Let [x] be the greatest integer less than or equal to x. Then, $f(x)=x \cos (\pi(x+[x]))$ is continuous at

x = 0

We have,

$f(x)=x \cos (\pi(x+[x])=x \cos (\pi[x]+\pi x)$

$\Rightarrow f(x)=\left\{\begin{array}{c}x \cos \pi x, & \text { for }-2 \leq x<-1 \\ -x \cos \pi x, & \text { for }-1 \leq x<0 \\ x \cos \pi x, & \text { for } ~~0 \leq x<1 \\ -x \cos \pi x, & \text { for } ~~1 \leq x<2 \\ x \cos \pi x, & \text { for } ~~2 \leq x<3\end{array}\right.$

Clearly,

$\lim\limits_{x \rightarrow 0^{-}} f(x)=\lim\limits_{x \rightarrow 0^{-}}-x \cos \pi x=0$

$\lim\limits_{x \rightarrow 0^{+}} f(x)=\lim\limits_{x \rightarrow 0^{+}} x \cos \pi x=0$,  and  f(0) = 0

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

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

$\lim\limits_{x \rightarrow 1^{-}} f(x)=\lim\limits_{x \rightarrow 1} x \cos \pi x=\cos \pi=-1$

$\lim\limits_{x \rightarrow 1^{+}} f(x)=\lim\limits_{x \rightarrow 1}-x \cos \pi x=-\cos \pi=1$

∴  $\lim\limits_{x \rightarrow 1^{-}} f(x) \neq \lim\limits_{x \rightarrow 1^{+}} f(x)$

So, f(x) is continuous at x = 1

$\lim\limits_{x \rightarrow 2^{-}} f(x)=\lim\limits_{x \rightarrow 2}-x \cos \pi x=-2 \cos 2 \pi=-2$

$\lim\limits_{x \rightarrow 2^{+}} f(x)=\lim\limits_{x \rightarrow 2} x \cos \pi x=2 \cos 2 \pi=2$

∴  $\lim\limits_{x \rightarrow 2^{-}} f(x) \neq \lim\limits_{x \rightarrow 2^{+}} f(x)$

So, f(x) is not continuous at x = 2

$\lim\limits_{x \rightarrow-1^{+}} f(x)=\lim\limits_{x \rightarrow-1}-x \cos \pi x=\cos (-\pi)=-1$

$\lim\limits_{x \rightarrow-1^{-}} f(x)=\lim\limits_{x \rightarrow-1} x \cos \pi x=-\cos (-\pi)=1$

So, f(x) is discontinuous at x = -1.