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

$\begin{array}{1 1}(a)\;\text{differentiable for all x}\\(b)\;\text{f is continuous everywhere}\\(c)\;\text{f is differentiable at x=-1}\\(d)\;\text{f is continuous at x=-1}\end{array}$

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$\lim\limits_{x\to 1^-}f(x)=1$
$\lim\limits_{x\to -1^+}=|-1|=1$
$\Rightarrow \lim\limits_{x\to{-1}}f(x)=f(-1)=\lim\limits_{x\to{-1^+}}f(x)$
$f$ is continuous at $x=-1$
It can be shown that it is not differentiable at $x=-1$
Hence (d) is the correct answer.
answered Dec 24, 2013 by sreemathi.v
 

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