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

 $$\begin{gathered} Given f\left(x\right)=\frac{2}{\mathrm{\pi }}\left({\sin }^{-1}\left[x\right]+co{t}^{-1}\left[x\right]+{\tan }^{-1}\left[x\right]\right) \\ {\sin }^{-1 }domain is \left[-1,1\right], co{t}^{-1} and {\tan }^{-1 }domain is \left(-\infty ,\infty \right). \\ \therefore The domain of f\left(x\right) is [-1,2). \left(\because if x\in [-1,2), then \left[x\right] can take -1,0,1 only \right) \\ \therefore A=[-1,2) \\ Now if x\in [-1,0), then \left[x\right]=-1 \\ \Rightarrow f\left(x\right)=\frac{2}{\mathrm{\pi }}\left(-\frac{\mathrm{\pi }}{2}+\frac{3\mathrm{\pi }}{4}-\frac{\mathrm{\pi }}{4}\right)=0 \\ for x\in [0,1), then \left[x\right]=0 \\ \Rightarrow f\left(x\right)=\frac{2}{\mathrm{\pi }}\left(0+\frac{\mathrm{\pi }}{2}+0\right)=1 \\ for x\in [1,2), then \left[x\right]=1 \\ \Rightarrow f\left(x\right)=\frac{2}{\mathrm{\pi }}\left(\frac{\mathrm{\pi }}{2}+\frac{\mathrm{\pi }}{4}+\frac{\mathrm{\pi }}{4}\right)=2 \\ \therefore range of f\left(x\right)=B=\{0,1,2\} \\ \therefore A\cup B=[-1,2)\cup \{0,1,2\}=\left[-1,2\right] \\ \therefore The number of integers in A\cup B is 4 \end{gathered}$$