The total number of distinct x ∈ 0 , 1 for which, ∫ 0 x t 2 1 + t 4 d t = 2 x - 1 is

Step-by-step solution

Let f x = ∫ 0 x t 2 1 + t 4 d t - 2 x + 1 ⇒ f ′ x = x 2 1 + x 4 - 2 As 1 + x 4 x 2   ≥ 2 ⇒ x 2 1 + x 4 ≤ 1 2 ⇒ f ′ x   ≤ - 3 2 ⇒ f x is continuous and decreasing f 0 = 1 and f 1 = ∫ 0 1 t 2 1 + t 4 d t - 1 ≤ -1 2 By intermediate value theorem (IVT), f x = 0 possesses exactly one solution in [0, 1]. As f ( 0 ) f ( 1 ) < 0