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(A) The function $f(x)$ is defined as:$f(x) = x - [x]$ if $[x]$ is odd,and $f(x) = 1 + [x] - x$ if $[x]$ is even.For $x \in [0, 1)$,$[x] = 0$ (even),s

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$4$

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(A) The function $f(x)$ is defined as:$f(x) = x - [x]$ if $[x]$ is odd,and $f(x) = 1 + [x] - x$ if $[x]$ is even.For $x \in [0, 1)$,$[x] = 0$ (even),so $f(x) = 1 + 0 - x = 1 - x$.For $x \in [1, 2)$,$[x] = 1$ (odd),so $f(x) = x - 1$.This function is periodic with period $T = 2$.Let $I = \int_{-10}^{10} f(x) \cos(\pi x) \, dx$.Since $f(x)$ is an even function $(f(-x) = f(x))$ and $\cos(\pi x)$ is an even function,their product is even.$I = 2 \int_{0}^{10} f(x) \cos(\pi x) \, dx = 2 	imes 5 \int_{0}^{2} f(x) \cos(\pi x) \, dx = 10 \left[ \int_{0}^{1} (1-x) \cos(\pi x) \, dx + \int_{1}^{2} (x-1) \cos(\pi x) \, dx ight]$.Let $I_1 = \int_{0}^{1} (1-x) \cos(\pi x) \, dx$. Using integration by parts:$I_1 = \left[ (1-x) \frac{\sin(\pi x)}{\pi} ight]_0^1 - \int_{0}^{1} (-1) \frac{\sin(\pi x)}{\pi} \, dx = 0 + \left[ -\frac{\cos(\pi x)}{\pi^2} ight]_0^1 = -\frac{1}{\pi^2} (-1 - 1) = \frac{2}{\pi^2}$.Let $I_2 = \int_{1}^{2} (x-1) \cos(\pi x) \, dx$. Let $t = x-1$,then $dt = dx$:$I_2 = \int_{0}^{1} t \cos(\pi(t+1)) \, dt = \int_{0}^{1} t \cos(\pi t + \pi) \, dt = -\int_{0}^{1} t \cos(\pi t) \, dt$.Using integration by parts for $\int t \cos(\pi t) \, dt = \frac{t \sin(\pi t)}{\pi} + \frac{\cos(\pi t)}{\pi^2}$:$I_2 = -\left[ \frac{t \sin(\pi t)}{\pi} + \frac{\cos(\pi t)}{\pi^2} ight]_0^1 = -\left( 0 + \frac{-1}{\pi^2} - (0 + \frac{1}{\pi^2}) ight) = \frac{2}{\pi^2}$.Thus,$I = 10 \left( \frac{2}{\pi^2} + \frac{2}{\pi^2} ight) = \frac{40}{\pi^2}$.Finally,$\frac{\pi^2}{10} I = \frac{\pi^2}{10} 	imes \frac{40}{\pi^2} = 4$.

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