Let f ⁡ : 1 2 1 ⟶ R (the set of all real numbers) be a positive, non-constant and differentiable function such that f ' x < 2 f ⁡ x and f ⁡ 1 2 = 1 . Then the value of ∫ 1 / 2 1 f ⁡ x dx lies in the interval:

Step-by-step solution

f ' x < 2 f x ⇒ f ' x f x < 2 ⇒ ∫ 1 2 x f ' x f x d x = ∫ 1 2 x 2 d x ⇒ ln f x < 2 x - 1 ⇒ f x < e 2 x - 1 ⇒ 0 < ∫ 1 2 1 f x d x < ∫ 1 2 1 e 2 x - 1 d x ⇒ 0 < ∫ 1 2 1 f x d x < e 2 x - 1 2 1 2 1 ⇒ 0 < ∫ 1 2 1 f x d x < e - 1 2