Let f: (0, π) → ℝ be a twice differentiable function such that _ t...

\lim_{t \to x}\dfrac{f\left(x\right)\sin t - f\left(t\right)\sin x}{t - x} = \sin^{2}x$<br >By using L'Hopital rule<br >$\lim_{t \to x}\dfrac{f\left(x\right)\cos t - {f}^{′}\left(t\right)\sin x}{1} = \sin^{2}x$<br >$⇒ f\left(x\right)\cos x - {f}^{′}\left(x\right)\sin x = \sin^{2}x$<br >$⇒ - \left(\dfrac{{f}^{′}\left(x\right)\sin x - f\left(x\right)\cos x}{\sin^{2}x}\right) = 1$<br >$⇒ - \dfrac{d \dfrac{f\left(x\right)}{\sin x}}{d x} = 1$<br >$⇒ \dfrac{f\left(x\right)}{\sin x} = - x + c$<br >Put$x = \dfrac{\pi}{6}$and$f\left(\dfrac{\pi}{6}\right) = - \dfrac{\pi}{12}$<br >$∴ c = 0 ⇒ f\left(x\right) = - xsinx$<br >\left(A\right) f\left(\dfrac{\pi}{4}\right) = - \dfrac{\pi}{4}\dfrac{1}{\sqrt{2}}$
\left(B\right) f\left(x\right) = - xsinx$<br >As$\sin x > x - \dfrac{{x}^{3}}{6}$,$- xsinx < - {x}^{2} + \dfrac{{x}^{4}}{6}$<br >$∴ f\left(x\right) < - {x}^{2} + \dfrac{{x}^{4}}{6} \forall x \in \left(0, \pi\right)$<br >\left(C\right) {f}^{′}\left(x\right) = - \sin x - xcosx$
${f}^{′}\left(x\right) = 0 ⇒ \tan x = - x ⇒$ there exist $\alpha \in \left(0, \pi\right)$ for which ${f}^{′}\left(\alpha\right) = 0$

$\left(D\right) {f}^{"}\left(x\right) = - 2\cos x + xsinx$
${f}^{"}\left(\dfrac{\pi}{2}\right) = \dfrac{\pi}{2}, f\left(\dfrac{\pi}{2}\right) = - \dfrac{\pi}{2}$
${f}^{"}\left(\dfrac{\pi}{2}\right) + f\left(\dfrac{\pi}{2}\right) = 0$