JEE Advanced Mathematics Question (2023) — Solution

Step-by-step solution

Given, f x = ∫ 0 x tan - 1 x e t - cos t 1 + t 2023 d t Now differentiating both side we get, f ' x = e x tan - 1 x - cos x tan - 1 x 1 + x tan - 1 x 2023 × x 1 + x 2 + tan - 1 x ⇒ f ' x = g x · h x  where  g x = e x tan − 1 ⁡ x − cos ⁡ x tan − 1 ⁡ x 1 + x tan − 1 ⁡ x 2023 > 0   ∀   x And  h x = x 1 + x 2 + tan − 1      x   which is   < 0  for  x < 0 = 0      x = 0 > 0      x > 0 ∴ f x has minimum at x = 0 And f x m i n = f 0 = 0