# Solve this

Question:

If $f(x)=\left\{\begin{array}{ll}m x+1, & x \leq \frac{\pi}{2} \\ \sin x+n, & x>\frac{\pi}{2}\end{array}\right.$ is continuous at $x=\frac{\pi}{2}$, then

(a) $m=1, n=0$

(b) $m=\frac{n \pi}{2}+1$

(c) $n=\frac{m \pi}{2}$

(d) $m=n=\frac{\pi}{2}$

Solution:

(c) $n=\frac{\mathrm{m} \pi}{2}$

Here,

$f\left(\frac{\pi}{2}\right)=\frac{m \pi}{2}+1$

We have

$\left(\mathrm{LHL}\right.$ at $\left.x=\frac{\pi}{2}\right)=\lim _{x \rightarrow \frac{\pi}{2}} f(x)=\lim _{h \rightarrow 0} f\left(\frac{\pi}{2}-h\right)=\lim _{h \rightarrow 0} m\left(\frac{\pi}{2}-h\right)+1=\frac{\mathrm{m} \pi}{2}+1$

$\left(\mathrm{RHL}\right.$ at $\left.x=\frac{\pi}{2}\right)=\lim _{x \rightarrow \frac{\pi}{2}^{+}} f(x)=\lim _{h \rightarrow 0} f\left(\frac{\pi}{2}+h\right)=\lim _{h \rightarrow 0}\left[\sin \left(\frac{\pi}{2}+h\right)+n\right]=n+1$

Thus,

If $f(x)$ is continuous at $x=\frac{\pi}{2}$, then

$\lim _{x \rightarrow \frac{\pi}{2}^{-}} f(x)=\lim _{x \rightarrow \frac{\pi}{2}^{+}} f(x)$

$\Rightarrow \frac{m \pi}{2}+1=n+1$

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