Advanced Higher Maths · Qualifications Scotland past paper

2026 Paper 2

Calculator · 16 questions
1

Question 1

Differentiation
2 Marks
2026 P2 Q1

Differentiate f(x)=3sin17x.\displaystyle f(x)=3\sin^{-1}7x.

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f(x)=3×11(7x)2×7=21149x2\displaystyle f'(x)=3\times\frac{1}{\sqrt{1-(7x)^{2}}}\times7=\frac{21}{\sqrt{1-49x^{2}}}
2

Question 2

Binomial Theorem
4 Marks
2026 P2 Q2

Write down the binomial expansion of

(x25x)4\displaystyle \left(x^{2}-\frac{5}{x}\right)^{4}

and simplify your answer.

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x820x5+150x2500x+625x4\displaystyle x^{8}-20x^{5}+150x^{2}-\frac{500}{x}+\frac{625}{x^{4}}
3

Question 3

Maclaurin Series
(2, 2, 2)6 Marks
2026 P2 Q3

(a)Find and simplify the Maclaurin expansion, up to and including the term in x3\displaystyle x^{3}, for:

(i) e3x\displaystyle e^{3x}

(ii) ln(1+x).\displaystyle \ln(1+x).

(b)Hence find and simplify the Maclaurin expansion, up to and including the term in x3\displaystyle x^{3}, for e3xln(11+x).\displaystyle e^{3x}\ln\left(\frac{1}{1+x}\right).

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(a)(i) e3x=1+3x+92x2+92x3\displaystyle e^{3x}=1+3x+\frac{9}{2}x^{2}+\frac{9}{2}x^{3}
(a)(ii) ln(1+x)=x12x2+13x3\displaystyle \ln(1+x)=x-\frac{1}{2}x^{2}+\frac{1}{3}x^{3}
(b) e3xln(11+x)=e3xln(1+x)=x52x2103x3\displaystyle e^{3x}\ln\left(\frac{1}{1+x}\right)=-e^{3x}\ln(1+x)=-x-\frac{5}{2}x^{2}-\frac{10}{3}x^{3}
4

Question 4

Number Theory
(1, 2)3 Marks
2026 P2 Q4

(a)Use the Euclidean algorithm to find d\displaystyle d, the greatest common divisor of 1428 and 567.

(b)Find integers a\displaystyle a and b\displaystyle b such that 1428a+567b=d.\displaystyle 1428a+567b=d.

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(a) 1428=2×567+294\displaystyle 1428=2\times567+294, 567=1×294+273\displaystyle 567=1\times294+273, 294=1×273+21\displaystyle 294=1\times273+21, 273=13×21\displaystyle 273=13\times21, so d=21\displaystyle d=21
(b) Working backwards, 21=2×14285×567\displaystyle 21=2\times1428-5\times567, so a=2\displaystyle a=2 and b=5\displaystyle b=-5
5

Question 5

Differentiation
5 Marks
2026 P2 Q5

Given y=x4x\displaystyle y=x^{4x}, use logarithmic differentiation to find dydx.\displaystyle \frac{dy}{dx}.
Write your answer in terms of x.\displaystyle x.

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lny=4xlnx\displaystyle \ln y=4x\ln x, so 1ydydx=4lnx+4.\displaystyle \frac{1}{y}\frac{dy}{dx}=4\ln x+4.
dydx=x4x(4lnx+4)=4x4x(lnx+1)\displaystyle \frac{dy}{dx}=x^{4x}(4\ln x+4)=4x^{4x}(\ln x+1)
6

Question 6

Sequences & Series
(1, 1, 1, 1, 1, 1, 1)7 Marks
2026 P2 Q6

(a)An arithmetic sequence has terms u3=6\displaystyle u_{3}=6 and u11=10.\displaystyle u_{11}=10.
For this sequence, find the:

(i) common difference

(ii) first term

(iii) sum of the first 109 terms.

(b)The terms v3=18\displaystyle v_{3}=18 and v4=27\displaystyle v_{4}=27 form part of a geometric sequence.
For this sequence, find:

(i) the common ratio

(ii) the first term

(iii) an expression, in terms of n\displaystyle n, for the sum of the first n\displaystyle n terms.

(c)Find algebraically the least value of n\displaystyle n such that the sum of the geometric series exceeds the sum of the first 109 terms in the arithmetic sequence.

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(a)(i) 8d=4\displaystyle 8d=4, so d=0.5\displaystyle d=0.5
(a)(ii) a+2(0.5)=6\displaystyle a+2(0.5)=6, so a=5\displaystyle a=5
(a)(iii) S109=1092(10+108×0.5)=3488\displaystyle S_{109}=\frac{109}{2}\left(10+108\times0.5\right)=3488
(b)(i) r=2718=1.5\displaystyle r=\frac{27}{18}=1.5
(b)(ii) a(1.5)2=18\displaystyle a(1.5)^{2}=18, so a=8\displaystyle a=8
(b)(iii) Sn=8(1.5n1)0.5=16(1.5n1)\displaystyle S_{n}=\frac{8(1.5^{n}-1)}{0.5}=16(1.5^{n}-1)
(c) 16(1.5n1)>34881.5n>219n>ln219ln1.5=13.29...\displaystyle 16(1.5^{n}-1)>3488\Rightarrow1.5^{n}>219\Rightarrow n>\frac{\ln219}{\ln1.5}=13.29..., so the least value is n=14\displaystyle n=14
7

Question 7

Differentiation
(2, 2)4 Marks
2026 P2 Q7

A curve is defined parametrically by

x=3ln(2t+1),y=t12t2,\displaystyle x=3\ln(2t+1),\quad y=t-\frac{1}{2}t^{2}, where t>0.\displaystyle t>0.

(a)Find an expression for dydx.\displaystyle \frac{dy}{dx}. Simplify your answer.

(b)Find the coordinates of the stationary point on the curve.

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(a) dxdt=62t+1\displaystyle \frac{dx}{dt}=\frac{6}{2t+1} and dydt=1t\displaystyle \frac{dy}{dt}=1-t, so dydx=(1t)(2t+1)6\displaystyle \frac{dy}{dx}=\frac{(1-t)(2t+1)}{6}
(b) dydx=0\displaystyle \frac{dy}{dx}=0 gives t=1\displaystyle t=1 (since t>0\displaystyle t>0), so the stationary point is (3ln3, 12)\displaystyle \left(3\ln 3,\ \frac{1}{2}\right)
8

Question 8

Differentiation
4 Marks
2026 P2 Q8

The volume, V\displaystyle V cubic metres, of water held in a reservoir is given by

V=18(2+h)61152\displaystyle V=18(2+\sqrt{h})^{6}-1152

where h\displaystyle h metres is the depth of water.
Water is pumped out of the reservoir at a constant rate of 0.6\displaystyle 0.6 m³s⁻¹.
Find the rate of change of the depth of water in the reservoir when the depth is 9 metres.

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dVdh=54(2+h)5h\displaystyle \frac{dV}{dh}=\frac{54(2+\sqrt{h})^{5}}{\sqrt{h}}, which at h=9\displaystyle h=9 is 54×553=56250.\displaystyle \frac{54\times5^{5}}{3}=56\,250.
dVdt=dVdh×dhdt\displaystyle \frac{dV}{dt}=\frac{dV}{dh}\times\frac{dh}{dt}, so 0.6=56250×dhdt\displaystyle -0.6=56\,250\times\frac{dh}{dt}
dhdt=193750\displaystyle \frac{dh}{dt}=-\frac{1}{93\,750} m/s (about 1.07×105\displaystyle -1.07\times10^{-5} m/s)
9

Question 9

Number Theory
3 Marks
2026 P2 Q9

Express 34425\displaystyle 3442_{5} in base 9.

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34425=3(125)+4(25)+4(5)+2=49710\displaystyle 3442_{5}=3(125)+4(25)+4(5)+2=497_{10}
497÷9=55\displaystyle 497\div9=55 r 2, 55÷9=6\displaystyle 55\div9=6 r 1, 6÷9=0\displaystyle 6\div9=0 r 6, so the answer is 6129\displaystyle 612_{9}
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Question 10

Differentiation
(4, 1)5 Marks
2026 P2 Q10

A curve is defined by the equation x2e6y+x2+y5=50.\displaystyle x^{2}e^{6y}+x^{2}+y^{5}=50.

(a)Find dydx\displaystyle \frac{dy}{dx} in terms of x\displaystyle x and y.\displaystyle y.

(b)Given x>0\displaystyle x>0, explain why the derivative is never zero.

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(a) 2xe6y+6x2e6ydydx+2x+5y4dydx=0\displaystyle 2xe^{6y}+6x^{2}e^{6y}\frac{dy}{dx}+2x+5y^{4}\frac{dy}{dx}=0, so dydx=2x(e6y+1)6x2e6y+5y4\displaystyle \frac{dy}{dx}=\frac{-2x(e^{6y}+1)}{6x^{2}e^{6y}+5y^{4}}
(b) For x>0\displaystyle x>0, 2x\displaystyle -2x is strictly negative and e6y+1\displaystyle e^{6y}+1 is strictly positive for every real y\displaystyle y, so the numerator is never zero. A fraction with a non-zero numerator cannot equal zero.
11

Question 11

Integration
4 Marks
2026 P2 Q11

Find

2x+4x2+4dx.\displaystyle \int\frac{2x+4}{x^{2}+4}\,dx.

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Splitting the integral, 2xx2+4dx=ln(x2+4)\displaystyle \int\frac{2x}{x^{2}+4}\,dx=\ln(x^{2}+4) and 4x2+4dx=2tan1(x2).\displaystyle \int\frac{4}{x^{2}+4}\,dx=2\tan^{-1}\left(\frac{x}{2}\right).
ln(x2+4)+2tan1(x2)+C\displaystyle \ln(x^{2}+4)+2\tan^{-1}\left(\frac{x}{2}\right)+C
12

Question 12

Methods of Proof
5 Marks
2026 P2 Q12

Prove by induction that 7n+2\displaystyle 7^{n}+2 is divisible by 3 for all nN.\displaystyle n\in\mathbb{N}.

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When n=1\displaystyle n=1, 71+2=9=3×3\displaystyle 7^{1}+2=9=3\times3, so the statement is true.
Assume true for n=k\displaystyle n=k, so 7k+2=3M\displaystyle 7^{k}+2=3M for some integer M\displaystyle M, giving 7k=3M2.\displaystyle 7^{k}=3M-2.
Then 7k+1+2=7(3M2)+2=21M12=3(7M4)\displaystyle 7^{k+1}+2=7(3M-2)+2=21M-12=3(7M-4), a multiple of 3.
True for n=1\displaystyle n=1, and true for n=k\displaystyle n=k implies true for n=k+1\displaystyle n=k+1, so by induction the statement holds for all nN.\displaystyle n\in\mathbb{N}.
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Question 13

Partial FractionsDifferential Equations
(2, 7)9 Marks
2026 P2 Q13

(a)Express using partial fractions x+3(x+7)(x+5).\displaystyle \frac{x+3}{(x+7)(x+5)}.

(b)Hence find the particular solution of the differential equation

dydx+2x+3y=1(x+7)(x+5)(x+3),\displaystyle \frac{dy}{dx}+\frac{2}{x+3}y=\frac{1}{(x+7)(x+5)(x+3)}, where x0,\displaystyle x\ge0,

given that y=0\displaystyle y=0 when x=3.\displaystyle x=3.

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(a) x+3(x+7)(x+5)=2x+71x+5\displaystyle \frac{x+3}{(x+7)(x+5)}=\frac{2}{x+7}-\frac{1}{x+5}
(b) Integrating factor e2x+3dx=(x+3)2\displaystyle e^{\int\frac{2}{x+3}dx}=(x+3)^{2}, so ddx(y(x+3)2)=x+3(x+7)(x+5).\displaystyle \frac{d}{dx}\left(y(x+3)^{2}\right)=\frac{x+3}{(x+7)(x+5)}.
y(x+3)2=2ln(x+7)ln(x+5)+C\displaystyle y(x+3)^{2}=2\ln(x+7)-\ln(x+5)+C
y=0\displaystyle y=0 when x=3\displaystyle x=3 gives C=ln252\displaystyle C=-\ln\frac{25}{2}, so y=1(x+3)2ln(2(x+7)225(x+5))\displaystyle y=\frac{1}{(x+3)^{2}}\ln\left(\frac{2(x+7)^{2}}{25(x+5)}\right)
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Question 14

Vectors
(4, 3, 3)10 Marks
2026 P2 Q14

The plane π\displaystyle \pi contains the points P(2,3,4)\displaystyle P(2,3,-4), Q(3,5,1)\displaystyle Q(3,5,1) and R(6,0,6).\displaystyle R(6,0,-6).

(a)Determine the Cartesian equation of π.\displaystyle \pi.

The line L\displaystyle L has symmetric equations

x82=y+21=z+23.\displaystyle \frac{x-8}{2}=\frac{y+2}{-1}=\frac{z+2}{3}.

L\displaystyle L intersects π\displaystyle \pi at the point S.\displaystyle S.

(b)Find the coordinates of S.\displaystyle S.

(c)Calculate the size of the acute angle between L\displaystyle L and π.\displaystyle \pi.

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(a) PQ=(1,2,5)\displaystyle \vec{PQ}=(1,2,5) and PR=(4,3,2)\displaystyle \vec{PR}=(4,-3,-2), so n=PQ×PR=(11,22,11)\displaystyle \mathbf{n}=\vec{PQ}\times\vec{PR}=(11,22,-11), or (1,2,1).\displaystyle (1,2,-1).
x+2yz=12\displaystyle x+2y-z=12
(b) With x=2t+8, y=t2, z=3t2\displaystyle x=2t+8,\ y=-t-2,\ z=3t-2: 3t+6=12\displaystyle -3t+6=12, so t=2\displaystyle t=-2 and S(4,0,8)\displaystyle S(4,0,-8)
(c) sinθ=dndn=3146\displaystyle \sin\theta=\frac{|\mathbf{d}\cdot\mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}=\frac{3}{\sqrt{14}\sqrt{6}}, so θ=19.1\displaystyle \theta=19.1^{\circ} (to 1 d.p.)
15

Question 15

Methods of Proof
(1, 3)4 Marks
2026 P2 Q15

Let r\displaystyle r be a positive real number and consider the following statement:

If r\displaystyle r is irrational then r\displaystyle \sqrt{r} is irrational.

(a)Write down the contrapositive of the statement.

(b)Hence prove that the statement is true.

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(a) If r\displaystyle \sqrt{r} is rational then r\displaystyle r is rational.
(b) Assume r\displaystyle \sqrt{r} is rational, so r=pq\displaystyle \sqrt{r}=\frac{p}{q} for integers p\displaystyle p and q\displaystyle q with q0.\displaystyle q\neq0.
Squaring gives r=p2q2\displaystyle r=\frac{p^{2}}{q^{2}}, a ratio of two integers with q20\displaystyle q^{2}\neq0, so r\displaystyle r is rational.
The contrapositive is true, so the original statement is true.
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Question 16

DifferentiationIntegration
(2, 2, 1)5 Marks
2026 P2 Q16

(a)Given y=ln(cosx), 0x<π2\displaystyle y=\ln(\cos x),\ 0\le x<\frac{\pi}{2}, show that dydx=tanx.\displaystyle \frac{dy}{dx}=-\tan x.

For a function g(x)\displaystyle g(x), it is known that

xg(x)dx=2xtan2x2tan2xdx.\displaystyle \int xg(x)\,dx=2x\tan 2x-\int2\tan 2x\,dx.

(b)

(i) Determine the exact value of 0π6xg(x)dx.\displaystyle \int_{0}^{\frac{\pi}{6}}xg(x)\,dx.

(ii) Find an expression for g(x)\displaystyle g(x) in terms of x.\displaystyle x.

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(a) dydx=1cosx×(sinx)=sinxcosx=tanx\displaystyle \frac{dy}{dx}=\frac{1}{\cos x}\times(-\sin x)=-\frac{\sin x}{\cos x}=-\tan x
(b)(i) From (a), 2tan2xdx=ln(cos2x)\displaystyle \int2\tan 2x\,dx=-\ln(\cos 2x), so 0π6xg(x)dx=[2xtan2x+ln(cos2x)]0π6\displaystyle \int_{0}^{\frac{\pi}{6}}xg(x)\,dx=\left[2x\tan 2x+\ln(\cos 2x)\right]_{0}^{\frac{\pi}{6}}
=π33+ln120=π33ln2\displaystyle =\frac{\pi}{3}\sqrt{3}+\ln\frac{1}{2}-0=\frac{\pi\sqrt{3}}{3}-\ln 2
(b)(ii) Comparing with udv=uvvdu\displaystyle \int u\,dv=uv-\int v\,du gives v=2tan2x\displaystyle v=2\tan 2x, so g(x)=ddx(2tan2x)=4sec22x\displaystyle g(x)=\frac{d}{dx}(2\tan 2x)=4\sec^{2}2x