$(a)\;R-\{0\}\qquad(b)\;R-\{1\}\qquad(c)\;R-\{2\}\qquad(d)\;None\;of\;these$

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We have

$f(x)=x|x|=\left\{\begin{array}{1 1}-x^2&x<0\\x^2&x\geq 0\end{array}\right.$

$f'(x)=\left\{\begin{array}{1 1}-2x&x<0\\2x&x\geq 0\end{array}\right.$

$f''(x)=\left\{\begin{array}{1 1}-2&x<0\\2&x\geq 0\end{array}\right.$

$\Rightarrow f''(x)$ exists at every point except at $x=0$

Thus $f(x)$ is twice differentiable on $R-\{0\}$.

Hence (a) is the correct answer.

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