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Let $[x]$ denote the greatest integer less than or equal to x. If $f(x)=[x\sin\pi x]$, then $f(x)$ is

$\begin{array}{1 1}(a)\;\text{continuous at x=0}\\(b)\;\text{continuous in (0,0)}\\(c)\;\text{differentiable at x=1}\\(d)\;\text{differentiable in (-1,1)}\end{array}$

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We have for $-1\leq x \leq 1$
$\Rightarrow 0\leq x\sin\pi x\leq \large\frac{1}{2}$
$\therefore f(x)=[x\sin\pi x]=0$
Also $x\sin \pi x$ becomes negative and numerically less than 1 and so by definition of [x]
$f(x)=[x\sin \pi x]=-1\;when 1 < x < 1+h$
Thus $f(x)$ is constant and equal to 0 in the closed interval [-1,1] and so f(x) is continuous and differentiable in the open interval (-1,1)
Hence (d) is the correct answer.
answered Jan 3, 2014 by sreemathi.v

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