Find the value of $k$ so that the function $f(x) = \begin{cases} \frac{2^{x+2} - 16}{4^x - 16}, & \text{if } x \neq 2 \\ k, & \text{if } x = 2 \end{cases}$ is continuous at $x=2$.

$\frac{1}{2}$

The correct answer is Option (1) → $\frac{1}{2}$ ##

We have, $f(x) = \begin{cases} \frac{2^{x+2} - 16}{4^x - 16}, & \text{if } x \neq 2 \\ k, & \text{if } x = 2 \end{cases}$ at $x = 2$

At $x = 2$,

$\lim\limits_{x \to 2} \frac{2^x \cdot 2^2 - 2^4}{4^x - 4^2} = \lim\limits_{x \to 2} \frac{4 \cdot (2^x - 4)}{(2^x)^2 - (4)^2}$

$= \lim\limits_{x \to 2} \frac{4 \cdot (2^x - 4)}{(2^x - 4)(2^x + 4)} \quad [∵a^2 - b^2 = (a+b)(a-b)]$

$= \lim\limits_{x \to 2} \frac{4}{2^x + 4} = \frac{4}{8} = \frac{1}{2}$

Since $f(x)$ is continuous at $x = 2$,

$∴\text{LHL} = \text{RHL} = f(2)$

But $f(2) = k$

$∴k = \frac{1}{2} \quad [∵f(x) \text{ is continuous at } x = 2]$