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TMUA 2019 · Paper 1 · Question 7 of 20

TMUA 2019 Paper 1 Question 7

Differentiation and integration — Differentiation · minimising a gradient over a parameter. Try it first; the answer and a full worked solution are below.

TMUA 2019 · Paper 1Differentiation and integrationDifferentiation · minimising a gradient over a parameter7 options
A curve has equation y = (2q − x2)(2qx + 3) The gradient of the curve at x = −1 is a function of q.

Find the value of q which minimises the gradient of the curve at x = −1.

  1. A−1
  2. B−34
  3. C−12
  4. D0
  5. E12
  6. F34
  7. G1
Show the answer and worked solution
answer · F
  1. A−1
  2. B−34
  3. C−12
  4. D0
  5. E12
  6. F34
  7. G1
Differentiate with the product rule, treating q as a constant: dydx = −2x(2qx + 3) + 2q(2q − x2). Substituting x = −1 gives 2(−2q + 3) + 2q(2q − 1) = 4q2 − 6q + 6. This is a quadratic in q opening upwards, so its least value is at the vertex q = 68 = 34. Expanding the brackets before differentiating also works, but is more error-prone.