Pure Mathematics 4

Question bank

Real questions, real marking

2 real questions this paper has actually set, drawn only from genuine Pearson question papers. Each one links to its real mark scheme, its real examiner report where one has been published, and our own level-by-level breakdown of how marks are actually earned — write your own answer first, then compare.

Oct 2024 · Q1

6 marks

Binomial expansion for rational n, then substitution into the series — the reciprocal trap

(a) Find the first 4 terms of the binomial expansion, in ascending powers of x, of (8 − 3x)^(−1/3), |x| < 8/3, giving each coefficient as a simplified fraction. (b) Use the answer from part (a) with x = 2/3 to find a rational approximation to ∛6.

Our analysis

(a) Find the first 4 terms of the binomial expansion, in ascending powers of x, of (8 − 3x)^(−1/3), |x| < 8/3, giving each coefficient as a simplified fraction. (b) Use the answer from part (a) with x = 2/3 to find a rational approximation to ∛6.

6 marks available

a — 4 marks

  1. 01

    (8 − 3x)^(−1/3) = (1/2)(1 − (3/8)x)^(−1/3)

    Takes out the correct factor of 8 so the bracket starts '1 ± …' — 8^(−1/3) must actually be evaluated to 1/2, not left as 8^(−1/3), before the (1+kx)^n expansion can be applied.

    B1
  2. 02

    (1 − (3/8)x)^(−1/3) = 1 + (−1/3)(−(3/8)x) + [(−1/3)(−4/3)/2!](−(3/8)x)² + [(−1/3)(−4/3)(−7/3)/3!](−(3/8)x)³ + …

    Attempts the binomial expansion of (1+kx)^n to reach the third or fourth term unsimplified, with an acceptable structure — the correct binomial coefficient for n = −1/3 combined with (−(3/8)x)² or (−(3/8)x)³, even if the −3/8 itself is mishandled.

    M1
  3. 03

    1/16 x, 1/64 x², 7/1536 x³ — any 2 of these 3 correctly simplified

    2 correct, fully simplified non-constant terms, which may be listed rather than combined into one expansion.

    A1
  4. 04

    (8 − 3x)^(−1/3) = 1/2 + 1/16 x + 1/64 x² + 7/1536 x³ + …

    All 4 terms correct and simplified, ignoring subsequent working once a correct answer is seen.

    A1

b — 2 marks

  1. 101

    1/2 + 1/16(2/3) + 1/64(2/3)² + 7/1536(2/3)³ + … = 2851/5184

    Attempts to substitute x = 2/3 into their expansion from part (a) to achieve a fraction, AND attempts to find the reciprocal of that fraction — both moves are required for this one mark, not just the substitution.

    M1
  2. 102

    ∛6 ≈ (2851/5184)⁻¹ = 5184/2851 (or 1 2333/2851)

    Correct final rational value. The real examiner report specifically flags that a decimal answer here does not satisfy the question's own request for a rational approximation.

    A1

Oct 2024 · Q2

4 marks

Proof by contradiction — a quartic that reduces to two impossible values of x²

The curve C₁ has equation y = x⁴ + 10x² + 8, x ∈ ℝ. The curve C₂ has equation y = 2x² − 7, x ∈ ℝ. Use algebra to prove by contradiction that C₁ and C₂ do not intersect.

Our analysis

The curve C₁ has equation y = x⁴ + 10x² + 8, x ∈ ℝ. The curve C₂ has equation y = 2x² − 7, x ∈ ℝ. Use algebra to prove by contradiction that C₁ and C₂ do not intersect.

4 marks available

  1. 01

    Assume that C₁ and C₂ do intersect, so x⁴ + 10x² + 8 = 2x² − 7 (and has real solutions).

    Sets up the contradiction by making a statement that the two curves intersect (or equivalently equating the curve equations) — the compulsory first move on a 'prove by contradiction' question. This mark is independent of the rest of the working, so it can still be earned even if the rest of the proof fails.

    B1
  2. 02

    x⁴ + 10x² + 8 = 2x² − 7 ⟹ x⁴ + 8x² + 15 = 0 ⟹ (x² + 5)(x² + 3) = 0 ⟹ x² = …

    Attempts to solve a three-term quadratic equation in x² by any valid non-calculator method — factorising, the quadratic formula, or completing the square are all credited equally. The question's own bold instruction ('solutions relying on calculator technology are not acceptable') rules out jumping straight to the roots with no method shown.

    M1
  3. 03

    x² = −5, −3

    Correct pair of values for x², reached from valid algebraic working rather than simply stated.

    A1
  4. 04

    Both values of x² are negative, so x⁴ + 8x² + 15 = 0 has no real roots — which is not possible, so C₁ and C₂ do not intersect.

    The concluding contradiction — explicitly ties the impossibility of a real x² back to the original assumption. Dependent on the preceding M1A1 having been scored.

    A1