Examine the continuity of the function $f(x) = \begin{cases} 3x + 5, & \text{if } x \geq 2 \\ x^2, & \text{if } x < 2 \end{cases}$ at $x = 2$

Discontinuous at $x = 2$ because $\text{LHL} \neq \text{RHL}$

The correct answer is Option (2) → Discontinuous at $x = 2$ because $\text{LHL} \neq \text{RHL}$ ##

We have, $f(x) = \begin{cases} 3x + 5, & \text{if } x \geq 2 \\ x^2, & \text{if } x < 2 \end{cases}$ at $x = 2$

At $x = 2$,

$\text{LHL} = \lim\limits_{x \to 2^-} (x)^2$

Put $x = 2 - h$,

$= \lim\limits_{h \to 0} (2 - h)^2 = \lim\limits_{h \to 0} (4 + h^2 - 4h) = 4$

and $\text{RHL} = \lim\limits_{x \to 2^+} (3x + 5) \quad [∵(a - b)^2 = a^2 + b^2 - 2ab]$

Put $x = 2 + h$,

$= \lim\limits_{h \to 0} [3(2 + h) + 5] = 11$

Since, $\text{LHL} \neq \text{RHL}$ at $x = 2$

So, $f(x)$ is discontinuous at $x = 2$.