A level Maths mock exam Free full practice test

Written by STEM-Tutor's tutors · Updated October 2026 · 30 questions and 4 exam-style questions · about 95 minutes

A full-length practice test built from our exam-style questions on every topic of A level Maths. Each answer is marked as you go, and every question has a worked solution to open afterwards.

On this page

Before you start
Section I
Section II
After the mock

Before you start

Set a timer for about 95 minutes and work without notes. A formula booklet is provided in the exam. Answers within 2.5% are marked correct, and you can type fractions such as 1/6.
There are no hints in the mock. Answer every question before you open its worked solution: opening it first counts the question as missed.
Write the exam-style questions on paper, then mark them yourself with the scoring guide.

Section I: 30 exam-style questions

Q1Exam-style

Rationalise 6/(3 − √3), writing it as a + √3. Find a.

Hint

Multiply top and bottom by (3 + √3).

Worked solution

6(3 + √3)/((3 − √3)(3 + √3)) = 6(3 + √3)/6.
= 3 + √3, so a = 3.

Q2Exam-style

x² + kx + 9 = 0 has equal roots, and k > 0. Find k.

Hint

Equal roots: the discriminant is zero.

Worked solution

k² − 36 = 0.
k = 6 (taking k > 0).

Q3Exam-style

f(x) = 2x + 1 and g(x) = x². Find fg(3).

Hint

Apply g first, then f.

Worked solution

g(3) = 9.
f(9) = 2(9) + 1 = 19.

Q4Exam-style

f(x) = (x + 4)/3. Find f⁻¹(5).

Hint

Find the inverse, or solve f(x) = 5.

Worked solution

(x + 4)/3 = 5 gives x + 4 = 15.
f⁻¹(5) = 11.

Q5ChallengeYear 2

(5x + 1)/((x + 1)(x − 1)) = A/(x + 1) + B/(x − 1). Find B.

Hint

Multiply through, then substitute x = 1.

Worked solution

5x + 1 = A(x − 1) + B(x + 1).
x = 1: 6 = 2B.
B = 3 (and A = 2, from x = −1).

Q6Exam-style

For the same arithmetic sequence (a = 5, d = 3), find the sum of the first 20 terms.

Hint

S_n = n/2(2a + (n − 1)d).

Worked solution

S₂₀ = 10 × (10 + 57).
= 670.

Q7Exam-style

A geometric series has first term 12 and common ratio 0.25. Find its sum to infinity.

Hint

S_∞ = a/(1 − r).

Worked solution

12/(1 − 0.25) = 12/0.75.
= 16.

Q8Exam-style

Find the coefficient of x³ in the expansion of (1 + 2x)⁵.

Hint

The x³ term is ⁵C₃(2x)³.

Worked solution

⁵C₃ = 10 and 2³ = 8.
Coefficient = 80.

Q9Exam-style

Find the distance between (1, 2) and (7, 10).

Hint

Use Pythagoras on the differences.

Worked solution

√(6² + 8²) = √100.
= 10.

Q10ChallengeYear 2

Find the coefficient of x² in the expansion of (1 + 3x)^1/2.

Hint

Use n(n − 1)/2! (3x)² with n = ½.

Worked solution

(½)(−½)/2 = −1/8.
−1/8 × 9 = −9/8.

Q11Exam-style

Find the area of a sector of radius 5 cm and angle 0.8 radians.

Hint

A = ½r²θ.

Worked solution

A = ½ × 25 × 0.8 = 10 cm².

Q12Exam-style

Solve sin x = 0.5 for 0° ≤ x ≤ 360°. Give the larger solution.

Hint

sin is positive in the first and second quadrants.

Worked solution

Principal value 30°; the other is 180° − 30°.
Larger solution: 150°.

Q13Exam-style

sin θ = 0.6 and θ is acute. Find tan θ.

Hint

Use sin²θ + cos²θ = 1 to find cos θ.

Worked solution

cos θ = √(1 − 0.36) = 0.8.
tan θ = 0.6/0.8 = 0.75.

Q14Exam-styleYear 2

Write 3 cos x + 4 sin x in the form R cos(x − α). Find R.

Hint

R = √(a² + b²).

Worked solution

R = √(9 + 16) = 5.

Q15Challenge

Solve 2cos²x − cos x − 1 = 0 for 0° < x < 360°. Give the smallest solution.

Hint

Treat it as a quadratic in cos x.

Worked solution

(2cos x + 1)(cos x − 1) = 0, so cos x = −½ or cos x = 1.
cos x = 1 gives x = 0° or 360°, both outside the range. cos x = −½ gives 120° and 240°.
Smallest: 120°.

Q16Exam-style

Solve ln x = 2, to 3 significant figures.

Hint

Undo ln with e.

Worked solution

x = e².
x = 7.39.

Q17Exam-style

Solve log₃ x + log₃ 9 = 4.

Hint

log₃ 9 = 2.

Worked solution

log₃ x = 4 − 2 = 2.
x = 3² = 9.

Q18Exam-style

A population is modelled by N = 500e^−0.2t. Find N when t = 5, to the nearest whole number.

Hint

Substitute t = 5.

Worked solution

N = 500e⁻¹ = 500 × 0.3679.
N = 184.

Q19Exam-style

For the same model, find the time taken for N to halve, to 3 significant figures.

Hint

Set e^−0.2t = ½.

Worked solution

−0.2t = ln ½, so t = ln 2/0.2.
t = 3.47.

Q20Challenge

Solve 2^2x − 6(2^x) + 8 = 0. Give the larger solution.

Hint

Let y = 2^x, so 2^2x = y².

Worked solution

y² − 6y + 8 = 0 gives y = 2 or y = 4.
2^x = 2 gives x = 1; 2^x = 4 gives x = 2.
Larger solution: 2.

Q21Exam-style

Find the x-coordinate of the stationary point of y = x² − 6x + 2.

Hint

Set dy/dx = 0.

Worked solution

dy/dx = 2x − 6 = 0.
x = 3.

Q22Exam-style

y = (2x + 1)⁴. Find dy/dx at x = 0.

Hint

Chain rule: 4(2x + 1)³ × 2.

Worked solution

dy/dx = 8(2x + 1)³.
At x = 0: 8.

Q23Exam-styleYear 2

y = xe^x. Find dy/dx at x = 1, to 3 significant figures.

Hint

Product rule with u = x and v = e^x.

Worked solution

dy/dx = e^x + xe^x.
At x = 1: 2e = 5.44.

Q24Exam-style

Find the gradient of the tangent to y = ln x at x = 4.

Hint

d/dx(ln x) = 1/x.

Worked solution

dy/dx = 1/4 = 0.25.

Q25ChallengeYear 2

The area of a circle increases at 10 cm²/s. Find the rate of increase of the radius when r = 5 cm, to 3 significant figures.

Hint

dA/dt = dA/dr × dr/dt, with A = πr².

Worked solution

dA/dr = 2πr = 10π at r = 5.
dr/dt = 10/(10π) = 1/π = 0.318 cm/s.

Q26Exam-style

Find the area under y = x² + 1 between x = 0 and x = 3.

Hint

Integrate between the limits.

Worked solution

[x³/3 + x]₀³ = 9 + 3.
Area = 12.

Q27Exam-style

Evaluate ∫₀¹ e^2x dx, to 3 significant figures.

Hint

∫ e^2x dx = ½e^2x.

Worked solution

[½e^2x]₀¹ = ½(e² − 1).
= 3.19.

Q28Exam-style

Evaluate ∫₀^π/2 cos x dx.

Hint

∫ cos x dx = sin x.

Worked solution

[sin x]₀^π/2 = 1 − 0 = 1.

Q29Exam-styleYear 2

Use the substitution u = x² + 1 to evaluate ∫₀¹ 2x(x² + 1)³ dx.

Hint

du = 2x dx, and the limits become u = 1 to u = 2.

Worked solution

∫₁² u³ du = [u⁴/4]₁².
= (16 − 1)/4 = 3.75.

Q30ChallengeYear 2

Use integration by parts to evaluate ∫₀¹ xe^x dx.

Hint

Let u = x and dv = e^x dx.

Worked solution

∫ xe^x dx = xe^x − ∫ e^x dx = xe^x − e^x.
[xe^x − e^x]₀¹ = (e − e) − (0 − 1) = 1.

Section II: 4 exam-style questions

Exam-style question 1: Algebra and functions

Exam-style question style · 6 points · about 8 minutes

f(x) = x³ − 7x − 6.

(a) Use the factor theorem to show that (x + 1) is a factor of f(x).
(b) Factorise f(x) completely.
(c) Hence solve f(x) = 0.

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

f(−1) = −1 + 7 − 6 = 0, so (x + 1) is a factor.

1 point for: M1: evaluating f(−1)
1 point for: A1: = 0 with a conclusion

Part (b)

f(x) = (x + 1)(x² − x − 6) = (x + 1)(x + 2)(x − 3).

1 point for: M1: dividing by (x + 1) or comparing coefficients
1 point for: A1: x² − x − 6
1 point for: A1: fully factorised

Part (c)

x = −1, x = −2 or x = 3.

1 point for: B1: all three solutions

Your mark: 0 of 6

Exam-style question 2: Trigonometry

Exam-style question style · 9 points · about 9 minutes

f(x) = 3 cos x + 4 sin x.

(a) Express f(x) in the form R cos(x − α), where R > 0 and 0° < α < 90°.
(b) Write down the maximum value of f(x) and the smallest positive value of x where it occurs.
(c) Solve f(x) = 2.5 for 0° ≤ x < 360°.

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

R = 5, tan α = 4/3, so α = 53.13°. f(x) = 5 cos(x − 53.13°).

1 point for: B1: R = 5
1 point for: M1: tan α = 4/3
1 point for: A1: α = 53.13°

Part (b)

Maximum 5, when x − 53.13° = 0, so x = 53.13°.

1 point for: B1: maximum 5
1 point for: B1: x = 53.13°

Part (c)

cos(x − 53.13°) = 0.5, so x − 53.13° = −60° or 60°. x = 113.13° or 353.13°.

1 point for: M1: cos(x − α) = 0.5
1 point for: M1: finding both angles in range
1 point for: A1: 113.1°
1 point for: A1: 353.1°

Your mark: 0 of 9

Exam-style question 3: Exponentials and logarithms

Exam-style question style · 8 points · about 8 minutes

The value of a machine, £V, t years after it is bought is modelled by V = 20 000e^−kt. After 2 years it is worth £15 000.

(a) Show that k = ½ ln(4/3).
(b) Find the value after 5 years, to the nearest pound.
(c) Find when the value first falls below £5000, to 1 decimal place.
(d) Give one limitation of this model.

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

15 000 = 20 000e^−2k, so e^−2k = 3/4 and −2k = ln(3/4), giving k = ½ ln(4/3).

1 point for: M1: substituting t = 2, V = 15 000
1 point for: A1: correct rearrangement to the given answer

Part (b)

k = 0.1438. V = 20 000e^−0.719 = £9743.

1 point for: M1: substituting t = 5
1 point for: A1: £9743

Part (c)

e^−kt < 1/4, so t > ln 4/k = 9.6 years.

1 point for: M1: setting up the inequality or equation
1 point for: M1: taking logs correctly
1 point for: A1: 9.6 years

Part (d)

It predicts the value never reaches zero, or it ignores factors such as condition and demand.

1 point for: B1: a sensible limitation

Your mark: 0 of 8

Exam-style question 4: Integration

Exam-style question style · 8 points · about 8 minutes

The curve C has equation y = 4x − x². The line L has equation y = x.

(a) Find the x-coordinates where C meets the x-axis.
(b) Find the area enclosed by C and the x-axis.
(c) Find the area of the region enclosed by C and L.

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

4x − x² = x(4 − x) = 0, so x = 0 and x = 4.

1 point for: B1: x = 0 and x = 4

Part (b)

∫₀⁴ (4x − x²) dx = [2x² − x³/3]₀⁴ = 32 − 64/3 = 32/3.

1 point for: M1: integrating
1 point for: M1: substituting the limits
1 point for: A1: 32/3

Part (c)

4x − x² = x gives x = 0 or 3. ∫₀³ (3x − x²) dx = 27/2 − 9 = 4.5.

1 point for: M1: finding the intersection x = 3
1 point for: M1: integrating curve minus line
1 point for: M1: correct limits
1 point for: A1: 4.5

Your mark: 0 of 8

After the mock

Note every question you missed and its topic, then go back to that topic's practice page: all A level Maths topics. Then try the real thing with official A level Maths past papers.