Explanation

(c)
$\because\lbrack\sin^{- 1}{x\rbrack > \lbrack\cos^{- 1}{x\rbrack}}$
$\Rightarrow x > 0$
Here, $\lbrack cos^{- 1}\ x\rbrack = \left\{ \begin{matrix}
0, & x \in (\cos{1,1}) \\
1, & x \in (0,\cos{1)}
\end{matrix} \right.\$
and, $\lbrack\sin^{- 1}\ x\rbrack = \left\{ \begin{matrix}
0, & x \in (0{,sin}{1\ }) \\
1, & x \in ({sin\ 1,}{1)}
\end{matrix} \right.\$
$\therefore x \in \lbrack\sin{1,1)}$
$\because\left\lbrack \frac{x}{2} \right\rbrack = 1$
Or we say that $x \in \lbrack\sin{1,1\rbrack}$