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TMUA 2021 · Paper 1 · Question 12 of 20

TMUA 2021 Paper 1 Question 12

Algebra and functions — Minimum of a quartic · substitution to a quadratic. Try it first; the answer and a full worked solution are below.

TMUA 2021 · Paper 1Algebra and functionsMinimum of a quartic · substitution to a quadratic6 options
The minimum value of the function x4 − p2x2 is −9, where p is a real number.

Find the minimum value of the function x2 − px + 6.

  1. A−3
  2. B6 − 3√22
  3. C32
  4. D3
  5. E92
  6. F6 + 3√22
Show the answer and worked solution
answer · E
  1. A−3
  2. B6 − 3√22
  3. C32
  4. D3
  5. E92
  6. F6 + 3√22
Put u = x2, so the quartic becomes u2 − p2 u = (u − p22)2 − p44, with minimum −p44 attained at u = p22, which is a legitimate value of x2. Hence p44 = 9, so p4 = 36 and p2 = 6. The second function completes the square as (x − p2)2 + 6 − p24, with minimum 6 − 64 = 92. Only p2 is needed, so the sign ambiguity in p does not matter.