Functions Question 686
Question: If $ f(x)= \begin{cases} & 1,,when0<x\le \frac{3\pi }{4} \\ & 2\sin \frac{2}{9}x,when,\frac{3\pi }{4}<x<\pi \\ \end{cases} . $ , then
[IIT 1991]
Options:
A) $ f(x) $ is continuous at $ x=0 $
B) $ f(x) $ is continuous at $ x=\pi $
C) $ f(x) $ is continuous at $ x=\frac{3\pi }{4} $
D) $ f(x) $ is discontinuous at $ x=\frac{3\pi }{4} $
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Answer:
Correct Answer: C
Solution:
Here $ f,( \frac{3\pi }{4} )=1 $ and $ \because f $ $ \underset{x\to 3\pi /4+}{\mathop{\lim }},f(x)=\underset{h\to 0}{\mathop{\lim }},2\sin \frac{2}{9},( \frac{3\pi }{4}+h )=2,\sin \frac{\pi }{6}=1 $ . Hence $ f(x) $ is continuous at $ \frac{\sin 2x}{\sin ,( \frac{x}{2} )} $ .
BETA
