Question

Let $f\left(\theta \right)=(1+{\sin }^2\theta )(2-{\sin }^2\theta )$. Then for all values of $\theta$

• A. $f\left(\theta \right)>\frac{9}{4}$
• B. $f\left(\theta \right)<2$
• C. $f\left(\theta \right)>\frac{11}{4}$
• D. $2\le f\left(\theta \right)\le \frac{9}{4}$

Answer 1

The correct option is D $2\le f\left(\theta \right)\le \frac{9}{4}$

Given

$f\left(\theta \right)=(1+{\sin }^2\theta )(2-{\sin }^2\theta )$

$=2+2{\sin }^2\theta -{\sin }^2\theta -{\sin }^4\theta$

$=-{\sin }^4\theta +{\sin }^2\theta +2$

$=-({\sin }^4\theta -{\sin }^2\theta -2)$

$=-\{{\sin }^4\theta -{\sin }^2\theta +\frac{1}{4}-\frac{9}{4}\}$

$=+\frac{9}{4}-{({\sin }^2\theta -\frac{1}{2})}^2...\left(i\right)$

$\because -1\le \sin \theta \le 1\Rightarrow 0{\sin }^2\theta \le 1$

$\Rightarrow -\frac{1}{2}\le {\sin }^2\theta -\frac{1}{2}\le \frac{1}{2}$

$\Rightarrow 0\le {({\sin }^2\theta -\frac{1}{2})}^2\le \frac{1}{4}$

$\Rightarrow 0\ge -{({\sin }^2\theta -\frac{1}{2})}^2\ge -\frac{1}{4}$

$\Rightarrow \frac{9}{4}\ge \frac{9}{4}-{({\sin }^2\theta -\frac{1}{2})}^2\ge \frac{9}{4}-\frac{1}{4}$

$\Rightarrow 2\le f\left(\theta \right)\le \frac{9}{4}$ (from eq. $i$)