Question

The range of the following function $f\left(x\right)={\cos }^2\frac{x}{4}+\sin \frac{x}{4},x\in R$ is

• A. $\left[0,\frac{5}{4}\right]$
• B. $\left[1,\frac{5}{4}\right]$
• C. $\left[1,-\frac{5}{4}\right]$
• D. $\left[-1,\frac{5}{4}\right]$

Answer 1

The correct option is D $\left[-1,\frac{5}{4}\right]$

$f\left(x\right)={\cos }^2\frac{x}{4}+\sin \frac{x}{4},x\in R$

Let $\frac{x}{4}=\theta \therefore \theta \in R$

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

$g^{{}^{\prime }}\theta =-2\sin \theta \cos \theta +\cos \theta =0$

$\therefore \sin \theta =\frac{1}{2}$ or $\cos \theta =0$

$g^{\prime \prime }\left(\theta \right)=-2({\cos }^2\theta -{\sin }^2\theta )-\sin \theta$

At $\sin \theta =\frac{1}{2},\cos \theta =\pm \frac{\sqrt{3}}{2}$

$\therefore g^{\prime \prime }\left(\theta \right)=-2(\frac{3}{4}-\frac{1}{4})-\frac{1}{2}=\frac{-3}{2}<0$

$\therefore$ at$\sin \theta =\sin \left(\frac{x}{4}\right)=\frac{1}{2}$, $f\left(x\right)$ has maximum value.

Maximum value$=\frac{3}{4}+\frac{1}{2}=\frac{5}{4}$

At $\cos \theta =0,\sin \theta =\pm 1$

$g^{\prime \prime }\left(\theta \right)=-2(0-(-{11}^2)-(-1)=3>0$ if $\sin \theta =-1$

$g^{\prime \prime }\left(\theta \right)=-2(0-(-{11}^2)-(1)=1>0$ if $\sin \theta =1$

$g\left(\theta \right)=f\left(x\right)=-1$ if $\sin \left(\frac{x}{4}\right)=-1$

$g\left(\theta \right)=f\left(x\right)=1$ if $\sin \left(\frac{x}{4}\right)=1$

$\therefore f\left(x\right)$ has local minimum at $\sin \left(\frac{x}{4}\right)=1$

$\therefore f\left(x\right)$ has gloal minimum at $\sin \left(\frac{x}{4}\right)=-1$

$\therefore$ Range of $f\left(x\right)$ is $[-1,\frac{5}{4}]$