Question

The range of the function $f\left(x\right)=\frac{1}{3-\sin 3x},x\in R$ is

• A. $\left(\frac{1}{4},\frac{1}{2}\right]$
• B. $\left(\frac{1}{4},1\right]$
• C. $\left(\frac{1}{4},\frac{1}{2}\right)$
• D. $\left(\frac{1}{4},1\right)$

Answer 1

The correct option is A $\left(\frac{1}{4},\frac{1}{2}\right]$

$f\left(x\right)=\frac{1}{3-\sin 3x},x\in R\mathrm{We}\mathrm{know}\mathrm{that},-1\le \sin 3x\le 1\mathrm{for}\mathrm{all}x\in R\Rightarrow -1\le -\sin 3x\le 1\mathrm{for}\mathrm{all}x\in R\Rightarrow 2\le 3-\sin 3x\le 4\mathrm{for}\mathrm{all}x\in R\left[\mathrm{Adding}3\mathrm{throughout}\right]\Rightarrow \frac{1}{4}\le \frac{1}{3-\sin 3x}\le \frac{1}{2}\mathrm{for}\mathrm{all}x\in R\Rightarrow \frac{1}{4}\le f\left(x\right)\le \frac{1}{2}\mathrm{for}\mathrm{all}x\in R\Rightarrow f\left(x\right)\in \left(\frac{1}{4},\frac{1}{2}\right]\mathrm{Hence},\mathrm{range}\mathrm{of}f=\left(\frac{1}{4},\frac{1}{2}\right]$