Explanation

(c)
Let $y = x^{1/x}$
On taking log on both sides, we get
$\log y = \frac{1}{x^{\ }}\log x$
$\Rightarrow \frac{1}{y}\frac{dy}{dx} = \frac{1}{x^{2}} - \frac{\log x}{x^{2}} = \frac{1 - \log x}{x}$
$\Rightarrow \frac{dy}{dx} = x^{1/x}\left( \frac{1 - \log x}{x} \right)$
Now, $x^{1/x} > 0$ for all positive values of $x$
And $\frac{1 - \log x}{x^{2}} > 0$ in $(1,\ e)$
And $\frac{1 - \log x}{x^{2}} < 0$in $(e,\ \infty)$
$\therefore\ \ f(x)$ is increasing in $(1,\ e)$ and decreasing in $(e,\ \infty)$