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TMUA 2019 · Paper 1 · Question 13 of 20

TMUA 2019 Paper 1 Question 13

Exponentials and logarithms — Hidden quadratic in 2sin x · range of the substitution. Try it first; the answer and a full worked solution are below.

TMUA 2019 · Paper 1Exponentials and logarithmsHidden quadratic in 2sin x · range of the substitution6 options
Find the maximum value of 4sin x − 4 × 2sin x + 174 for real x.
  1. A14
  2. B52
  3. C132
  4. D212
  5. E654
  6. FThere is no maximum value.
Show the answer and worked solution
answer · B
  1. A14
  2. B52
  3. C132
  4. D212
  5. E654
  6. FThere is no maximum value.
Substitute u = 2sin x, so that 4sin x = u2 and the expression becomes g(u) = u2 − 4u + 174. The crucial step is the range of u: since sin x ∈ [−1,  1], u runs over [12,  2], not all of ℝ. The parabola g has its vertex at u = 2, so on this interval it is decreasing and the maximum is at the left end: g!(12) = 14 − 2 + 174 = 52.