Since A $\subseteq$ A . $\therefore$ Relation ' $\subseteq$ ' is relfexive 

Since A $\subseteq$ B , B $\subseteq$ C $\Rightarrow$ A $\subseteq$ C 

$\therefore$ Relation ' $\subseteq$ ' is transitive. 

But A ' $\subseteq$ ' B , $\Rightarrow$ B  $\subseteq$ A . $\therefore$ relation is not symmetric.

Here n(A × B) = 3 × 3 = 9 

Since every subset of A × B defines a relation from A to B, the number of relations from A to B is equal to the number of subsets of A $\times$ B = $2^{\text{n(A}\times \text{B)}}$
                                               = $2^9$
                                               = 512

$R=\left\{\right(a,b):a,b\in N,a-b=3\}=\left\{\right((n+3),n):n\in N\}$ 

={(4,1),(5,2),(6,3), .......} . 

n ( A × A) = n(A)n(A) = $3^2$ = 9 

So, the total number of subsets of A × A is $2^9$ 

and a subset of A × A is a relation over the set A . 

$Wehave,f\left(x\right)=\frac{x^2+x+2}{x^2+x+1}=\frac{(x^2+x+1)}{x^2+x+1}=1+\frac{1}{{(x+\frac{1}{2})}^2+\frac{3}{4}}Wecanseeherethatasx\to \infty ,f\left(x\right)\to 1whichistheminvalueoff\left(x\right).Alsof\left(x\right)ismaxwhen{(x+\frac{1}{2})}^2+\frac{3}{4}isminwhichissowhenx=-\frac{1}{2}andthen\frac{3}{4}.\therefore f_{max}=1+\frac{1}{\frac{3}{4}}=\frac{7}{3}\therefore R_i=(1,\frac{7}{3}]$

A relation from P to Q is a subset of P $\times$ Q.

$f\left(x\right)+2f(1-x)=x^2+1$   .........(i)
Replacing x by 1 – x
$f(1-x)+2f\left(x\right)=(1-x^2)+1$  .......(ii)
multiplying (ii) by 2 and subtracting it from (1), we get
$-3f\left(x\right)=x^2-2(1-x{)}^2-1$
$3f\left(x\right)=2(1-x{)}^2+1-x^2=(1-x\left)\right(2-2x+1+x)=(1-x\left)\right(3-x)=x^2-4x+3$
$f\left(x\right)=\frac{1}{3}(x^2-4x+3)$
 

(a, b)
g(x) = f(x) + f(–x)
g(–x) = f(–x) + f(x) = g(x)
therefore g(x) is even

Let f(x), g(x) be odd
Let F(x) = f(x)g(x)
F(–x) = f(–x)g(–x) = F(x)
therefore F(x) is even