A) one
B) two
C) three
D) none of these
Correct Answer: C
Solution :
| \[f(x)\,=\left\{ \begin{align} & {{\cot }^{-1}}x,\,\,\,\left| x \right|\ge 1 \\ & -\frac{1}{2}x+\frac{\pi }{4}-\frac{1}{2},-1<x<0 \\ & \frac{1}{2}x+\frac{\pi }{4}-\frac{1}{2},0\le x<1 \\ \end{align} \right.\] |
| \[f(1-0)=\frac{1}{2}+\frac{\pi }{4}-\frac{1}{2}=\frac{\pi }{4}\] |
| and \[f(1+0)=co{{t}^{-1}}(1)\,=\frac{\pi }{4}\] |
| \[f(0-0)=f(0+0)=\frac{\pi }{4}-\frac{1}{2}\]\[f(-1-0)={{\cot }^{-1}}(-1)=\frac{3\pi }{2}\] |
| \[and\,f(-1+0)=\frac{\pi }{4}\] |
| \[\therefore \,f(x)\]Is continuous at x=1 and 0 but discontinuous at x=-1\[\begin{align} & Now\,f'(x)\,=\left\{ \begin{align} & -\frac{1}{1+{{x}^{2}}}\,\,\,\,\,\,\,\,\left| x \right|>1 \\ & -\frac{1}{2},-1<x<0 \\ & \frac{1}{2},0<x<1 \\ \end{align} \right. \\ & \\ \end{align}\] |
| \[f'(1-0)=\frac{1}{2}and\,f'(1+0)=-\frac{1}{2}\] |
| \[f'(0-0)=\frac{-1}{2}and\,f'(0+0)=-\frac{1}{2}\] |
| Thus \[f\left( x \right)\]Is not differentiable at \[x=0\]and 1 further \[f\left( x \right)\]is discontinuous\[x=-1\], so not differentiable. |
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