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Q. The function $\frac{\log x}{x}$ is increasing in

Solution:

Let $y=\frac{\log x}{x} $
$\Rightarrow \frac{d y}{d x}=\frac{x(1 / x)-\log x}{x^{2}} $
$=\frac{1-\log x}{x^{2}}$
$y$ is increasing if $\frac{1-\log x}{x^{2}}>0$
$\Rightarrow 1-\log x>0$
$\Rightarrow \log x< 1$
$\Rightarrow x< e$
$\therefore y$ is increasing in $(0, e)$.