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(B) Given $f(x) = f(1-x)$. Differentiating with respect to $x$,we get $f^{\prime}(x) = -f^{\prime}(1-x)$.At $x = 1/2$,$f^{\prime}(1/2) = -f^{\prime}(1

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(B) Given $f(x) = f(1-x)$. Differentiating with respect to $x$,we get $f^{\prime}(x) = -f^{\prime}(1-x)$.At $x = 1/2$,$f^{\prime}(1/2) = -f^{\prime}(1-1/2) = -f^{\prime}(1/2)$,which implies $2f^{\prime}(1/2) = 0$,so $f^{\prime}(1/2) = 0$. Thus,$(B)$ is correct.Since $f(x) = f(1-x)$,the graph of $f(x)$ is symmetric about $x = 1/2$. Let $x = 1/2 + t$,then $f(1/2 + t) = f(1/2 - t)$,so $g(t) = f(1/2 + t)$ is an even function.For $(C)$,$\int_{-1/2}^{1/2} f(x+1/2) \sin x dx$. Since $f(x+1/2)$ is even and $\sin x$ is odd,their product is an odd function. The integral of an odd function over a symmetric interval $[-a, a]$ is $0$. Thus,$(C)$ is correct.For $(A)$,we have $f^{\prime}(1/4) = 0$. Since $f^{\prime}(x) = -f^{\prime}(1-x)$,$f^{\prime}(3/4) = -f^{\prime}(1/4) = 0$. Also $f^{\prime}(1/2) = 0$. By Rolle's Theorem,$f^{\prime\prime}(x)$ vanishes at least once in $(1/4, 1/2)$ and at least once in $(1/2, 3/4)$. Thus,$f^{\prime\prime}(x)$ vanishes at least twice in $[0, 1]$. Thus,$(A)$ is correct.For $(D)$,let $I = \int_{1/2}^1 f(1-t) e^{\sin \pi t} dt$. Let $1-t = u$,then $dt = -du$. When $t=1/2, u=1/2$; when $t=1, u=0$. $I = \int_{1/2}^0 f(u) e^{\sin \pi (1-u)} (-du) = \int_0^{1/2} f(u) e^{\sin \pi u} du$. Thus,$(D)$ is correct.

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