Let A=R-{3}, B=R-{1}. If f: A$\to$B be defined by $f\left(x\right)=\frac{x-2}{x-3}, \forall x\in A$. Then, show that f is bijective.                                                     

OR

Let A={1, 2, 3} and R={(a, b):a, b $\in$A and $\left|a^2-b^2\right|\le 5$}. Write R as set of ordered pairs. Mention whether R is

i.                 Reflexive

ii.               Symmetric

iii.             Transitive

              Given reason in each case.   

Given,  $\mathrm{A}=\mathrm{R}-\left\{3\right\},\mathrm{B}=\mathrm{R}-\left\{1\right\}$

$\mathrm{f}:\mathrm{A}\to \mathrm{B}$ is defined by  $f\left(x\right)=\frac{x-2}{x-3}\forall x\in A$

For injectivity 

$\begin{array}{rr} & \text{ Let }f\left(x_1\right)=f\left(x_2\right)\Rightarrow \frac{x_1-2}{x_1-3}=\frac{x_2-2}{x_2-3}\\ & \Rightarrow \left({\mathrm{x}}_1-2\right)\left({\mathrm{x}}_2-3\right)=\left({\mathrm{x}}_2-2\right)\left({\mathrm{x}}_1-3\right)\\ & \Rightarrow {\mathrm{x}}_1{\mathrm{x}}_2-3{\mathrm{x}}_1-2{\mathrm{x}}_2+6={\mathrm{x}}_1{\mathrm{x}}_2-3{\mathrm{x}}_2-2{\mathrm{x}}_1+6\\ & \Rightarrow -3{\mathrm{x}}_1-2{\mathrm{x}}_2=-3{\mathrm{x}}_2-2{\mathrm{x}}_1\\ & \Rightarrow -{\mathrm{x}}_1=-{\mathrm{x}}_2\Rightarrow {\mathrm{x}}_1={\mathrm{x}}_2\end{array}$

So, $f\left(\mathrm{x}\right)$ is an injective function

For surjectivity

Let $\mathrm{y}=\frac{\mathrm{x}-2}{\mathrm{x}-3}\Rightarrow \mathrm{x}-2=\mathrm{xy}-3\mathrm{y}$

$\Rightarrow \mathrm{x}(1-\mathrm{y})=2-3\mathrm{y}\Rightarrow x=\frac{2-3y}{1-y}$

$\Rightarrow \mathrm{x}=\frac{3\mathrm{y}-2}{\mathrm{y}-1}\in \mathrm{A},\forall \mathrm{y}\in \mathrm{B}\left[\mathrm{codomain}\right]$

So, $f\left(\mathrm{x}\right)$ is surjective function.

Therefore, we can say that $f\left(\mathrm{x}\right)$ is a bijective function.

OR

Given: $$\begin{gathered} \mathrm{A}=\{1,2,3\} \mathrm{and} \mathrm{R}=\left\{(\mathrm{a},\mathrm{b}):\mathrm{a},\mathrm{b}\in \mathrm{A}\right.\mathrm{and} \left.\left|{\mathrm{a}}^2-{\mathrm{b}}^2\right|\le 5\right\} . \\ \mathrm{Put} \mathrm{a}=1,\mathrm{b}=1,\Rightarrow \left|1^2-1^2\right|\le 5,\left(\mathrm{1,1}\right) is an order pair \end{gathered}$$

Put $\mathrm{a}=1,\mathrm{b}=2,\left|1^2-2^2\right|\le 5,(1,2)$ is an ordered pair.

Put $\mathrm{a}=1,\mathrm{b}=3,\left|1^2-3^2\right|>5,\left(\mathrm{1,3}\right)$ is not an ordered pair.

Put $\mathrm{a}=2,\mathrm{b}=1,\left|2^2-1^2\right|\le 5,\left(\mathrm{2,1}\right)$ is  an ordered pair.

Put $\mathrm{a}=2,\mathrm{b}=2,\left|2^2-2^2\right|\le 5,\left(\mathrm{2,2}\right)$ is an ordered pair.

Put $\mathrm{a}=2,\mathrm{b}=3,\left|2^2-3^2\right|\le 5,\left(\mathrm{2,3}\right)$ is an ordered pair.

Put $\mathrm{a}=3,\mathrm{b}=1,\left|3^2-1^2\right|>5,\left(\mathrm{3,1}\right)$ is not an ordered pair.

Put $\mathrm{a}=3,\mathrm{b}=2,\left|3^2-2^2\right|\le 5,\left(\mathrm{3,2}\right)$ is not an ordered pair.

Put $\mathrm{a}=3,\mathrm{b}=3,\left|3^2-3^2\right|\le 5,\left(\mathrm{3,3}\right)$ is an ordered pair.

$\mathrm{R}=\left\{\right(\mathrm{1,1}),(\mathrm{1,2}),(\mathrm{2,1}),(\mathrm{2,2}),(\mathrm{2,3}),(\mathrm{3,2}),(\mathrm{3,3}\left)\right\}$

i. For  $(a,a)\in R$

$\left|a^2-a^2\right|=0\le 5$. Thus, it is reflexive.

ii. Let  $(a,b)\in R$

$(a,b)\in R,\left|a^2-b^2\right|\le 5$

$\left|{\mathrm{b}}^2-{\mathrm{a}}^2\right|\le 5$

(b, a) $\in \mathrm{R}$

Hence, it is symmetric

iii. Put $\mathrm{a}=1,\mathrm{b}=2,\mathrm{c}=3$

$$\begin{gathered} \left|1^2-2^2\right|\le 5 \\ \left|2^2-3^2\right|\le 5 \end{gathered}$$

But $\left|1^2-3^2\right|>5$

Thus, it is not transitive

Therefore, (i) R is reflexive. (ii) R is symmetric but (iii) R is not transitive.