Let for a differentiable function , Then is equal to

2890 Explanation: f(x)-f(y)\geq\log_e\left(\frac{x}{y}\right)+x-y \begin{aligned} & \Rightarrow f(x)-f(y)\geq\log x-\log x+x-y \\ & \Rightarrow f(x)-\log x-x\geq f(y)-\log y-y \\ & \text{Interchanging x and y} \\ & f(y)-\log y-y\geq f(x)-\log x-x \\ & \mathrm{So},f(x)=\log x+x+\lambda \\ & \Rightarrow f^{\prime}(x)=\frac{1}{x}+1 & & ...(\mathrm{i}) \\ & \therefore\sum_{n=1}^{20}f^{\prime}\left(\frac{1}{x^2}\right) & & \text{[Using (i)]} \\ & =\sum_{n=1}^{20}x^{2}+=\sum_{n=1}^{20}1 \\ & =\frac{20\times21\times4}{6}=20 \\ & =2870+20=2890 \end{aligned}

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JEE 2024 — Mathematics PYQ

JEE | Mathematics | 2024

Let for a differentiable function $f : (0, \infty) \to R$ , $f(x)\:-\:f(y)\:>\:\log_{\mathrm{e}}\biggl(\frac{x}{y}\biggr)+x-y,\forall x,y\in(0,\infty)$ Then $\sum_{n = 1}^{20} f^{'} (\frac{1}{n ^{2}})$ is equal to

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