A Level Mathematics: Pure Mathematics 3 Algebra and Calculus — Practice Questions (Cambridge 9709)

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 Pure Mathematics 3, most of them skills tested in the June 2025 Paper 32. After each answer there is a tip and, where the paper has a matching question, the real question to try next.

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Questions

1. Solve the inequality |3x − 2| < |x + 4|. [3]

2. Solve the equation (2eˣ + 3e⁻ˣ)/(eˣ − 1) = 5, giving x to 3 decimal places. [4]

3. (a) Find the first three terms, in ascending powers of x, of the expansion of (2 − x)(1 + 4x)^(−½), with each coefficient in its simplest form. (b) For which values of x is this expansion valid? [4]

4. Express (5x + 4)/((x − 1)(x + 2)²) in partial fractions. [4]

5. A curve is defined by the parametric equations x = t² + 1 and y = t³ − 2t. Find the gradient of the curve at the point where t = 2. [2]

6. Find the exact value of ∫₁ᵉ x² ln x dx. [4]

7. The variables x and y satisfy the differential equation dy/dx = (2y + 1) tan x, and y = 1 when x = 0. Solve the differential equation, giving y in terms of x. [5]

8. By using the substitution u = sin x, find the exact value of ∫₀^(π/2) sin 2x/(1 + sin x) dx. [4]

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Answers

1. Both sides are non-negative, so square: 9x² − 12x + 4 < x² + 8x + 16 [1]. This gives 8x² − 20x − 12 < 0, so 2x² − 5x − 3 < 0, which factorises as (2x + 1)(x − 3) < 0, with critical values −½ and 3 [1]. The quadratic is negative between its roots: −½ < x < 3 [1].

Tip: squaring both sides is safe here because both sides are moduli, so both are non-negative. Always check which side of the critical values you need by testing a value such as x = 0.

2. Multiply up: 2eˣ + 3e⁻ˣ = 5eˣ − 5, so 3eˣ − 5 − 3e⁻ˣ = 0 [1]. Multiply by eˣ and let u = eˣ: 3u² − 5u − 3 = 0 [1]. u = (5 ± √61)/6; eˣ must be positive, so u = (5 + √61)/6 = 2.1350… [1]. x = ln 2.1350… = 0.758 [1].

Tip: multiplying through by eˣ turns e⁻ˣ into a constant and gives a quadratic in eˣ. Reject any negative value of eˣ and say why.

Try the real question next: Cambridge International AS & A Level Mathematics 9709, June 2025, Paper 32, Question 1.

3. (a) (1 + 4x)^(−½) = 1 + (−½)(4x) + [(−½)(−3/2)/2](4x)² + … [1] = 1 − 2x + 6x² + … [1]. Then (2 − x)(1 − 2x + 6x²) = 2 − 4x + 12x² − x + 2x² + … = 2 − 5x + 14x² [1]. (b) Valid for |4x| < 1, that is |x| < ¼ [1].

Tip: with a fractional power, square the whole term (4x)², not just x, and multiply by the extra bracket only after the series is simplified.

Try the real question next: Cambridge International AS & A Level Mathematics 9709, June 2025, Paper 32, Question 2.

4. Write (5x + 4)/((x − 1)(x + 2)²) ≡ A/(x − 1) + B/(x + 2) + C/(x + 2)², so 5x + 4 ≡ A(x + 2)² + B(x − 1)(x + 2) + C(x − 1) [1]. x = 1 gives 9 = 9A, so A = 1; x = −2 gives −6 = −3C, so C = 2 [1]. x = 0 gives 4 = 4A − 2B − C = 2 − 2B, so B = −1 [1]. Answer: 1/(x − 1) − 1/(x + 2) + 2/(x + 2)² [1].

Tip: a repeated factor (x + 2)² needs two fractions, one over (x + 2) and one over (x + 2)². Leaving one out makes the identity impossible to satisfy.

5. dx/dt = 2t and dy/dt = 3t² − 2, so dy/dx = (3t² − 2)/(2t) [1]. When t = 2, dy/dx = 10/4 = 5/2 [1].

Tip: divide dy/dt by dx/dt, not the other way round.

6. Take u = ln x and dv/dx = x², so du/dx = 1/x and v = x³/3 [1]. ∫ x² ln x dx = (x³/3) ln x − ∫ x²/3 dx = (x³/3) ln x − x³/9 [1]. Limits: (e³/3 − e³/9) − (0 − 1/9) [1] = (2e³ + 1)/9 [1].

Tip: when one factor is a logarithm (or an inverse trig function), let u be that factor, because it becomes simpler when you differentiate it.

Try the real question next: Cambridge International AS & A Level Mathematics 9709, June 2025, Paper 32, Question 10(b).

7. Separate: ∫ 1/(2y + 1) dy = ∫ tan x dx [1]. ½ ln(2y + 1) = −ln(cos x) + c [1]. So ln(2y + 1) = −2 ln(cos x) + k. When x = 0, y = 1: ln 3 = k [1]. Then ln(2y + 1) = ln(3 sec² x), so 2y + 1 = 3 sec² x [1], giving y = (3 sec² x − 1)/2 [1].

Tip: remember the ½ when you integrate 1/(2y + 1), and find the constant before you remove the logarithms.

Try the real question next: Cambridge International AS & A Level Mathematics 9709, June 2025, Paper 32, Question 8.

8. sin 2x = 2 sin x cos x and du = cos x dx, so the integral becomes ∫ 2u/(1 + u) du [1]. Limits: x = 0 gives u = 0, x = π/2 gives u = 1 [1]. 2u/(1 + u) = 2 − 2/(1 + u), so the integral is [2u − 2 ln(1 + u)] from 0 to 1 [1] = 2 − 2 ln 2 [1].

Tip: change the limits to values of u as soon as you substitute, then you never need to go back to x.

Try the real question next: Cambridge International AS & A Level Mathematics 9709, June 2025, Paper 32, Question 11(b).

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Where marks are usually lost

• Negative values of eˣ kept instead of being rejected.
• The x in a binomial bracket not multiplied by its coefficient, for example writing x² instead of (4x)².
• A repeated factor in partial fractions given only one fraction.
• The constant of integration found after the logarithms have been removed, or missed altogether.
• Limits left in terms of x after a substitution.