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The ratio of the maximum and minimum values attained by the function $f(x)=1+2 \sin x+3 \cos ^2 x, 0 \leq x \leq \frac{2 \pi}{3}$ is
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Verified Answer
The correct answer is:
13 : 9
Given,
$\begin{aligned}
& f(x)=1+2 \sin x+3 \cos ^2 x \\
& f(x)=1+2 \sin x+3-3 \sin ^2 x \\
& f(x)=4-3\left(\sin ^2 x-\frac{2}{3} \sin x+\frac{1}{9}\right)+\frac{1}{3}
\end{aligned}$
$f(x)=\frac{13}{3}-3\left(\sin x-\frac{1}{3}\right)^2$
$f(x)=\frac{13}{3}-3\left(\sin x-\frac{1}{3}\right)^2$
Maximum value of $f(x)=\frac{13}{3}$ at $\sin x=\frac{1}{3}$
Minimum value of $f(x)=\frac{9}{3}$ at $\sin x=1 x \in\left[0, \frac{2 \pi}{3}\right]$
$\therefore$ Ratio of Maximum to Minimum 13:9.
$\begin{aligned}
& f(x)=1+2 \sin x+3 \cos ^2 x \\
& f(x)=1+2 \sin x+3-3 \sin ^2 x \\
& f(x)=4-3\left(\sin ^2 x-\frac{2}{3} \sin x+\frac{1}{9}\right)+\frac{1}{3}
\end{aligned}$
$f(x)=\frac{13}{3}-3\left(\sin x-\frac{1}{3}\right)^2$
$f(x)=\frac{13}{3}-3\left(\sin x-\frac{1}{3}\right)^2$
Maximum value of $f(x)=\frac{13}{3}$ at $\sin x=\frac{1}{3}$
Minimum value of $f(x)=\frac{9}{3}$ at $\sin x=1 x \in\left[0, \frac{2 \pi}{3}\right]$
$\therefore$ Ratio of Maximum to Minimum 13:9.
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