Pure Mathematics 2

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 2021 · Q10(i)

5 marks

tan² equation solved by taking the square root then arctan, across a full symmetric interval

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. Solve, for −π/2 < x < π/2, tan²(2x + π/4) = 3.

Our analysis

Solve, for −π/2 < x < π/2, tan²(2x + π/4) = 3.

5 marks available

  1. 01

    tan²(2x+π/4) = 3 ⟹ tan(2x+π/4) = ±√3 ⟹ 2x+π/4 = ±π/3, ±2π/3

    Correct order of operations — takes the square root of the whole equation FIRST (producing the ± that the rest of the question depends on), then arctan, rather than any other order.

    M1
  2. 02

    x = (2x+π/4 value − π/4) ÷ 2, e.g. x = (π/3 − π/4) ÷ 2

    Dependent on the first M1 — a complete attempt to isolate x by moving the π/4 across before dividing by 2.

    dM1
  3. 03

    x = π/24 [from 2x+π/4 = π/3]

    One correct value from the first solution pairing (π/24 or −11π/24).

    A1
  4. 04

    x = 5π/24 [from 2x+π/4 = 2π/3]

    One correct value from the second solution pairing (5π/24 or −7π/24).

    A1
  5. 05

    x = π/24, 5π/24, −7π/24, −11π/24

    All four values found, and no extra values outside the given range −π/2 < x < π/2 included.

    A1

Oct 2019 · Q10

7 marks

Finding cubic coefficients from a point and gradient, then a discriminant-based 'prove no turning points'

The curve C has equation y = ax³ − 3x² + 3x + b, where a and b are constants. Given that: the point (2, 5) lies on C; the gradient of the curve at (2, 5) is 7; (a) find the value of a and the value of b. (4) (b) Prove that C has no turning points. (3)

Our analysis

The curve C has equation y = ax³ − 3x² + 3x + b. Given (2,5) lies on C with gradient 7 there: (a) find a and b. (b) Prove that C has no turning points.

7 marks available

a — 4 marks

  1. 01

    x=2, y=5 into y=ax³−3x²+3x+b: 5 = 8a − 12 + 6 + b ⟹ 8a + b = 11

    Substitutes the given point into the curve's equation to form an equation in a and b — slips in the arithmetic are condoned for this mark.

    M1
  2. 02

    dy/dx = 3ax² − 6x + 3; x=2, dy/dx=7: 7 = 12a − 12 + 3 ⟹ 12a − 9 = 7

    Differentiates correctly and substitutes x=2 with the given gradient of 7 to form an equation purely in a — this equation itself must be correct, unlike the first M1.

    M1
  3. 03

    12a = 16 ⟹ a = 4/3

    Correct value of a from solving the two simultaneous equations.

    A1
  4. 04

    8(4/3) + b = 11 ⟹ 32/3 + b = 11 ⟹ b = 1/3

    Correct value of b, following from a correct a.

    A1

b — 3 marks

  1. 101

    dy/dx = 3(4/3)x² − 6x + 3 = 0 ⟹ 4x² − 6x + 3 = 0

    Sets dy/dx = 3ax² − 6x + 3 equal to zero using their own value of a — this explicit '= 0' step is what the examiner report calls the rigour a 'prove that' question specifically requires, not just an implied step.

    M1
  2. 102

    b² − 4ac = (−6)² − 4(4)(3) = 36 − 48 = −12

    Dependent on the first M1 — attempts the discriminant (or the roots via the quadratic formula) of the quadratic just formed, not of the original unset expression.

    dM1
  3. 103

    b² − 4ac = −12 < 0, so 4x² − 6x + 3 = 0 has no real roots, hence dy/dx is never zero, hence C has no turning points

    States the numerical discriminant, that it is negative, and draws the explicit conclusion (no real roots ⟹ dy/dx never zero ⟹ no turning points) — the starred mark, since the result to reach was printed on the question paper.

    A1*