Q33 / 90 · JEE Main 2024 (04 Apr Shift 2) · Mathematics
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
- Correct answerA42
- B38
- C24
- D44
Solution
$\begin{aligned}
& f(x)=3 \sqrt{x-2}+\sqrt{4-x} \\
& x-2 \geq 0 \& 4-x \geq 0 \\
& \therefore x \in[2,4]
\end{aligned}$
Let $x=2 \sin ^2 \theta+4 \cos ^2 \theta$ $\begin{aligned} & \therefore \mathrm{f}(\mathrm{x})=3 \sqrt{2}|\cos \theta|+\sqrt{2}|\sin \theta| \\ & \therefore \sqrt{2} \leq 3 \sqrt{2}|\cos \theta|+\sqrt{2}|\sin \theta| \leq \sqrt{9 \times 2+2} \\ & \sqrt{2} \leq 3 \sqrt{2}|\cos \theta|+\sqrt{2}|\sin \theta| \leq \sqrt{20} \\ & \therefore \alpha=\sqrt{2} \quad \beta=\sqrt{20} \\ & \alpha^2+2 \beta^2=2+40=42 \end{aligned}$
Let $x=2 \sin ^2 \theta+4 \cos ^2 \theta$ $\begin{aligned} & \therefore \mathrm{f}(\mathrm{x})=3 \sqrt{2}|\cos \theta|+\sqrt{2}|\sin \theta| \\ & \therefore \sqrt{2} \leq 3 \sqrt{2}|\cos \theta|+\sqrt{2}|\sin \theta| \leq \sqrt{9 \times 2+2} \\ & \sqrt{2} \leq 3 \sqrt{2}|\cos \theta|+\sqrt{2}|\sin \theta| \leq \sqrt{20} \\ & \therefore \alpha=\sqrt{2} \quad \beta=\sqrt{20} \\ & \alpha^2+2 \beta^2=2+40=42 \end{aligned}$
Hence, the correct answer is option: 'A'.