Let f(x)=3x-2+√4-x be a real valued function.

Question

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

Accepted Answer

$$\begin{aligned}
& f(x)=3 \sqrt{x-2}+\sqrt{4-x} \
& \text { Let } x=2 \sin ^2 \theta+4 \cos ^2 \theta \
& =3 \sqrt{2 \sin ^2 \theta+4 \cos ^2 \theta-2}+\sqrt{4-2 \sin ^2 \theta-4 \cos ^2 \theta} \
& =3 \sqrt{2 \cos ^2 \theta}+\sqrt{2 \sin ^2 \theta} \
& =3 \sqrt{2}|\cos \theta|+\sqrt{2}|\sin \theta|
\end{aligned}$$

$$\begin{aligned}
& \Rightarrow 3 \sqrt{2} \cos \theta+\sqrt{2} \sin \theta \leq \sqrt{18+2} \
& \Rightarrow 3 \sqrt{2} \cos \theta+\sqrt{2} \sin \theta \leq \sqrt{20}
\end{aligned}$$

Minimum value exists when $$\theta=\frac{\pi}{2}$$

$$\begin{aligned}
& \text { So, minimum value }=\sqrt{2} \
& \Rightarrow \alpha=\sqrt{2} \text { and } \beta=\sqrt{20} \
& \Rightarrow \alpha^2+2 \beta^2=2+40 \
& =42
\end{aligned}$$

Solutions · 1