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)

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) $