Pure Mathematics 3

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 · Q7(a)

4 marks

Product rule + chain rule on a square-root factor, simplified to one fraction

f(x) = x³(4x + 7)^(1/2). Show that f'(x) = 7x²(2x + 3) / √(4x + 7).

Our analysis

f(x) = x³(4x + 7)^(1/2). Show that f'(x) = 7x²(2x + 3) / √(4x + 7).

4 marks available

  1. 01

    f'(x) = 3x²(4x+7)^(1/2) + x³ · (1/2)(4x+7)^(-1/2) · 4

    Product rule applied: (uv)' = u'v + uv', with the chain rule correctly applied inside v' = (4x+7)^(1/2).

    M1
  2. 02

    f'(x) = 3x²(4x+7)^(1/2) + 2x³(4x+7)^(-1/2)

    Correct unsimplified derivative — both terms present with correct coefficients, before any combining.

    A1
  3. 03

    f'(x) = [3x²(4x+7) + 2x³] / (4x+7)^(1/2) = [12x³ + 21x² + 2x³] / (4x+7)^(1/2)

    Dependent on the first M1 — combines both terms into a single fraction over a common denominator of √(4x+7). This is the step examiners report most candidates lose: getting two separate correct-looking terms is not the same as combining them.

    dM1
  4. 04

    f'(x) = (14x³ + 21x²) / √(4x+7) = 7x²(2x+3) / √(4x+7)

    Fully factorised final form. The examiner report specifically flags this factorisation step, not the differentiation itself, as the more discriminating mark.

    A1

Oct 2024 · Q8(c)

5 marks

Maximum of a sum of two exponential decay terms with different rate constants

H = 32 + 40e^(-0.2t) - 20e^(-0.9t), where H is a horse's heart rate (beats per minute) t minutes after exercise. Find the time at which H is a minimum, and find this minimum value.

Our analysis

H = 32 + 40e^(-0.2t) - 20e^(-0.9t). Find the time at which H is a minimum, and find this minimum value.

5 marks available

  1. 01

    dH/dt = 40(-0.2)e^(-0.2t) - 20(-0.9)e^(-0.9t) = -8e^(-0.2t) + 18e^(-0.9t)

    Differentiates a sum of two exponential terms with different rate constants term-by-term — the same chain rule move as a single exponential term, just applied twice and kept separate, not combined yet.

    M1
  2. 02

    -8e^(-0.2t) + 18e^(-0.9t) [confirmed exact coefficients]

    Correct derivative with both real coefficients.

    A1
  3. 03

    -8e^(-0.2t) + 18e^(-0.9t) = 0 ⟹ 18e^(-0.9t) = 8e^(-0.2t) ⟹ e^(-0.7t) = 8/18

    Dependent on the first M1 — sets the derivative to zero and combines the two DIFFERENT exponential terms into a single exponential equation by dividing one side by the other, collapsing e^(-0.2t) and e^(-0.9t) into one e^(-0.7t) term. This combining step is the specific technique a single-exponential-term model never requires.

    dM1
  4. 04

    -0.7t = ln(8/18) ⟹ t = ln(8/18) / -0.7

    Solves the single exponential equation using correct log work — taking ln of both sides once it's down to one exponential term.

    M1
  5. 05

    t = awrt 1.158 (minutes)

    Correct final value to the accuracy the mark scheme requires.

    A1