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 \[x=0,\,\pi \]
Correct Answer: A
Solution :
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.
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