Question

The range of the function $f\left(x\right)=\frac{1}{2-\cos 3x}$ is

• A. $(-2,\infty )$
• B. $[-2,3]$
• C. $(\frac{1}{3},2)$
• D. $(\frac{1}{2},1)$
• E. $[\frac{1}{3},1]$

Answer 1

Answer: (E)

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

Clearly, domain of $f\left(x\right)$ is $R$ and $f\left(x\right)>0$, $\forall x\in R$
Let $y=f\left(x\right)$
Then, $y=\frac{1}{2-\cos 3x}$
$\Rightarrow \cos 3x=\frac{2y-1}{y}$
$\Rightarrow x=\frac{1}{3}{\cos }^{-1}\left(\frac{2y-1}{y}\right)$
For $x$ to be real, we must have
$-1\le \frac{2y-1}{y}\le 1$
$\Rightarrow -y\le 2y-1\le y$
$\Rightarrow -y\le 2y-1$ and $2y-1\le y$
$\Rightarrow 3y\ge 1$ and $y\le 1$
$\Rightarrow y\ge \frac{1}{3}$ and $y\le 1$
$\Rightarrow y\in [\frac{1}{3},1]$
Therefore, range $\left(f\right)=[\frac{1}{3},1]$