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TMUA 2020 · Paper 1 · Question 8 of 20

TMUA 2020 Paper 1 Question 8

Algebra and functions — Maximum of a quadratic · range of a parameter. Try it first; the answer and a full worked solution are below.

TMUA 2020 · Paper 1Algebra and functionsMaximum of a quadratic · range of a parameter6 options
The function f is defined for all real x as f(x) = (p − x)(x + 2) Find the complete set of values of p for which the maximum value of f(x) is less than 4.
  1. A−2 − 4√2 < p < −2 + 4√2
  2. B−2 − 2√2 < p < −2 + 2√2
  3. C−2√5 < p < 2√5
  4. D−6 < p < 2
  5. E−4 < p < 0
  6. F−2 < p < 2
Show the answer and worked solution
answer · D
  1. A−2 − 4√2 < p < −2 + 4√2
  2. B−2 − 2√2 < p < −2 + 2√2
  3. C−2√5 < p < 2√5
  4. D−6 < p < 2
  5. E−4 < p < 0
  6. F−2 < p < 2
The roots are x = p and x = −2, and the x2 coefficient is −1, so the parabola opens downwards and its maximum sits midway between the roots, at x = p−22. The value there is (p − p−22)(p−22 + 2) = p+22⋅p+22 = (p+2)24. Requiring (p+2)24 < 4 gives |p+2| < 4, so −6 < p < 2.