Explanation

(c)
Given, $y = x^{1/x}$
$\log{y = \frac{1}{x}\log x =}f(x)$ [say]
$\therefore\ \ \ \ \ \ \ f^{'}(x) = \frac{1 - \log x}{x^{2}}$
For maxima and minima, $\frac{1 - \log x}{x^{2}} = 0$
$\Rightarrow \ \ \ \ \ 1 - \log x = 0$
$\Rightarrow \ \ \ \ \ \ x = e$
Now, $f^{''}(x) = \frac{- x - 2(1 - \log x)x}{\left( x^{2} \right)^{2}}$
At $x = e$
$f^{''}(x) = - ve < 0$
$f(x)$ is maximum and maximum value of $x^{1/x} = e^{1/e}$.