$\mathrm{If} \mathrm{f}\left(\mathrm{x}\right)=\sqrt{4-{\mathrm{x}}^2}+\sqrt{{\mathrm{x}}^2-1},$ then the maximum value of ${\left(\mathrm{f}\left(\mathrm{x}\right)\right)}^2$  is __________

$\begin{array}{l}\mathrm{Let} {\mathrm{x}}^2=4 {\cos }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta }. \mathrm{Then} \\ \left(4-{\mathrm{x}}^2\right)=3{\sin }^2\mathrm{\theta } \mathrm{and} \left({\mathrm{x}}^2-1\right)=3{\cos }^2\mathrm{\theta }\\ \therefore \mathrm{f}\left(\mathrm{x}\right)=\sqrt{3} \left|\mathrm{sin\theta }\right|+\sqrt{3}\left|\mathrm{cos\theta }\right|\\ \mathrm{or} {\mathrm{y}}_{\min }=\sqrt{3} \mathrm{and} {\mathrm{y}}_{\max } =\sqrt{3}\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)=\sqrt{6}\\ \mathrm{Hence}, \mathrm{range} \mathrm{of} \mathrm{f}\left(\mathrm{x}\right) \mathrm{is} \left[\sqrt{3},\sqrt{6}\right].\\ \mathrm{Hence}, \mathrm{maximum} \mathrm{value} \mathrm{of} {\left( \mathrm{f}\left(\mathrm{x}\right)\right)}^2 \mathrm{is} 6.\end{array}$