A) \[(-\infty ,\infty )\]
B) \[(-\infty ,0)\cup (0,\infty )\]
C) \[(0,\infty )\]
D) \[[0,\infty )\]
Correct Answer: A
Solution :
We have, \[f(x)=\left\{ \begin{matrix} {{x}^{2}}, & x\ge 0 \\ -{{x}^{2}}, & x<0 \\ \end{matrix} \right.\]Clearly\[f(x)\]is differentiable for all\[x>0\]and for all\[x<0\]. So, we check the differentiable at \[x=0\]. Now,(RHD at\[x=0\]) \[{{\left( \frac{d}{dx}{{(x)}^{2}} \right)}_{x=0}}={{(2x)}_{x=0}}=0\] and (LHD at \[x=0\]) \[{{\left( \frac{d}{dx}(-{{x}^{2}}) \right)}_{x=0}}={{(-2x)}_{x=0}}=0\] \[\therefore \]\[(LHD\]at\[x=0\])\[=(RHD\]at\[x=0)\] So,\[f(x)\]is diffierendable for all\[x\]ie, the set of all points where\[f(x)\]is differentiable is\[(-\infty ,\infty )\].
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