A) m = 1, n = 0
B) \[m=\frac{n\pi }{2}+1\]
C) \[n=m\left( \frac{\pi }{2} \right)\]
D) \[m=n=\frac{\pi }{2}\]
Correct Answer: C
Solution :
| [c] Given function is |
| \[f(x)=\left\{ \begin{matrix} mx+1, & x\le \frac{\pi }{2} \\ \sin x+n, & x>\frac{\pi }{2} \\ \end{matrix} \right.\] |
| As given this function is continuous at \[x=\frac{\pi }{2}\]. |
| So, limit of function when \[x\to \frac{\pi }{2}=f\left( \frac{\pi }{2} \right)\] |
| \[\Rightarrow \underset{x\to \frac{\pi }{2}+}{\mathop{\lim }}\,(\sin \,x+n)=f\left( \frac{\pi }{2} \right)\] |
| \[\Rightarrow \underset{h\to 0}{\mathop{\lim }}\,\left( \sin \left( \frac{\pi }{2}+h \right)+n \right)=\frac{m\pi }{2}+1\] |
| \[\Rightarrow \sin \frac{\pi }{2}+n=\frac{m\pi }{2}+1\] |
| \[\Rightarrow 1+n=\frac{m\pi }{2}+1\] |
| \[\Rightarrow n=\frac{m\pi }{2}\] |
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