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

Question

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, 4, 2, 0, - 1

MARKS App · Accepted Answer

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

Search any question & find its solution

Looking for more such questions to practice?