JEE Main 2024 — 4 April, Shift 2 — Question 12

$\mathrm{f}(\mathrm{x})=3 \sqrt{\mathrm{x}-2}+\sqrt{4-\mathrm{x}}$

$x-2 \geq 0 \& 4-x \geq 0$

$\therefore \mathrm{x} \in[2,4]$

Let $x=2 \sin ^{2} \theta+4 \cos ^{2} \theta$

$\therefore \mathrm{f}(\mathrm{x})=3 \sqrt{2}|\cos \theta|+\sqrt{2}|\sin \theta|$

$\therefore \sqrt{2} \leq 3 \sqrt{2}|\cos \theta|+\sqrt{2}|\sin \theta| \leq \sqrt{9 \times 2+2}$

$\sqrt{2} \leq 3 \sqrt{2}|\cos \theta|+\sqrt{2}|\sin \theta| \leq \sqrt{20}$

$\therefore \alpha=\sqrt{2} \quad \beta=\sqrt{20}$

$\alpha^{2}+2 \beta^{2}=2+40=42$