QJEEM24Maf999: JEE Main 2024 Mathematics - Class 12 JEEM24Maf999 Solution: Let for a differentiable function . Then is equal to…

$\begin{aligned}& f(x)-f(y) \geq \ln x-\ln y+x-y \\& \frac{f(x)-f(y)}{x-y} \geq \frac{\ln x-\ln y}{x-y}+1\end{aligned}$

Let x>y

$$\underset{y\to x}{lim}{f}^{\prime }\left(x^{-}\right)\ge \frac{1}{x}+1\text{.... (1)}$$

Let x< y

$$\underset{y\to x}{lim}{f}^{\prime }\left(x^{+}\right)\le \frac{1}{x}+1\text{.... (2)}$$

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

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