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# RD Sharma Solution for Class 12 Humanities Mathematics Chapter 1 - Relations

Exercise/Page

## RD Sharma Solution for Class 12 Humanities Mathematics Chapter 1 - Relations MCQ

Solution 35

Given: R = {(1, 2)}

R is not reflexive as (1, 1), (2, 2), (3, 3) R.

Since (1, 2) R but (2, 1) R

Therefore, R is not symmetric.

Also, R is not transitive.

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

Solution 26

Solution 27

Solution 28

Solution 29

Solution 30

Solution 31

Solution 32

Solution 33

Solution 34

## RD Sharma Solution for Class 12 Humanities Mathematics Chapter 1 - Relations Ex. 1.1

Solution 1(i)

Solution 1(ii)

Solution 1(iii)

Solution 1(iv)

Solution 2

Solution 3(i)

Solution 3(ii)

Solution 3(iii)

Solution 4

Solution 5(i)

Solution 5(ii)

Solution 5(iii)

Solution 6

Solution 7(i)

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 16

Solution 17

A relation R in A is said to be reflexive if aRa for all aA

R is said to be transitive if aRb and bRc aRc

for all a, b, c A.

Hence for R to be reflexive (b, b) and (c, c) must be there in the set R.

Also for R to be transitive (a, c) must be in R because (a, b) R and (b, c) R so (a, c) must be in R.

So at least 3 ordered pairs must be added for R to be reflexive and transitive.

Solution 18

A relation R in A is said to be reflexive if aRa for all aA, R is symmetric if aRb bRa, for all a, b A and it is said to be transitive if aRb and bRc aRc for all a, b, c A.

x > y,  x, y ϵ N

(x, y) ϵ {(2, 1), (3, 1).......(3, 2), (4, 2)....}

This is not reflexive as (1, 1), (2, 2)....are absent.

This is not symmetric as (2,1) is present but (1,2) is absent.

This is transitive as (3, 2) ϵ R and (2,1) ϵ R also (3,1) ϵ R ,similarly this property satisfies all cases.

x + y = 10,  x, y ϵ N

(x, y)ϵ {(1, 9), (9, 1), (2, 8), (8, 2), (3, 7), (7, 3), (4, 6), (6, 4), (5, 5)}

This is not reflexive as (1, 1),(2, 2)..... are absent.

This only follows the condition of symmetric set as  (1, 9)ϵR also (9, 1)ϵR similarly other cases are also satisfy the condition.

This is not transitive because {(1, 9),(9, 1)}ϵR but (1, 1) is absent.

xy is square of an integer, x, y ϵ N

(x, y) ϵ {(1, 1), (2, 2), (4, 1), (1, 4), (3, 3), (9, 1),  (1, 9), (4, 4), (2, 8), (8, 2), (16, 1), (1, 16)...........}

This is reflexive as (1,1),(2,2)..... are present.

This is also symmetric because if aRb bRa, for all a,bϵN.

This is transitive also because if aRb and bRc aRc for all a, b, c ϵ N.

x + 4y = 10, x, y ϵ N

(x, y) ϵ {(6, 1), (2, 2)}

This is not reflexive as (1, 1), (2, 2).....are absent.

This is not symmetric because (6,1) ϵ R but (1,6) is absent.

This is not transitive as there are only two elements in the set having no element common.

Solution 7(ii)

Given: R = {(a, b) : a < b}

Clearly, 1 < 2

That is, (1, 2) R

But 2 1

So, (2, 1) R

Therefore, R is not symmetric.

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

a < b and b < c

a < c

Therefore, (a, c) R

Thus, R is transitive.

Solution 15

Given: R = {(x, y) W × W such that x and y have at least one letter in common}

Clearly, (x, x) R as x and x have all the letters common.

Therefore, R is reflexive.

Now suppose, x and y have at least one letter in common.

So, y and x will also have at least one letter in common.

Therefore, R is symmetric.

Now take, x = Sun, y = Moon and z = Mars

Here, (x, y) R and (y, z) R as x and y have the letter "n" common whereas y and z have the letter "m" common.

But, x and z do not have any letter common.

Therefore, (x, z) R.

Thus, R is not transitive.

## RD Sharma Solution for Class 12 Humanities Mathematics Chapter 1 - Relations Ex. 1.2

Solution 13

Given: R = {(a, b): a is divisor of b}

Any natural number "a" divides itself.

So, a is divisor of a.

Therefore, R is reflexive.

We know that 2 is divisor of 4.

So, (2, 4) R

But 4 does not divide 2, then (4, 2) R

Therefore, R is not symmetric.

As R is not symmetric.

Thus, R is not an equivalence relation.

Solution 18

To check reflexive:

Now,  which is real.

So, z1 R z1.

Therefore, R is reflexive.

To check symmetric:

Suppose z1 R z2

is real

Let  where k is real.

Now,

Therefore,  is real as -k is real.

Therefore, R is symmetric.

To check transitive:

Suppose z1 R z2 and z2 R z3

are real

Let

… (i)

Now,  which is real.

Thus, R is transitive.

Hence, R is an equivalence relation.

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7