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The maximum value of the expression $\frac{1}{\sin ^2 \theta+3 \sin \theta \cos \theta+5 \cos ^2 \theta}$ is
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Verified Answer
The correct answer is:
2
Let
$$
\begin{aligned}
& f(\theta)=\frac{1}{\sin ^2 \theta+3 \sin \theta \cos \theta+5 \cos ^2 \theta} \\
& \text { Again let } \\
& g(\theta)=\sin ^2 \theta+3 \sin \theta \cos \theta+5 \cos ^2 \theta \\
& =\frac{1-\cos 2 \theta}{2}+5\left(\frac{1+\cos 2 \theta}{2}\right)+\frac{3}{2} \sin 2 \theta \\
& =3+2 \cos 2 \theta+\frac{3}{2} \sin 2 \theta \\
& \therefore \quad g(\theta)_{\min }=3-\sqrt{4+\frac{9}{4}}=3-\frac{5}{2}=\frac{1}{2} \\
& \therefore \quad f(\theta)=\frac{1}{g(\theta)_{\min }}=2
\end{aligned}
$$
$$
\begin{aligned}
& f(\theta)=\frac{1}{\sin ^2 \theta+3 \sin \theta \cos \theta+5 \cos ^2 \theta} \\
& \text { Again let } \\
& g(\theta)=\sin ^2 \theta+3 \sin \theta \cos \theta+5 \cos ^2 \theta \\
& =\frac{1-\cos 2 \theta}{2}+5\left(\frac{1+\cos 2 \theta}{2}\right)+\frac{3}{2} \sin 2 \theta \\
& =3+2 \cos 2 \theta+\frac{3}{2} \sin 2 \theta \\
& \therefore \quad g(\theta)_{\min }=3-\sqrt{4+\frac{9}{4}}=3-\frac{5}{2}=\frac{1}{2} \\
& \therefore \quad f(\theta)=\frac{1}{g(\theta)_{\min }}=2
\end{aligned}
$$
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