Explanation

(d)
$f(x) = x + \frac{1}{x}$
On differentiating w.r.t. $x$, we get
$f^{'}(x) = 1 - \frac{1}{x^{2}}$
For maximum put $f^{'}(x)$=0
$\therefore\ \ x^{2} - 1 = 0\ \Rightarrow x = \pm 1$
Again differentiating, we get
$f^{'}(x) = \frac{2}{x^{3}}$
Since, $x > 0$, so we can check only at $x = 1$
At y $x = 1,\ \ f^{''}(x) > 0$
$\Rightarrow \ \ f(c)$ is minimum at $x = 1$
At $x = 1,\ \ f^{''}(x) > 0$
$\Rightarrow \ \ \ f(x)$ is minimum at $x = 1$
$\therefore$ No maximum value can be found