Ncert Solutions Maths class 12th

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&

The given relation in set A of points in a plane is

R= $\{\left(P,Q\right):$ distance of point P from origin=distance of point Q from $origin\}$

If O is the point of origin

R= $\left\{\left(P,Q\right):PO=QO\right\}$

Then, for $P\in A$ we have PO=PO

So,  $\left(P,P\right)\in R$

i.e., P is reflexive

for,  $P,Q\in A$ and $\left(P,Q\right)\in R$ we have

PO=QO

QO=PO i.e.,  $\left(Q,P\right)\in R$

i.e., R is s

Let A= $\left\{a,b,c\right\}$

(i) R= $\left\{\left(a,b\right),\left(b,a\right)\right\}$ is a relation in set A

So, $\left(a,b\right)\in R$ and $\left(b,a\right)\in R\to$ Symmetric

$\left(a,a\right)\notin R\to$ not reflexive

$\left(a,b\right)\in R,\left(b,a\right)\in R$ but $\left(a,a\right)\notin R$ $\to$ not transitive

(ii) R= $\left\{\left(a,b\right),\left(b,c\right),\left(a,c\right)\right\}$ is a relation in set A

So, $\left(a,a\right)\notin R\to$ not reflexive

$\left(a,b\right)\in R$ but $\left(b,a\right)\notin R\to$ not symmetric

$\left(a,b\right)\in R\&\left(b,c\right)\in R$ and also $\left(a,c\right)\in R\to$ transitive

(iii) R= $\left\{\left(a,a\right),\left(b,b\right),\left(c,c\right),\left(a,b\right),\left(b,a\right),\left(a,c\right),\left(c,a\right)\right\}$

So, $\left(a,a\right),\left(b,b\right),\left(c,c\right)\in R\to$&n

We have,

A= $\{x\in 2,0\le x\le 12\}$

The relation in set A is defined by

R= { $\left(a,b\right):\left|a-b\right|$ is a multiple of 4}

For all $a\in A$ ,

$\left|a-a\right|=0$ is a multiple of 4

So, $\left(a,a\right)\in R$ i.e., R is reflexive

For $a,b\in A\&\left(a,b\right)\in R$ we have,

$\left|a-b\right|$ is multiple of 4

$\left|-\left(b-a\right)\right|$ is multiple of 4

$\left|b-a\right|$ is multiple of 4

So, $\left(b,a\right)\in R$

i.e., R is symmetric

for $a,b,c\in A$ $\&\left(a,b\right)\in R\&\left(b,c\right)\in R$

$\left|a-b\right|$&n

We have,

R= $\{\left(a,b\right)\therefore \left|a-b\right|$ is even $\}$ is a relation in set A= $\{1,2,3,4,5\}$

For all $a\in A$ , $\left|a-a\right|=0$ is even.

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

For $a,b\in A$ and $\left(a,b\right)\in R$

$\left|a-b\right|$ is even

$\left|-b+a\right|$ is even $|$

$\left|-\left(b-a\right)\right|$ is even

$\left|b-a\right|$ is even

i.e., $\left(b,a\right)\in R$

Hence, R is symmetric.

For $a,b,c\in A$ and $\left(a,b\right)\in R$ and 

We have,

R= $\{\left(x,y\right):x\&y$ have same number of pages $\}$ is a relation in set of A of all books in

For $\left(x,y\right)\in R\&x,y\in A$

As x=y=same no. of pages

Then,  $\left(x,x\right)\in R$

Hence, R is reflexive.

For $\left(x,y\right)\in R$ and $x,y\in A$

Also,  $\left(y,x\right)\in R$ ,  $\therefore x=y$

Hence, R is symmetric.

For $x,y,z\in A$ and $\left(x,y\right)\in R$ and $\left(y,z\right)\in R$

x=y and y=z

x=z

i.e., &

We have,

R= $\{\left(1,2\right),\left(2,1\right)\}$ is a relation in set $\{1,2,3\}$

Then, as $\left(1,1\right)\notin R$ and $\left(2,2\right)\notin R$

So, R is not reflective

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

So, R is symmetric

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

So, R is not transitive.

We have,

R= $\{\left(a,b\right):a\le b^3\}$ is a relation in R.

For, $\left(a,b\right)\in R$ and $a=\frac{1}{2}$ we can write

$a\le a^3$ => $\frac{1}{2}\le {\left(\frac{1}{2}\right)}^3$ => $\frac{1}{2}\le \frac{1}{8}$ which is not true.

So, R is not reflexive.

For $\left(a,b\right)=\left(1,2\right)\in R$ we have,

$a\le b^3$ => $1\le 2^3$ => $1\le 8$ is true.

So, $\left(1,2\right)\in R$

But $2\le 1^3$ => $2\le 1$ is not true

So, $\left(2,1\right)\notin R$ and $\left(b,a\right)\notin R$

Hen

We have, R= $\{\left(a,b\right):a\le b\}$ is a relation in R.

For,  $a\in R$ ,

$a\le b$ but $b\le a$ is not possible i.e.,  $\left(b,a\right)\notin R$

Hence, R is not symmetric.

For $\left(a,b\right)\in R\&\left(b,c\right)\in R$ and $a,b,c\in R$

$a\le b$ and $b\le c$

So,  $a\le c$

i.e.,  $\left(a,c\right)\in R$

$\therefore$ R is transitive.

We have,

R= $\{\left(a,b\right):b=a+1\}$ is a relation in set $\{1,2,3,4,5,6\}$

So, R= $\{\left(1,2\right),\left(2,3\right),\left(3,4\right),\left(4,5\right),\left(5,6\right),\left(6,7\right)\}$

As,  $\left(1,1\right)\notin R$ , R is not reflexive

As,  $\left(1,2\right)\in R$ but $\left(2,1\right)\notin R$ , R is not symmetric

And as $\left(1,2\right)$ & $\left(2,3\right)\in R$ but $\left(1,3\right)\notin R$

Hence, R is not transitive.