Let $\mathrm{f}\left(\mathrm{x}\right)=\frac{2}{\mathrm{\pi }}\left({\sin }^{-1}\left[\mathrm{x}\right]+{\tan }^{-1}\left[\mathrm{x}\right]+{\cot }^{-1}\left[\mathrm{x}\right]\right)\text{ where }\left[\mathrm{x}\right]$ denotes greatest integer less than or equal to x. If A and B denote the domain and range of f (x) respectively, then the number of integers in $(\mathrm{A}\cup \mathrm{B})\text{, is }$

$\mathrm{f}\left(\mathrm{x}\right)=\frac{2}{\mathrm{\pi }}\left({\sin }^{-1}\left[\mathrm{x}\right]+{\tan }^{-1}\left[\mathrm{x}\right]+{\cot }^{-1}\left[\mathrm{x}\right]\right)$
For domain of f(x), we must have –1 ≤ [x] ≤ 1
$\Rightarrow -1\le \mathrm{x}<2, \mathrm{so} \mathrm{set} \mathrm{A} = [–1, 2)$
$\mathrm{f}\left(\mathrm{x}\right)=\left({\sin }^{-1}\left[\mathrm{x}\right]+\frac{\mathrm{\pi }}{2}\right)$
$\left(\text{ As }{\tan }^{-1}\left[\mathrm{x}\right]+{\cot }^{-1}\left[\mathrm{x}\right]=\frac{\mathrm{\pi }}{2}\mathrm{\forall }\mathrm{x}\in \mathrm{A}\right)$
So, set B = {0, 1, 2} = Range of f(x).
$\mathrm{Now} \mathrm{A}\cup \mathrm{B}=[-1,2)\cup \{0,1,2\}=[-1,2]$
Hence number of integers in $(\mathrm{A}\cup \mathrm{B})=4$