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Let $f(x)=x|x|$. The set of points where $f(x)$ is twice differentiable is

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

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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.
answered Dec 20, 2013 by sreemathi.v
 

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