Limits Continuity And Differentiability Question 4
Question: If $ f(x)= \begin{cases} mx+1x\le \frac{\pi }{2} \\ \sin x+nx>\frac{\pi }{2} \\ \end{cases}$is continuous at . $ x=\frac{\pi }{2} $ , then which one of the following is correct?
Options:
A) m = 1, n = 0
B) $ m=\frac{n\pi }{2}+1 $
C) $ n=m( \frac{\pi }{2} ) $
D) $ m=n=\frac{\pi }{2} $
Show Answer
Answer:
Correct Answer: C
Solution:
Given function is $ f(x)= \begin{cases} mx+1, & x\le \frac{\pi }{2} \\ \sin x+n, & x>\frac{\pi }{2} \\ \end{cases} . $
As given this function is continuous at $ x=\frac{\pi }{2} $ .
So, limit of function when $ x\to \frac{\pi }{2}=f( \frac{\pi }{2} ) $
$ \Rightarrow \underset{x\to \frac{\pi }{2}+}{\mathop{\lim }}(\sin x+n)=f( \frac{\pi }{2} ) $
$ \Rightarrow \underset{h\to 0}{\mathop{\lim }}( \sin ( \frac{\pi }{2}+h )+n )=\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} $
BETA
