Question

The function is $f(x)=\frac{1}{2-\cos \,3x},\,x\,\in \left[ 0,\frac{\pi }{3} \right],$ is

• A. one-one, but not onto
• B. onto, but not one-one
• C. one-one as well as onto
• D. neither one-one nor onto

Answer 1

Answer: (C)

one-one as well as onto

Answer 2

Given, $f(x)=\frac{1}{2-\cos \,3x},\,x\,\in \left[ 0,\frac{\pi }{3} \right]$ For one - one Let $f({{x}_{1}})=f({{x}_{2}})$
$\Rightarrow$ $\frac{1}{2-\cos \,3{{x}_{1}}}=\frac{1}{2-\cos \,3\,{{x}_{2}}}$
$\Rightarrow$ $2-\cos \,3{{x}_{1}}=2-\cos \,3{{x}_{2}}$
$\Rightarrow$ $\cos \,3{{x}_{1}}=\cos \,3{{x}_{2}}\,\,\Rightarrow \,\,{{x}_{1}}={{x}_{2}}$
$\Rightarrow$ f is one-one For onto Let $y=f(x),\,\,y\,\,\in$ codomain
$\Rightarrow$ $y=\frac{1}{2-\cos \,3x}$
$\Rightarrow$ $y(2-\cos \,3x)=1$
$\Rightarrow$ $2-\cos \,3x=\frac{1}{y}$
$\Rightarrow$ $\cos \,3x=2-\frac{1}{y}$
$\Rightarrow$ $x=\frac{1}{3}\,{{\cos }^{-1}}\left( 2-\frac{1}{y} \right)$
Here, for all $y\,\in$ codomain there exist $x\,\in$ domain. so $f(x)$ is onto.
Here for all $y\,E$ codomain there exist $x\,\in$ domain. so is on to. $f(x)$