Question

A derivable function $f:R^{+}\to R$ satisfies the condition $f\left(x\right)-f\left(y\right)\ge \ln \frac{x}{y}+x-y;\forall x,y\in R^{+}$. If $g$ denotes the derivative of $f$ then the value of the sum ${\sum }_{n=1}^{100}g\left(\frac{1}{n}\right)$ is $1030k$.Find the value of $k$.

Answer 1

$RHD{f}^{\prime }\left(x\right)=\underset{h\to 0^{+}}{lim}\frac{f(x+h)-f\left(x\right)}{h}$
$\ge \underset{h\to 0^{+}}{lim}\frac{\ln \left(\frac{x+h}{x}\right)+x+h-x}{h}$
$\ge \underset{h\to 0^{+}}{lim}\frac{\ln (1+\frac{h}{x})}{h}+1\ge \frac{1}{x}+\mathrm{1.......}\left(i\right)$
$LHD{f}^{\prime }\left(x\right)=\underset{h\to 0^{-}}{lim}\frac{f(x-h)-f\left(x\right)}{-h}$
$\le \underset{h\to 0^{-}}{lim}\frac{\ln \left(\frac{x+h}{x}\right)+x-h-x}{h}$
$\le \underset{h\to 0^{-}}{lim}\frac{\ln (1-\frac{h}{x})}{-h}+1\le \frac{1}{x}+\mathrm{1.......}\left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$
$\Rightarrow f^{\prime }\left(x\right)=\frac{1}{x}+1$
$\therefore {\sum }_{n=1}^{100}g\left(\frac{1}{n}\right)=g\left(\frac{1}{1}\right)+g\left(\frac{1}{2}\right)+......g\left(\frac{1}{100}\right)=(1+2+3+....+100)+100=5150$
$\Rightarrow k=5$