Explanation

(d)
Given, $f(x) = ax + \frac{b}{x}$
On differentiating w.r.t. $x$, we get
$f^{'}(x) = a - \frac{b}{x^{2}}$
For maxima or minima, put $f^{'}(x) = 0$
$\Rightarrow \ \ \ x = \sqrt{\frac{b}{a}}$
Again, differentiating w.r.t. $x$, we get
$f^{''}(x) = \frac{2b}{x^{3}}$
At $x = \sqrt{\frac{b}{a}},\ \ f^{''}(x) = + ve$
$\Rightarrow \ \ f(x)$ is minimum at $x = \sqrt{\frac{b}{a}}$
$\therefore\ \ f(x)$ has the least value at $x = \sqrt{\frac{b}{a}}$