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

$$\begin{gathered} {\text{Let x}}^2=4co{s}^2\theta +si{n}^2\theta then 4-x^2=3 si{n}^2 \theta and x^2-1=3 co{s}^2 \theta \\ now f(x)=\sqrt{4-x^2}+\sqrt{x^2-1} \\ \therefore f(x) =\sqrt{3}\left|sin\theta \right| +\sqrt{3}\left|cos \theta \right| \end{gathered}$$

$$\begin{gathered} (f{\left(x\right))}^2=3si{n}^2\theta +3co{s}^2\theta +2(3)\left|sin \theta cos\theta \right| \\ \Rightarrow (f{\left(x\right))}^2=3+3\left|sin2 \theta \right| \\ \Rightarrow max of \left|sin2\theta \right| =1 \\ Hence, maximum value of (f{\left(x\right))}^2 is 6 \end{gathered}$$