If $f(x)=[x][\sin \pi x], x \in(-1,1)$. Then, f(x) is

continuous on (-1, 0)

We have,

$f(x)=[x \sin \pi x]=\left\{\begin{aligned} -1 \times-1=1, & \text { if }-1<x<0 \\ 0 \times 0=0, & \text { if } ~0 \leq x<1 \end{aligned}\right.$

Clearly, f(x) is not continuous at x = 0.

Consequently, it is not differentiable at x = 0.

Since f(x) is a constant function on $(-1,0) \cup(0,1)$. So, it is continuous and differentiable on $(-1,0) \cup(0,1)$.