Let a differentiable function f satisfy f x + ∫ 3 x f t t d t = x + 1 , x ≥ 3 . Then 12 f 8 is equal to:

Step-by-step solution

Given: f x + ∫ 3 x f t t d t = x + 1 ⇒ f ' x + f x x = 1 2 x + 1 Put y = f x , then d y d x + y x = 1 2 x + 1 So, I . F . = e ∫ d x x = e ln x = x Hence, solution is x y = 1 2 ∫ x x + 1 d x ⇒ x y = 1 2 ∫ x + 1 - 1 x + 1 d x ⇒ x y = 1 2 ∫ x + 1 - 1 x + 1 d x ⇒ x y = 1 2 2 3 x + 1 3 2 - 2 x + 1 + C ⇒ x y = 1 3 x + 1 3 2 - x + 1 + C Put x = 3 , then f 3 = 2 So, 6 = 8 3 - 2 + C ⇒ C = 16 3 Hence, x y = 1 3 x + 1 3 2 - x + 1 + 16 3 So, put x = 8 , then