MCQ
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 $\mathrm{f}$, then $\alpha^2+2 \beta^2$ is equal to
- A$44$
- ✓$42$
- C$24$
- D$38$
$ \mathrm{x}-2 \geq 0 \& 4-\mathrm{x} \geq 0 $
$ \therefore \mathrm{x} \in[2,4] $
$ \text { Let } \mathrm{x}=2 \sin ^2 \theta+4 \cos ^2 \theta $
$ \therefore \mathrm{f}(\mathrm{x})=3 \sqrt{2}|\cos \theta|+\sqrt{2}|\sin \theta| $
$ \therefore \sqrt{2} \leq 3 \sqrt{2}|\cos \theta|+\sqrt{2}|\sin \theta| \leq \sqrt{9 \times 2+2} $
$ \sqrt{2} \leq 3 \sqrt{2}|\cos \theta|+\sqrt{2}|\sin \theta| \leq \sqrt{20} $
$ \therefore \alpha=\sqrt{2} \quad \beta=\sqrt{20} $
$ \alpha^2+2 \beta^2=2+40=42$
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