(a)OR(b)

Let $\displaystyle \mathrm{A}=\mathrm{R}-\{5\}$ and $\displaystyle \mathrm{B}=\mathrm{R}-\{1\}$. Consider the function $\displaystyle \mathrm{f}: \mathrm{A} \rightarrow \mathrm{B}$, defined by $\displaystyle \mathrm{f}(x)=\frac{x-3}{x-5}$. Show that f is one-one and onto.

$$\begin{aligned} & \text { Let } \mathrm{f}\left(x_{1}\right)=\mathrm{f}\left(x_{2}\right), \text { for some } x_{1}, x_{2} \in \mathrm{A} & \Rightarrow \frac{x_{1}-3}{x_{1}-5}=\frac{x_{2}-3}{x_{2}-5} & \Rightarrow\left(x_{1}-3\right)\left(x_{2}-5\right)=\left(x_{2}-3\right)\left(x_{1}-5\right) & \Rightarrow x_{1}=x_{2}, \text { So } \mathrm{f} \text { is one }- \text { one Function } . \end{aligned}Let $\displaystyle y=\mathrm{f}(x)=\frac{x-3}{x-5} \Rightarrow y(x-5)=x-3$ $\displaystyle \Rightarrow y x-5 y=x-3$ $\displaystyle \Rightarrow x=\frac{5 y-3}{y-1}$, We observe that $\displaystyle x$ is defined for all values of y except $\displaystyle y=1$,\} So, Range $\displaystyle =\mathrm{R}-\{1\}$ and Co-domain is Given $\displaystyle \mathrm{R}-\{1\} \quad$ [As, $\displaystyle \mathrm{f}: \mathrm{A} \rightarrow \mathrm{B}$ ] Since, Range $\displaystyle =\mathrm{Co}$-domain, $\displaystyle \mathrm{f}$ is onto Function. Thus, $\displaystyle \mathrm{f}$ is one-one & onto function. Reflexive: For $\displaystyle \mathrm{a} \in \mathrm{S}$ $\displaystyle \left.\begin{array}{l}\Rightarrow \mathrm{a}-\mathrm{a}+\sqrt{2} \text { is irrational number } \\ \Rightarrow \sqrt{2} \text { is irrational number } \\ \Rightarrow(\mathrm{a}, \mathrm{a}) \in \mathrm{S} \\ \text { Thus, } \mathrm{S} \text { is Reflexive Relation. }\end{array}\right\}$ Symmetric : Let $\displaystyle (\mathrm{a}, \mathrm{b}) \in \mathrm{S} \Rightarrow \mathrm{a}-\mathrm{b}+\sqrt{2}$ is irrational number but $\displaystyle \mathrm{b}-\mathrm{a}+\sqrt{2}$ may not be irrational number For example, $\displaystyle (\sqrt{2}, 1) \in \mathrm{S} \Rightarrow \sqrt{2}-1+\sqrt{2}=2 \sqrt{2}-1$ is irrational number\} $\displaystyle (1, \sqrt{2}) \notin \mathrm{S}$ as $\displaystyle 1-\sqrt{2}+\sqrt{2}=1$ is not irrational number $\displaystyle \therefore(\mathrm{b}, \mathrm{a}) \notin \mathrm{S}$, So S is NOT Symmetric Relation.