Consider,$f\left(x\right)={\sin }^{-1}\left(1-2\sqrt{x}\right)+{\sec }^{-1}\left(\frac{1}{2\sqrt{\sqrt{x}-x}}\right)+{\tan }^{-1}\left(\frac{\sqrt{2}-1-\sqrt{x}}{1+\sqrt{2x}-\sqrt{x}}\right),x\in (0,1)$   Identify incorrect statement (s) about  $f\left(x\right)$

Note:    $f\text{'}\left(x\right)$ means derivative of   $f\left(x\right) w.r.t x$
 

  $f\left(x\right)={\sin }^{-1}(1-2\sqrt{x})+{\cos }^{-1}\left(2\sqrt{\sqrt{x}-x}\right)$
Domain:   $x\in (0,1)$
$f\left(x\right)={\sin }^{-1}(1-2\sqrt{x})+{\sin }^{-1}\left|2\sqrt{x-1}\right|+{\tan }^{-1}(\sqrt{2}-1)$
$-{\tan }^{-1}\left(\sqrt{x}\right)$
$\therefore f\left(x\right)=\{\begin{array}{c}\frac{\pi }{8}-{\tan }^{-1}\sqrt{x};forx\in (\frac{1}{4},1)\\ 2{\sin }^1(1-2\sqrt{x})+\frac{\pi }{8}-{\tan }^{-1}\sqrt{x};forx\in (0,\frac{1}{4})\end{array}$
$\therefore f\text{'}\left(\frac{1^{+}}{4}\right)=\frac{-1}{1}.\frac{1}{1}=\frac{-4}{5}$
And   $f\text{'}\left(\frac{1}{4}\right)=\frac{2}{\sqrt{1-0}}(-2)-\frac{4}{5}=\frac{-24}{5}$
Also,   $f\text{'}\left(x\right)<0$
$\therefore f\left(x\right)$ is decreasing in   (0,1)
Also minimum value of  $f\left(x\right)$   does not exist