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.

$$\begin{array}{r} \text { (i) } x \text { is greater than } y, x, y \in N \\ (x, x) \in R \end{array}$$

For $x R x \quad x>x$ is not true for any $x \in N$.

Therefore, $R$ is not reflexive.

$$\begin{aligned} \text{Let}\quad (x, y) & \in R \Rightarrow x R y \\ x & >y \end{aligned}$$

but $y>x$ is not true for any $x, y \in N$

Thus, $R$ is not symmetric.

Let $x R y$ and $y R z$

$x>y$ and $y>z \Rightarrow x>z$

$x R z$

So, R is transitive.

(ii) $x+y=10, x, y \in N$

$$\begin{aligned} & R=\{(x, y) ; x+y=10, x, y \in \mathcal{N}\} \\ & R=\{(1,9),(2,8),(3,7),(4,6),(5,5),(6,4),(7,3),(8,2),(9,1)\}(1,1) \notin R \end{aligned}$$

So, $R$ is not reflexive.

$$(x, y) \in R \quad \Rightarrow \quad(y, x) \in R$$

Therefore, $R$ is symmetric.

$$(1,9) \in R,(9,1) \in R \quad \Rightarrow \quad(1,1) \notin R$$

Hence, $R$ is not transitive.

(iii) Given $x y$, is square of an integer $x, y \in N$.

$$\Rightarrow \quad \begin{aligned} R & =\{(x, y): x y \text { is a square of an integer } x, y \in N\} \\ (x, x) & \in R, \forall x \in N \end{aligned}$$

As $x^2$ is square of an integer for any $x \in N$.

Hence, $R$ is reflexive.

If $$(x, y) \in R \Rightarrow(y, x) \in R$$

Therefore, $R$ is symmetric.

If $$(x, y) \in R,(y, z) \in R$$

So, $x y$ is square of an integer and $y z$ is square of an integer.

Let $x y=m^2$ and $y z=n^2$ for some $m, n \in Z$

$$\begin{aligned} & x=\frac{m^2}{y} \text { and } z=\frac{x^2}{y} \\ & x z=\frac{m^2 n^2}{y^2}, \text { which is square of an integer. } \end{aligned}$$

So, $R$ is transitive.

$$\begin{aligned} &\begin{aligned} \text { (iv) } & x+4 y=10, x, y \in N \\ & R=\{(x, y): x+4 y=10, x, y \in N\} \\ & R=\{(2,2),(6,1)\} \\ &(1,1),(3,3), \ldots, \notin R \end{aligned} \end{aligned}$$

Thus, $R$ is not reflexive.

$$(6,1) \in R \text { but }(1,6) \notin R$$

Hence, $R$ is not symmetric.

$$\begin{aligned} (x, y) \in R & \Rightarrow x+4 y=10 \text { but }(y, z) \in R \\ y+4 z=10 & \Rightarrow(x, z) \in R \end{aligned}$$

So, $R$ is transitive.