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TMUA 2021 · Paper 2 · Question 20 of 20

TMUA 2021 Paper 2 Question 20

Differentiation and integration — Iterated modulus functions · integrating the result. Try it first; the answer and a full worked solution are below.

TMUA 2021 · Paper 2Differentiation and integrationIterated modulus functions · integrating the result8 optionshard
A sequence of functions f1, f2, f3, … is defined by f1(x) = |x| fn+1(x) = |fn(x) + x|   for  n ≥ 1 Find the value of ∫−11 f99(x) dx
  1. A0
  2. B0.5
  3. C1
  4. D49.5
  5. E50
  6. F99
  7. G99.5
  8. H100
Show the answer and worked solution
answer · E
  1. A0
  2. B0.5
  3. C1
  4. D49.5
  5. E50
  6. F99
  7. G99.5
  8. H100
Split at zero and iterate on each side. For x ≥ 0, f1(x) = x and each step adds another x with nothing to cancel, so fn(x) = nx and f99(x) = 99x. For x < 0 the pattern collapses instead of growing: f1(x) = −x, then f2(x) = |−x + x| = 0, then f3(x) = |0 + x| = −x, and so on alternating. Since 99 is odd, f99(x) = −x there. So the integral is ∫−10 (−x) dx + ∫01 99x dx = 12 + 992 = 50.