Question

If f(x) is differentiable function satisfying f(x) – f(y) ≥ log \(\frac {x}{y}+x-y,\) then find \(\displaystyle\sum_{N=1}^{20}f'\) \(\left(\frac {1}{N^2}\right)\)

Answer 1

Correct answer is 2890

⇒ f'(x^-) = f'(x⁺) as f(x) is differentiable function

\(f'(x) = \frac {1}{x} + 1\)

\(f'\left(\frac {1}{N^2}\right) =N^2 + 1\)

\(\displaystyle\sum_{N=1}^{20}f'\left(\frac {1}{N^2}\right )\) \(=\sum (N^2 + 1) = \frac {20 \times 21 \times 41 }{6}\) + 20 = 2890