Let F x =   ∫ x x 2 + π 6 2 cos 2 ⁡ t   d t for all x   ∈ R   a n d   f   :   0 , 1 2 → 0 ,   ∞ be a continuous function. For a ∈ 0 , 1 2 , if F ′ a + 2 is the area of the region bounded by x = 0 ,   y = 0 ,   y = f x   a n d   x = a , then f ( 0 ) is

Step-by-step solution

F ′ a + 2 = ∫ 0 a f x d x ⇒ F ′ ′ a = f a ...(i) ∵ F x = ∫ x x 2 + π 6 2 cos 2 ⁡ t   d t ∴ F ′ x = 2 cos 2 ⁡ x 2 + π 6 . 2 x - 2 cos 2 ⁡ x F ′ ′ x = 4 cos 2 ⁡ x 2 + π 6 - 8 x cos ⁡ x 2 + π 6 . sin ⁡ x 2 + π 6 2 x + 4 cos ⁡ x sin ⁡ x f 0 = F ′ ′ 0 = 4 × 3 4 = 3