Let  $g\left(x\right)=2f\left(\frac{x}{2}\right)+f(2-x)$  and  $f\text{'}\text{'}\left(x\right)>0$  for  $x\in (0,3)$. If g is decreasing in $(0,\alpha )$  and increasing in  $(\alpha ,3)$, then $\underset{\theta \to \frac{3}{4}\alpha -1}{\lim }\frac{1-\cos 2\theta }{\theta \sin 2\theta }$  is

$g\left(x\right)=2.f\left(\frac{x}{2}\right)+f(2-x)$ 
 $g\text{'}\left(x\right)=2.f\text{'}\left(\frac{x}{2}\right)+f\text{'}(2-x)(-1)$
 $g\text{'}\left(x\right)=f\text{'}\left(\frac{x}{2}\right)-f\text{'}(2-x)\mathrm{.....}\left(1\right)$
$g\left(x\right)$  is decreasing
 $\Rightarrow g\text{'}\left(x\right)<0$
By eqn (1)  $f\text{'}\left(\frac{x}{2}\right)-f\text{'}(2-x)<0$
 $$\begin{gathered} \Rightarrow f\text{'}\left(\frac{x}{2}\right)<f\text{'}(2-x) \\ \Rightarrow \frac{x}{2}<2-x\left[\because f\text{'}\text{'}\left(x\right)>0\Rightarrow f\text{'}\left(x\right)isincrea\sin gin(0,3)\right] \\ \Rightarrow \frac{x}{2}+x<2\Rightarrow \frac{3x}{2}<2\Rightarrow x<\frac{4}{3} \\ \Rightarrow \alpha =\frac{4}{3} \end{gathered}$$