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
Correct Answer: C
Solution :
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.\[f(x)\]
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