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Let $f(x)=\left\{\small\begin{array}{1 1}(x-1)^2\sin\large\frac{1}{(x-1)}-\normalsize |x|&ifx\neq 1\\-1&if\;x=1\end{array}\right.$ be a real valued function. Then the set of points where $f(x)$ is not differentiable is


1 Answer

Given :
$f(x)=\left\{\begin{array}{1 1}(x-1)^2\sin\large\frac{1}{(x-1)}-\normalsize |x|&ifx\neq 1\\-1&if\;x=1\end{array}\right.$
We know that $|x|$ is not differentiable at $x=0$
$\therefore (x-1)^2\sin\large\frac{1}{x-1}$$-|x|$ is not differentiable at $x=0$.
At all other values of $x,f(x)$ is differentiable .
The required set of points is $\{0\}$.
Hence (c) is the correct answer.
answered Dec 20, 2013 by sreemathi.v

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