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(D) Given $f(x) = [x] \sin(\pi x)$.$1$. Differentiability: $f(x)$ is a step function multiplied by a trigonometric function. It is discontinuous at al

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For each real $\alpha$,the equation $f(x) - \alpha = 0$ has infinitely many roots.

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(D) Given $f(x) = [x] \sin(\pi x)$.$1$. Differentiability: $f(x)$ is a step function multiplied by a trigonometric function. It is discontinuous at all integers $n \in \mathbb{Z}$ (except $n=0$). Since differentiability implies continuity,$f(x)$ is not differentiable on $\mathbb{R}$.$2$. Symmetry: $f(-x) = [-x] \sin(-\pi x) = -[-x] \sin(\pi x)$. Since $[-x] = -[x] - 1$ for $x 
otin \mathbb{Z}$,$f(-x) = ([x] + 1) \sin(\pi x) 
eq f(x)$. Thus,it is not symmetric about $x=0$.$3$. Integral: $\int_{-3}^{3} [x] \sin(\pi x) \, dx = \sum_{k=-3}^{2} \int_{k}^{k+1} k \sin(\pi x) \, dx = \sum_{k=-3}^{2} k \left[ -\frac{\cos(\pi x)}{\pi} ight]_{k}^{k+1} = \sum_{k=-3}^{2} \frac{k}{\pi} ((-1)^k - (-1)^{k+1}) = \sum_{k=-3}^{2} \frac{2k(-1)^k}{\pi} = \frac{2}{\pi} [3 - 2 + 1 - 0 - 1 + 2] = \frac{6}{\pi} 
eq 0$.$4$. Roots: For any $\alpha$,the function $f(x)$ oscillates between values determined by $[x]$. Since $\sin(\pi x)$ is periodic and $[x]$ takes integer values,the function $f(x)$ takes values in intervals that repeat. For any $\alpha$ within the range of the function,the horizontal line $y = \alpha$ will intersect the graph of $f(x)$ infinitely many times due to the periodic nature of $\sin(\pi x)$ in each interval $[n, n+1)$. Thus,option $(D)$ is correct.

For a real number $x$,let $[x]$ denote the greatest integer less than or equal to $x$. For $x \in \mathbb{R}$,let $f(x) = [x] \sin(\pi x)$. Then,

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