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(A) The function $f(x)$ is periodic with period $T = 2$. Given the integral $I = \frac{\pi^2}{10} \int_{-10}^{10} f(x) \cos(\pi x) dx$. Since $f(x)$ i

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$4$

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(A) The function $f(x)$ is periodic with period $T = 2$. Given the integral $I = \frac{\pi^2}{10} \int_{-10}^{10} f(x) \cos(\pi x) dx$. Since $f(x)$ is periodic with period $2$ and $\cos(\pi x)$ is also periodic with period $2$,the integral over $[-10, 10]$ is $10$ times the integral over one period $[0, 2]$. $I = \frac{\pi^2}{10} 	imes 10 \int_{0}^{2} f(x) \cos(\pi x) dx = \pi^2 \int_{0}^{2} f(x) \cos(\pi x) dx$. 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 $f(x) = x - 1$. Thus,$I = \pi^2 \left( \int_{0}^{1} (1 - x) \cos(\pi x) dx + \int_{1}^{2} (x - 1) \cos(\pi x) dx ight)$. Evaluating the first integral: $\int_{0}^{1} (1 - x) \cos(\pi x) dx = \left[ (1 - x) \frac{\sin(\pi x)}{\pi} ight]_0^1 - \int_0^1 (-1) \frac{\sin(\pi x)}{\pi} dx = 0 + \left[ -\frac{\cos(\pi x)}{\pi^2} ight]_0^1 = -\frac{1}{\pi^2} (-1 - 1) = \frac{2}{\pi^2}$. Evaluating the second integral: $\int_{1}^{2} (x - 1) \cos(\pi x) dx = \left[ (x - 1) \frac{\sin(\pi x)}{\pi} ight]_1^2 - \int_1^2 (1) \frac{\sin(\pi x)}{\pi} dx = 0 - \left[ -\frac{\cos(\pi x)}{\pi^2} ight]_1^2 = \frac{1}{\pi^2} (\cos(2\pi) - \cos(\pi)) = \frac{1}{\pi^2} (1 - (-1)) = \frac{2}{\pi^2}$. Summing these,$I = \pi^2 \left( \frac{2}{\pi^2} + \frac{2}{\pi^2} ight) = \pi^2 \left( \frac{4}{\pi^2} ight) = 4$.

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