Question

The maximum value of the expression $\frac{1}{\left({\sin }^2\theta +3\sin \theta \cos \theta +5{\cos }^2\theta \right)}$ is

• A. $2$
• B. $1$
• C. $3$
• D. $4$

Answer 1

The correct option is A

$2$

Explanation for the correct option:

To solve for the maximum value of the expression $\frac{1}{\left({\sin }^2\theta +3\sin \theta \cos \theta +5{\cos }^2\theta \right)}$

Let y$=\left({\sin }^2\theta +3\sin \theta \cos \theta +5{\cos }^2\theta \right)$

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

$=1+\left(\frac{3}{2}\right)\left(\sin 2\theta \right)+2(2\cos 2\theta )$ $\left[\because \left({\sin }^{2}A+{\cos }^{2}A\right)=1\right]$$\left[\because \sin 2A=2\mathrm{sinA}\mathrm{cosA}\right]$

$$\begin{gathered} =1+\left(\frac{3}{2}\right)\left(\sin 2\theta \right)+2(1+\cos 2\theta )\left[\because \cos 2A=2{\cos }^{2}A-1\right] \\ =2\cos 2\theta +\left(\frac{3}{2}\right)\left(\sin 2\theta \right)+3 \end{gathered}$$

We know that, $c-\sqrt{a^2+b^2}\le a\cos \theta +b\sin \theta +c\le c+\sqrt{a^2+b^2}$

Substituting $a=2,b=\left(\frac{3}{2}\right),c=3,$

$3-\sqrt{\frac{25}{4}}\le 2\cos 2\theta +\left(\frac{3}{2}\right)\sin 2\theta +3\le 3+\sqrt{\frac{25}{4}}$

The minimum value of y$=3-\left(\frac{5}{2}\right)=\frac{1}{2}$

Thus, the maximum value of $\frac{1}{y}=2.$

Hence the maximum value of the expression $\frac{1}{\left({\sin }^2\theta +3\sin \theta \cos \theta +5{\cos }^2\theta \right)}$ is $2.$

Therefore, option (A) is correct answer.