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Mathematics Question on Differentiability

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

A

Continuous at x=0, πx=0,\,\pi

B

Discontinuous at x=0x=0 but continuous at x=πx=\pi

C

Continuous at x=0x=0 but discontinuous at x=πx=\pi

D

Discontinuous at x=0, πx=0,\,\pi

Answer

Continuous at x=0, πx=0,\,\pi

Explanation

Solution

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