A) not differentiable at one point
B) differentiable at all points
C) not continuous
D) not differentiable at two points
Correct Answer: A
Solution :
Given, \[f(x)=\left\{ \begin{matrix} -1, & -2\le x<0 \\ {{x}^{2}}-1, & 0\le x\le 2 \\ \end{matrix} \right.\] \[\therefore \]\[|f(x)|=\left\{ \begin{matrix} 1, & -2\le x<0 \\ 1-{{x}^{2}}, & 0\le x<1 \\ {{x}^{2}}-1, & 1\le x\le 2 \\ \end{matrix} \right.\] and\[f(|x|)={{x}^{2}}-1,x\in [-2,2]\] \[\therefore \]\[g(x)=\left\{ \begin{matrix} {{x}^{2}}, & -2\le x<0 \\ 0, & 0\le x<1 \\ 2{{x}^{2}}-1, & 1\le x\le 2 \\ \end{matrix} \right.\] \[g'(x)=\left\{ \begin{matrix} 2x & -2<x<0 \\ 0, & 0<x<1 \\ 4x & 1<x<2 \\ \end{matrix} \right.\] \[g'({{0}^{-}})=0=g'({{0}^{+}})\]and\[g'({{1}^{-}})=0,g'({{1}^{+}})=4\] So, g(x) is not differentiable at x = 1.
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