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TMUA 2021 · Paper 1 · Question 14 of 20

TMUA 2021 Paper 1 Question 14

Trigonometry — Counting solutions · bounding both sides of an equation. Try it first; the answer and a full worked solution are below.

TMUA 2021 · Paper 1TrigonometryCounting solutions · bounding both sides of an equation6 optionshard
This question uses radians.

Find the number of distinct values of x that satisfy the equation (x+1)(3−x) = 2(1 − cos(π x))

  1. A2
  2. B3
  3. C4
  4. D5
  5. E6
  6. F7
Show the answer and worked solution
answer · B
  1. A2
  2. B3
  3. C4
  4. D5
  5. E6
  6. F7
Complete the square on the left: (x+1)(3−x) = 4 − (x−1)2, a downward parabola peaking at 4 when x = 1. The right side is 2 − 2cos(π x), which oscillates between 0 and 4 and reaches 4 exactly at odd integers. Substituting t = x − 1 and using cos(π x) = −cos(π t) turns the equation into 2 − t2 − 2cos(π t) = 0, which is even in t. At t = 0 both sides equal 4, so x = 1 is a solution; for small t the expression behaves like (π2 − 1)t2 > 0, and it stays positive up to t = 1 (where it is 3) before turning negative by t = 1.5. So there is exactly one further root with t > 0, near t ≈ 1.43, and by symmetry one with t < 0: three distinct values of x in all.