OxfordAQA A Level Mathematics: Pure Maths 1 — Practice Questions

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 — 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, differentiation and integration for Unit P1 (with circle and trigonometry sections marked as Unit PSM1 content).

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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

Questions 6 and 7 test trigonometry (sine/cosine rule, area of a triangle, solving a trig equation). OxfordAQA’s 9660 specification places all trigonometry in Unit PSM1’s pure maths strand (PP1.2), not in Unit P1 – P1 contains no trigonometry at all, including the sin/cos/tan graphs. They’re included in this P1 practice set because every P1 candidate also sits PSM1, but they are PSM1-syllabus questions, not P1 ones.

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. (PSM1 content) 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. (PSM1 content) 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 lies entirely below the x-axis on the interval: if it does, the signed integral comes out negative and the area is its magnitude. (P1 will not set a region that is partially above and partially below the x-axis, so the further technique of splitting at the roots and taking each part’s absolute value separately – needed only for that mixed case – is background knowledge rather than something a P1 paper requires.) For the underlying rules, including how the second derivative classifies a stationary point, see the Pure Maths 1 revision notes.