Limits Continuity And Differentiability Question 79
Question: If $ f(x)= \begin{cases} x-1,x<0 \\ x^{2}-2x,x\ge 0 \\ \end{cases} ., $ , then
Options:
A) $ f(| x |) $ is discontinuous at x=0
B) $ f(| x |) $ is differentiable at x=0
C) $ |f(x)| $ is non-differentiable at x=0, 2
D) $ |f(x)| $ is continuous at x=0
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Answer:
Correct Answer: C
Solution:
$ f(x)= \begin{cases} | x |-1, \ {| x |}^{2}-2| x |, \ \end{cases} \begin{matrix} | x |<0 \ | x |\ge 0 \ \end{matrix} $ Where $ | x |\le 0 $ is not possible.
Thus neglecting, we obtain $ f(| x |)=\begin{cases} {{| x |}^{2}}-2| x |,| x |\ge 0 \\ \end{cases} $
$ f(| x |)= \begin{cases} x^{2}+2x, & x<0 \\ x^{2}-2x & x\ge 0 \\ \end{cases} ….(i) $
$ \therefore f’(| x |)= \begin{cases} -x, & x<0 \ x, & x\ge 0 \end{cases} . $
Clearly, $ | f(x) | $ is continuous at $ x=0 $ , but non-differentiable at $ x=0 $ . $ f(x)= \begin{cases} | x |-1, \ {| x |}^{2}-2| x |, \ \end{cases} \begin{matrix} | x |<0 \ | x |\ge 0 \ \end{matrix} $
$ g(x)=| f(x) |= \begin{cases} 1-x, & x<0 \\ -x^{2}+2x & 0\le x<2 \\ x^{2}-2x & x\ge 2 \\ \end{cases} ….(ii) $
Clearly, $ | f(x) | $ is discontinuous at $ x=0 $ but differentiable at $ x=2. $
Also, $ g’(x)= \begin{cases} -1, & x<0 \\ 2x+2, & 0<x<2. \\ 2x-2, & x>2 \\ \end{cases} . $
$ | f(x) | $ is non-differentiable at $ x=0 $ and $ x=2. $
BETA
