Advanced Higher Maths · SQA past paper

2025 Paper 2

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Question 1

Product, quotient, chain rules
2 Marks
2025 P2 Q1

A function is defined by f(x)=cos⁡−14x.\displaystyle f(x) = \cos^{-1} 4x.
Find f′(x).\displaystyle f'(x).

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−41−(4x)2\displaystyle -\frac{4}{\sqrt{1 - (4x)^2}} (or −41−16x2\displaystyle -\frac{4}{\sqrt{1 - 16x^2}})

Question 2

Implicit differentiation
3 Marks
2025 P2 Q2

A curve is defined by the equation 2y2+4xe2y=3x.\displaystyle 2y^2 + 4xe^{2y} = 3x.
Find an expression for dydx\displaystyle \frac{dy}{dx} in terms of x\displaystyle x and y.\displaystyle y.

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3−4e2y4y+8xe2y\displaystyle \frac{3 - 4e^{2y}}{4y + 8xe^{2y}}

Question 3

Partial fractions
3 Marks
2025 P2 Q3

Express 2x2−18x+4(x−1)(x−3)(x+5)\displaystyle \frac{2x^2 - 18x + 4}{(x - 1)(x - 3)(x + 5)} in partial fractions.

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1x−1−2x−3+3x+5\displaystyle \frac{1}{x - 1} - \frac{2}{x - 3} + \frac{3}{x + 5}

Question 4

Euclidean algorithm
(1, 2)3 Marks
2025 P2 Q4

(a)Use the Euclidean algorithm to find d\displaystyle d, the greatest common divisor of 1118 and 416.

(b)Hence find integers a\displaystyle a and b\displaystyle b such that 1118a+416b=d.\displaystyle 1118a + 416b = d.

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(a) 26
(b) a=3,b=−8\displaystyle a = 3, b = -8

Question 5

Logarithmic differentiation
5 Marks
2025 P2 Q5

A curve is defined by y=xcot⁡x.\displaystyle y = x^{\cot x}.
Use logarithmic differentiation to find dydx.\displaystyle \frac{dy}{dx}.
Write your answer in terms of x.\displaystyle x.

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xcot⁡x(−cosec2xln⁡x+cot⁡xx)\displaystyle x^{\cot x} \left( -\text{cosec}^2 x \ln x + \frac{\cot x}{x} \right)

Question 6

Maclaurin Series
(2, 2)4 Marks
2025 P2 Q6

(a)Find and simplify the Maclaurin expansion, up to and including the term in x4\displaystyle x^4, for cos⁡3x.\displaystyle \cos 3x.

(b)Hence find and simplify the Maclaurin expansion, up to and including the term in x4\displaystyle x^4, for cos⁡23x.\displaystyle \cos^2 3x.

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(a) 1−92x2+278x4\displaystyle 1 - \frac{9}{2}x^2 + \frac{27}{8}x^4
(b) 1−9x2+27x4\displaystyle 1 - 9x^2 + 27x^4

Question 7

Parametric differentiation
(2, 3)5 Marks
2025 P2 Q7

A curve is defined on a suitable domain by the equations x=t2\displaystyle x = t^2 and y=tan⁡t.\displaystyle y = \tan t.
Find in terms of t\displaystyle t:

(a)dydx\displaystyle \frac{dy}{dx}

(b)d2ydx2\displaystyle \frac{d^2y}{dx^2}

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(a) sec⁡2t2t\displaystyle \frac{\sec^2 t}{2t}
(b) 2tsec⁡2ttan⁡t−sec⁡2t4t3\displaystyle \frac{2t \sec^2 t \tan t - \sec^2 t}{4t^3}

Question 8

Matrix operationsDeterminant and inverse
(2, 2)4 Marks
2025 P2 Q8

The matrix A\displaystyle A has the following property:

A2=6A−I\displaystyle A^2 = 6A - I, where I\displaystyle I is the identity matrix.

(a)Express A3\displaystyle A^3 in the form pA+qI\displaystyle pA + qI, where p,q∈R.\displaystyle p, q \in \mathbb{R}.

Matrix A\displaystyle A is non-singular.

(b)Find a similar expression for A−1\displaystyle A^{-1} in terms of A\displaystyle A and I.\displaystyle I.

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(a) 35A−6I\displaystyle 35A - 6I
(b) 6I−A\displaystyle 6I - A

Question 9

Rectilinear motion
(2, 2)4 Marks
2025 P2 Q9

Relative to a fixed origin, the velocity, v\displaystyle v metres per second, of an object at time t\displaystyle t seconds is given by v=2t+e5t.\displaystyle v = 2t + e^{5t}.

(a)Find an expression for the displacement of the object, s\displaystyle s metres, in terms of t\displaystyle t, given that when t=0,s=0.\displaystyle t = 0, s = 0.

(b)Show that the acceleration of the object is always positive.

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(a) t2+15e5t−15\displaystyle t^2 + \frac{1}{5}e^{5t} - \frac{1}{5}
(b) a=2+5e5t.\displaystyle a = 2 + 5e^{5t}. Since e5t>0\displaystyle e^{5t} > 0 for all t\displaystyle t, a>0.\displaystyle a > 0.

Question 10

Summation formulae
2 Marks
2025 P2 Q10

Find and fully factorise an expression for ∑r=1n(r3−3r).\displaystyle \sum_{r=1}^{n} (r^3 - 3r).

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14n(n+1)(n−2)(n+3)\displaystyle \frac{1}{4}n(n + 1)(n - 2)(n + 3)

Question 11

Integration by substitutionVolume of revolution
(2, 3)5 Marks
2025 P2 Q11

(a)Using the substitution u=2x2\displaystyle u = 2x^2, or otherwise, find ∫xe−2x2dx.\displaystyle \int xe^{-2x^2} dx.

The diagram shows part of the curve with equation y=4xex2.\displaystyle y = \frac{4\sqrt{x}}{e^{x^2}}.

Curve of y = 4*sqrt(x) / e^(x^2) rotated about x-axis

A solid is generated by rotating the curve through 2π\displaystyle 2\pi radians about the x\displaystyle x-axis from x=0\displaystyle x = 0 to x=1.\displaystyle x = 1.

(b)Calculate the exact value of the volume generated.

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(a) −14e−2x2+c\displaystyle -\frac{1}{4}e^{-2x^2} + c
(b) 4π(1−e−2)\displaystyle 4\pi(1 - e^{-2})

Question 12

Second order non-homogeneous DE
9 Marks
2025 P2 Q12

Solve the differential equation

d2ydx2−8dydx+15y=15x2−31x+40\displaystyle \frac{d^2y}{dx^2} - 8\frac{dy}{dx} + 15y = 15x^2 - 31x + 40

given that y=4\displaystyle y = 4 and dydx=13\displaystyle \frac{dy}{dx} = 13 when x=0.\displaystyle x = 0.

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y=−2e3x+4e5x+x2−x+2\displaystyle y = -2e^{3x} + 4e^{5x} + x^2 - x + 2

Question 13

Geometric sequences and series
(2, 1, 1, 1, 1, 1)7 Marks
2025 P2 Q13

An infinite geometric sequence of positive numbers has second term 100 and fourth term 16.

(a)Determine:

(i) the common ratio
(ii) the first term of this sequence.

(b)Explain why the associated geometric series has a sum to infinity.

(c)Determine this sum to infinity.

A new geometric sequence is formed by multiplying each term in the sequence above by the real number k\displaystyle k, where k≠0.\displaystyle k \neq 0.

(d)State the effect that this will have on:

(i) the common ratio
(ii) the sum to infinity of the associated series.

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(a)(i) 25\displaystyle \frac{2}{5}
(a)(ii) 250\displaystyle 250
(b) A sum to infinity exists because −1<25<1\displaystyle -1 < \frac{2}{5} < 1
(c) 12503\displaystyle \frac{1250}{3}
(d)(i) The common ratio is unchanged.
(d)(ii) The sum to infinity is multiplied by k.\displaystyle k.

Question 14

First order linear DE
5 Marks
2025 P2 Q14

Find the general solution of the differential equation

dydx−2xy=x2sec⁡23x.\displaystyle \frac{dy}{dx} - \frac{2}{x}y = x^2 \sec^2 3x.

Give your answer in the form y=f(x).\displaystyle y = f(x).

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y=x2(13tan⁡3x+c)\displaystyle y = x^2 \left(\frac{1}{3} \tan 3x + c\right)

Question 15

Proof by induction
5 Marks
2025 P2 Q15

Prove by induction that, for all positive integers n\displaystyle n,

∑r=1n1(2r+1)(2r−1)=n2n+1.\displaystyle \sum_{r=1}^{n} \frac{1}{(2r + 1)(2r - 1)} = \frac{n}{2n + 1}.

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When n=1\displaystyle n=1, LHS =13×1=13\displaystyle =\frac{1}{3\times1}=\frac{1}{3} and RHS =12+1=13\displaystyle =\frac{1}{2+1}=\frac{1}{3}, so the statement is true.
Assume true for n=k\displaystyle n=k: ∑r=1k1(2r+1)(2r−1)=k2k+1.\displaystyle \sum_{r=1}^{k}\frac{1}{(2r+1)(2r-1)}=\frac{k}{2k+1}.
Then ∑r=1k+11(2r+1)(2r−1)=k2k+1+1(2k+3)(2k+1)=(2k+3)k+1(2k+1)(2k+3)=(k+1)(2k+1)(2k+1)(2k+3)=k+12(k+1)+1.\displaystyle \sum_{r=1}^{k+1}\frac{1}{(2r+1)(2r-1)}=\frac{k}{2k+1}+\frac{1}{(2k+3)(2k+1)}=\frac{(2k+3)k+1}{(2k+1)(2k+3)}=\frac{(k+1)(2k+1)}{(2k+1)(2k+3)}=\frac{k+1}{2(k+1)+1}.
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 positive integers n.\displaystyle n.

Question 16

Integration by parts
5 Marks
2025 P2 Q16

Use integration by parts to find

∫e2xsin⁡5x dx.\displaystyle \int e^{2x} \sin 5x \,dx.

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229e2xsin⁡5x−529e2xcos⁡5x+c\displaystyle \frac{2}{29}e^{2x} \sin 5x - \frac{5}{29}e^{2x} \cos 5x + c

Question 17

Related rates of change
4 Marks
2025 P2 Q17

The volume, V\displaystyle V cm3\displaystyle ^3, of water in a tank is given by

V=15h3\displaystyle V = \frac{1}{5}h^3, where h\displaystyle h cm is the depth of water in the tank.

Water is being piped into the tank at a rate of 6 cm3\displaystyle ^3/second.
Water is leaking from the bottom of the tank at a rate of 110h\displaystyle \frac{1}{10}\sqrt{h} cm3\displaystyle ^3/second.

Calculate the rate of change of the depth of water when h=400.\displaystyle h = 400.

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124000\displaystyle \frac{1}{24000} cm/sec

Question 18

Complex number operationsde Moivre's theorem
(2, 1, 2)5 Marks
2025 P2 Q18

Let z=x+iy\displaystyle z = x + iy be a complex number, where x,y∈R.\displaystyle x, y \in \mathbb{R}.

(a)(i) Express zˉ+iz\displaystyle \bar{z} + iz in Cartesian form, where zˉ\displaystyle \bar{z} is the complex conjugate of z.\displaystyle z.
(ii) Given x>y\displaystyle x > y, find the argument of zˉ+iz.\displaystyle \bar{z} + iz.

When x<y\displaystyle x < y, zˉ+iz=r(cos⁡(−3π4)+isin⁡(−3π4))\displaystyle \bar{z} + iz = r\left(\cos\left(-\frac{3\pi}{4}\right) + i\sin\left(-\frac{3\pi}{4}\right)\right) where r\displaystyle r is the modulus of zˉ+iz.\displaystyle \bar{z} + iz.

(b)Use de Moivre's theorem to find, in polar form, both square roots of zˉ+iz.\displaystyle \bar{z} + iz.

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(a)(i) (x−y)+i(x−y)\displaystyle (x - y) + i(x - y)
(a)(ii) π4\displaystyle \frac{\pi}{4} (or 45∘\displaystyle 45^\circ)
(b) r(cos⁡(−3π8)+isin⁡(−3π8))\displaystyle \sqrt{r}\left(\cos\left(-\frac{3\pi}{8}\right) + i\sin\left(-\frac{3\pi}{8}\right)\right) and r(cos⁡(5π8)+isin⁡(5π8))\displaystyle \sqrt{r}\left(\cos\left(\frac{5\pi}{8}\right) + i\sin\left(\frac{5\pi}{8}\right)\right)