Let $f(x)=\left\{\begin{array}{l}-1,-2 \leq x<0 \\ x^{2}-1,0 \leq x \leq 2\end{array}\right.$ and
$\mathrm{g}(\mathrm{x})=|\mathrm{f}(\mathrm{x})|+\mathrm{f}(|\mathrm{x}|)$. Then, in the interval $(-2,2), \mathrm{g}$ is :-
(–2, 2), g is :-
Correct Option: , 4
$|f(\mathrm{x})|=\left\{\begin{array}{cc}1, & -2 \leq \mathrm{x}<0 \\ 1-\mathrm{x}^{2}, & 0 \leq \mathrm{x}<1 \\ \mathrm{x}^{2}-1, & 1 \leq \mathrm{x} \leq 2\end{array}\right.$
and $f(|\mathrm{x}|)=\mathrm{x}^{2}-1, \mathrm{x} \in[-2,2]$
Hence $g(x)=\left\{\begin{array}{ccc}x^{2} & , x \in[-2,0) \\ 0 & , \quad x \in[0,1) \\ 2\left(x^{2}-1\right) & , \quad x \in[1,2]\end{array}\right.$
It is not differentiable at x = 1
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