JEE Advanced Mathematics Question (2023) — Solution

Step-by-step solution

Given, f : 0 ,   1 → ℝ be the function defined as f x = n if x ∈ [ 1 n + 1 ,   1 n ) where n ∈ ℕ And  g : 0 ,   1 → ℝ be a function such that ∫ x 2 x 1 - t t d t < g x < 2 x for all x ∈ 0 ,   1 .  Now we need to solve 1 sided limit here to get some answer, as   lim x → 0 -  doesn’t exist here (not in domain) As  1 n + 1 ≤ x <   1 n ⇒ n + 1 ≥ 1 x >   n ⇒ n ≥ 1 x - 1 ⇒ n ≥ 1 x - 1 So, let  f x = 1 x - 1  where  · =  least integer function Now multiplying  f x  in  ∫ x 2 x 1 - t t d t < g x < 2 x  and taking limit we get, lim x → 0 + ∫ x 2 x 1 - t t d t · 1 x - 1 ≤ lim x → 0 + f x · g x ≤ lim x → 0 + 1 x - 1 × 2 x Now  lim x → 0 + 1 x - 1 × 2 x = lim x → 0 + 2 x 1 x    1 x ∉ Z = lim x → 0 + 2 x 1 x - 1 x = 2 = lim x → 0 + 2 1 - x 1 x = 2 ;   1 x ∉ Z So,  lim x → 0 + ∫ x 2 x 1 - t t d t · 1 x - 1 x = ∫ x 2 x 1 - t t d t · 1 - x 1 x x And  lim x → 0 + ∫ x 2 x 1 - t t d t x = lim x → 0 + 1 - x x - 2 x 1 - x 2 x 2 1 2 x Using L-hospital rule = lim x → 0 + 2 1 - x - 4 x · 1 - x 2 = 2 Similarly for  1 x ∈ Z  is equal to 2 .