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 $f,$ then $\alpha^2+2 \beta^2$ is equal to
  • A
    44
  • B
    42
  • C
    24
  • D
    38

Answer

$ f ( x )=3 \sqrt{ x -2}+\sqrt{4- x }$
$x -2 \geq 0 \ 4- x \geq 0$
$\therefore x \in[2,4]$
Let $ x =2 \sin ^2 \theta+4 \cos ^2 \theta$
$\therefore f ( x )=3 \sqrt{2}|\cos \theta|+\sqrt{2}|\sin \theta|$
$\therefore \sqrt{2} \leq 3 \sqrt{2}|\cos \theta|+\sqrt{2}|\sin \theta| \leq \sqrt{9 \times 2+2}$
$\sqrt{2} \leq 3 \sqrt{2}|\cos \theta|+\sqrt{2}|\sin \theta| \leq \sqrt{20}$
$\therefore \alpha=\sqrt{2} \beta=\sqrt{20}$
$\alpha^2+2 \beta^2$
$=2+40=42$

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