Let [x] denote the greatest integer less than or equal to x. If $\mathrm{f}\left(\mathrm{x}\right)=[\mathrm{xsin} \mathrm{\pi x}]$ then f(x) is:

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)=\left[\mathrm{xsin\pi x}\right] \\ \mathrm{f}\left(0\right)=0, \\ \mathrm{f}\left(0^{-}\right)={\lim }_{\mathrm{h}\to 0^{-}} [-\mathrm{hsin} \mathrm{\pi }(-\mathrm{h}\left)\right]=0; \\ \mathrm{f}\left(0^{+}\right)={\lim }_{\mathrm{h}\to 0^{+}} [\mathrm{hsin} \mathrm{\pi h}]=0 \end{gathered}$$

Here $f\left(0^{-}\right)=f\left(0^{+}\right)=f\left(0\right)=0$ So f(x) is continuous at x = 0

Thus f(x) is constant and equal to 0 in the interval [−1;1] and
so it is continuous and differentiable in (−1,1). In particular, f is not
continuous at x=1 and, not differentiable at x=1.

Since graph of f(x) is as shown in the
figure

Now all options can be checked from graph