Advanced Higher Maths · SQA past paper

2021 Paper 1

Non-Calculator · 9 questions
1(a)

Question 1(a)

Differentiation
2 Marks
2021 P1 Q1(a)

Differentiate y=x3e5x.\displaystyle y = x^3e^{5x}.

Show answer
3x2e5x+5x3e5x\displaystyle 3x^2e^{5x} + 5x^3e^{5x}
Show marking instructions

Question 1(a)

•¹ evidence use of product rule
3x2(...)+...\displaystyle 3x^2(...) + ... OR ...+5e5x(...)\displaystyle ... + 5e^{5x}(...)

•² complete differentiation
3x2e5x+5x3e5x\displaystyle 3x^2e^{5x} + 5x^3e^{5x}

1(b)

Question 1(b)

Differentiation
2 Marks
2021 P1 Q1(b)

Given y=tanxx6+1\displaystyle y = \frac{\tan x}{x^6 + 1}
find dydx.\displaystyle \frac{dy}{dx}.

Show answer
(x6+1)sec2x6x5tanx(x6+1)2\displaystyle \frac{(x^6 + 1)\sec^2 x - 6x^5\tan x}{(x^6 + 1)^2}
Show marking instructions

Question 1(b)

•³ evidence use of quotient rule with denominator and one term of the numerator correct
(x6+1)sec2x...(x6+1)2\displaystyle \frac{(x^6+1)\sec^2 x - ...}{(x^6+1)^2} OR ...6x5tanx(x6+1)2\displaystyle \frac{... - 6x^5\tan x}{(x^6+1)^2}

•⁴ complete differentiation
(x6+1)sec2x6x5tanx(x6+1)2\displaystyle \frac{(x^6+1)\sec^2 x - 6x^5\tan x}{(x^6+1)^2}

2

Question 2

Matrices
(2, 2)4 Marks
2021 P1 Q2

Matrices A\displaystyle A and B\displaystyle B are defined as follows

A=(2437)\displaystyle A = \begin{pmatrix} -2 & 4 \\ -3 & 7 \end{pmatrix},
B=(402321).\displaystyle B = \begin{pmatrix} 4 & 0 \\ 2 & 3 \\ -2 & 1 \end{pmatrix}.

(a)Find AB\displaystyle AB', where B\displaystyle B' is the transpose of B.\displaystyle B.

(b)Find A1.\displaystyle A^{-1}.

Show answer
(a) (888121513)\displaystyle \begin{pmatrix} -8 & 8 & 8 \\ -12 & 15 & 13 \end{pmatrix}
(b) 12(7432)\displaystyle \frac{1}{2}\begin{pmatrix} -7 & 4 \\ -3 & 2 \end{pmatrix}
Show marking instructions

Question 2(a)

•¹ state the transpose of B
B=(422031)\displaystyle B' = \begin{pmatrix} 4 & 2 & -2 \\ 0 & 3 & 1 \end{pmatrix}

•² calculate AB'
(888121513)\displaystyle \begin{pmatrix} -8 & 8 & 8 \\ -12 & 15 & 13 \end{pmatrix}

Question 2(b)

•³ calculate the determinant of A
detA=2\displaystyle \det A = -2

•⁴ find the inverse of A
12(7432)\displaystyle \frac{1}{2}\begin{pmatrix} -7 & 4 \\ -3 & 2 \end{pmatrix}

3

Question 3

Integration
2 Marks
2021 P1 Q3

Use the substitution u=sinθ\displaystyle u = \sin \theta to find cosθsin3θdθ.\displaystyle \int \cos \theta \sin^3 \theta \, d\theta.
Write your answer in terms of θ.\displaystyle \theta.

Show answer
14sin4θ+c\displaystyle \frac{1}{4}\sin^4 \theta + c
Show marking instructions

Question 3

•¹ rewrite integral
u3du\displaystyle \int u^3 \, du

•² integrate and rewrite in terms of θ\displaystyle \theta, including constant of integration
14sin4θ+c\displaystyle \frac{1}{4}\sin^4 \theta + c

4

Question 4

Systems of Equations
4 Marks
2021 P1 Q4

A system of equations is given by

x+2y+z=5\displaystyle x + 2y + z = 5
3xy+2z=4\displaystyle 3x - y + 2z = 4
2x+3y+λz=8\displaystyle -2x + 3y + \lambda z = -8

where λR.\displaystyle \lambda \in \mathbb{R}.

Use Gaussian elimination to determine the value of λ\displaystyle \lambda for which this system of equations has no solution.

Show answer
1\displaystyle -1
Show marking instructions

Question 4

•¹ set up augmented matrix
(1215312423λ8)\displaystyle \begin{pmatrix} 1 & 2 & 1 & 5 \\ 3 & -1 & 2 & 4 \\ -2 & 3 & \lambda & -8 \end{pmatrix}

•² obtain two zeros
e.g. (121507111072+λ2)\displaystyle \begin{pmatrix} 1 & 2 & 1 & 5 \\ 0 & -7 & -1 & -11 \\ 0 & 7 & 2+\lambda & 2 \end{pmatrix}

•³ complete row operations
e.g. (121507111001+λ9)\displaystyle \begin{pmatrix} 1 & 2 & 1 & 5 \\ 0 & -7 & -1 & -11 \\ 0 & 0 & 1+\lambda & -9 \end{pmatrix}

•⁴ write down value of λ\displaystyle \lambda
1\displaystyle -1

5

Question 5

Integration
2 Marks
2021 P1 Q5

A solid is formed by rotating the curve with equation y=2x\displaystyle y = 2\sqrt{x} between x=3\displaystyle x = 3 and x=5\displaystyle x = 5 through 2π\displaystyle 2\pi radians about the x\displaystyle x-axis.
Calculate the exact value of the volume of this solid.

Show answer
32π\displaystyle 32\pi
Show marking instructions

Question 5

•¹ correct form of integral
π35y2dx\displaystyle \pi\int_{3}^{5} y^2 \, dx

•² evaluate
32π\displaystyle 32\pi (cubic units)

6

Question 6

DifferentiationIntegration
(2, 2)4 Marks
2021 P1 Q6

The velocity, v\displaystyle v ms1\displaystyle ^{-1}, of a particle after t\displaystyle t seconds is given by v=3t2e2t.\displaystyle v = 3t^2 - e^{-2t}. At time t=0\displaystyle t = 0 the displacement of the particle is zero.

(a)Find an expression for the displacement of the particle.

(b)Calculate the acceleration of the particle when t=0.\displaystyle t = 0.

Show answer
(a) t3+12e2t12\displaystyle t^3 + \frac{1}{2}e^{-2t} - \frac{1}{2}
(b) 2\displaystyle 2 ms2\displaystyle ^{-2}
Show marking instructions

Question 6(a)

•¹ start to integrate
t3...\displaystyle t^3 ... or ...12e2t\displaystyle ... - \frac{1}{-2}e^{-2t}

•² find expression
t3+12e2t12\displaystyle t^3 + \frac{1}{2}e^{-2t} - \frac{1}{2}

Question 6(b)

•³ differentiate
6t+2e2t\displaystyle 6t + 2e^{-2t}

•⁴ calculate acceleration
2 ms2\displaystyle 2 \text{ ms}^{-2}

7

Question 7

Functions & Graphs
(1, 2, 1, 1, 1)6 Marks
2021 P1 Q7

A function is defined on a suitable domain by f(x)=x2x2.\displaystyle f(x) = \frac{x^2}{x - 2}.

(a)For the graph of y=f(x)\displaystyle y = f(x)

(i) state the equation of the vertical asymptote
(ii) find the equation of the non-vertical asymptote. Justify your answer.

The turning points on the graph are (0, 0) and (4, 8). There are no other stationary points.

(b)On the diagram provided, sketch the graph of y=f(x).\displaystyle y = f(x).

(c)(i) On the diagram provided, sketch the graph of y=f(x).\displaystyle y = |f(x)|. Show all asymptotes.

(c)(ii) State the values of k\displaystyle k for which f(x)=k\displaystyle |f(x)| = k has exactly two distinct solutions.

Show answer
(a)(i) x=2\displaystyle x = 2
(a)(ii) y=x+2\displaystyle y = x + 2 (Justification: As x±\displaystyle x \rightarrow \pm \infty, 4x20\displaystyle \frac{4}{x - 2} \rightarrow 0)
(b) A sketch showing the shape of the curve approaching the asymptotes (x=2\displaystyle x = 2 and y=x+2\displaystyle y = x + 2), with turning points at (0, 0) and (4, 8).
(c)(i) A sketch showing the shape of the curve approaching the asymptotes, with the negative section reflected above the x\displaystyle x-axis.
(c)(ii) 0<k<8\displaystyle 0 < k < 8
Show marking instructions

Question 7(a)(i)

•¹ state vertical asymptote
x=2\displaystyle x = 2

Question 7(a)(ii)

•² complete algebraic division and restate function
x+2+4x2\displaystyle x + 2 + \frac{4}{x - 2}

•³ state non-vertical asymptote with justification
e.g. As x±\displaystyle x \rightarrow \pm \infty, 4x20\displaystyle \frac{4}{x - 2} \rightarrow 0
y=x+2\displaystyle y = x + 2

Question 7(b)

•⁴ sketch showing shape of curve with approach to asymptotes
Valid sketch showing asymptotes x=2\displaystyle x = 2 and y=x+2\displaystyle y = x + 2 with turning points at (0,0)\displaystyle (0, 0) and (4,8).\displaystyle (4, 8).

Question 7(c)(i)

•⁵ sketch showing shape of curve with approach to asymptotes
Valid sketch showing the negative section of the curve reflected above the x\displaystyle x-axis.

Question 7(c)(ii)

•⁶ state values of k
0<k<8\displaystyle 0 < k < 8

8

Question 8

Differential Equations
9 Marks
2021 P1 Q8

Find the particular solution of the differential equation

d2ydx2+dydx6y=35e2x\displaystyle \frac{d^2y}{dx^2} + \frac{dy}{dx} - 6y = 35e^{2x}

given y=5\displaystyle y = 5 and dydx=12\displaystyle \frac{dy}{dx} = 12 when x=0.\displaystyle x = 0.

Show answer
y=4e2x+e3x+7xe2x\displaystyle y = 4e^{2x} + e^{-3x} + 7xe^{2x}
Show marking instructions

Question 8

•¹ solve auxiliary equation
m=2,m=3\displaystyle m = 2, m = -3

•² state complementary function
y=Ae2x+Be3x\displaystyle y = Ae^{2x} + Be^{-3x}

•³ state particular integral
y=Cxe2x\displaystyle y = Cxe^{2x}

•⁴ differentiate complementary function
dydx=Ce2x+2Cxe2x\displaystyle \frac{dy}{dx} = Ce^{2x} + 2Cxe^{2x}
d2ydx2=4Ce2x+4Cxe2x\displaystyle \frac{d^2y}{dx^2} = 4Ce^{2x} + 4Cxe^{2x}

•⁵ evaluate C
C=7\displaystyle C = 7

•⁶ general solution stated or implied
y=Ae2x+Be3x+7xe2x\displaystyle y = Ae^{2x} + Be^{-3x} + 7xe^{2x}

•⁷ differentiate
dydx=2Ae2x3Be3x+7e2x+14xe2x\displaystyle \frac{dy}{dx} = 2Ae^{2x} - 3Be^{-3x} + 7e^{2x} + 14xe^{2x}

•⁸ form equations and solve for one constant
A=4\displaystyle A = 4 or B=1\displaystyle B = 1

•⁹ give particular solution
y=4e2x+e3x+7xe2x\displaystyle y = 4e^{2x} + e^{-3x} + 7xe^{2x}