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TMUA 2021 · Paper 1 · Question 11 of 20

TMUA 2021 Paper 1 Question 11

Differentiation and integration — Differentiation with fractional powers · where a function decreases. Try it first; the answer and a full worked solution are below.

TMUA 2021 · Paper 1Differentiation and integrationDifferentiation with fractional powers · where a function decreases7 options
The function f is given by f(x) = x17(x2 − x + 1) Find the fraction of the interval 0 < x < 1 for which f(x) is decreasing.
  1. A215
  2. B15
  3. C13
  4. D12
  5. E23
  6. F45
  7. G1315
Show the answer and worked solution
answer · A
  1. A215
  2. B15
  3. C13
  4. D12
  5. E23
  6. F45
  7. G1315
Multiply out before differentiating: f(x) = x157 − x87 + x17, so f'(x) = 157x87 − 87 x17 + 17 x−67. Take out the common factor 17 x−67, which is positive for x > 0, leaving f'(x) = 17x−67(15x2 − 8x + 1). The sign is therefore the sign of (5x − 1)(3x − 1), which is negative exactly for 15 < x < 13. That interval has length 13 − 15 = 215, and the whole interval has length 1.