Question

Discuss the continuity of the function $ f(x)=\sin 2x-1 $ at the point $ x=0, $ and $ x=\pi $ .

• A. Continuous at $ x=0,\,\pi $
• B. Discontinuous at $ x=0 $ but continuous at $ x=\pi $
• C. Continuous at $ x=0 $ but discontinuous at $ x=\pi $
• D. Discontinuous at $ x=0,\,\pi $

Answer 1

The Correct Option is A

Given, $ f(x)=\sin \,2x-1 $ At $ x=0, $ $ \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\,\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\,f(0-h) $
$=\underset{h\to 0}{\mathop{\lim }}\,\,[-\sin \,2h-1] $
$=0-1=-1 $ $ \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\,f(0+h) $
$=\underset{h\to 0}{\mathop{\lim }}\,\,\,\sin \,2h-1 $
$=0-1=-1 $ and $ f(0)=sin0-1=-1 $ $ \because $ $ \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\,\,f(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\,f(x)=f(0) $
$ \therefore $ $ f(x) $ is continuous at $ x=0 $ Now, at $ x=\pi $ $ \underset{x\to {{\pi }^{-}}}{\mathop{\lim }}\,\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,f(\pi -h) $
$=\underset{h\to 0}{\mathop{\lim }}\,\,\sin 2(\pi -h)-1 $
$=\underset{h\to 0}{\mathop{\lim }}\,\,\,-\sin 1h-1=-1 $ $ \underset{x\to {{\pi }^{+}}}{\mathop{\lim }}\,\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\,f(\pi +h) $
$=\underset{h\to 0}{\mathop{\lim }}\,\sin 2(\pi +h)-1 $
$=\underset{h\to 0}{\mathop{\lim }}\,\,\,\sin 2h-1 $
$=0-1=-1 $ and $ f(\pi )=\sin \,2\pi -1=-1 $ $ \because $ $ \underset{x\to {{\pi }^{-}}}{\mathop{\lim }}\,\,f(x)=\underset{x\to {{\pi }^{+}}}{\mathop{\lim }}\,\,\,f(x)=f(\pi ) $
$ \therefore $ $ f(x) $ is continuous at $ x=\pi $ also.