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TMUA 2017 · Paper 1 · Question 10 of 20

TMUA 2017 Paper 1 Question 10

Differentiation and integration — Gradient of a normal · optimising over a parameter. Try it first; the answer and a full worked solution are below.

TMUA 2017 · Paper 1Differentiation and integrationGradient of a normal · optimising over a parameter6 optionshard
A curve C has equation y = f(x) where f(x) = p3 − 6p2x + 3px2 − x3 and p is real.

The gradient of the normal to the curve C at the point where x = −1 is M.

What is the greatest possible value of M as p varies?

  1. A−32
  2. B−23
  3. C−12
  4. D14
  5. E23
  6. F32
Show the answer and worked solution
answer · E
  1. A−32
  2. B−23
  3. C−12
  4. D14
  5. E23
  6. F32
Differentiate with respect to x, treating p as a constant: f'(x) = −6p2 + 6px − 3x2. At x = −1 the tangent gradient is −6p2 − 6p − 3, so M = −1−6p2−6p−3 = 16p2 + 6p + 3. Completing the square, 6p2 + 6p + 3 = 6(p + 12)2 + 32, which is always positive with least value 32 at p = −12. A positive denominator at its smallest makes M largest, so the greatest value of M is 13/2 = 23.