Q76 / 120 · VITEEE 2012 · Mathematics
If $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{l}\mathrm{mx}+1, \quad \mathrm{x} \leq \frac{\pi}{2} \\ \sin \mathrm{x}+\mathrm{n}, \mathrm{x}>\frac{\pi}{2}\end{array}\right.$ is continuous at $x=\frac{\pi}{2}$, then
- A$m=1, n=0$
- B$\mathrm{m}=\frac{\mathrm{n} \pi}{2}+1$
- Correct answerC$\mathrm{n}=\mathrm{m} \frac{\pi}{2}$
- D$\mathrm{m}=\mathrm{n}=\frac{\pi}{2}$
Solution
$\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=\frac{\pi}{2}$.
$$
\begin{array}{l}
\text { So, } \lim _{x \rightarrow \frac{\pi^{-}}{2}} f(x)=\lim _{x \rightarrow \frac{\pi^{+}}{2}} f(x) \\
\Rightarrow m \frac{\pi}{2}+1=\sin \frac{\pi}{2}+n \\
\Rightarrow m \frac{\pi}{2}+1=1+n \Rightarrow \frac{m \pi}{2}=n
\end{array}
$$
Hence, the correct answer is option: 'C'.