For x in mathbb{R},let f(x) = |sin x| and g(x | vedclass

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(A) Given $f(x) = |\sin x|$ and $g(x) = \int_0^x |\sin t| \, dt$.$p(x) = g(x) - \frac{2}{\pi} x$.We evaluate $p(x + \pi) = g(x + \pi) - \frac{2}{\pi}(

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$p(x + \pi) = p(x)$ for all $x$

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(A) Given $f(x) = |\sin x|$ and $g(x) = \int_0^x |\sin t| \, dt$.$p(x) = g(x) - \frac{2}{\pi} x$.We evaluate $p(x + \pi) = g(x + \pi) - \frac{2}{\pi}(x + \pi)$.$g(x + \pi) = \int_0^{x + \pi} |\sin t| \, dt = \int_0^{\pi} |\sin t| \, dt + \int_{\pi}^{x + \pi} |\sin t| \, dt$.Since $|\sin t|$ is periodic with period $\pi$,$\int_{\pi}^{x + \pi} |\sin t| \, dt = \int_0^x |\sin t| \, dt = g(x)$.Also,$\int_0^{\pi} |\sin t| \, dt = 2 \int_0^{\pi/2} \sin t \, dt = 2[-\cos t]_0^{\pi/2} = 2(0 - (-1)) = 2$.Thus,$g(x + \pi) = 2 + g(x)$.Substituting this into $p(x + \pi)$:$p(x + \pi) = (2 + g(x)) - \frac{2}{\pi} x - \frac{2}{\pi} \cdot \pi = 2 + g(x) - \frac{2}{\pi} x - 2 = g(x) - \frac{2}{\pi} x = p(x)$.Therefore,$p(x + \pi) = p(x)$ for all $x$.

For $x \in \mathbb{R}$,let $f(x) = |\sin x|$ and $g(x) = \int_0^x f(t) \, dt$. Let $p(x) = g(x) - \frac{2}{\pi} x$. Then:

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