$\text{ Let }f:\left[0,\frac{\pi }{2}\right]\to [0,1]\text{ be a differentiable function such that }f\left(0\right)=0,f\left(\frac{\pi }{2}\right)=1\text{ then }$

$\text{ Consider }g\left(x\right)={\sin }^{-1}f\left(x\right)-x$

$g\left(0\right)=0,g\left(\frac{\pi }{2}\right)=0$

$\text{ A) By Rolle's theorem there is atleast one value of }\alpha \in \left(0,\frac{\pi }{2}\right)\text{ such that }$

$g^{\text{'}}\left(\alpha \right)=0\Rightarrow \frac{f^{\text{'}}\left(\alpha \right)}{{\sqrt{1-f\left(\alpha \right)}}^2}-1=0\Rightarrow f^{\text{'}}\left(\alpha \right)=\sqrt{1-\left(f\right(\alpha ){)}^2}$

$\text{ It is not true for all values of }a$

$\text{ B) Consider }g\left(x\right)=f\left(x\right)-\frac{2x}{\pi }$

$g\left(0\right)=0,g\left(\frac{\pi }{2}\right)=0$

$g^{\text{'}}\left(x\right)=0\text{ at least for one value of }\alpha \in \left(0,\frac{\pi }{2}\right)$

$\Rightarrow f^{\text{'}}\left(\alpha \right)=\frac{2}{\pi }\text{ for atleast one value of }\alpha \in \left(0,\frac{\pi }{2}\right)$

$\text{ C) Consider }g\left(x\right)=\left[f\right(x){]}^2-\frac{4x^2}{{\pi }^2}$

$g\left(0\right)=0,g\left(\frac{\pi }{2}\right)=0$

$\Rightarrow g^{\text{'}}\left(x\right)=0\text{ for atleast one value of }\alpha \in \left(0,\frac{\pi }{2}\right)$

$f\left(\alpha \right)f^{\text{'}}\left(\alpha \right)=\frac{1}{\pi }\left(\text{ true }\right)$

$\text{ D) Consider }g\left(x\right)=f\left(x\right)-\frac{4x^2}{{\pi }^2},g\left(0\right)=0\text{ and }g\left(\frac{\pi }{2}\right)=0$

$f^{\text{'}}\left(\alpha \right)=\frac{8\alpha }{{\pi }^2}\text{ atleast for one value of }\alpha \in \left(0,\frac{\pi }{2}\right)$