Let for a differentiable function f:(0, infty | vedclass

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(B) Given $f(x)-f(y) \geq \ln x - \ln y + x - y$.Rearranging the terms,we get $f(x) - x - \ln x \geq f(y) - y - \ln y$.Let $g(x) = f(x) - x - \ln x$.

Vedclass · Accepted Answer

$2890$

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(B) Given $f(x)-f(y) \geq \ln x - \ln y + x - y$.Rearranging the terms,we get $f(x) - x - \ln x \geq f(y) - y - \ln y$.Let $g(x) = f(x) - x - \ln x$. Then $g(x) \geq g(y)$ for all $x, y \in (0, \infty)$.This implies that $g(x)$ is a constant function,say $C$.So,$f(x) - x - \ln x = C$,which means $f(x) = x + \ln x + C$.Differentiating with respect to $x$,we get $f^{\prime}(x) = 1 + \frac{1}{x}$.Now,we need to calculate $\sum_{n=1}^{20} f^{\prime}\left(\frac{1}{n^2}ight)$.$f^{\prime}\left(\frac{1}{n^2}ight) = 1 + \frac{1}{1/n^2} = 1 + n^2$.Therefore,$\sum_{n=1}^{20} (1 + n^2) = \sum_{n=1}^{20} 1 + \sum_{n=1}^{20} n^2$.$= 20 + \frac{20(20+1)(2 	imes 20 + 1)}{6} = 20 + \frac{20 	imes 21 	imes 41}{6}$.$= 20 + 10 	imes 7 	imes 41 = 20 + 2870 = 2890$.

Let for a differentiable function $f:(0, \infty) \rightarrow \mathbb{R}$,$f(x)-f(y) \geq \log_e\left(\frac{x}{y}\right)+x-y, \forall x, y \in(0, \infty)$. Then $\sum_{n=1}^{20} f^{\prime}\left(\frac{1}{n^2}\right)$ is equal to

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