Advanced Higher Maths · SQA past paper

2023 Paper 2

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Question 1

Product, quotient, chain rules
2 Marks
2023 P2 Q1

The function f\displaystyle f is defined by f(x)=2sin⁡−13x.\displaystyle f(x)=2 \sin^{-1} 3x.
Find f′(x).\displaystyle f'(x).

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61−9x2\displaystyle \frac{6}{\sqrt{1-9x^2}} (or 61−(3x)2\displaystyle \frac{6}{\sqrt{1-(3x)^2}})

Question 2

Integration by substitution
2 Marks
2023 P2 Q2

Find ∫x2x3+10dx.\displaystyle \int \frac{x^2}{x^3+10}dx.

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13ln⁡∣x3+10∣+c\displaystyle \frac{1}{3}\ln|x^3+10|+c

Question 3

Determinant and inverse
(2, 1)3 Marks
2023 P2 Q3

Matrix A\displaystyle A is defined by

A=(22x4x−1010−2)\displaystyle A = \begin{pmatrix} 2 & 2x & 4 \\ x & -1 & 0 \\ 1 & 0 & -2 \end{pmatrix}

where x∈R.\displaystyle x \in \mathbb{R}.

(a)Find a simplified expression for the determinant of A.\displaystyle A.

(b)Hence, determine whether A−1\displaystyle A^{-1} exists for all values of x.\displaystyle x.

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(a) 4x2+8\displaystyle 4x^2+8
(b) Since 4x2+8>0\displaystyle 4x^2+8 > 0 for all real x\displaystyle x, A−1\displaystyle A^{-1} always exists.

Question 4

Implicit differentiation
3 Marks
2023 P2 Q4

Calculate the gradient of the tangent to the curve with equation x2y2−2y=sin⁡3x\displaystyle x^2y^2-2y=\sin 3x at the point (0,0).

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−32\displaystyle -\frac{3}{2}

Question 5

Binomial terms
(3, 2)5 Marks
2023 P2 Q5

(a)Write down and simplify the general term in the binomial expansion of (3x−2x2)8.\displaystyle \left(3x-\frac{2}{x^2}\right)^8.

(b)Hence, or otherwise, determine the coefficient of x−1.\displaystyle x^{-1}.

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(a) (8r)38−r(−2)rx8−3r\displaystyle \binom{8}{r}3^{8-r}(-2)^r x^{8-3r}
(b) −108864\displaystyle -108864

Question 6

Euclidean algorithm
(1, 2, 1)4 Marks
2023 P2 Q6

(a)Use the Euclidean algorithm to find d\displaystyle d, the greatest common divisor of 703 and 399.

(b)Find integers a\displaystyle a and b\displaystyle b such that d=703a+399b.\displaystyle d=703a+399b.

(c)Hence find integers p\displaystyle p and q\displaystyle q such that 76=703p+399q.\displaystyle 76=703p+399q.

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(a) 19\displaystyle 19
(b) a=4,b=−7\displaystyle a=4, b=-7
(c) p=16,q=−28\displaystyle p=16, q=-28

Question 7

First order linear DE
(4, 2)6 Marks
2023 P2 Q7

(a)Solve the differential equation

dydx−2y=6e5x\displaystyle \frac{dy}{dx}-2y=6e^{5x}

given that when x=0,y=−1.\displaystyle x=0, y=-1. Express y\displaystyle y in terms of x.\displaystyle x.

(b)The solution of the differential equation in (a) is also a solution of

d3ydx3−5d2ydx2=ke2x\displaystyle \frac{d^3y}{dx^3}-5\frac{d^2y}{dx^2}=ke^{2x}, k∈R.\displaystyle k \in \mathbb{R}.

Find the value of k.\displaystyle k.

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(a) y=2e5x−3e2x\displaystyle y=2e^{5x}-3e^{2x}
(b) k=36\displaystyle k=36

Question 8

Geometric sequences and series
(1, 1, 2)4 Marks
2023 P2 Q8

The fourth and seventh terms of a geometric sequence are 9 and 243 respectively.

(a)Find the:
(i) common ratio
(ii) first term.

(b)Show that S2nSn=1+3n\displaystyle \frac{S_{2n}}{S_n} = 1+3^n where Sn\displaystyle S_n represents the sum of the first n\displaystyle n terms of this geometric sequence.

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(a)(i) 3\displaystyle 3
(a)(ii) 13\displaystyle \frac{1}{3}
(b) Sn=13(1−3n)1−3\displaystyle S_n = \frac{\frac{1}{3}(1-3^n)}{1-3} and S2n=13(1−32n)1−3.\displaystyle S_{2n} = \frac{\frac{1}{3}(1-3^{2n})}{1-3}. Dividing gives 1−32n1−3n=(1−3n)(1+3n)1−3n=1+3n.\displaystyle \frac{1-3^{2n}}{1-3^n} = \frac{(1-3^n)(1+3^n)}{1-3^n} = 1+3^n.

Question 9

Number bases
2 Marks
2023 P2 Q9

Express 57210\displaystyle 572_{10} in base 9.

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7059\displaystyle 705_9

Question 10

Logarithmic differentiation
5 Marks
2023 P2 Q10

A curve is defined by y=x5x2\displaystyle y=x^{5x^2}, where x>0.\displaystyle x>0.
Find dydx\displaystyle \frac{dy}{dx} in terms of x.\displaystyle x.

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x5x2(10xln⁡x+5x)\displaystyle x^{5x^2}(10x \ln x + 5x)

Question 11

Related rates of change
(1, 5)6 Marks
2023 P2 Q11

On a building site, water is stored in a container. The container is a cone with diameter 180 cm at its widest point and height of 150 cm.

Cross section of a cone with diameter 180cm and height 150cm

(a)Show that when the water level is at a height of h\displaystyle h cm, 0≤h≤150\displaystyle 0 \le h \le 150, the volume of water in the container can be written as

V=3πh325\displaystyle V = \frac{3\pi h^3}{25}

[The volume of a cone is given by V=13πr2h.\displaystyle V = \frac{1}{3}\pi r^2 h.]

Water is pumped into the container at a constant rate of 10 litres per second.

(b)Find the rate at which the height is increasing when h=125.\displaystyle h = 125.

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(a) From similar triangles, rh=90150  ⟹  r=35h.\displaystyle \frac{r}{h} = \frac{90}{150} \implies r = \frac{3}{5}h. Substituting: V=13π(35h)2h=3πh325.\displaystyle V = \frac{1}{3}\pi \left(\frac{3}{5}h\right)^2 h = \frac{3\pi h^3}{25}.
(b) 169π\displaystyle \frac{16}{9\pi} cm/s (or approx 0.57\displaystyle 0.57 cm/s)

Question 12

Proof by induction
5 Marks
2023 P2 Q12

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

∑r=1n2r−1r=2n(n−1)+1.\displaystyle \sum_{r=1}^{n} 2^{r-1}r = 2^n(n-1)+1.

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When n=1\displaystyle n=1, LHS =20×1=1\displaystyle =2^{0}\times1=1 and RHS =21(1−1)+1=1\displaystyle =2^{1}(1-1)+1=1, so the statement is true.
Assume true for n=k\displaystyle n=k: ∑r=1k2r−1r=2k(k−1)+1.\displaystyle \sum_{r=1}^{k}2^{r-1}r=2^{k}(k-1)+1.
Then ∑r=1k+12r−1r=2k(k−1)+1+2k(k+1)=2k⋅2k+1=2k+1((k+1)−1)+1.\displaystyle \sum_{r=1}^{k+1}2^{r-1}r=2^{k}(k-1)+1+2^{k}(k+1)=2^{k}\cdot2k+1=2^{k+1}\left((k+1)-1\right)+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 13

Separable DE in context
6 Marks
2023 P2 Q13

Points scored in the long jump element of the decathlon can be calculated using a solution of the differential equation

dPdm=1.4Pm−220\displaystyle \frac{dP}{dm} = \frac{1.4P}{m-220}, m>220\displaystyle m > 220

where m\displaystyle m is the distance jumped in centimetres and P\displaystyle P the points scored.

Given that a jump of 807 centimetres scores 1079 points, find an expression for P\displaystyle P in terms of m.\displaystyle m.

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P=0.14(m−220)1.4\displaystyle P = 0.14(m-220)^{1.4} (or P=e1.4ln⁡(m−220)−1.94\displaystyle P = e^{1.4\ln(m-220)-1.94})

Question 14

Equations with complex roots
4 Marks
2023 P2 Q14

A complex number is defined by w=a+ib\displaystyle w=a+ib, where a\displaystyle a and b\displaystyle b are positive real numbers.

Given w2=8+6i\displaystyle w^2=8+6i, determine the values of a\displaystyle a and b.\displaystyle b.

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a=3,b=1\displaystyle a=3, b=1

Question 15

Integration by substitution
(3, 3, 2)8 Marks
2023 P2 Q15

A function f(x)\displaystyle f(x) has the following properties:

f′(x)=x+11+(x+1)4\displaystyle f'(x) = \frac{x+1}{1+(x+1)^4}

the first term in the Maclaurin expansion of f(x)\displaystyle f(x) is 1.

(a)Find the Maclaurin expansion of f(x)\displaystyle f(x) up to and including the term in x2.\displaystyle x^2.

(b)Use the substitution u=(x+1)2\displaystyle u=(x+1)^2 to find ∫x+11+(x+1)4dx.\displaystyle \int \frac{x+1}{1+(x+1)^4}dx.

(c)Determine an expression for f(x).\displaystyle f(x).

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(a) 1+12x−14x2\displaystyle 1 + \frac{1}{2}x - \frac{1}{4}x^2
(b) 12tan⁡−1((x+1)2)+c\displaystyle \frac{1}{2}\tan^{-1}((x+1)^2)+c
(c) 12tan⁡−1((x+1)2)+1−π8\displaystyle \frac{1}{2}\tan^{-1}((x+1)^2) + 1 - \frac{\pi}{8}