Question

Let $f(θ) = (1+sin^2θ)(2-sin^2θ)$. Then for all values of $θ$

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

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 \leq \sin \theta \leq 1 $
$\Rightarrow 0 \leq \sin ^{2} \theta \leq 1 $
$\Rightarrow -\frac{1}{2} \leq \sin ^{2} \theta-\frac{1}{2} \leq \frac{1}{2}$
$\Rightarrow 0 \leq\left(\sin ^{2} \theta-\frac{1}{2}\right)^{2} \leq \frac{1}{4}$
$\Rightarrow 0 \geq-\left(\sin ^{2} \theta-\frac{1}{2}\right)^{2} \geq-\frac{1}{4}$
$\Rightarrow \frac{9}{4} \geq \frac{9}{4}-\left(\sin ^{2} \theta-\frac{1}{2}\right)^{2} \geq \frac{9}{4}-\frac{1}{4}$
$\Rightarrow 2 \leq f(\theta) \leq \frac{9}{4} $ [from Eq.(i)]