The function f(x) = x/2 + 2/x has a local minima at x equal to:

The correct answer is Option (1) → $2$ ##

We have,

$f(x) = \frac{x}{2} + \frac{2}{x}$

$f'(x) = \frac{1}{2} - \frac{2}{x^2}$

Put, $f'(x) = 0 \Rightarrow \frac{1}{2} - \frac{2}{x^2} = 0$

$\Rightarrow \frac{1}{2} = \frac{2}{x^2}$

$\Rightarrow x^2 = 4$

$\Rightarrow x = \pm 2$

Now, $f''(x) = \frac{4}{x^3}$

At $x = 2$, $f''(2) = \frac{4}{8} = \frac{1}{2} > 0$

At $x = -2$, $f''(-2) = \frac{4}{-8} = -\frac{1}{2} < 0$

Thus, $f(x)$ has local minima at $x = 2$.