Let for a differentiable function f : ( 0 , ∞ ) → R , f x - f y ≥ log e x y + x - y , ∀ x , y ∈ 0 , ∞ . Then ∑ n = 1 20 f ' 1 n 2 is equal to

Step-by-step solution

Given: f x - f y ≥ log e x y + x - y Now, taking x = x + h & y = x we get, f x + h - f x ≥ log e x + h x + x + h - x f x + h - f x ≥ log e x + h x + h Now, using f ' x = lim h → 0 f x + h - f x h ⇒ f ' x = lim h → 0 log e x + h x + h h ⇒ f ' x = lim h → 0 log e 1 + h x x · h x + 1 ⇒ f ' x = 1 x + 1 ⇒ f ' 1 x 2 = x 2 + 1 ⇒ ∑ x = 1 20 f ' 1 x 2 = ∑ x = 1 20 x 2 + 1 ⇒ ∑ x = 1 20 f ' 1 x 2 = ∑ x = 1 20 x 2 + 20 ⇒ ∑ x = 1 20 f ' 1 x 2 = 20 × 21 × 41 6 + 20 ⇒ ∑ x = 1 20 f ' 1 x 2 = 2870 + 20 ⇒ ∑ x = 1 20 f ' 1 x 2 = 2890