Application of Derivatives 4 Question 65
68. The maximum value of the expression $\frac{1}{\sin ^{2} \theta+3 \sin \theta \cos \theta+5 \cos ^{2} \theta}$ is ……
(2010)
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Solution:
- Let
$$ f(\theta)=\frac{1}{\sin ^{2} \theta+3 \sin \theta \cos \theta+5 \cos ^{2} \theta} $$
Again let, $g(\theta)=\sin ^{2} \theta+3 \sin \theta \cos \theta+5 \cos ^{2} \theta$
$$ \begin{aligned} & =\frac{1-\cos 2 \theta}{2}+5 \frac{1+\cos 2 \theta}{2}+\frac{3}{2} \sin 2 \theta \\ & =3+2 \cos 2 \theta+\frac{3}{2} \sin 2 \theta \\ \therefore \quad g(\theta) _{\min } & =3-\sqrt{4+\frac{9}{4}} \\ & =3-\frac{5}{2}=\frac{1}{2} \end{aligned} $$
$\therefore \quad$ Maximum value of $f(\theta)=\frac{1}{1 / 2}=2$
