(A) We know $\sin ^{-1}\left(\frac{\sin x+\cos x}{2}\right)=\sin ^{-1}\left(\frac{\sqrt{2} \sin (x+(\pi / 4))}{2}\right)$ lie between $-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}$ for all $x \in R$
$\therefore $ domain is $x \in R$
$\sin ^{-1} x$ is increasing function and max. value of given function is $\sin ^{-1} \sqrt{\frac{1}{2}}$ which is $\pi / 4$ and minimum value $\sin ^{-1}(-\sqrt{1 / 2})$ which is $-\pi / 4$, so range contain only one integer which is zero
$\Rightarrow ( Q )$ is correct
$\therefore f ( x )=\sin ^{-1}\left(\sin \left( x +\frac{\pi}{3}\right)\right) \rightarrow$ not odd $\Rightarrow$
(R) is not correct
vertical tangent exist for $\sin ^{-1} x$ is $x = \pm 1$ but this is not possible so (S) is also correct.
(B)$g ( x )=\sin ^{-1}\left(\frac{2}{\pi} \tan ^{-1} x \right)$
$-1<\frac{2}{\pi} \tan ^{-1} x <1$ for all $x \in R . \Rightarrow$
(P) is correct
$\therefore $ range is $(-\pi / 2, \pi / 2) \Rightarrow ( Q )$ is wrong $g(x)$ is odd $\Rightarrow( R )$ is correct and no vertical tangent $\Rightarrow$
(S) is also correct
(C)$h(x) =\tan ^{-1}\left(\frac{2}{\pi}\left(2 \tan ^{-1} x+\cot ^{-1} x-\left(\sin ^{-1} x+\cos ^{-1} x\right)\right)\right) $
$ =\tan ^{-1}\left[\frac{2}{\pi}\left(\tan ^{-1} x+\frac{\pi}{2}-\frac{\pi}{2}\right)\right]=\tan ^{-1}\left(\frac{2}{\pi} \tan ^{-1} x\right)$
It is easy to analyse domain is $[-1,1]$
$-\frac{1}{2} \leq \frac{2}{\pi} \tan ^{-1} x \leq \frac{1}{2} \text { for } x \in[-1,1]$
$\therefore \tan ^{-1} \frac{1}{2} \leq 1 \Rightarrow h ( x )$ contain only one integer which is zero $h ( x )$ is odd no vertical tangent.
(D) domain is $R$ (cubic polynomial) range is $(-\pi / 2, \pi / 2)$ $x ^3+ x$ is odd
$\therefore \tan ^{-1}\left( x ^3+ x \right)$ is also odd no vertical tangent