Let f be a differentiable function in 0 , π 2 . If ∫ cos x 1 t 2 f t d t = sin 3 x + cos x , then 1 3 f ' 1 3 is equal to

Step-by-step solution

∫ cos x 1 t 2 f t d t = sin 3 x + cos x On differentiating, we get f cos x sin x · cos 2 x = 3 sin 2 x cos x - sin x f cos x = 3 tan x - sec 2 x Again differentiating, we get - sin x f ' cos x = 3 sec 2 x - 2 sec 2 x tan x When cos x = 1 3 then sec x = 3 , tan x = 2   &   sin x = 2 3 Then - 2 3 f ' 1 3 = 3 × 3 - 2 × 3 2 ⇒ 1 3 f ' 1 3 = 6 - 9 2