Question

The range of the function $f\left(x\right)=\frac{1}{(2-sin3x)}$ is

• A. $]\frac{1}{3},1[$
• B. $[\frac{1}{3},1]$
• C. $[\frac{1}{3},1[$
• D. $]1,\frac{1}{3}]$

Answer 1

The correct option is B $[\frac{1}{3},1]$

Given function is :

$f\left(x\right)=\frac{1}{(2-sin3x)}$

Function f(x) is not defined when

$=>2-sin3x=0$

so ,$sin3x=2---------\left(1\right)$

but we know that range of $sinx€[-1,1]$

so maximum value of $sin3x=1$

therefore$sin3x\ne 2$ ( not possible)

$=>2-sin3x\ne 0------\left(2\right)$

so from equation (2) we can see tha function is defined for all values of $x$

=> $Domain\in R$

FOR RANGE :

now multiplying by -1 we get,

$=>1\ge -sin3x\ge -$1

adding 2 we get,

$=>2+1\ge 2-sin3x\ge 2-1$

$=>3\ge 2-sin3x\ge 1$

now taking inverse we get,

so , $Range\in [1/3,1]$

Hence, Option $B$ is correct answer.