Let f : (0, pi) rightarrow mathbb{R} be a twi | vedclass

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(C) Given $\lim _{t \rightarrow x} \frac{f(x) \sin t - f(t) \sin x}{t-x} = \sin^2 x$.Applying $L$'$H$ôpital's rule with respect to $t$:$\lim _{t \righ

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$B, C, D$

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(C) Given $\lim _{t ightarrow x} \frac{f(x) \sin t - f(t) \sin x}{t-x} = \sin^2 x$.Applying $L$'$H$ôpital's rule with respect to $t$:$\lim _{t ightarrow x} \frac{f(x) \cos t - f^{\prime}(t) \sin x}{1} = \sin^2 x$.Substituting $t = x$:$f(x) \cos x - f^{\prime}(x) \sin x = \sin^2 x$.Dividing by $\sin^2 x$:$\frac{f^{\prime}(x) \sin x - f(x) \cos x}{\sin^2 x} = -1$.This is the derivative of $\frac{f(x)}{\sin x}$:$\frac{d}{dx} \left( \frac{f(x)}{\sin x} ight) = -1$.Integrating both sides:$\frac{f(x)}{\sin x} = -x + C$.Given $f \left(\frac{\pi}{6}ight) = -\frac{\pi}{12}$,we have $\frac{-\pi/12}{1/2} = -\frac{\pi}{6} = -\frac{\pi}{6} + C$,so $C = 0$.Thus,$f(x) = -x \sin x$.Checking options:$(A) f \left(\frac{\pi}{4}ight) = -\frac{\pi}{4} \cdot \frac{1}{\sqrt{2}} = -\frac{\pi}{4\sqrt{2}} 
eq \frac{\pi}{4\sqrt{2}}$. (False)$(B) f(x) = -x \sin x$. Since $\sin x > x - \frac{x^3}{6}$,then $-x \sin x $(C) f^{\prime}(x) = -\sin x - x \cos x$. $f^{\prime}(x) = 0 \Rightarrow 	an x = -x$. There exists $\alpha \in (0, \pi)$ such that $	an \alpha = -\alpha$. (True)$(D) f^{\prime \prime}(x) = -\cos x - \cos x + x \sin x = x \sin x - 2 \cos x$. $f^{\prime \prime}\left(\frac{\pi}{2}ight) + f\left(\frac{\pi}{2}ight) = (\frac{\pi}{2} - 0) + (-\frac{\pi}{2}) = 0$. (True)

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