing to be zero. So we just need to examine the top bit. So when x is greater than 0 - 2x is less than zero.

[00:37:01]What about the other bit though? Well for all y a member of r e to the 6 y is greater than zero because it's e to the think of a positive number. It's it's a number. e to the 0 is 1 and e to the negative number is still a number because it's just one over e to the positive number. Again, if e 6 y is greater than 0, then e to the e 6 y + 1 would be greater than zero and therefore positive.

[00:37:34]And that means that we've got a negative time a positive. So - 2x e 6 y + 1 is strictly less than zero when x is greater than zero. and for all y if that is true just say therefore d y by dx is never zero when x is greater than zero now some you might have went further than that and tried to look at the denominator if you do look at the denominator I'll just do it quickly there by the way that's positive that's always positive so that's a positive anything to the power four h is positive because it's an even rooe. So the denominator is strictly positive anyway which means the whole thing is strictly negative because a negative plus it's a negative but you don't need to look at the denominator I don't think because it's a fraction so it's just a numerator that counts. Qualifications Scotland advance math paper 2 question 11.

[00:38:39]Integrate 2x + 4x^2 + 4. Well, don't think so. I'm going to integrate instead 2x^2 + 4 and 4x^2 + 4. Just simplify things out as much as you can. You can always simplify by splitting the numerator up. Now, this is actually a standard result, but you don't need to know that. You can just use a substitution. So let u = x^2 + 4 for that bit. Then d u by dx is 2x and that means that 2x dx is du.

[00:39:16]So our integral becomes the integral of du over u which I know you know is a log now. See plus in the other one 4x^2 + 4 is just a standard result from the start of the exam paper. So standard results if we look at the start of the exam paper we get this 1 / a^2 + x^2 integrates to 1 / a inverse tan x a. So that means that our integration becomes log u which change back in a minute plus well the four just comes along for the ride times now what is a? Well x2 + 4 a 2 + x 2 a is 2.

[00:39:55]So it's 1 / 2 inverse tan of x / 2. And we've not got any limits, so it's plus c.

[00:40:04]But u is way back up here, x2 + 4. So I've got log x^2 + 4. 4 / 2 is 2 inverse tan x / 2 plus c. And we're done there.

[00:40:20]Not absolute values. We're not strictly necessary for this because the log of x2 + 4 is always positive anyway because the minimum it can possibly be is four.

[00:40:30]So you could have just normal brackets and probably not lose any marks on that and I probably could just put normal brackets there. Quif as maths 2026 paper 2 question 12 proof by induction 7 n + 2 is divided by three for all n. So let's check the base case n= 1. We've got 7 1 + 2 which = 9 and 9 = 3 * 3. Therefore divisible by 3.

[00:41:02]Okay. Assume true for n= k.

[00:41:05]Assume true for n= k.

[00:41:12]So we're trying to say that in other words 7 the k + 2 is divisible by 3. But we want to be able to say that in a way that makes sense in maths. So 7 k + 2 equals 3 times some number a say where a is a member of z.

[00:41:38]Now let's prove true for n= k + 1.

[00:41:45]So, we're going to try and prove that 7 k + 1 + 2 is divisible by three. Well, if we can make this look like something time 7 k + 2, then we know that 7 k + 2 can be swapped out for 3 a. So, let's see if we can do that.

[00:42:08]And if it could be swapped out for 3 a, it doesn't matter what times what's left because it'll be three times something.

[00:42:14]which is divisible by three by definition because it's three times something. So let's see 7 the k + 1 + 2 well 7 the k + 1 is 7 * 7 the k and I've still got plus two. Now this is a little bit trick this one we know that 7 the k + 2 = 3 a. So can we replace 7 the k? Well yes we can.

[00:42:43]70 the k + 2 = 3 amp implies 7 the k is 3 a minus 2. So that means we can see that this is 7 * 3 a - 2 straight away plus another two and then we can expand the bracket. 7 3es is 21 a 7 2's is 14 plus another two 21 a -4 add 2 is -2 and all we're doing is seen as that is we a common factor that's 3 * 7 a - 4 a is a member of z therefore 7 a is a member of z and therefore 7 a minus 4 is a member of Z.

[00:43:36]So since 7 A - 4 is an integer of N that means that therefore by um 7 7 K + 1 + 2 is divisible by 3.

[00:43:52]So now our statement true for n= to 1 and assuming true for n= to k implies true for n= to k + one by induction true for all n a member of the natural numbers as required and we're done there Scotland of maths 2026 paper to question 13. Express using partial fractions that as partial fractions then it's a differential equation. It looks like a first order linear not separate variables. So yeah first order line differential equation shouldn't be too bad. Let's hope part a x + 3 x + 7 x + 5 are just linear. So it's a / x + 7 + b / x + 5.

[00:44:54]That implies I can do a whatever that be do a * x + 5 plus b * x + 7 equals our numerator.

[00:45:07]Now when x = -5, I can just get rid of this. -5 + 5 is 0. -5 + 7 is 2 b.

[00:45:16]-5 + 3 is -2. So b is -1.

[00:45:21]And then when x is -7 - 7 add 5 is -2. A b is gone. -7 add 3 is -4.

[00:45:33]So a is 2. So our partial fractions are 2x + 7 minus 1 /x + 5.

[00:45:48]Right? Part A done. Part B. Okay. Part B. Hence find a particular solution of a differential equation d of xy + 2x + 3 y = that. Now that is a first order linear differential equation. So that's dy by dx plus pxy = qx where q is the right hand side and p of x is a bit in front of y which is 2x + 3 and then we're going to sub in initial conditions at the end. So our integrating factor we do first a lot of people call that mu of x and it's just e to the integral of p of x dx.

[00:46:32]So that is e to the integral of 2x + 3 dx. But that's just 2 log. So that's just e to the 2 log x + 3. That's e to the log x + 3 all^ 2. e and the log cancel. So mu of x is just x + 3 all squared for integrating factor. Now the next step mu of x y = the integral of mu of x q of x dx.

[00:47:12]So mu of x is x + 3^ 2 * y = integral of x + 3 2 again* 1 / x + 7 x + 5 x + 3.

[00:47:33]Now that looks nasty. However, we can cancel out some of that to get the integral of an x + 3 cancel x + 3 x + 3 over x + 7 x + 5. I'm pretty sure part was that part a is x + 3 x + 7 x + 5.

[00:47:52]We've already done the partial fractions. 2 x + 7 - 1 x + 5. So, we'll put that in instead.

[00:48:00]So x + 3^ 2 y equals finally the integral of 2 x + 7 - 1 x + 5 dx.

[00:48:18]So that equals 2 log x + 7 - log x + 5 plus a c So now we can just use the rules of logs to combine the logs log x + 7^ 2 which is always positive actually over x + 5 + c. So you probably don't need the four bars because that's always positive.

[00:48:52]Now we've got some initial conditions. Y is zero and x is three. So we can get this c back. So you can sub it now or after the same y equals I think now would be easier. So what did we say x is free and y is zero.

[00:49:09]So our left hand side then gives you 3 + 3 * 0 equ= log 3 + 7^ 2. So that's 3 + 7 over 3 + 5 + c. So 0 is log 10^ 2ar is 100 / 8 + c. c is minus log 4 25s is 100, four twos is 8. Or you can say that c is log 2 over 25 because you can take the minus up to the power position which means you just flip the fraction upside down. Okay. So we've got our c i a2 as long as you put a minus in. So back up to what we started with, we knew that this was true here. So we got x + 3 2 y = log x + 7 2 x + 5.

[00:50:12]So I used to put the bars in it. I just do it. plus log 2 over 25 or minus log 2 / 2.

[00:50:22]Finally, my y is all divided by x + 3 2 then. So log x + 7^ 2 over x + 5 plus log 2 over 25 / x + 3 2. But I've not actually combined my logs. I just go y = log.

[00:50:43]It's 2 over 25. I can put 2 x + 7^ 2 for neatness over 25 x + 5 all over x + 3^ 2.

[00:51:00]And we're done there. For neatness, quite a lot of people would write 1 / x + 3 2 like that and then just put the log to the side. I don't think you'll not lose a mark for it anyway.

[00:51:10]It just looks a little bit neater if you do that instead.

[00:51:17]But either way, it's the same answer.

[00:51:20]Scam math 26 paper 2 question 14.

[00:51:23]Vectors a plane pi has points P, Q, and R. Determine the equation of the plane.

[00:51:29]So for equation of a plane, I need two vectors on the plane. So imagine we have got a plane. There we are. draw a bad picture for you and you've want two vectors. So we can imagine PQ and P R and then we can find the normal to the plane. So P to Q I'm not going to do the math for this because it is just higher mass here. 3 - 2 is 1. 5 - 3 is 2. 1 - - 4 is 5.

[00:52:04]PTR 6 - 2 is 4 0 - 3 is - 3 - 6 - -4 - 6 add 4 is 2 well minus the normal is got by taking the cross product say that in my last minute live stream so our normal we do the cross productduct of our vectors which is formal exam paper the determinant of II K 125 and 4 - 3 - 2 doesn't really matter the order of 125 and four or top or bottom so start exam paper says that that equals I times the determinant of you just delete this to get 25 - 3 - 2 minus J delete your I row and column 1 54 minus 2 and K delete your row and your column 1 2 4 - 3.

[00:53:20]Okay, so that is -4 + 15 because it's just that times that minus that times that. So -4 + 15 is 11 I - 2 - 20 - 22 * -1 is + 22 J - 3 - 8 is - 11 K. So there's a normal vector or you can take 11 out as a common factor. I'll just write it as a column 1 2 - 1. So you can actually just use 1 2 - 1 instead of 11 * it because 1 2 - 1 is parallel to 11 * 1 2 - 1 n dot r equ= n do a so n dot r we're going to use as 1 2 -1 dotted with xyz = 1 2 -1 point p which is 2 3 - 4 so that's x + 2 y - 2 1 z = 1 * 2 is 2. 2 3's is 6 - 1 * - 4 is 4.

[00:54:40]So x + 2 y - z = 12 is the equation of our plane. And we're done there. Okay.

[00:54:49]The line has symmetrical equations. x - 8 / 2 + y + 2 - 1 = z + 2 3. The line p is x s. Find the coordinates of s. To find the point of intersection, we just need to do simultaneous equations. We substitute the equation of a line into our plane. But our line is in symmetrical form. So we'll need to make that equal to some lambda say where it's a constant. And then you just rearrange for lambda. So that means that x= 2 lambda + 8 you just times by the denominator then sign changes on the top y = - lambda - 2 and z = 3 lambda - 2. We've got the equation of a plane then which was x + 2 y - z = 12. So subbing n we get 2 lambda + 8 + 2 - lambda - 2 - 3 lambda - 2 = 12 2 lambda + 8 - 2 lambda - 4 - 3 lambda + 2. Watch your signs.

[00:56:06]2 - 2 is 0. You got - 3 lambda.

[00:56:10]8 - 4 is 4 + 2 is 6 - 3 lambda of n is 6. So lambda must be min -2. So we know a lambda now we need to get a point to get a point sub it into our xy z to get three numbers. So when lambda is -2, x is 2 * -2 + 8 - 4 + 8 is 4. y = - lambda - 2. So - - 2 - another 2.

[00:56:46]That's 2 - 2 is 0. And zed equals 3 lambda - 2.

[00:56:55]That's -6 - 8. So our point of intersection is finally x y z and we're done there. The size of acute angle between l and pi. Okay. So if we want to calculate the size of angle between two planes, we can use cos theta u dob over uv.

[00:57:22]So we need our normal vector our normal to the plane 2 minus one.

[00:57:34]And then we need the direction of the line. So the direction of a line is given by the bottom numbers here or the numbers in front of lambda. In other words, 2 - 1 3.

[00:57:50]So n do d is 1 * 2 is 2 2 * -2 is -2 - 1 * 3 is -3 that's - 3. The size of n is the square root of 1 2 + 2^ 2 + - 1 2 1 1's is 1 2's is four that's five 5 + 1 is 6. And for D we've got 2^ 2 + - 1^ 2 + 3 2 4 5 5 + 9 is 14. So we can say that cos of some theta equ= -3 over<unk> 6 <unk>14.

[00:58:37]So we can just do inverse cos of 3 /<unk> 6 - 3 over<unk> 6<unk>4 which is 109.1°.

[00:58:54]But that's not acute. We want acute.

[00:58:58]So we can just do 180 minus 109.1° which is 70.9. But because we're doing a line against a plane and the line the plane is you've just done it to the normal which is at right angles to the plane to get the actual angle between the line and the plane you need to do 90 minus. So our final answer is 90 minus 70.9 - 70.9 as with 19.1° and we're done there. Scotland maths 2026 paper 2 question 15 let r be positive real number and consider the following statement is if r is irrational then root r is irrational.

[00:59:51]Write down the contraositive.

[00:59:53]If not Q not P. If root R is rational then R is rational.

[01:00:09]There's your contraositive statement there. Part B. Hence prove the statement is true. Again I did a prove root something is rational. I was rational in the last minute live stream. So it shouldn't be too difficult for me to do again. Prove the statement is true. So we prove the contraositive which means you prove the original statement is true. So we're going to prove the contraositive. So let's do that. If R is rational then we can write a rational number as root r= a / b and they are co-p has no common factor. So we can write that side as A and B are member of Z and obviously B does not divide A. Right?

[01:00:57]Obviously B as well doesn't equal zero because you can't divide by a fraction.

[01:01:02]So that's that. Now we just square both sides. R= A^ 2 over B^ 2.

[01:01:08]And that's it. A squ is a member of Z.

[01:01:11]Obviously B^ square is a member of Z as well because A and B were a member of Z.

[01:01:17]B square is definitely not equal to zero cuz B was equal to zero and so R is rational and it's a given that a squ over b square share no common factor because a and b didn't share any common factors but fundamentally.

[01:01:36]So therefore we've done that. So contraositive true therefore original statement true and so so if R is irrational then root R is irrational and we're done there Scotland as 226 paper 3 question 16 given y = log cos x show that d y by d x is - tan x and then part b I don't know integration by partial for it seems so let's do part a y is log cos x so we're going to use the chain rule of course d y by dx then must be 1 / cos x times cos x differentiates to negative sin x that's e of n that's - tan x. We're done. Part B, determine the exact value of the integral of 0 to 5 of x gx. And it tells us that x gx is this.

[01:02:45]But in part a, we differentiated log cos x and got minus tan x. So we can integrate minus tan x to get back to log cos of x, which means we just use the chain rule on this bit. So for part B, so the integral between 0 and pi / 6 of x gx must equal 2x tan 2x - 2 * the integral take the 2 out of tan 2x dx between 0 and pi / 6.

[01:03:27]So that must be 2x tan 2x and - 2 - tan 2x goes to log cos 2x. So + 2 log cos 2x but then the chain will divide by 2 and that's between 0 and p<unk> / 6.

[01:03:51]So that is 2<unk>i / 6 tan 2<unk> / 6 plus the twos cancel log cos 2<unk> / 6 take away 0 + log cos of 0. Exact value triangle 60 30 1 2<unk>3.

[01:04:20]So 2<unk> / 6 is p<unk> / 3 tan p<unk> / 3 then is <unk>3 plus log cos / 3 is a half - 0 - log cos 0 is 1.

[01:04:41]So finally that's <unk>3<unk>i over 3 plus log just a half because rules of logs or log 1 is equal to 0. Or if you want another neater way which I don't think you'll need to do roo<unk>3 pi over 3 minus log 2 and that's because log a half is log 2us1 which is minus log 2 but I don't think you need to do that. Okay, so that's a got exact value. Part two, find an expression for g of x in terms of x.

[01:05:15]And we know that integral of x gx dx equals all this. So let's just take a note of that and we'll just do something to get g of x. So if we got that in part two, then that looks like integration by parts. So we just need to identify our u and our v dash because that equals u v minus integral of v u dash. Now u and v dash. So let's have a look.

[01:05:39]X and GX. So one of MU and one of them V dash. You just do it trial and error and try or you can just look at the right hand side. You've got a tan 2x appear twice here and here. So V must be part of tan 2x.

[01:05:53]But is it just tan 2x?

[01:05:56]So u if u was 2x then we' have a 2x here because that's u and u. So u must be actually just the x which means the v must be 2 tan 2x.

[01:06:10]Let's just check that then. UV v 2x tan 2x minus the integral of v 2 tan 2x u dash is 1. Yep. That works very well.

[01:06:23]That means that g of x is just v dash.

[01:06:26]So I just need to work out v dash 2 tan 2x. Go to the start of the exam paper.

[01:06:31]Start exam paper says tan differentiates to sex 2. So 2 se 2x times differentiate 2 x to 2. So that's 4 sex^2 2x. Therefore g of x is 4 x^2 x and we're done there. This has been Mic Maths. We did qualification Scotland as high maths 2026 paper two. Hopefully you found that useful and you've done well in it. Take care. Stay safe.