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

The correct option is C $[\frac{3}{4},1]$

Given,

$f\left(\theta \right)=si{n}^4\theta +co{s}^2\theta$

$ExtremePointsof{\sin }^4\left(\theta \right)+{\cos }^2\left(\theta \right):Maximum(\frac{\pi }{2}n,1),Minimum(\frac{\pi }{4}+\frac{\pi }{2}n,\frac{3}{4})$

${lim}_{\theta \to \frac{\pi }{2}+\frac{\pi }{2}n-}({\sin }^4\left(\theta \right)+{\cos }^2\left(\theta \right))=1$

$Theintervalhasamaximumpointat\theta =\frac{\pi }{2}n+\frac{\pi }{2}nwithvaluef(\frac{\pi }{2}n+\frac{\pi }{2}n)=1$

$Theintervalhasaminimumpointat\theta =\frac{\pi }{4}+\frac{\pi }{2}n+\frac{\pi }{2}nwithvaluef(\frac{\pi }{4}+\frac{\pi }{2}n+\frac{\pi }{2}n)=\frac{3}{4}$

$Minimumfunctionvalueatthedomaininterval-\infty <\theta <\infty is\frac{3}{4}$

$Maximumfunctionvalueatthedomaininterval-\infty <\theta <\infty is1$

$Thereforetherangeof{\sin }^4\left(\theta \right)+{\cos }^2\left(\theta \right)atthedomaininterval-\infty <\theta <\infty is$

$\frac{3}{4}\le f\left(\theta \right)\le 1$

$\left[\begin{array}{cc}Solution:& \frac{3}{4}\le f\left(\theta \right)\le 1\\ IntervalNotation:& [\frac{3}{4},1]\end{array}\right]$