Let f(x) = [x]sin left( frac{pi}{[x + 1]} rig | vedclass

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(C) The function $f(x) = [x]\sin \left( \frac{\pi}{[x + 1]} ight)$ is defined if $[x + 1] 
eq 0$. Since $[x + 1] = 0$ for $0 \le x + 1 < 1$,which i

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$\left\{ x \in R \mid x 
otin [ - 1, 0) ight\}, I - \{ 0 \}$

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(C) The function $f(x) = [x]\sin \left( \frac{\pi}{[x + 1]} ight)$ is defined if $[x + 1] 
eq 0$. Since $[x + 1] = 0$ for $0 \le x + 1 Thus,the domain is $\left\{ x \in R \mid x 
otin [-1, 0) ight\}$. The function $[x]$ is discontinuous at all integers $I$. For $x \in I$,$f(x)$ is discontinuous unless the product $[x]\sin \left( \frac{\pi}{[x + 1]} ight)$ is continuous. At $x = 0$,$f(0) = [0]\sin \left( \frac{\pi}{[1]} ight) = 0$. For $x 	o 0^+$,$[x] = 0$,so $f(x) = 0 \cdot \sin \left( \frac{\pi}{1} ight) = 0$. Thus,$f$ is continuous at $x = 0$. For other integers $n \in I \setminus \{0\}$,the function remains discontinuous. Therefore,the set of points of discontinuity is $I - \{0\}$.

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

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