A Level Mathematics: Pure Mathematics 1 Mixed Practice (Cambridge 9709 Paper 1)

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs — Cambridge International holds copyright in its own papers. Use these alongside the official past papers available from your board.

Each question practises a skill tested in the June 2025 Paper 1 series. After each answer there is a tip and the real question to try next.

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Questions

1. Find the coordinates of the points where the curve y² + 4x = 13 meets the line 2x + y = 5. [4]

2. The coefficient of x in the expansion of (2x + k/x²)⁷ is 2688. Find the value of the positive constant k. [4]

3. A point moves along the curve y = 2x³ − 5x so that its x-coordinate increases at a constant rate of 0.3 units per second. Find the rate at which its y-coordinate is changing when x = 2. [3]

4. The equation of a curve is y = 3 sin 2x − 1 for 0 ≤ x ≤ π. (a) State the greatest and least values of y. (b) State the number of solutions of 3 sin 2x − 1 = 0 for 0 ≤ x ≤ π. [3]

5. A circle has equation x² + y² − 6x + 4y − 12 = 0. (a) Find the coordinates of the centre C and the radius of the circle. (b) The line y = 2 meets the circle at A and B. Find the angle ACB in radians, correct to 3 significant figures. [4]

6. A curve is such that d²y/dx² = 16/x³. The curve has a stationary point where x = 2. (a) Find an expression for dy/dx. (b) Find the x-coordinate of the other stationary point, and determine its nature. [4]

7. The first three terms of an arithmetic progression are 3k, k² and 5k, where k is a non-zero constant. (a) Find the value of k. (b) Find the sum of the first 20 terms. [4]

8. The third and fifth terms of a geometric progression are 48 and 12. The common ratio is positive. Find the sum to infinity. [3]

9. The function f is defined by f(x) = x² − 6x + 5 for x ≥ 3. (a) Express f(x) in the form (x − a)² + b. (b) Find an expression for f⁻¹(x) and state its domain. [4]

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Answers

1. From the line, y = 5 − 2x [1]. Substitute: (5 − 2x)² + 4x = 13, so 4x² − 16x + 12 = 0, so x² − 4x + 3 = 0 [1]. (x − 1)(x − 3) = 0, so x = 1 or x = 3 [1]. The points are (1, 3) and (3, −1) [1].

Tip: substitute the linear equation into the curve, not the other way round, and give each answer as a coordinate pair.

Try the real question next: Cambridge International AS & A Level Mathematics 9709, June 2025, Paper 12, Question 2.

2. General term: ⁷Cᵣ(2x)⁷⁻ʳ(k/x²)ʳ, whose power of x is 7 − 3r [1]. For x¹, 7 − 3r = 1, so r = 2 [1]. Coefficient = ⁷C₂ × 2⁵ × k² = 21 × 32 × k² = 672k² [1]. 672k² = 2688, so k² = 4 and k = 2 (k is positive) [1].

Tip: find which term you need from the power of x before you calculate any coefficients.

Try the real question next: Cambridge International AS & A Level Mathematics 9709, June 2025, Paper 12, Question 3.

3. dy/dx = 6x² − 5, which is 19 when x = 2 [1]. dy/dt = dy/dx × dx/dt [1] = 19 × 0.3 = 5.7 units per second [1].

Tip: write the chain rule out in full before substituting numbers.

Try the real question next: Cambridge International AS & A Level Mathematics 9709, June 2025, Paper 12, Question 4.

4. (a) sin 2x lies between −1 and 1, so the greatest value is 3 − 1 = 2 [1] and the least is −3 − 1 = −4 [1]. (b) sin 2x = 1/3. As x goes from 0 to π, 2x goes from 0 to 2π, and sin 2x = 1/3 happens twice: 2 solutions [1].

Tip: the range of a sin k x + b is from b − a to b + a, whatever the value of k.

Try the real question next: Cambridge International AS & A Level Mathematics 9709, June 2025, Paper 12, Question 5.

5. (a) Completing the square: (x − 3)² + (y + 2)² = 12 + 9 + 4 = 25 [1]. Centre C(3, −2), radius 5 [1]. (b) With y = 2: (x − 3)² + 16 = 25, so x = 0 or 6, and AB = 6 [1]. The distance from C to AB is 4, and half of AB is 3, so sin(½ACB) = 3/5 and ACB = 2 sin⁻¹(0.6) = 1.29 radians [1].

Tip: halve the triangle ACB to get a right angle, and keep your calculator in radians.

Try the real question next: Cambridge International AS & A Level Mathematics 9709, June 2025, Paper 12, Question 8.

6. (a) dy/dx = ∫16x⁻³ dx = −8x⁻² + c [1]. At x = 2, dy/dx = 0: −2 + c = 0, so c = 2, and dy/dx = 2 − 8/x² [1]. (b) 2 − 8/x² = 0 gives x² = 4, so the other stationary point is at x = −2 [1]. At x = −2, d²y/dx² = 16/(−8) = −2 < 0, so it is a maximum [1].

Tip: use the given stationary point to find the constant of integration. Without it you cannot find the other point.

Try the real question next: Cambridge International AS & A Level Mathematics 9709, June 2025, Paper 12, Question 9.

7. (a) In an AP, the middle term is the mean of its neighbours: 2k² = 3k + 5k = 8k [1]. As k ≠ 0, k = 4 [1]. (b) The terms are 12, 16, 20, …, so a = 12 and d = 4 [1]. S₂₀ = 20/2 × (2 × 12 + 19 × 4) = 10 × 100 = 1000 [1].

Tip: state why you can divide by k (it is non-zero), so you don’t lose the k = 0 case without reason.

Try the real question next: Cambridge International AS & A Level Mathematics 9709, June 2025, Paper 12, Question 10(a).

8. ar⁴ ÷ ar² = r² = 12/48 = 1/4, so r = 1/2 (positive) [1]. a = 48 ÷ (1/2)² = 192 [1]. S∞ = 192 ÷ (1 − 1/2) = 384 [1].

Tip: dividing one term by another removes a and leaves a power of r.

Try the real question next: Cambridge International AS & A Level Mathematics 9709, June 2025, Paper 12, Question 10(b).

9. (a) x² − 6x + 5 = (x − 3)² − 9 + 5 = (x − 3)² − 4 [2]. (b) y = (x − 3)² − 4, so x − 3 = ±√(y + 4) [1]. Since x ≥ 3, take the positive root: f⁻¹(x) = 3 + √(x + 4), for x ≥ −4 [1].

Tip: the domain of f decides the sign of the square root, and the range of f is the domain of f⁻¹.

Try the real question next: Cambridge International AS & A Level Mathematics 9709, June 2025, Paper 12, Question 11.

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Where marks are usually lost

• Answers left as x-values when coordinates are asked for.
• The constant of integration missing, or not found from the given point.
• Degrees used instead of radians in circle and arc questions.
• The wrong sign of square root chosen for an inverse function.