OCR A Level Mathematics: Pure Mathematics — 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 Mathematics revision notes

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Section A

1. Solve 3^(2x) = 40, giving your answer to 3 significant figures. [3]

2. Write as a single logarithm: 2 log a + log b − 3 log c. [3]

Section B

3. An arithmetic series has first term 7 and common difference 4.

(a) Find the 20th term. [2] (b) Find the sum of the first 20 terms. [3] (c) Find the least n for which the sum exceeds 1000. [4]

4. Differentiate with respect to x:

(a) y = (3x² + 1)⁵ [2] (b) y = x² ln x [3] (c) y = (2x + 1) ÷ (x − 3) [3]

5. Evaluate ∫₀^(π/2) sin 2x dx. [3]

6. Prove by contradiction that √2 is irrational. [5]

7. Prove that the sum of any two consecutive odd numbers is a multiple of 4. [3]

8. The functions f and g are defined for all real x by f(x) = 3x + 2 and g(x) = x² − 1.

(a) Find gf(x) in the form ax² + bx + c. [2] (b) Find f⁻¹(x). [2]

9. A sector of a circle has radius 8 cm and angle 1.2 radians.

(a) Find the arc length of the sector. [1] (b) Find the area of the sector. [2]

10. The vectors a = 2i + 3j − 6k and b = 4i + 6j − 12k are given.

(a) Find |a|. [2] (b) State, with a reason, whether a and b are parallel. [2]

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Answers

1. 2x log 3 = log 40 [1]; x = log 40 ÷ (2 log 3) [1] = 1.68 [1].

2. log a² + log b − log c³ [1] [1] = log(a²b ÷ c³) [1].

3. (a) u₂₀ = 7 + 19 × 4 [1] = 83 [1]. (b) S₂₀ = 20/2 (7 + 83) [1] [1] = 900 [1]. (c) Sₙ = n/2 [14 + 4(n − 1)] = n(2n + 5) [1]. Solve 2n² + 5n − 1000 > 0 [1]; n = (−5 + √(25 + 8000)) ÷ 4 = 21.1 [1]; least n = 22 [1].

4. (a) dy/dx = 5(3x² + 1)⁴ × 6x [1] = 30x(3x² + 1)⁴ [1]. (b) Product rule: 2x ln x + x² × (1/x) [1] [1] = 2x ln x + x [1]. (c) Quotient rule: [2(x − 3) − (2x + 1)] ÷ (x − 3)² [1] [1] = −7 ÷ (x − 3)² [1].

5. ∫ sin 2x dx = −½ cos 2x [1]. At π/2: −½ cos π = ½ [1]. At 0: −½ cos 0 = −½. Value = ½ − (−½) = 1 [1].

6. Assume √2 is rational, so √2 = a/b in lowest terms with a, b integers and no common factor [1]. Then 2 = a²/b², so a² = 2b², meaning a² is even and therefore a is even [1]. Write a = 2k; then 4k² = 2b², so b² = 2k², meaning b is also even [1]. So a and b share a factor of 2, contradicting the assumption that the fraction was in lowest terms [1]. The assumption must be false, so √2 is irrational [1].

7. Let the odd numbers be 2n + 1 and 2n + 3 [1]. Their sum is 4n + 4 [1] = 4(n + 1), which is a multiple of 4 for all integers n [1].

8. (a) gf(x) = (3x + 2)² − 1 [1] = 9x² + 12x + 4 − 1 = 9x² + 12x + 3 [1]. (b) Let y = 3x + 2, so x = (y − 2) ÷ 3 [1]; f⁻¹(x) = (x − 2) ÷ 3 [1].

9. (a) s = rθ = 8 × 1.2 = 9.6 cm [1]. (b) Area = ½r²θ = ½ × 8² × 1.2 = ½ × 64 × 1.2 [1] = 38.4 cm² [1].

10. (a) |a| = √(2² + 3² + (−6)²) = √(4 + 9 + 36) [1] = √49 = 7 [1]. (b) b = 2a [1], so a and b are parallel — every component of b is exactly twice the corresponding component of a, confirming b is a scalar multiple of a [1].

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Where marks are usually lost

• Forgetting to divide by 2 as well as by log 3.
• Using the wrong sign in the quotient rule numerator.
• Omitting the factor ½ when integrating sin 2x.
• Not stating the “in lowest terms” assumption, which is the key to the contradiction.
• Applying gf(x) as fg(x) instead — always substitute the inner function’s whole expression into the outer function, working from the inside out.
• Swapping x and y the wrong way round when finding an inverse function, or forgetting to relabel the result in terms of x.
• Using a calculator in degree mode for arc length and sector area — s = rθ and Area = ½r²θ only hold when θ is in radians.
• Checking vectors for a parallel relationship by comparing only one or two components instead of all three — a genuine scalar multiple must hold across every component.