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 \[{{f}^{-1}}(x)=\frac{3x-2}{x-1}\]
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