A) \[f(\left| x \right|)\]is discontinuous at x=0
B) \[f(\left| x \right|)\]is differentiable at x=0
C) \[|f(x)|\]is non-differentiable at x=0, 2
D) \[|f(x)|\]is continuous at x=0
Correct Answer: C
Solution :
[c]| \[f(x)=\left\{ \begin{matrix} \left| x \right|-1,\, \\ {{\left| x \right|}^{2}}-2\left| x \right|, \\ \end{matrix} \right.\,\,\,\,\,\,\,\,\,\begin{matrix} \left| x \right|<0 \\ \left| x \right|\ge 0 \\ \end{matrix}\] |
| Where \[\left| x \right|\le 0\] is not possible. Thus neglecting, we get |
| \[f(\left| x \right|)=\begin{matrix} {{\left| x \right|}^{2}}-2\left| x \right|,\,\,\left| x \right|\ge 0 \\ \end{matrix}\] |
| \[f(\left| x \right|)=\left\{ \begin{matrix} {{x}^{2}}+2x, & x<0 \\ {{x}^{2}}-2x & x\ge 0 \\ \end{matrix} \right....(i)\] |
| \[\therefore f'(\left| x \right|)=\left\{ \begin{matrix} x+2x, & x<0 \\ x-2x & x\ge 0 \\ \end{matrix} \right.\] |
| Clearly, \[\left| f(x) \right|\] is continuous at \[x=0\], but non-differentiable at\[x=0\]. |
| \[f(x)=\left\{ \begin{matrix} \left| x \right|-1,\, \\ {{\left| x \right|}^{2}}-2\left| x \right|, \\ \end{matrix} \right.\,\,\,\,\,\,\,\,\,\begin{matrix} \left| x \right|<0 \\ \left| x \right|\ge 0 \\ \end{matrix}\] |
| \[g(x)=\left| f(x) \right|=\left\{ \begin{matrix} 1-x, & x<0 \\ -{{x}^{2}}+2x & 0\le x<2 \\ {{x}^{2}}-2x & x\ge 2 \\ \end{matrix} \right....(ii)\] |
| Clearly, \[\left| f(x) \right|\] is discontinuous at \[x=0\] but non-differentiable at \[x=2.\] |
| Also, \[g'(x)=\left\{ \begin{matrix} -1, & x<0 \\ 2x+2, & 0<x<2. \\ 2x-2, & x>2 \\ \end{matrix} \right.\] |
| \[\left| f(x) \right|\] is non-differentiable at \[x=0\] and \[x=2.\] |
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