Question

The maximum value of $\frac{1}{{\sin }^2\theta +3\sin \theta \cos \theta +5{\cos }^2\theta }$ is

• A. $4$
• B. $2$
• C. $\frac{3}{2}$
• D. $1$

Answer 1

The correct option is B $2$

$f\left(\theta \right)=\frac{1}{{\sin }^2\theta +3\sin \theta \cos \theta +5{\cos }^2\theta }=\frac{1}{4{\cos }^2\theta +1+\frac{3}{2}\sin 2\theta }=\frac{1}{2(1+\cos 2\theta )+1+\frac{3}{2}\sin 2\theta }=\frac{1}{3+2\cos 2\theta +\frac{3}{2}\sin 2\theta }$

$f\left(\theta \right)$ is maximum when denominator is minimum and denominator is minimum when $2\cos 2\theta +\frac{3}{2}\sin 2\theta$ is minimum.

We know that,
$-\sqrt{4+\frac{9}{4}}\le 2\cos 2\theta +\frac{3}{2}\sin 2\theta \le \sqrt{4+\frac{9}{4}}$
$\Rightarrow -\frac{5}{2}\le 2\cos 2\theta +\frac{3}{2}\sin 2\theta \le \frac{5}{2}$

$\therefore f(\theta {)}_{max}=\frac{1}{3-\frac{5}{2}}=2$