Let f : R → R be a differentiable function such that f π 4 = 2 , f π 2 = 0 and f ' π 2 = 1 and let g x = ∫ x π 4 f ' t sec t + tan t sec t f t d t for x ∈ π 4 , π 2 . Then lim x → π 2 - g x is equal to

Step-by-step solution

Given, g x = ∫ x π 4 f ' t sec t + tan t sec t f t d t ⇒ g x = ∫ x π 4 d f t · sec t ⇒ g x = f t sec t x π 4 g x = f π 4 sec π 4 - f x · sec x g x = 2 - f x sec x = 2 - f x cos x Now taking limit both side, we get lim x → π 2 - g x = 2 - lim x → π 2 - f x cos x Using L'Hospital Rule = 2 - lim x → π 2 - f ' x - sin x = 2 + f ' π 2 sin π 2 = 2 + 1 1 = 3