Let f be a differentiable function defined on 0 , π 2 such that f x > 0 and f x + ∫ 0 x f t 1 - log e f t 2 d t = e ∀ x ∈ 0 , π 2 , then 6 log e f π 6 2 is equal to

Step-by-step solution

Given: f x + ∫ 0 x f t 1 - log e f t 2 d t = e       . . . 1 Put x = 0 , then f 0 = e . Differentiating 1 w.r.t. x , we get f ' x + f x 1 - log e f x 2 = 0 Put y = f x . d y d x + y 1 - log e y 2 = 0 ⇒ ∫ d y y 1 - log e y 2 = - ∫ d x ⇒ ∫ d log e y 1 - log e y 2 = - ∫ d x ⇒ sin - 1 log e y = - x + C Put x = 0 ,   y = e sin - 1 log e e = - 0 + C ⇒ C = π 2 So, sin - 1 log e y = - x + π 2 ⇒ log e y = sin - x + π 2 ⇒