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Let $f(x)=[x]\sin\bigg[\large\frac{\pi}{[x+1]}\bigg]$ where [.] denotes the greatest integer function. The domain of $f$ is and the point of discontinuity of $f$ in the domain are

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

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Thus function is not defined for those values of x for which $[x+1]=0$.In other words it means that $0\leq x+1 < 1$ or $-1 \leq x < 0$-------(1)
Hence the function is defined outside the region given by (1)
Required domain is $[-\infty,-1]\cup [0,\infty]$
Now consider integral values of x say $x=0$
RHL=$\lim\limits_{h\to 0}[n+h]\sin\large\frac{\pi}{[n+1+h]}$$=n\sin\large\frac{\pi}{n+1}$
LHL=$\lim\limits_{h\to 0}[n-h]\sin\large\frac{\pi}{[n+1-h]}$$=n-h\sin\large\frac{\pi( n-1)}{n}$
Clearly LHL $\neq$ RHL.Hence the given function is not continuous for integral values of $n(n\neq 0,-1)$
At $x=0,f(0)=0$
$\lim\limits_{h\to 0}f(0+h)=\lim\limits_{h\to 0}[h]\sin\large\frac{\pi}{[h+1]}$$=0$
The function is not defined for $x < 0$.Hence we cannot find $\lim\limits_{h\to 0}f(0-h)$.Thus f(x) is continuous at x=0.Hence the points of discontinuity are given by $I-\{0\}$ where $I$ is set of integers n except n=-1.
Hence (a) is the correct answer.
answered Jan 3, 2014 by sreemathi.v
 

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