Determine the value of the constant '$k$' so that the function $f(x) = \begin{cases} \frac{kx}{|x|}, & \text{if } x < 0 \\ 3, & \text{if } x \geq 0 \end{cases}$ is continuous at $x = 0$.

$k = -3$

The correct answer is Option (3) → $k = -3$ ##

Since, $f$ is continuous at $x = 0$,

$\lim\limits_{x \to 0^-} f(x) = \lim\limits_{x \to 0^+} f(x) = f(0)$

Here, $f(0) = 3,$

$LHL = \lim\limits_{x \to 0^-} f(x)$

$= \lim\limits_{x \to 0^-} \frac{kx}{|x|} = \lim\limits_{x \to 0^-} \frac{kx}{-x} = -k$

$∴-k = 3 \text{ or } k = -3.$