Relations and Functions

7. Given,

A= {1, 2}, B = {1,2,3,4}, C= {5,6} and D= {5,6,7,8}

(i) L.H. S = A × (B∩ C) = {1,2} [ {1,2,3,4} ∩ {5,6}]

= {1,2}× 

=  .

R.H.S = (A× B)∩ (A ×C)= [ {1,2}× {1,2,3,4}]∩ [ {1,2} {5,6}]

= [ { (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)]∩ [ {1,5), (1,6), (2,5), (

6. Given,

A ×B = { (a, x), (a, y), (b, x), (b, y)}

We know that,

A ×B = { (p, q); p ∈ A and q ∈ B}

So, A = {a, b} and B = {x, y}.

5. Given, A = {1,1}

So, A* A = { (1,1), (1,1), (1,1), (1,1)}

A *A *A = { (1,1), (1,1), (1,1), (1,1)} * {1,1}

= { (1,1.1), (1, 1), (1, ), (1,1,1), (1,1,1), (1,1), (1,1), (1,1,1)}

The $fx$ n is $f(x)=\frac{1}{x}$ , which is a $f:R_{*}$ $\to$ $R_{*}$ and $R_{*}$ is set of all non-zero real numbers

For, $x_1,x_2\in R_{*},f(x_1)=f(x_2)$

$\Rightarrow \frac{1}{x_1}=\frac{1}{x_2}$

$\Rightarrow x_1=x_2$ So, $f$ is one-one

For, $y\in R_{*},$ $x=\frac{1}{f(x)}=\frac{1}{y}$ such that

So, $f(x)=y$

So, every element in the co-domain has a pre-image in $f$

So, $f$ is onto

If $f:N\to R_{*}$ such th

4. (i) False. Here P = {m, n}, n (p)=2

Q = {n, m}, n (Q)=2

n (P× Q) = n (P)× n (Q) = 2× 2 = 4.

So, P ×Q = { (m, n), (m, m), (n, n), (n, m)}

(ii) True.

(iii) True. { A × (B ∩ ?) = A× ? . {∴ B ∩ ? = ?  }

= n (A) ×0 {? is empty set}

= ? 

The given relation in set N defined by

$R=\left\{\left(a,b\right):a=b-2,b>6\right\}$

For (2,4),        4>6 is not true

For (3,8),     8>6  but  3= 8-2 ⇒3=6 is not true

For (6,8),      8>6 and 6= 8-2 ⇒6=6 is true

And for (8,7), 7>6 but 8= 7-2 ⇒8=5 is not true

Hence, option (C) is correct

The set in $A=\left\{1,2,3,4\right\}$

The relation in this set $A$ is given by

$R=\left\{\left(1,2\right),\left(2,2\right),\left(1,1\right),\left(4,4\right),\left(1,3\right),\left(3,3\right),\left(3,2\right)\right\}$

$R$ is reflexive as $\left(1,1\right),\left(2,2\right),\left(3,3\right),\left(4,4\right)\in R$

As, $\left(1,2\right)\in R$ but $\left(2,1\right)\notin R$

$R$ is not symmetric

For $\left(1,2\right)\in R$ and $\left(2,2\right)\in R;\left(1,2\right)\in R$

And for $\left(1,3\right)\in R$ and $\left(3,2\right)\in R;\left(1,3\right)\in R$

∴ $R$ is transitive

Hence, option (B) is correct

The given relation in the set $L=$ all lines in $XY-$ plane is defined as

$R=\{\left(L_1,L_2\right):L_1$ is parallel to $L_2\}$

Let $L_1\in A$ then as $L_1$ is parallel to $L_1$ ,

$\left(L_1,L_1\right)\in R$

So, $R$ is reflexive

Let $L_1,L_2\in A$ and $\left(L_1,L_1\right)\in R$

Then, $L_1$ is parallel to $L_2$

$L_2$ is parallel to $L_1$

So, $\left(L_2,L_1\right)\in R$

i.e.,&

The given relation in set A of all polygons is defined as

R= $\{\left(P_1,P_2\right):P_1$ and $P_2$ have same number of sides $\}$

Let $P_1\in A$ ,

As number of sides $\left(P_1\right)$ = number of sides $\left(P_1\right)$

$\left(P_1,P_1\right)\in R$

So, R is reflexive.

Let $P_1,P_2\in A$ and $\left(P_1,P_2\right)\in R$

Then, number of sides of $P_1$ = number of sides of $P_2$

Number of

The given relation to set A of all triangles is defined as

R= $\{\left(T_1,T_2\right):T_1$ is similar to $T_2\}$

For $T_1\in A$ ,

$T_1$ is always similar to $T_1$

So, $\left(T_1,T_1\right)\in R$ . Hence R is reflexive.

For $T_{1,}{T}_2\in A$ and $\left(T_1,T_2\right)\in R$ we have

$T_1\sim T_2\left(similar\right)$

$T_2\sim T_1$ i.e., $\left(T_2,T_1\right)\in R$

so, R is symmetric.

for, $T_1,T_2,T_3\in A$ and $\left(T_1,T_2\right)\in R$ and $\left(T_2,T_3\right)\in R$

$T_1\sim T_2$ and&