JEE MAIN - Mathematics (2024 - 27th January Morning Shift - No. 25)
Let for a differentiable function $f:(0, \infty) \rightarrow \mathbf{R}, f(x)-f(y) \geqslant \log _{\mathrm{e}}\left(\frac{x}{y}\right)+x-y, \forall x, y \in(0, \infty)$. Then $\sum\limits_{n=1}^{20} f^{\prime}\left(\frac{1}{n^2}\right)$ is equal to ____________.
Answer
2890
Explanation
$$\begin{aligned} & f(x)-f(y) \geq \ln x-\ln y+x-y \\ & \frac{f(x)-f(y)}{x-y} \geq \frac{\ln x-\ln y}{x-y}+1 \end{aligned}$$
Let $$x>y$$
$$\lim _\limits{y \rightarrow x} f^{\prime}\left(x^{-}\right) \geq \frac{1}{x}+1\quad\text{.... (1)}$$
Let $$x< y$$
$$\lim _\limits{y \rightarrow x} f^{\prime}\left(x^{+}\right) \leq \frac{1}{x}+1 \quad\text{.... (2)}$$
$$\begin{aligned} & \mathrm{f}^1\left(\mathrm{x}^{-}\right)=\mathrm{f}^1\left(\mathrm{x}^{+}\right) \\ & \mathrm{f}^1(\mathrm{x})=\frac{1}{\mathrm{x}}+1 \\ & \mathrm{f}^{\prime}\left(\frac{1}{\mathrm{x}^2}\right)=\mathrm{x}^2+1 \end{aligned}$$
$$\begin{aligned} & \sum_{x=1}^{20}\left(x^2+1\right)=\sum_{x-1}^{20} x^2+20 \\ & =\frac{20 \times 21 \times 41}{6}+20 \\ & =2890 \end{aligned}$$
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