Q107 / 200 · WBJEE 2013 · Mathematics
Let $f(\theta)=\left(1+\sin ^{2} \theta\right)\left(2-\sin ^{2} \theta\right) .$ Then, for all
values of $\theta$
- A$f(\theta)>\frac{9}{4}$
- B$f(\theta) < 2$
- C$f(\theta)>\frac{11}{4}$
- Correct answerD$2 \leq f(\theta) \leq \frac{9}{4}$
Solution
Given
$$
\begin{aligned}
f(\theta)=\left(1+\sin ^{2} \theta\right)\left(2-\sin ^{2} \theta\right) \\
&=2+2 \sin ^{2} \theta-\sin ^{2} \theta-\sin ^{4} \theta \\
&=-\sin ^{4} \theta+\sin ^{2} \theta+2 \\
&=-\left(\sin ^{4} \theta-\sin ^{2} \theta-2\right) \\
&=-\left\{\sin ^{4} \theta-\sin ^{2} \theta+\frac{1}{4}-\frac{9}{4}\right\} \\
&=+\frac{9}{4}-\left(\sin ^{2} \theta-\frac{1}{2}\right)^{2} \\
\because \quad-1 \leq \sin \theta \leq 1 \Rightarrow 0 \leq \sin ^{2} \theta \leq 1\\
\Rightarrow & 0 \geq-\left(\sin ^{2} \theta-\frac{1}{2}\right)^{2} \geq-\frac{1}{4} \\
\Rightarrow \quad \quad-\frac{1}{2} \leq \sin ^{2} \theta-\frac{1}{2} \leq \frac{1}{2}\\
\Rightarrow \quad 0 \leq\left(\sin ^{2} \theta-\frac{1}{2}\right)^{2} \leq \frac{1}{4}
\Rightarrow & \frac{9}{4} \geq \frac{9}{4}-\left(\sin ^{2} \theta-\frac{1}{2}\right)^{2} \geq \frac{9}{4}-\frac{1}{4} \\
\Rightarrow \quad 2 \leq f(\theta) \leq \frac{9}{4} & \quad[\text { from Eq. }(0)]
\end{aligned}
$$
Hence, the correct answer is option: 'D'.