Advanced Higher Maths · SQA past paper

2025 Paper 2

Calculator · 18 questions
1

Question 1

Differentiation
2 Marks
2025 P2 Q1

A function is defined by f(x)=cos14x.\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 4116x2\displaystyle -\frac{4}{\sqrt{1 - 16x^2}})
2

Question 2

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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34e2y4y+8xe2y\displaystyle \frac{3 - 4e^{2y}}{4y + 8xe^{2y}}
3

Question 3

Partial Fractions
3 Marks
2025 P2 Q3

Express 2x218x+4(x1)(x3)(x+5)\displaystyle \frac{2x^2 - 18x + 4}{(x - 1)(x - 3)(x + 5)} in partial fractions.

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1x12x3+3x+5\displaystyle \frac{1}{x - 1} - \frac{2}{x - 3} + \frac{3}{x + 5}
4

Question 4

Number Theory
(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
5

Question 5

Differentiation
5 Marks
2025 P2 Q5

A curve is defined by y=xcotx.\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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xcotx(cosec2xlnx+cotxx)\displaystyle x^{\cot x} \left( -\text{cosec}^2 x \ln x + \frac{\cot x}{x} \right)
6

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 cos3x.\displaystyle \cos 3x.

(b)Hence find and simplify the Maclaurin expansion, up to and including the term in x4\displaystyle x^4, for cos23x.\displaystyle \cos^2 3x.

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(a) 192x2+278x4\displaystyle 1 - \frac{9}{2}x^2 + \frac{27}{8}x^4
(b) 19x2+27x4\displaystyle 1 - 9x^2 + 27x^4
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Question 7

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=tant.\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) sec2t2t\displaystyle \frac{\sec^2 t}{2t}
(b) 2tsec2ttantsec2t4t3\displaystyle \frac{2t \sec^2 t \tan t - \sec^2 t}{4t^3}
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Question 8

Matrices
(2, 2)4 Marks
2025 P2 Q8

The matrix A\displaystyle A has the following property:

A2=6AI\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,qR.\displaystyle p, q \in \mathbb{R}.

Matrix A\displaystyle A is non-singular.

(b)Find a similar expression for A1\displaystyle A^{-1} in terms of A\displaystyle A and I.\displaystyle I.

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(a) 35A6I\displaystyle 35A - 6I
(b) 6IA\displaystyle 6I - A
9

Question 9

DifferentiationIntegration
(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+15e5t15\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.
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Question 10

Sequences & Series
2 Marks
2025 P2 Q10

Find and fully factorise an expression for r=1n(r33r).\displaystyle \sum_{r=1}^{n} (r^3 - 3r).

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14n(n+1)(n2)(n+3)\displaystyle \frac{1}{4}n(n + 1)(n - 2)(n + 3)
11

Question 11

Integration
(2, 3)5 Marks
2025 P2 Q11

(a)Using the substitution u=2x2\displaystyle u = 2x^2, or otherwise, find xe2x2dx.\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) 14e2x2+c\displaystyle -\frac{1}{4}e^{-2x^2} + c
(b) 4π(1e2)\displaystyle 4\pi(1 - e^{-2})
12

Question 12

Differential Equations
9 Marks
2025 P2 Q12

Solve the differential equation

d2ydx28dydx+15y=15x231x+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+x2x+2\displaystyle y = -2e^{3x} + 4e^{5x} + x^2 - x + 2
13

Question 13

Sequences & 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 k0.\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.
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Question 14

Differential Equations
5 Marks
2025 P2 Q14

Find the general solution of the differential equation

dydx2xy=x2sec23x.\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(13tan3x+c)\displaystyle y = x^2 \left(\frac{1}{3} \tan 3x + c\right)
15

Question 15

Methods of Proof
5 Marks
2025 P2 Q15

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

r=1n1(2r+1)(2r1)=n2n+1.\displaystyle \sum_{r=1}^{n} \frac{1}{(2r + 1)(2r - 1)} = \frac{n}{2n + 1}.

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Proof by induction showing true for n=1\displaystyle n=1 (LHS = RHS = 13\displaystyle \frac{1}{3}), assuming true for n=k\displaystyle n=k, and showing the sum to k+1\displaystyle k+1 simplifies to k+12(k+1)+1\displaystyle \frac{k+1}{2(k+1)+1}, concluding the proof for all positive integers n.\displaystyle n.
16

Question 16

Integration
5 Marks
2025 P2 Q16

Use integration by parts to find

e2xsin5xdx.\displaystyle \int e^{2x} \sin 5x \,dx.

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229e2xsin5x529e2xcos5x+c\displaystyle \frac{2}{29}e^{2x} \sin 5x - \frac{5}{29}e^{2x} \cos 5x + c
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Question 17

Differentiation
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
18

Question 18

Complex Numbers
(2, 1, 2)5 Marks
2025 P2 Q18

Let z=x+iy\displaystyle z = x + iy be a complex number, where x,yR.\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) (xy)+i(xy)\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)