Find the value of $k$ so that the function $f(x) = \begin{cases} 3x - 8, & \text{if } x \leq 5 \\ 2k, & \text{if } x > 5 \end{cases}$ is continuous at $x = 5$.

$\frac{7}{2}$

The correct answer is Option (2) → $\frac{7}{2}$ ##

We have, $f(x) = \begin{cases} 3x - 8, & \text{if } x \leq 5 \\ 2k, & \text{if } x > 5 \end{cases}$ at $x = 5$

Now at $x = 5$,

$\text{LHL} = \lim\limits_{x \to 5^-} (3x - 8) = \lim\limits_{h \to 0} [3(5 - h) - 8]$

Put $x = 5 - h$,

$= \lim\limits_{h \to 0} [15 - 3h - 8] = 7$

$\text{RHL} = \lim\limits_{x \to 5^+} 2k = 2k \text{ and } f(5) = 3 \times 5 - 8 = 7$

Since, $f(x)$ is continuous at $x = 5$.

$∴\text{LHL} = \text{RHL} = f(5)$

$∴2k = 7$

$⇒k = \frac{7}{2}$