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

Z - {-1, 0}

Clearly, f(x) is defined for all $x \in R-[-1,0)$. For any integer $k \neq-1$, we have

$f(x)=k \sin \frac{\pi}{k+1}, k \leq x<k+1$

So, f(x), being a constant function is continuous at all points other than integers in its domain.

Also,

$\lim\limits_{x \rightarrow k^{-}} f(x)=\lim\limits_{h \rightarrow 0} f(k-h)$

$\Rightarrow \lim\limits_{x \rightarrow k^{-}} f(x)=\lim\limits_{h \rightarrow 0}[k-h] \sin \frac{\pi}{[k-h+1]}=(k-1) \sin \frac{\pi}{k}, k \neq 0$

$\lim\limits_{x \rightarrow k^{+}} f(x)=\lim\limits_{h \rightarrow 0} f(k+h)$

$\Rightarrow \lim\limits_{x \rightarrow k^{+}} f(x)=\lim\limits_{h \rightarrow 0}[k+h] \sin \frac{\pi}{[k+h+1]}$

$=k \sin \left(\frac{\pi}{k+1}\right), k \neq-1$

∴ $\lim\limits_{x \rightarrow k^{-}} f(x) \neq \lim\limits_{x \rightarrow k^{+}} f(x), k \neq 0$

Thus, f(x) is discontinuous at all non-zero integer points in its domain.

When k = 0, we have

f(x) = 0 sin π = 0 for all x ∈ [0, 1)

$\Rightarrow \lim\limits_{x \rightarrow 0^{+}} f(x)=f(0)$

So, f(x) is right continuous at x = 0

Hence, the set of points of discontinuity in its domain is Z - {-1, 0}.