Step 1 (Formula / Governing Law): (1/2 Mark)

To find the local minima of a function $f(x)$, we use the first derivative test. We find the critical points by setting the first derivative $f'(x) = 0$ and then verify the nature of the point using the second derivative test, where $f''(x) > 0$ indicates a local minimum.

Step 2 (Step-by-Step Working): (1/4 Mark)

Given the function:

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

Differentiating with respect to $x$:

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

Setting $f'(x) = 0$ for critical points:

$$\frac{1}{2} - \frac{2}{x^2} = 0 \implies \frac{1}{2} = \frac{2}{x^2} \implies x^2 = 4 \implies x = \pm 2$$

Now, find the second derivative $f''(x)$:

$$f''(x) = \frac{d}{dx}\left(\frac{1}{2} - 2x^{-2}\right) = 0 - 2(-2)x^{-3} = \frac{4}{x^3}$$

Testing $x = 2$:

$$f''(2) = \frac{4}{2^3} = \frac{4}{8} = \frac{1}{2} > 0$$

Since $f''(2) > 0$, the function has a local minima at $x = 2$.

Step 3 (Final Answer): (1/4 Mark)

The function has a local minima at $x = 2$.

Correct Option: (A)