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) $m=n=\frac{\pi}{2}$
(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
$\lim _{x \rightarrow \frac{\pi}{2}^{-}} f(x)=\lim _{x \rightarrow \frac{\pi}{2}^{+}} f(x)$
$\Rightarrow \frac{m \pi}{2}+1=n+1$
$\Rightarrow \frac{m \pi}{2}=n$
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