Question

Let $f(\theta )=(1+si{n}^2\theta )(2-si{n}^2\theta )$. 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

Answer: (D)

Given $f(\theta )=\left(1+{\sin }^2\theta \right)\left(2-{\sin }^2\theta \right)$
$=2+2{\sin }^2\theta -{\sin }^2\theta -{\sin }^4\theta$
$=-{\sin }^4\theta +{\sin }^2\theta +2$
$=-\left({\sin }^4\theta -{\sin }^2\theta -2\right)$
$=-\left\{{\sin }^4\theta -{\sin }^2\theta +\frac{1}{4}-\frac{9}{4}\right\}$
$=+\frac{9}{4}-{\left({\sin }^2\theta -\frac{1}{2}\right)}^2$ ....(i)
$\because -1\le \sin \theta \le 1$
$\Rightarrow 0\le {\sin }^2\theta \le 1$
$\Rightarrow -\frac{1}{2}\le {\sin }^2\theta -\frac{1}{2}\le \frac{1}{2}$
$\Rightarrow 0\le {\left({\sin }^2\theta -\frac{1}{2}\right)}^2\le \frac{1}{4}$
$\Rightarrow 0\ge -{\left({\sin }^2\theta -\frac{1}{2}\right)}^2\ge -\frac{1}{4}$
$\Rightarrow \frac{9}{4}\ge \frac{9}{4}-{\left({\sin }^2\theta -\frac{1}{2}\right)}^2\ge \frac{9}{4}-\frac{1}{4}$
$\Rightarrow 2\le f(\theta )\le \frac{9}{4}$ [from Eq.(i)]