OxfordAQA A Level Mathematics: Pure Maths 1 — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Pure Maths 1 revision notes, covering algebra, quadratics, coordinate geometry, trigonometry, differentiation and integration for International AS Mathematics.

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Section A

1. Simplify (2x³y⁻²)³ ÷ (4x⁵y⁻¹). [3]

2. Rationalise the denominator of 5 ÷ (3 − √2). [3]

Section B

3. The quadratic kx² + (k + 3)x + 4 = 0 has equal roots.

(a) Show that k² − 10k + 9 = 0. [3] (b) Hence find the possible values of k. [2]

4. Find the first four terms of the expansion of (1 + 2x)⁸ in ascending powers of x. [4]

5. Solve the simultaneous equations y = x² − 3x + 4 and y = 2x − 2. [5]

6. In triangle ABC, AB = 8 cm, AC = 11 cm and angle BAC = 52°.

(a) Calculate BC. [3] (b) Calculate the area of the triangle. [2] (c) Calculate angle ABC. [3]

7. Solve 2 tan x = 3 for 0 ≤ x ≤ 2π, giving answers to 3 significant figures in radians. [3]

8. A curve has equation y = x³ − 3x² + 2.

(a) Find dy/dx. [2] (b) Find the equation of the tangent to the curve at the point where x = 1. [3] (c) Find the equation of the normal to the curve at the same point. [2]

9. Find the area enclosed between the curve y = 4 − x² and the x-axis. [5]

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Answers

1. (2x³y⁻²)³ = 8x⁹y⁻⁶ [1]; ÷ 4x⁵y⁻¹ [1] = 2x⁴y⁻⁵ [1].

2. Multiply by (3 + √2) over itself [1]; numerator 5(3 + √2) = 15 + 5√2, denominator 9 − 2 = 7 [1]; = (15 + 5√2) ÷ 7 [1].

3. (a) Equal roots means the discriminant is zero: (k + 3)² − 4 × k × 4 = 0 [1]; k² + 6k + 9 − 16k = 0 [1]; k² − 10k + 9 = 0 [1]. (b) (k − 1)(k − 9) = 0 [1]; k = 1 or 9 [1].

4. (1 + 2x)⁸ = 1 + 8(2x) + 28(2x)² + 56(2x)³ [1] [1] = 1 + 16x + 112x² + 448x³ [1] [1].

5. x² − 3x + 4 = 2x − 2 [1]; x² − 5x + 6 = 0 [1]; (x − 2)(x − 3) = 0, so x = 2 or 3 [1]. When x = 2, y = 2; when x = 3, y = 4 [1]. Solutions (2, 2) and (3, 4) [1].

6. (a) Cosine rule: BC² = 8² + 11² − 2(8)(11)cos 52° [1] = 64 + 121 − 108.4 [1]; BC = √76.6 = 8.75 cm [1]. (b) Area = ½ × 8 × 11 × sin 52° [1] = 34.7 cm² [1]. (c) Cosine rule for angle B, using the unrounded BC² = 76.6436 from part (a) rather than the rounded 8.75 cm (rounding the side before using it again compounds the error): cos B = (BC² + AB² − AC²) ÷ (2 × BC × AB) [1] = (76.6436 + 64 − 121) ÷ (2 × 8.7546 × 8) = 0.1403 [1]; B = 81.9° [1].

7. tan x = 1.5 [1]; x = 0.983 rad [1]; the second solution is 0.983 + π = 4.12 rad [1].

8. (a) dy/dx = 3x² − 6x [1] [1]. (b) At x = 1: y = 1 − 3 + 2 = 0, and gradient = 3 − 6 = −3 [1]. Tangent: y − 0 = −3(x − 1) [1], so y = −3x + 3 [1]. (c) Normal gradient = 1/3 (negative reciprocal of −3) [1]; y − 0 = (1/3)(x − 1), so y = (x − 1)/3 [1].

9. The curve meets the x-axis where 4 − x² = 0, so x = ±2 [1]. Area = ∫₋₂² (4 − x²) dx [1] = [4x − x³/3]₋₂² [1] = (8 − 8/3) − (−8 + 8/3) [1] = 16/3 − (−16/3) = 32/3 (≈ 10.7) [1].

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Where marks are usually lost

• Forgetting to cube the coefficient as well as the variables.
• Using b² − 4ac > 0 instead of = 0 for equal roots.
• Not squaring the 2 in (2x)² in the binomial expansion.
• Giving only one solution to a trigonometric equation in the stated range.
• Using the tangent gradient instead of its negative reciprocal when finding the equation of a normal.
• Simply integrating from one root to the other without checking the sign of the function — here y = 4 − x² is entirely non-negative on [−2, 2], so the signed integral does equal the true area, but this must be checked rather than assumed for every region.

Differentiation and integration — the two mark-heavy checks

Two checks are worth making automatic whenever calculus appears in this unit. First, a tangent takes the gradient found by differentiating directly, while a normal takes its negative reciprocal — mixing these up is the single most common calculus error in P1. Second, before integrating to find an area, check whether the curve dips below the x-axis anywhere in the interval: if it does, the signed integral will understate the true area unless the region is split at the roots and each part’s absolute value is taken separately. For the underlying rules, including how the second derivative classifies a stationary point, see the Pure Maths 1 revision notes.