Explanation

(d)
Let $y = \frac{\log x}{x}$
$\frac{dy}{dx} = \frac{x.\frac{1}{x} - \log{x.1}}{x^{2}} = \frac{1 - \log x}{x^{2}}$
For maxima, put $\frac{dy}{dx} = 0$
$\Rightarrow \ \ \ \frac{1 - \log x}{x^{2}} = 0\ \ \Rightarrow \ \ \ x = e$
Now, $\frac{d^{2}y}{dx^{2}} = \frac{x^{2}\left( - \frac{1}{x} \right) - (1 - \log x)2x}{\left( x^{2} \right)^{2}}$
At $x = e,\ \frac{d^{2}y}{dx^{2}} \leq 0$, maxima
$\therefore$ The maximum value at $x = e$ is $y = \frac{1}{e}$