Given that, $$A=\{1,2,3,4\}$$

(i) Let $$R_1=\{(1,1),(1,2),(2,3),(2,2),(1,3),(3,3)\}$$

$R_1$ is reflexive, since, $(1,1)(2,2)(3,3)$ lie in $R_1$.

Now, $$(1,2) \in R_1,(2,3) \in R_1 \Rightarrow(1,3) \in R_1$$

Hence, $R_1$ is also transitive but $(1,2) \in R_1 \Rightarrow(2,1) \notin R_1$.

So, it is not symmetric.

(ii) Let $$R_2=\{(1,2),(2,1)\}$$

Now, $$(1,2) \in R_2,(2,1) \in R_2$$

So, it is symmetric.

(iii) Let $$R_3=\{(1,2),(2,1),(1,1),(2,2),(3,3),(1,3),(3,1),(2,3)\}$$

Hence, $R_3$ is reflexive, symmetric and transitive.

$$\begin{aligned} \text{Given that,}\quad R & =\{(x, y): x \in N, y \in N, 2 x+y=41\} \\ \text { Domain } & =\{1,2,3, \ldots, 20\} \\ \text { Range } & =\{1,3,5,7, \ldots, 39\} \\ R & =\{(1,39),(2,37),(3,35), \ldots,(19,3),(20,1)\} \end{aligned}$$

$R$ is not reflexive as $(2,2) \notin R$

$$2 \times 2+2 \neq 41$$

So, $R$ is not symmetric.

As $$(1,39) \in R \text { but }(39,1) \notin R $$

So, $R$ is not transitive.

As $$(11,19) \in R,(19,3) \in R$$

But $(11,3) \notin R$

Hence, $R$ is neither reflexive, nor symmetric and nor transitive.

Given that, $$A=\{2,3,4\}, B=\{2,5,6,7\}$$

(i) Let $f: A \rightarrow B$ denote a mapping

$$\begin{aligned} & f=\{(x, y): y=x+3\} \\ \text{i.e.,}\quad & f=\{(2,5),(3,-6),(4,7)\}, \text { which is an injective mapping. } \end{aligned}$$

(ii) Let $g: A \rightarrow B$ denote a mapping such that $g=\{(2,2),(3,5),(4,5)\}$, which is not an injective mapping.

(iii) Let $h: B \rightarrow A$ denote a mapping such that $h=\{(2,2),(5,3),(6,4),(7,4)\}$, which is a mapping from $B$ to $A$.

(i) Let $f: N \rightarrow N$, be a mapping defined by $f(x)=2 x$ which is one-one.

$$\begin{aligned} \text{For}\quad f\left(x_1\right) & =f\left(x_2\right) \\ \Rightarrow\quad 2 x_1 & =2 x_2 \\ x_1 & =x_2 \end{aligned}$$

Further $f$ is not onto, as for $1 \in N$, there does not exist any $x$ in $N$ such that $f(x)=2 x+1$.

(ii) Let $f: N \rightarrow N$, given by $f(1)=f(2)=1$ and $f(x)=x-1$ for every $x>2$ is onto but not one-one. $f$ is not one-one as $f(1)=f(2)=1$. But $f$ is onto.

(iii) The mapping $f: R \rightarrow R$ defined as $f(x)=x^2$, is neither one-one nor onto.

Given that, $$A=R-\{3\}, B=R-\{1\} .$$

$f: A \rightarrow B$ is defined by $f(x)=\frac{x-2}{x-3}, \forall x \in A$

For injectivity

Let $$f\left(x_1\right)=f\left(x_2\right) \Rightarrow \frac{x_1-2}{x_1-3}=\frac{x_2-2}{x_2-3}$$

$$\begin{aligned} &\begin{array}{rrr} \Rightarrow & \left(x_1-2\right)\left(x_2-3\right) =\left(x_2-2\right)\left(x_1-3\right) \\ \Rightarrow & x_1 x_2-3 x_1-2 x_2+6 =x_1 x_2-3 x_2-2 x_1+6 \\ \Rightarrow & -3 x_1-2 x_2 =-3 x_2-2 x_1 \\ \Rightarrow & -x_1 =-x_2 \Rightarrow x_1=x_2 \end{array}\\ &\text { So, } f(x) \text { is an injective function. } \end{aligned}$$

For surjectivity

$$\begin{aligned} \text{Let}\quad y & =\frac{x-2}{x-3} \Rightarrow x-2=x y-3 y \\ \Rightarrow\quad x(1-y) & =2-3 y \Rightarrow x=\frac{2-3 y}{1-y} \\ \Rightarrow\quad x & =\frac{3 y-2}{y-1} \in A, \forall y \in B\quad\text{[codomain]} \end{aligned}$$

So, $f(x)$ is surjective function.

Hence, $f(x)$ is a bijective function.