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TMUA 2023 · Paper 1 · Question 1 of 20

TMUA 2023 Paper 1 Question 1

Differentiation and integration — Definite integrals · simultaneous conditions. Try it first; the answer and a full worked solution are below.

TMUA 2023 · Paper 1Differentiation and integrationDefinite integrals · simultaneous conditions7 options
Given that ∫01 (ax + b) dx = 1 and ∫01 x(ax + b) dx = 1 find the value of a + b.
  1. A−1
  2. B0
  3. C1
  4. D2
  5. E3
  6. F4
  7. G5
Show the answer and worked solution
answer · F
  1. A−1
  2. B0
  3. C1
  4. D2
  5. E3
  6. F4
  7. G5
The first integral gives [ax22 + bx]01 = a2 + b = 1. The second is ∫01 (ax2 + bx) dx = a3 + b2 = 1. Substituting b = 1 − a2 into the second gives a3 + 12 − a4 = 1, so a12 = 12 and a = 6. Then b = −2, and a + b = 4.