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

$R$ 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.

Given that, $$A=[-1,1]$$

(i) $f(x)=\frac{x}{2}$

$$\begin{array}{ll} \text { Let } & f\left(x_1\right)=f\left(x_2\right) \\ \Rightarrow & \frac{x_1}{2}=\frac{x_2}{2} \Rightarrow x_1=x_2 \end{array}$$

So, $f(x)$ is one-one.

Now, let $$y=\frac{x}{2}$$

$$\begin{array}{ll} \Rightarrow & x=2 y \notin A, \forall y \in A \\ \text { As for } & y=1 \in A, x=2 \notin A \end{array}$$

So, $f(x)$ is not onto.

Also, $f(x)$ is not bijective as it is not onto.

(ii) $g(x)=|x|$

Let $$g\left(x_1\right)=g\left(x_2\right)$$

So, $g(x)$ is not one-one.

Now, $\quad y=|x| \Rightarrow x= \pm y \notin A, \forall y \in A$

So, $g(x)$ is not onto, also, $g(x)$ is not bijective.

(iii) $h(x)=x|x|$

$$\begin{aligned} &\begin{aligned} \text { Let } \quad h\left(x_1\right) & =h\left(x_2\right) \\ \Rightarrow\quad x_1\left|x_1\right| & =x_2\left|x_2\right| \quad \Rightarrow x_1=x_2 \end{aligned} \end{aligned}$$

So, $h(x)$ is one-one.

$$\begin{array}{ll} \text { Now, let } & y=x|x| \\ \Rightarrow & y=x^2 \in A, \forall x \in A \end{array}$$

So, $h(x)$ is onto also, $h(x)$ is a bijective.

(iv) $k(x)=x^2$

$$\begin{aligned} &\begin{aligned} \text { Let }\quad k\left(x_1\right) & =k\left(x_2\right) \\ x_1^2 & =x_2^2 \Rightarrow x_1= \pm x_2 \end{aligned} \end{aligned}$$

Thus, $k(x)$ is not one-one.

Now, let $$y=x^2$$

$$\Rightarrow \quad x=\sqrt{y} \notin A, \forall y \in A$$

As for $y=-1, x=\sqrt{-1} \notin A$

Hence, $k(x)$ is neither one-one nor onto.