Target Exam

CUET

Subject

Section B1

Chapter

Applications of Derivatives

Question:

The function $f(x) = \frac{x}{2} + \frac{2}{x}$ has a local minima at $x$ equal to:

Options:

$2$

$1$

$0$

$-2$

Correct Answer:

$2$

Explanation:

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 $x = 2$.