A) 0
B) \[\pm 1,\,0\]
C) 1
D) \[\pm \,1\]
Correct Answer: B
Solution :
\[=\left\{ \begin{align} & |x|-1,\,\,\,\,\,\,\,|x|-1\ge 0 \\ & -|x|+1,\,\,\,|x|-1<0 \\ \end{align} \right.\] \[=\left\{ \begin{align} & |x|-1,\,\,\,x\le -1\,\,\text{or}\,x\ge 1 \\ & -|x|+1,\,\,\,\,\,\,\,-1<x<1 \\ \end{align} \right.\] \[=\left\{ \begin{align} & -x-1,\,\,\,\,\,x\le -1 \\ & x+1,\,\,\,\,\,\,-1<x<0 \\ & -x+1,\,\,\,\,\,0\le x<1 \\ & \,\,x-1,\,\,\,\,\,\,\,x\ge 1 \\ \end{align} \right.\]
From the graph. It is clear that \[f(x)\] is not differentiable at \[x=-1,\,0\] and 1.
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