The domain of the function $\mathrm{f}\left(\mathrm{x}\right)={\log }_{\mathrm{e}} \left\{\mathrm{sgn} \left(9-{\mathrm{x}}^2\right)\right\}+\sqrt{[\mathrm{x}{]}^3-4[\mathrm{x}]}$ where [.] = G.I.F

Given $\mathrm{f}\left(\mathrm{x}\right)={\log }_{\mathrm{e}} \left\{\mathrm{sgn} \left(9-{\mathrm{x}}^2\right)\right\}+\sqrt{[\mathrm{x}{]}^3-4[\mathrm{x}]}={\mathrm{y}}_1+{\mathrm{y}}_2$(say) Now, ${\mathrm{y}}_1$ is defined if sgn$\left(9-{\mathrm{x}}^2\right)>0$

But $\mathrm{sgn} \mathrm{x}=1(\mathrm{i}.\mathrm{e}. >\hspace{0.17em}0) \mathrm{if} \mathrm{x}>0$

$\therefore \mathrm{sgn} \left(9-{\mathrm{x}}^2\right)>0\Rightarrow 9-{\mathrm{x}}^2>0\Rightarrow {\mathrm{x}}^2-9<0\Rightarrow (\mathrm{x}-3)(\mathrm{x}+3)<0\Rightarrow -3<\mathrm{x}<3$Again ${\mathrm{y}}_2$ is defined if $[\mathrm{x}{]}^3-4[\mathrm{x}]\ge 0\Rightarrow [\mathrm{x}]\left\{[\mathrm{x}{]}^2-4\right\}\ge 0\Rightarrow [\mathrm{x}\left]\right(\left[\mathrm{x}\right]-2\left)\right(\left[\mathrm{x}\right]+2)\ge 0$Following the wavy curve method, we find

Thus $\left[\mathrm{x}\right]\ge 2$ or [x] lies between – 2 and 0, i.e.[x] = -2,  -1 or 0
Now, $\left[\mathrm{x}\right]\ge 2\Rightarrow \mathrm{x}\ge 2$

$\left[\mathrm{x}\right]=-2\Rightarrow -2\le \mathrm{x}<1\left[\mathrm{x}\right]=-1\Rightarrow -1\le \mathrm{x}<0\left[\mathrm{x}\right]=0\Rightarrow 0\le \mathrm{x}<1$Hence $\left[\mathrm{x}\right]=-2,-1,0\Rightarrow -2\le \mathrm{x}<1\therefore \left(\mathrm{B}\right)\cup \left(\mathrm{C}\right)=(\mathrm{x}\ge 2)$...(C)Hence $D_{\mathit{f}}\mathit{=}\mathit{\left(}A\mathit{\right)}\mathit{\cup }\mathit{\left(}C\mathit{\right)}\mathit{=}\mathit{[}\mathit{-}\mathit{2}\mathit{,1}\mathit{)}\mathit{\cup }\mathit{[}\mathit{2}\mathit{,3}\mathit{)}$