If $f\left(x\right)={\sin }^{-1}\left(x\right).{\cos }^{-1}\left(x\right).{\tan }^{-1}\left(x\right).{\cot }^{-1}\left(x\right).{\sec }^{-1}\left(x\right).\cos e{c}^{-1}\left(x\right)$ then which of the following statement is correct?

Domain of ${\sin }^{-1}x$  and  ${\cos }^{-1}x,$ each is  $[-1,1]$ and that of  ${\sec }^{-1}x$ and $\cos e{c}^{-1}x,$ each is  
  Domain of $f\left(x\right)$  must be  $\{-1,1\}$
  Range of $f\left(x\right)$  will be  $\{f(-1),f\left(1\right)\}$
Where,  $f(-1)={\sin }^{-1}(-1).{\cos }^{-1}(-1).{\tan }^{-1}(-1)$
 . ${\cot }^{-1}(-1).{\sec }^{-1}(-1).\cos e{c}^{-1}(-1)$
 $=\left(\frac{-\pi }{2}\right).\left(\pi \right).\left(\frac{-\pi }{4}\right).\left(\frac{3\pi }{4}\right).\left(\pi \right).\left(\frac{-\pi }{2}\right)$
 $=\frac{-3{\pi }^6}{64}$ and  $f\left(1\right)=0\{as{\cos }^{-1}1=0\}$
(i)  Thus, the graph of $f\left(x\right)$  is a two point graph which doesn’t lie above  $x-axis.$
  (a) is correct statement.
ii) $f{\left(x\right)}_{\max }=0$  and  $f{\left(x\right)}_{\min }=\frac{-3{\pi }^6}{64}$
 $\therefore |f{\left(x\right)}_{\max }-f{\left(x\right)}_{\min }|=\frac{3{\pi }^6}{64}$
 
  (b) is correct statement .
(iii) $f\left(x\right)$  is one-one hence, injective 
  (c) is incorrect statement .
(iv) Domain is  $\{-1,1\}$
  Number of non-negative integers in the domain of  $f\left(x\right)$ is one.
  (d) is incorrect statement.