JEE Main 2024 (Online) 27th January Morning ShiftQuestion 1
Let for a differentiable function $f:(0,\infty)\to\mathbf{R}$, $f(x)-f(y)\ge\log_e\left(\frac{x}{y}\right)+x-y$, $\forall x,y\in(0,\infty)$. Then $\sum_{n=1}^{20}f'\left(\frac{1}{n^{2}}\right)$ is equal to $\_\_\_\_\_\_$ .
Differentiation and its Application
Answer$2890$
Solution
Concept. With $g(x)=f(x)-\ln x-x$ the condition says $g(x)\ge g(y)$ for all $x,y$; swapping $x,y$ gives equality, so $g$ is constant. Formulas used. (1) $\sum_{n=1}^{N}n^2=\frac{N(N+1)(2N+1)}{6}$Step 1 - $g$ is constant. $g(x)\ge g(y)$ and $g(y)\ge g(x)$, so $g(x)=g(y)$ for all $x,y$. Step 2 - $f'$. $f(x)=\ln x+x+C$, so $f'(x)=\frac1x+1$ and $f'\left(\frac1{n^2}\right)=n^2+1$. Step 3 - Sum. By (1), $$\sum_{n=1}^{20}\left(n^2+1\right)=\frac{20\cdot21\cdot41}{6}+20=2870+20=2890.$$ Answer: 2890