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

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.

Given that, $A=\{1,2,3, \ldots, 9\}$ and $(a, b) R(c, d)$ if $a+d=b+c$ for $(a, b) \in A \times A$ and $(c, d) \in A \times A$.

Let $(a, b) R(a, b)$

$$\Rightarrow \quad a+b=b+a, \forall a, b \in A$$

which is true for any $a, b \in A$.

Hence, $R$ is reflexive.

Let $(a, b) R(c, d)$

$$\begin{aligned} & a+d=b+c \\ & c+b=d+a \quad \Rightarrow \quad(c, d) R(a, b) \end{aligned}$$

So, $R$ is symmetric.

$$\begin{aligned} &\text { Let }\\ &\begin{aligned} (a, b) R(c, d) \text { and } & (c, d) R(e, f) \\ a+d & =b+c \text { and } c+f=d+e \\ a+d & =b+c \text { and } d+e=c+f \\ (a+d)-(d+e) & =(b+c)-(c+f) \\ (a-e) & =b-f \\ a+f & =b+e \\ (a, b) & R(e, f) \end{aligned} \end{aligned}$$

So, $R$ is transitive.

Hence, $R$ is an equivalence relation.

Now, equivalence class containing $[(2,5)]$ is $\{(1,4),(2,5),(3,6),(4,7),(5,8),(6,9)\}$.

A function $f: X \rightarrow Y$ is defined to be invertible, if there exist a function $g=Y \rightarrow X$ such that gof $=I_X$ and fog $=I_Y$. The function is called the inverse of $f$ and is denoted by $f^{-1}$. A function $f=X \rightarrow Y$ is invertible iff $f$ is a bijective function.

Given that,

$$f(x)=x^2+3 x+1, g(x)=2 x-3$$

(i)

$$\begin{aligned} f \circ g & =f\{g(x)\}=f(2 x-3) \\ & =(2 x-3)^2+3(2 x-3)+1 \\ & =4 x^2+9-12 x+6 x-9+1=4 x^2-6 x+1 \end{aligned}$$

(ii)

$$\begin{aligned} g \circ f & =g\{f(x)\}=g\left(x^2+3 x+1\right) \\ & =2\left(x^2+3 x+1\right)-3 \\ & =2 x^2+6 x+2-3=2 x^2+6 x-1 \end{aligned}$$

(iii)

$$\begin{aligned} f \circ f & =f\{f(x)\}=f\left(x^2+3 x+1\right) \\ & =\left(x^2+3 x+1\right)^2+3\left(x^2+3 x+1\right)+1 \\ & =x^4+9 x^2+1+6 x^3+6 x+2 x^2+3 x^2+9 x+3+1 \\ & =x^4+6 x^3+14 x^2+15 x+5 \end{aligned}$$

(iv)

$$\begin{aligned} g \circ g & =g\{g(x)\}=g(2 x-3) \\ & =2(2 x-3)-3 \\ & =4 x-6-3=4 x-9 \end{aligned}$$

Given that * be the binary operation defined on $Q$.

(i) $a * b=a-b, \forall a, b \in Q$ and $b * a=b-a$

So, $$a * b \neq b * a\quad$$ $$[\because b-a \neq a-b]$$

Hence, * is not commutative.

(ii)

$$\begin{aligned} & a * b=a^2+b^2 \\ & b * a=b^2+a^2 \end{aligned}$$

So, * is commutative.

[since, ' + ' is on rational is commutative]

$$\begin{aligned} &\text { (iii) }\\ &\begin{aligned} a * b & =a+a b \\ b * a & =b+a b \\ \text { Clearly, } \quad a+a b & \neq b+a b \end{aligned}\\ &\text { So, * is not commutative. } \end{aligned}$$

(iv)

$$\begin{aligned} a * b & =(a-b)^2, \forall a, b \in Q \\ b * a & =(b-a)^2 \\ \because \quad(a-b)^2 & =(b-a)^2 \end{aligned}$$

Hence, * is commutative.