Let f x be a continuous and positive function, such that the area bounded by y = f x ,   x -axis and the lines x 2 = 2 a x is 6 a 2 + sin ⁡ a   ∀ a > 0 sq. units. If f π = k π , then the value of k is equal to

Step-by-step solution

x 2 - 2 a x = 0 ⇒ x = 0 , 2 a So, the given area is ∫ 0 2 a f x d x ∴ ∫ 0 2 a f x d x = 6 a 2 + sin ⁡ a Differentiating with respect to a , we get, 2 f 2 a = 12 a + cos ⁡ a ⇒ f x = 3 x + 1 2 cos x 2 ⇒ f π = 3 π ⇒ k = 3