Let f : 0 , 1 → 0 , 1 be a continuous function such that x 2 + f x 2 ≤ 1 for all x ∈ 0 , 1 and ∫ 0 1 f x d x = π 4 Then, ∫ 1 2 1 2 f x 1 - x 2 d x equals

Step-by-step solution

For a continuous function f : 0 , 1 ⟶ 0 , 1 it is given that x 2 + f x 2 ≤ 1 ⇒ f x 2 ≤ 1 - x 2 ⇒ f x ≤ 1 - x 2    ∵ f x ∈ 0 , 1 ∴ ∫ 0 1 f x d x ≤ ∫ 0 1 1 - x 2 d x ⇒ ∫ 0 1 f x d x ≤ x 2 1 - x 2 + 1 2 sin - 1 x 0 1 ⇒ ∫ 0 1 f x d x ≤ π 4 So, f x = 1 - x 2 , because it is given that ∫ 0 1 f x = π 4 . Now, ∫ 1 / 2 1 2 f x 1 - x 2 d x = ∫ 1 / 2 1 2 d x 1 - x 2 = sin - 1 x 1 /