Let $\displaystyle \mathrm{f}: \mathrm{A} \rightarrow \mathrm{B}$ be defined by $\displaystyle \mathrm{f}(\mathrm{x})=\frac{\mathrm{x}-2}{\mathrm{x}-3}$, where $\displaystyle \mathrm{A}=\mathrm{R}-\{3\}$ and $\displaystyle \mathrm{B}=\mathrm{R}-\{1\}$. Discuss the bijectivity of the function.

Let $\displaystyle \mathrm{x}_{1}, \mathrm{x}_{2} \in \mathrm{A}$ such that $\displaystyle \mathrm{f}\left(\mathrm{x}_{1}\right)=\mathrm{f}\left(\mathrm{x}_{2}\right) \Rightarrow \frac{\mathrm{x}_{1}-2}{\mathrm{x}_{1}-3}=\frac{\mathrm{x}_{2}-2}{\mathrm{x}_{2}-3} \Rightarrow \mathrm{x}_{1}=\mathrm{x}_{2}, \therefore$ ' $\displaystyle \mathrm{f}$ ' is one-one. For each $\displaystyle y \in \mathrm{B}$, there exists $\displaystyle \mathrm{x}=\frac{3 y-2}{y-1} \in \mathrm{R}-\{3\}$, such that $\displaystyle \mathrm{f}(\mathrm{x})=y, \therefore$ ' $\displaystyle \mathrm{f}$ ' is onto ⇒ ' $\displaystyle \mathbf{f}$ ' is one-one $\displaystyle \boldsymbol{\&}$ onto, or ' $\displaystyle \mathbf{f}$ ' is a bijective function.