Let f(x) = frac{sin pi x}{x^2}, x 0. Let x_ | vedclass

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(C) Given $f(x) = \frac{\sin \pi x}{x^2}$.Taking the derivative,$f'(x) = \frac{\pi x^2 \cos \pi x - 2x \sin \pi x}{x^4} = \frac{x(\pi x \cos \pi x - 2

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$1, 3, 4$

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(C) Given $f(x) = \frac{\sin \pi x}{x^2}$.Taking the derivative,$f'(x) = \frac{\pi x^2 \cos \pi x - 2x \sin \pi x}{x^4} = \frac{x(\pi x \cos \pi x - 2 \sin \pi x)}{x^4} = \frac{\pi x \cos \pi x - 2 \sin \pi x}{x^3}$.For local extrema,$f'(x) = 0 \implies \pi x \cos \pi x = 2 \sin \pi x \implies 	an \pi x = \frac{\pi x}{2}$.Let $g(x) = 	an \pi x$ and $h(x) = \frac{\pi x}{2}$. The points of intersection of these curves give the extrema.From the graph,the points of local maximum $x_n$ occur in the intervals $(2n, 2n + 1/2)$ and points of local minimum $y_n$ occur in the intervals $(2n-1/2, 2n)$.$(1)$ $|x_n - y_n| > 1$ is true as the distance between consecutive extrema is greater than $1$.$(2)$ $x_1$ is in $(2, 2.5)$ and $y_1$ is in $(0.5, 1)$,so $x_1 > y_1$. Thus,statement $(2)$ is false.$(3)$ $x_n \in (2n, 2n + 1/2)$ is true from the analysis of $	an \pi x = \frac{\pi x}{2}$.$(4)$ $x_{n+1} - x_n > 2$ is true because the roots of $	an \pi x = \frac{\pi x}{2}$ are separated by at least $1$,and specifically for local maxima,the gap is greater than $2$.Therefore,statements $(1)$,$(3)$,and $(4)$ are true.

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