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 $2\le f\left(\theta \right)\le \frac{9}{4}$

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

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

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

Now the maximum value of $f\left(\theta \right)$ is obtained when $({\sin }^2\theta -\frac{1}{2}{)}^2=0$. Hence maximum value is $\frac{9}{4}$.

The minimum value of $f\left(\theta \right)$ is obtained when $({\sin }^2\theta -\frac{1}{2}{)}^2$ is maximum.

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

Differentiating $({\sin }^2\theta -\frac{1}{2}{)}^2$ with respect to $\theta$ and equating to zero gives

$4{\sin }^3\theta \cos \theta -2\sin \theta \cos \theta =0$

$2\sin \theta \cos \theta (2{\sin }^2\theta -1)=0⟹\sin 2\theta (2{\sin }^2\theta -1)=0$

$\sin 2\theta (-\cos 2\theta )=0⟹-\sin 4\theta =0$

Either $\theta =\frac{\pi }{4}$ or $\theta =0$.

For $\theta =\frac{\pi }{4}$ we got $({\sin }^2\theta -\frac{1}{2}{)}^2=0$ and maximum value of $f\left(\theta \right)$.

For $\theta =0$ ,$({\sin }^2\theta -\frac{1}{2}{)}^2=\frac{1}{4}$ and minimum value of $f\left(\theta \right)=\frac{9}{4}-\frac{1}{4}=2$.

Hence $2\le f\left(\theta \right)\le \frac{9}{4}$