Question
If \[f\left( x \right) = \left\{ \begin{array}{l}
\,\,\,mx + 1,\,\,\,\,\,\,\,x \le \frac{\pi }{2}\\
\sin \,x + n,\,\,\,\,\,x > \frac{\pi }{2}
\end{array} \right.\] is continuous at $$x = \frac{\pi }{2},$$ then which one of the following is correct ?
A.
$$m = 1,\,n = 0$$
B.
$$m = \frac{{n\pi }}{2} + 1$$
C.
$$n = m\left( {\frac{\pi }{2}} \right)$$
D.
$$m = n = \frac{\pi }{2}$$
Answer :
$$n = m\left( {\frac{\pi }{2}} \right)$$
Solution :
Given function is \[f\left( x \right) = \left\{ \begin{array}{l}
\,\,\,mx + 1,\,\,\,\,\,\,\,x \le \frac{\pi }{2}\\
\sin \,x + n,\,\,\,\,\,x > \frac{\pi }{2}
\end{array} \right.\]
As given this function is continuous at $$x = \frac{\pi }{2}$$
So, limit of function when $$x \to \frac{\pi }{2} = f\left( {\frac{\pi }{2}} \right)$$
$$\eqalign{
& \Rightarrow \mathop {\lim }\limits_{x \to \frac{\pi }{2} + } \left( {\sin \,x + n} \right) = f\left( {\frac{\pi }{2}} \right) \cr
& \Rightarrow \mathop {\lim }\limits_{h \to 0} \left( {\sin \left( {\frac{\pi }{2} + h} \right) + n} \right) = \frac{{mx}}{2} + 1 \cr
& \Rightarrow \sin \frac{\pi }{2} + n = \frac{{m\pi }}{2} + 1 \cr
& \Rightarrow 1 + n = \frac{{m\pi }}{2} + 1 \cr
& \Rightarrow n = \frac{{m\pi }}{2} \cr} $$