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TMUA 2016 · Paper 1 · Question 15 of 20

TMUA 2016 Paper 1 Question 15

Differentiation and integration — Gradient as a quadratic in a parameter · completing the square. Try it first; the answer and a full worked solution are below.

TMUA 2016 · Paper 1Differentiation and integrationGradient as a quadratic in a parameter · completing the square5 options
The least possible value of the gradient of the curve y = (2x + a)(x − 2a)2 at the point where x = 1, as a varies, is
  1. A−494
  2. B−8
  3. C−254
  4. D74
  5. E4716
Show the answer and worked solution
answer · C
  1. A−494
  2. B−8
  3. C−254
  4. D74
  5. E4716
Differentiate with the product rule, treating a as a constant: dydx = 2(x−2a)2 + (2x+a)⋅ 2(x−2a) = 2(x−2a)[(x−2a) + (2x+a)] = 2(x−2a)(3x−a). At x = 1 this is 2(1−2a)(3−a) = 4a2 − 14a + 6. That is now a quadratic in a, minimised at a = 74, where its value is 4⋅4916 − 492 + 6 = −494 + 6 = −254.