The minimum and maximum values of $\cos \left(x+\frac{\pi}{3}\right)+2 \sqrt{2} \sin…
The minimum and maximum values of $\cos \left(x+\frac{\pi}{3}\right)+2 \sqrt{2} \sin \left(x+\frac{\pi}{3}\right)$ are respectively
- $-(2 \sqrt{3}-1)$ and $2 \sqrt{3}-1$
- $-(1+2 \sqrt{2})$ and $1+2 \sqrt{2}$
- -3 and 3
- -2 and 2
Solution
Let
$
\begin{aligned}
f(x)= & \cos \left(x+\frac{\pi}{3}\right)+2 \sqrt{2} \sin \left(x+\frac{\pi}{3}\right) \\
= & \left(\cos x \cdot \cos \frac{\pi}{3}-\sin x \cdot \sin \frac{\pi}{3}\right) \\
& +2 \sqrt{2}\left(\sin x \cos \frac{\pi}{3}+\cos x \sin \frac{\pi}{3}\right)
\end{aligned}
$
$
\begin{aligned}
&=\left(\cos x \cdot \frac{1}{2}-\sin x \frac{\sqrt{3}}{2}\right) \\
&+2 \sqrt{2}\left(\sin x \cdot \frac{1}{2}+\cos x \cdot \frac{\sqrt{3}}{2}\right) \\
&= \frac{\cos x}{2}-\frac{\sqrt{3} \sin x}{2}+\sqrt{2} \sin x+\sqrt{6} \cos x \\
& f(x)= \cos x\left(\frac{1}{2}+\sqrt{6}\right)+\sin x\left(\sqrt{2}-\frac{\sqrt{3}}{2}\right) \\
& \therefore \quad a= \frac{1}{2}+\sqrt{6} \text { and } b=\sqrt{2}-\frac{\sqrt{3}}{2} \\
& \sqrt{a^2+b^2}= \sqrt{\left(\frac{1}{2}+\sqrt{6}\right)^2+\left(\sqrt{2}-\frac{\sqrt{3}}{2}\right)^2}=\sqrt{9} \\
& \sqrt{a^2+b^2}= 3
\end{aligned}
$
$\therefore$ Maximum value of $f(x)=\sqrt{a^2+b^2}=3$
Minimum value of $f(x)=-\sqrt{a^2+b^2}=-3$
Hence, option (3) is correct
Asked in: AP EAMCET 2020 (22 Sep Shift 2)
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