A) \[R-\{0\}\]
B) \[R-\{1\}\]
C) \[R-\{-1\}\]
D) \[R-\{-1,\ 1\}\]
Correct Answer: C
Solution :
\[f(x)=\left\{ \begin{align} & \frac{1}{2}(-x-1),\,\,\,x<-1 \\ & {{\tan }^{-1}}x,\,\,\,\,-1\le x\le 1 \\ & \frac{1}{2}(x+1),\,\,\,x>1 \\ \end{align} \right.\]; \[{f}'(x)=\left\{ \begin{align} & -\frac{1}{2},\,\,\,\,\,\,\,\,x<-1 \\ & \frac{1}{1+{{x}^{2}}},\,\,-1<x<1 \\ & \frac{1}{2},\,\,\,\,\,\,\,\,\,\,\,\,x>1 \\ \end{align} \right.\] \[{f}'(-1-0)=-\frac{1}{2};\,\,{f}'(-1+0)=\frac{1}{1+{{(-1+0)}^{2}}}=\frac{1}{2}\] \[{f}'(1-0)=\frac{1}{1+{{(1-0)}^{2}}}=\frac{1}{2};\,\,{f}'(1+0)=\frac{1}{2}\] \\[{f}'(-1)\] Does not exist; \ domain of \[{f}'(x)=R-\{-1\}\].
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