Question

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

• A. $\left(\frac{1}{3},1\right)$
• B. $\left[\frac{1}{3},1\right)$
• C. $\left[\frac{1}{3},1\right]$
• D. None of these

Answer 1

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

$y=\frac{1}{(2-\sin 3x)}$ --------- ( 1 )

Let $f\left(x\right)=\frac{1}{(2-\sin 3x)}$

Domain of $f\left(x\right)$ is $(-\infty ,\infty )$

From ( 1 ),

$2-\sin 3x=\frac{1}{y}$

$\Rightarrow$ $\sin 3x=2-\frac{1}{y}$

$\Rightarrow$ $\sin 3x=\frac{2y-1}{y}$

$\Rightarrow$ $3x={\sin }^{-1}\left(\frac{2y-1}{y}\right)$

$\Rightarrow$ $x=\frac{1}{3}{\sin }^{-1}\left(\frac{2y-1}{y}\right)$ [ $\because$ $y=\frac{1}{2-\sin 3x}\therefore y>0as-1\le \sin 3x\le 1$ ]

For $x$ to be real

$-1\le \frac{2y-1}{y}\le 1$

$\Rightarrow$ $-y\le 2y-1\le y$ [ $y>0$ ]

$\Rightarrow$ $2y-1\ge -y$ and $2y-1\le y$

$\Rightarrow$ $y\ge \frac{1}{3}$, $y\le 1$

$\therefore$ Range of $f\left(x\right)=[\frac{1}{3},1]$