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(D) For the interval $-1 \le x \le 1$,consider the function $g(x) = x \sin \pi x$.Since $\sin \pi x$ has the same sign as $x$ for $x \in (-1, 1)$,the

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All the above

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(D) For the interval $-1 \le x \le 1$,consider the function $g(x) = x \sin \pi x$.Since $\sin \pi x$ has the same sign as $x$ for $x \in (-1, 1)$,the product $x \sin \pi x$ is always non-negative.Specifically,for $x \in (-1, 1)$,$0 \le x \sin \pi x Therefore,the greatest integer function $[x \sin \pi x] = 0$ for all $x \in [-1, 1]$.Since $f(x) = 0$ is a constant function on the interval $[-1, 1]$,it is continuous everywhere in this interval,including at $x = 0$ and in the sub-interval $(-1, 0)$.Furthermore,a constant function is differentiable everywhere,so $f(x)$ is differentiable in $(-1, 1)$.Thus,all the given statements are correct.

Let $[x]$ denote the greatest integer less than or equal to $x$. If $f(x) = [x \sin \pi x]$,then $f(x)$ is

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