1. Find the domain of $f(x)$

Given $f(x)=3\sqrt{x-2}+\sqrt{4-x}$ for real values, we need: $x-2\ge 0 \quad \text{and} \quad 4-x\ge 0$ So, $2\le x\le 4$

Thus, the function is defined on the closed interval $[2,4]$.

2. Find extrema on $[2,4]$

Since $f$ is continuous on $[2,4]$, its minimum and maximum occur either at critical points or at endpoints.

Differentiate: $f'(x)=3\cdot \frac{1}{2\sqrt{x-2}}+\frac{d}{dx}(4-x)^{1/2}$ $f'(x)=\frac{3}{2\sqrt{x-2}}-\frac{1}{2\sqrt{4-x}}$

Set $f'(x)=0$: $\frac{3}{2\sqrt{x-2}}-\frac{1}{2\sqrt{4-x}}=0$ $\frac{3}{\sqrt{x-2}}=\frac{1}{\sqrt{4-x}}$ $3\sqrt{4-x}=\sqrt{x-2}$

Squaring both sides: $9(4-x)=x-2$ $36-9x=x-2$ $38=10x$ $x=\frac{19}{5}$

This lies in $[2,4]$, so it is a critical point.

3. Evaluate $f(x)$ at endpoints and critical point

• At $x=2$: $f(2)=3\sqrt{0}+\sqrt{2}=\sqrt{2}$

• At $x=4$: $f(4)=3\sqrt{2}+\sqrt{0}=3\sqrt{2}$

• At $x=\frac{19}{5}$: $x-2=\frac{19}{5}-\frac{10}{5}=\frac{9}{5}, \qquad 4-x=\frac{20}{5}-\frac{19}{5}=\frac{1}{5}$ So, $f\left(\frac{19}{5}\right)=3\sqrt{\frac{9}{5}}+\sqrt{\frac{1}{5}}$ $=3\cdot \frac{3}{\sqrt{5}}+\frac{1}{\sqrt{5}}=\frac{10}{\sqrt{5}}=2\sqrt{5}$

Now compare: $\sqrt{2},\quad 3\sqrt{2},\quad 2\sqrt{5}$ Numerically, $\sqrt{2}\approx 1.414,\quad 3\sqrt{2}\approx 4.243,\quad 2\sqrt{5}\approx 4.472$ Hence, $\alpha=\min f(x)=\sqrt{2}, \qquad \beta=\max f(x)=2\sqrt{5}$

4. Compute $\alpha^2+2\beta^2$

$\alpha^2=(\sqrt{2})^2=2$ $\beta^2=(2\sqrt{5})^2=20$ Therefore, $\alpha^2+2\beta^2=2+2(20)=2+40=42$

5. Match with options

$42$ So the correct option is A.