If $f(x) = \begin{cases} mx + 1, & \text{if } x \leq \frac{\pi}{2} \\ \sin x + n, & \text{if } x > \frac{\pi}{2} \end{cases}$ is continuous at $x = \frac{\pi}{2}$, then

$n = \frac{m\pi}{2}$

The correct answer is Option (3) → $n = \frac{m\pi}{2}$ ##

We have, $f(x) = \begin{cases} mx + 1, & \text{if } x \leq \frac{\pi}{2} \\ (\sin x + n), & \text{if } x > \frac{\pi}{2} \end{cases}$ is continuous at $x = \frac{\pi}{2}$

$∴\text{LHL} = \lim\limits_{x \to \frac{\pi}{2}^-} (mx + 1) = \lim\limits_{h \to 0} \left[ m \left( \frac{\pi}{2} - h \right) + 1 \right] = \frac{m\pi}{2} + 1$

$\left[ ∵\text{put } x = \frac{\pi}{2} - h \right]$

and

$\text{RHL} = \lim\limits_{x \to \frac{\pi}{2}^+} (\sin x + n) = \lim\limits_{h \to 0} \left[ \sin \left( \frac{\pi}{2} + h \right) + n \right] \quad \left[ ∵\text{put } x = \frac{\pi}{2} + h \right]$

$= \lim\limits_{h \to 0} \cos h + n = 1 + n$

$∴\text{LHL} = \text{RHL} \quad \left[ \text{to be continuous at } x = \frac{\pi}{2} \right]$

$⇒m \cdot \frac{\pi}{2} + 1 = n + 1$

$∴n = m \cdot \frac{\pi}{2} = \frac{m\pi}{2}$