Functions Question 8
Question: If $ f(x)= \begin{cases} & x\sin x,,\text{when }0<x\le \frac{\pi }{2} \\ & \frac{\pi }{2}\sin (\pi +x),when\frac{\pi }{2}<x<\pi \\ \end{cases} . $ , then
[IIT 1991]
Options:
A) $ f(x) $ is discontinuous at $ x=\pi /2 $
B) $ f(x) $ is continuous at $ x=\pi /2 $
C) $ f(x) $ is continuous at $ x=0 $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ \underset{x\to \infty }{\mathop{\lim }}\frac{4x}{(\sqrt{x^{2}+8x+3}+\sqrt{x^{2}+4x+3}} $ and $ f( \frac{\pi }{2} )=\frac{\pi }{2}. $
BETA
