Let $f(x)=3 \sqrt{x-2}+\sqrt{4-x}$ be a real valued function. If $\alpha$ and $\beta$ are respectively the minimum and the maximum values of $f$, then $\alpha^2+2 \beta^2$ is equal to

<p>$$\begin{aligned} & f(x)=3 \sqrt{x-2}+\sqrt{4-x} \\ & \text { Let } x=2 \sin ^2 \theta+4 \cos ^2 \theta \\ & =3 \sqrt{2 \sin ^2 \theta+4 \cos ^2 \theta-2}+\sqrt{4-2 \sin ^2 \theta-4 \cos ^2 \theta} \\ & =3 \sqrt{2 \cos ^2 \theta}+\sqrt{2 \sin ^2 \theta} \\ & =3 \sqrt{2}|\cos \theta|+\sqrt{2}|\sin \theta| \end{aligned}$$</p> <p>$$\begin{aligned} & \Rightarrow 3 \sqrt{2} \cos \theta+\sqrt{2} \sin \theta \leq \sqrt{18+2} \\ & \Rightarrow 3 \sqrt{2} \cos \theta+\sqrt{2} \sin \theta \leq \sqrt{20} \end{aligned}$$</p> <p>Minimum value exists when $\theta=\frac{\pi}{2}$</p> <p>$$\begin{aligned} & \text { So, minimum value }=\sqrt{2} \\ & \Rightarrow \alpha=\sqrt{2} \text { and } \beta=\sqrt{20} \\ & \Rightarrow \alpha^2+2 \beta^2=2+40 \\ & =42 \end{aligned}$$</p>