A) \[f(x)\]is discontinuous everywhere
B) \[f(x)\]is continuous everywhere
C) \[f'(x)\]exists in \[(-1,1)\]
D) \[f'(x)\]exists in \[(-2,2)\]
Correct Answer: B
Solution :
We have \[f(x)=\left\{ \begin{array}{*{35}{r}} \frac{{{x}^{2}}}{|x|}, & x\ne 0 \\ 0, & x=0 \\ \end{array}=\left\{ \begin{array}{*{35}{r}} \frac{{{x}^{2}}}{x}=x, & x>0 \\ 0, & x=0 \\ \frac{{{x}^{2}}}{-x}=-x, & x<0 \\ \end{array} \right. \right.\] We have \[\underset{x\to 0-}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,\,-x=0,\,\,\underset{x\to 0+}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,x=0\] and \[f(0)=0.\] So \[f(x)\] is continuous at \[x=0.\] Also \[f(x)\] is continuous for all other values of x. Hence, \[f(x)\] is continuous everywhere. Clearly, \[Lf'(0)=-1\] and \[Rf'(0)=1.\] therefore \[f(x)\] is not differentiable at \[x=0.\]
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