$\begin{array}{l}\text{ Let }f:(0,\pi )\to \mathrm{ℝ}\text{ be a twice differentiable function such that }\underset{t\to \infty }{\lim }\frac{f\left(x\right)\sin t-f\left(t\right)\sin x}{t-x}={\sin }^{2}x\\ \text{ for all }x\in (0,\pi )\text{ . If }f\left(\frac{\pi }{6}\right)=-\frac{\pi }{12},\text{ then which of the following statement (s) is (are) TRUE? }\end{array}$

$\begin{array}{l}\underset{t\to x}{\lim }\frac{f\left(x\right)\sin t-f\left(t\right)\sin x}{t-x}={\sin }^{2}x \text{ By using L'Hopital }\\ \underset{t\to x}{\lim }\frac{f\left(x\right)\cos t-f^1\left(t\right)\sin x}{1}={\sin }^{2}x\\ \Rightarrow f\left(x\right)\cos x-f\left(x\right)\sin x={\sin }^{2}x \Rightarrow -\left(\frac{f^1\left(x\right)\sin x-f\left(x\right)\cos x}{{\sin }^{2}x}\right)=1\\ \Rightarrow -d\left(\frac{f\left(x\right)}{\sin x}\right)=1 \Rightarrow \frac{f\left(x\right)}{\sin x}=x+c\end{array}$

$\text{ Put }x=\frac{\pi }{6}\&f\left(\frac{\pi }{6}\right)=-\frac{\pi }{12} \therefore c=0\Rightarrow f\left(x\right)=-x\sin x$

$\text{ A) }f\left(\frac{\pi }{4}\right)=\frac{-\pi }{4}\frac{1}{\sqrt{2}}$

$\text{ B) }f\left(x\right)=-x\sin x\text{ as } \sin x>x-\frac{x^3}{6},-x\sin x<-x^2+\frac{x^4}{6} \therefore f\left(x\right)<-x^2+\frac{x^4}{6}\forall x\in (0,\pi )$

$\text{ (c) }{f}^{1}x=-\sin x-x\cos x{f}^1\left(x\right)=0\Rightarrow \tan x=-x\Rightarrow \text{ there exist }\alpha \in (0,\pi )\text{ for which }{f}^1\left(\alpha \right)=0$

$\text{ D) }{f}^{11}\left(x\right)=-2\cos x+x\sin x f^{11}\left(\frac{\pi }{2}\right)=\frac{\pi }{2},f\left(\frac{\pi }{2}\right)=-\frac{\pi }{2} f^{11}\left(\frac{\pi }{2}\right)+f\left(\frac{\pi }{2}\right)=0$