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If I = ∫ d x x 2 - 2 x + 5 = 1 2 tan ⁡ - 1 f x + C (where, C is the constant of integration) and f 2 = 1 2 , then the maximum value of y = f sin ⁡ x   ∀ x ∈ R is

Options

  1. A4
  2. B2
  3. C0
  4. D- 1

Correct answer

C. 0

Step-by-step solution

The given integral is I = ∫ d x x - 1 2 + 4 = 1 2 tan - 1 ⁡ x - 1 2 + c ∴ f x = x - 1 2 i.e. f sin ⁡ x = sin ⁡ x - 1 2 Hence, max f sin ⁡ x = 0

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