Let f be a twice differentiable function on R . If f ' 0 = 4 and f x + ∫ 0 x x - t f ' t d t = e 2 x + e - 2 x cos 2 x + 2 a x , then 2 a + 1 5 a 2 is equal to _______.

Step-by-step solution

Given, f x + ∫ 0 x x - t f ' t d t = e 2 x + e - 2 x cos 2 x + 2 x a           ⋯ i Putting x = 0 both side we get, f 0 = 2           ⋯ ii On differentiating equation i w.r.t. x we get: f ' x + ∫ 0 x f ' t d t + x f ' x - x f ' x = 2 e 2 x - e - 2 x cos 2 x - 2 e 2 x + e - 2 x sin 2 x + 2 a ⇒ f ' x + f x - f 0 = 2 e 2 x - e - 2 x cos 2 x - 2 e 2 x + e - 2 x sin 2 x + 2 a Replace x by 0 we get: ⇒ 4 = 2 a  &#