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TMUA 2021 · Paper 1 · Question 4 of 20

TMUA 2021 Paper 1 Question 4

Exponentials and logarithms — Exponentials · hidden quadratic and its minimum. Try it first; the answer and a full worked solution are below.

TMUA 2021 · Paper 1Exponentials and logarithmsExponentials · hidden quadratic and its minimum6 options
Find the minimum value of the function 22x − 2x+3 + 4
  1. A−16
  2. B−12
  3. C−8
  4. D0
  5. E4
  6. F20
Show the answer and worked solution
answer · B
  1. A−16
  2. B−12
  3. C−8
  4. D0
  5. E4
  6. F20
Put t = 2x, so 22x = t2 and 2x+3 = 8t. The function becomes t2 − 8t + 4 = (t−4)2 − 12, a quadratic in t with minimum −12 at t = 4. The value t = 4 is attainable, since t = 2x ranges over all positive reals and x = 2 gives it. So the minimum is −12; the restriction t > 0 only matters when the vertex sits at a non-positive t, which it does not here.