JEE Main 2024 — Application Of Derivatives Question with Solution

$$\begin{array}{rl}& f(x)-f(y)\ge \ln x-\ln y+x-y\\ & \frac{f(x)-f(y)}{x-y}\ge \frac{\ln x-\ln y}{x-y}+1\end{array}$$

Let $$x>y$$

\lim _\limits{y \rightarrow x} f^{\prime}\left(x^{-}\right) \geq \frac{1}{x}+1\quad\text{.... (1)}

Let $$x<y$$

\lim _\limits{y \rightarrow x} f^{\prime}\left(x^{+}\right) \leq \frac{1}{x}+1 \quad\text{.... (2)}

$$\begin{array}{rl}& {\mathrm{f}}^1\left({\mathrm{x}}^{-}\right)={\mathrm{f}}^1\left({\mathrm{x}}^{+}\right)\\ & {\mathrm{f}}^1(\mathrm{x})=\frac{1}{\mathrm{x}}+1\\ & {\mathrm{f}}^{\prime }\left(\frac{1}{{\mathrm{x}}^2}\right)={\mathrm{x}}^2+1\end{array}$$

$$\begin{array}{rl}& \sum _{x=1}^{20}\left(x^2+1\right)=\sum _{x-1}^{20}{x}^2+20\\ & =\frac{20\times 21\times 41}{6}+20\\ & =2890\end{array}$$