Advanced Higher Maths · SQA past paper

2025 Paper 1

Non-Calculator · 8 questions
1

Question 1

Binomial Theorem
4 Marks
2025 P1 Q1

Use the binomial theorem to expand (1x3x)4.\displaystyle \left(\frac{1}{x} - 3x\right)^4.
Simplify your answer.

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1x412x2+54108x2+81x4\displaystyle \frac{1}{x^4} - \frac{12}{x^2} + 54 - 108x^2 + 81x^4
2

Question 2

Differentiation
3 Marks
2025 P1 Q2

Given f(x)=2x3+x3+2x\displaystyle f(x) = \frac{2x^3 + x}{3 + 2x}, find f(x).\displaystyle f'(x). Simplify your answer.

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8x3+18x2+3(3+2x)2\displaystyle \frac{8x^3 + 18x^2 + 3}{(3 + 2x)^2}
3

Question 3

Complex Numbers
2 Marks
2025 P1 Q3

Two complex numbers are defined as z=11+10i\displaystyle z = 11 + 10i and w=32i.\displaystyle w = 3 - 2i.
Find zw\displaystyle \frac{z}{w} in the form a+bi\displaystyle a + bi, where a,bR.\displaystyle a, b \in \mathbb{R}.

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1+4i\displaystyle 1 + 4i
4

Question 4

Matrices
(1, 2, 1, 2)6 Marks
2025 P1 Q4

Matrices A\displaystyle A and B\displaystyle B are defined by A=(32 01)\displaystyle A = \begin{pmatrix} -3 & 2 \\\ 0 & 1 \end{pmatrix} and B=(22 5λ)\displaystyle B = \begin{pmatrix} 2 & 2 \\\ 5 & \lambda \end{pmatrix} where λR.\displaystyle \lambda \in \mathbb{R}.

(a)Find 3A+2B.\displaystyle 3A + 2B.

(b)(i) Find AB\displaystyle A'B, where A\displaystyle A' is the transpose of A.\displaystyle A.

(b)(ii) Find an expression for the determinant of AB.\displaystyle A'B.

(b)(iii) Determine the value of λ\displaystyle \lambda such that AB\displaystyle A'B is singular.

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(a) (510 103+2λ)\displaystyle \begin{pmatrix} -5 & 10 \\\ 10 & 3 + 2\lambda \end{pmatrix}
(b)(i) (66 94+λ)\displaystyle \begin{pmatrix} -6 & -6 \\\ 9 & 4 + \lambda \end{pmatrix}
(b)(ii) 6(4+λ)(6)(9)\displaystyle -6(4 + \lambda) - (-6)(9) or 306λ\displaystyle 30 - 6\lambda
(b)(iii) λ=5\displaystyle \lambda = 5
5

Question 5

Integration
2 Marks
2025 P1 Q5

Find 11+4x2dx.\displaystyle \int \frac{1}{1 + 4x^2} dx.

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12tan1(2x)+c\displaystyle \frac{1}{2} \tan^{-1}(2x) + c
6

Question 6

Functions & GraphsPartial Fractions
(2, 2)4 Marks
2025 P1 Q6

On a suitable domain a curve is given by the equation y=f(x)\displaystyle y = f(x), where
f(x)=x2+x+5x2.\displaystyle f(x) = \frac{x^2 + x + 5}{x - 2}.

(a)Express f(x)\displaystyle f(x) in the form Ax+B+Cx2\displaystyle Ax + B + \frac{C}{x - 2}, where A,B\displaystyle A, B and C\displaystyle C are constants.

(b)State the equations of the vertical and non-vertical asymptotes of the curve.

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(a) x+3+11x2\displaystyle x + 3 + \frac{11}{x - 2}
(b) Vertical asymptote: x=2\displaystyle x = 2
Non-vertical asymptote: y=x+3\displaystyle y = x + 3
7

Question 7

Differential Equations
5 Marks
2025 P1 Q7

Solve the differential equation

dydx=y2x1\displaystyle \frac{dy}{dx} = \frac{y}{2x - 1}, x,y>1\displaystyle x, y > 1,

given that y=12\displaystyle y = 12 when x=5.\displaystyle x = 5. Express y\displaystyle y in terms of x.\displaystyle x.

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y=4(2x1)12\displaystyle y = 4(2x - 1)^{\frac{1}{2}} (or y=42x1\displaystyle y = 4\sqrt{2x - 1})
8

Question 8

Vectors
(4, 3, 2)9 Marks
2025 P1 Q8

Three planes are defined by

π1:xy+4z=6\displaystyle \pi_1:\quad x - y + 4z = -6
π2:2x+3y+z=15\displaystyle \pi_2:\quad 2x + 3y + z = 15
π3:3x+2y2z=16\displaystyle \pi_3:\quad 3x + 2y - 2z = 16

(a)Use Gaussian elimination to find T, the point of intersection of the three planes.

The line L1\displaystyle L_1 is defined by x+43=y72=z41.\displaystyle \frac{x + 4}{3} = \frac{y - 7}{2} = \frac{z - 4}{1}.

(b)Find P, the point of intersection of the line L1\displaystyle L_1 and the plane π3.\displaystyle \pi_3.

The line L2\displaystyle L_2 passes through points T (the point of intersection of the three planes) and P (the point of intersection of the line L1\displaystyle L_1 and the plane π3\displaystyle \pi_3).

(c)Find, in parametric form, the equations of the line L2.\displaystyle L_2.

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(a) x=2,y=4,z=1\displaystyle x = 2, y = 4, z = -1
(b) P(2,11,6)\displaystyle P(2, 11, 6)
(c) x=2,y=11+7λ,z=6+7λ\displaystyle x = 2, y = 11 + 7\lambda, z = 6 + 7\lambda (or equivalent)