Question

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

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

Answer 1

Answer: (C)

Clearly, domain of $f(x)$ is $R$ and $f(x)>0,\forall x\in R$
Let $y=f(x)$
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
$-1\le \frac{2y-1}{y}\le 1$
$\Rightarrow -y\le 2y-1\le y$
$\Rightarrow -y\le 2y-1\text{ and }2y-1\le y$
$\Rightarrow 3y\ge 1\text{ and }y\le 1$
$\Rightarrow y\ge \frac{1}{3}\text{ and }y\le 1$
$\Rightarrow y\in \left[\frac{1}{3},1\right]$
Therefore, range $(f)=\left[\frac{1}{3},1\right]$