Advanced Higher Maths · SQA past paper

2023 Paper 1

Non-Calculator · 9 questions
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Question 1

Product, quotient, chain rules
2 Marks
2023 P1 Q1

Given y=7xtan⁡2x\displaystyle y = 7x \tan 2x, find dydx.\displaystyle \frac{dy}{dx}.

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7tan⁡2x+14xsec⁡22x\displaystyle 7 \tan 2x + 14x \sec^2 2x

Question 2

Partial fractions
3 Marks
2023 P1 Q2

Express 3x2−x−14(x+3)(x−1)2\displaystyle \frac{3x^2 - x - 14}{(x + 3)(x - 1)^2} in partial fractions.

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1x+3+2x−1−3(x−1)2\displaystyle \frac{1}{x + 3} + \frac{2}{x - 1} - \frac{3}{(x - 1)^2}

Question 3

Systems of Equations
3 Marks
2023 P1 Q3

A system of equations is defined by
x−3y+z=−1\displaystyle x - 3y + z = -1
3x−2y+4z=11\displaystyle 3x - 2y + 4z = 11
x+4y+2z=15\displaystyle x + 4y + 2z = 15
Use Gaussian elimination to determine whether the system shows redundancy, inconsistency or has a unique solution.

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Inconsistent (as the final row of the augmented matrix leads to an impossible equation, e.g., 0=2\displaystyle 0 = 2 or 14≠16\displaystyle 14 \neq 16)

Question 4

Integration by parts
3 Marks
2023 P1 Q4

Use integration by parts to find ∫x4ln⁡x dx\displaystyle \int x^4 \ln x \, dx, x>0.\displaystyle x > 0.

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15x5ln⁡x−x525+c\displaystyle \frac{1}{5}x^5 \ln x - \frac{x^5}{25} + c

Question 5

Second order non-homogeneous DE
9 Marks
2023 P1 Q5

Find the particular solution of the differential equation
d2ydx2−4dydx−5y=10x2+11x−23\displaystyle \frac{d^2y}{dx^2} - 4\frac{dy}{dx} - 5y = 10x^2 + 11x - 23
given that y=2\displaystyle y = 2, dydx=14\displaystyle \frac{dy}{dx} = 14 when x=0.\displaystyle x = 0.

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

Question 6

Argand diagramde Moivre's theorem
(2, 2)4 Marks
2023 P1 Q6

(a)Express z=1+3i\displaystyle z = 1 + \sqrt{3}i in polar form.

(b)Hence, or otherwise, show that z3\displaystyle z^3 is real.

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(a) 2(cos⁡π3+isin⁡π3)\displaystyle 2\left(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}\right)
(b) Using de Moivre's Theorem, z3=23(cos⁡3π3+isin⁡3π3)=8(cos⁡π+isin⁡π)=−8.\displaystyle z^3 = 2^3 \left(\cos \frac{3\pi}{3} + i \sin \frac{3\pi}{3}\right) = 8(\cos \pi + i \sin \pi) = -8. Since the imaginary part is zero, z3\displaystyle z^3 is real.

Question 7

Summation formulae
(2, 2)4 Marks
2023 P1 Q7

(a)Find an expression for ∑r=1n(r2+3r)\displaystyle \sum_{r=1}^{n} (r^2 + 3r) in terms of n.\displaystyle n.
Express your answer in the form 13n(n+a)(n+b).\displaystyle \frac{1}{3}n(n + a)(n + b).

(b)Hence, or otherwise, find ∑r=1120(r2+3r).\displaystyle \sum_{r=11}^{20} (r^2 + 3r).

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(a) 13n(n+1)(n+5)\displaystyle \frac{1}{3}n(n + 1)(n + 5)
(b) 2950\displaystyle 2950

Question 8

Disproof by counterexampleDirect proof
(1, 2)3 Marks
2023 P1 Q8

(a)Consider the statement:
For all integers a\displaystyle a and b\displaystyle b, if a<b\displaystyle a < b then a2<b2.\displaystyle a^2 < b^2.
Find a counterexample to show that the statement is false.

(b)Let n\displaystyle n be an odd integer.
Prove directly that n2−1\displaystyle n^2 - 1 is divisible by 4.

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(a) e.g., let a=−2\displaystyle a = -2 and b=1.\displaystyle b = 1. −2<1\displaystyle -2 < 1 is true, but (−2)2<12⇒4<1\displaystyle (-2)^2 < 1^2 \Rightarrow 4 < 1 is false.
(b) Let n=2k+1\displaystyle n = 2k + 1 for k∈Z.\displaystyle k \in \mathbb{Z}. Then n2−1=(2k+1)2−1=4k2+4k=4(k2+k)\displaystyle n^2 - 1 = (2k+1)^2 - 1 = 4k^2 + 4k = 4(k^2 + k), which is divisible by 4.

Question 9

Transformation matrices
(1, 1, 1, 1)4 Marks
2023 P1 Q9

(a)State the matrix A\displaystyle A, associated with an anti-clockwise rotation of π2\displaystyle \frac{\pi}{2} radians about the origin.

The matrix B\displaystyle B is given by B=(−3212−12−32).\displaystyle B = \begin{pmatrix} -\dfrac{\sqrt{3}}{2} & \dfrac{1}{2} \\[7pt] -\dfrac{1}{2} & -\dfrac{\sqrt{3}}{2} \end{pmatrix}.
The matrix AB\displaystyle AB is associated with an anti-clockwise rotation of α\displaystyle \alpha radians about the origin.

(b)(i) Determine AB.\displaystyle AB.

(b)(ii) Find the value of α.\displaystyle \alpha.

(c)Determine the least positive integer value of n\displaystyle n such that (AB)n=I\displaystyle (AB)^n = I, where I\displaystyle I is the 2×2\displaystyle 2 \times 2 identity matrix.

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(a) (0−110)\displaystyle \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}
(b)(i) (1232−3212)\displaystyle \begin{pmatrix} \dfrac{1}{2} & \dfrac{\sqrt{3}}{2} \\[7pt] -\dfrac{\sqrt{3}}{2} & \dfrac{1}{2} \end{pmatrix}
(b)(ii) 5π3\displaystyle \frac{5\pi}{3} (or −π3\displaystyle -\frac{\pi}{3})
(c) 6\displaystyle 6