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(B) Given $f(x) = \sin(2 \pi x)$.For Rolle's theorem,$f'(C) = 0$ for some $C \in (-1, 1)$.$f'(x) = 2 \pi \cos(2 \pi x)$.Setting $f'(C) = 0$ gives $2 \

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$4$

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(B) Given $f(x) = \sin(2 \pi x)$.For Rolle's theorem,$f'(C) = 0$ for some $C \in (-1, 1)$.$f'(x) = 2 \pi \cos(2 \pi x)$.Setting $f'(C) = 0$ gives $2 \pi \cos(2 \pi C) = 0$,which implies $\cos(2 \pi C) = 0$.Since $C \in (-1, 1)$,we have $2 \pi C \in (-2 \pi, 2 \pi)$.The values of $2 \pi C$ in the interval $(-2 \pi, 2 \pi)$ where $\cos(2 \pi C) = 0$ are $\frac{-3 \pi}{2}, \frac{-\pi}{2}, \frac{\pi}{2}, \frac{3 \pi}{2}$.Dividing by $2 \pi$,we get $C = \frac{-3}{4}, \frac{-1}{4}, \frac{1}{4}, \frac{3}{4}$.Thus,there are $4$ such values of $C$.

The number of values of $C$ that satisfy the conclusion of Rolle's theorem for the function $f(x) = \sin(2 \pi x)$ on the interval $x \in [-1, 1]$ is:

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