Question

If $f\left(\theta \right)={\sin }^4\theta +{\cos }^2\theta$, then range of $f\left(\theta \right)$ is

• A. $[\frac{1}{2},1]$
• B. $[\frac{1}{2},\frac{3}{4}]$
• C. $[\frac{3}{4},1]$
• D. None of these

Answer 1

Answer: (C)

$[\frac{3}{4},1]$

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

Put ${\sin }^2\theta =t$, we get

$f\left(t\right)=t^2-t+1$

$f\left(t\right)$ is quadratic equation with $D<0$

$\therefore$ minimum value of $f\left(t\right)$, when

$\frac{df\left(t\right)}{dt}=0\Rightarrow t=\frac{1}{2}$

$f(t{)}_{min}=\frac{1}{4}-\frac{1}{2}+1=\frac{3}{4}$

and the maximum value of $t=1$, therefore the maximum value of $f\left(t\right)$

$\Rightarrow f\left(t\right)=1-1+1=1$

$\therefore$ range of $f\left(t\right)\in [\frac{3}{4},1]$