The function $f(x) = \frac{x}{2} + \frac{2}{x}, x \neq 0$ is increasing on

(A) $(-\infty, -2)$

(B) $(-2, 2)$

(C) $(2, \infty)$

(D) $(-1, 1)$

Choose the correct answer from the options given below:

A function is increasing where its derivative $f'(x) > 0$.

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

Differentiating term by term:

Derivative of $\frac{x}{2}$ is $\frac{1}{2}$

Derivative of $\frac{2}{x} = 2x^{-1}$ is $2 \times (-1) \times x^{-2} = -\frac{2}{x^2}$

Therefore:

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

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For the function to be increasing:

$\frac{1}{2} - \frac{2}{x^2} > 0$

$\frac{1}{2} > \frac{2}{x^2}$

$\frac{x^2}{2} > 2$

$x^2 > 4$

$|x| > 2$

This means: $x > 2$ or $x < -2$

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Checking the options:

(A) $(-\infty, -2)$ represents $x < -2$ — the function is increasing here

(B) $(-2, 2)$ represents $-2 < x < 2$ — the function is not increasing here

(C) $(2, \infty)$ represents $x > 2$ — the function is increasing here

(D) $(-1, 1)$ represents $-1 < x < 1$ — the function is not increasing here

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The function is increasing on $(-\infty, -2)$ and $(2, \infty)$.

The correct answer is (A) and (C).