$\begin{array}{1 1}(a)\;\text{continuous at x=0}\\(b)\;\text{continuous in (0,0)}\\(c)\;\text{differentiable at x=1}\\(d)\;\text{differentiable in (-1,1)}\end{array}$

We have for $-1\leq x \leq 1$

$\Rightarrow 0\leq x\sin\pi x\leq \large\frac{1}{2}$

$\therefore f(x)=[x\sin\pi x]=0$

Also $x\sin \pi x$ becomes negative and numerically less than 1 and so by definition of [x]

$f(x)=[x\sin \pi x]=-1\;when 1 < x < 1+h$

Thus $f(x)$ is constant and equal to 0 in the closed interval [-1,1] and so f(x) is continuous and differentiable in the open interval (-1,1)

Hence (d) is the correct answer.

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