question_answer

Let \[A=R-\{3\},\] \[B=R-\{1\}.\] If \[f\,\,:\,\,A\to B\] be defined by \[f(x)=\frac{x-2}{x-3},\] \[\forall \,\,x\in A.\] Show that f is bijective and find the inverse of f.

Answer:

Given, \[A=R-\{3\},\] \[B=R-\{1\}.\] and             \[f\,\,:\,\,A\to B\] defined by \[f(x)=\frac{x-2}{x-3},\] \[\forall \,\,x\in A.\]             For one-one             Let \[f({{x}_{1}})=f({{x}_{2}})\] \[\Rightarrow \] \[\frac{{{x}_{1}}-2}{{{x}_{1}}-3}=\frac{{{x}_{2}}-2}{{{x}_{2}}-3}\]             \[\Rightarrow \]   \[({{x}_{1}}-2)\,({{x}_{2}}-3)=({{x}_{2}}-2)\,({{x}_{1}}-3)\]             \[\Rightarrow \]   \[{{x}_{1}}{{x}_{2}}-3{{x}_{1}}-2{{x}_{2}}+6={{x}_{1}}{{x}_{2}}-3{{x}_{2}}-2{{x}_{1}}+6\]             \[\Rightarrow \]   \[-\,3{{x}_{1}}-2{{x}_{2}}=-\,3{{x}_{2}}-2{{x}_{1}}\]             \[\Rightarrow \]   \[-{{x}_{1}}=-{{x}_{2}}\]\[\Rightarrow \] \[{{x}_{1}}={{x}_{2}}\]             So, f(x) is onto.             For onto             Let        \[y=\frac{x-2}{x-3}\] \[\Rightarrow \] \[x-2=xy-3y\]             \[\Rightarrow \]   \[x(1-y)=2-3y\]   \[\Rightarrow \] \[x=\frac{2-3y}{1-y}\]      \[\because \]       \[x=\frac{3y-2}{y-1},\] \[\in A,\] \[\forall \,\,y\in B\] (codomain)             So, f(x) is onto.             Thus, f(x) is bijective.             Hence, \[{{f}^{-1}}:B\to A\] defined by             \[{{f}^{-1}}(y)=\frac{3y-2}{y-1}\] or \[{{f}^{-1}}(x)=\frac{3x-2}{x-1}\]