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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 Page/Excercise 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

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

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.

## RD Sharma Solution for Class 12 Humanities Mathematics Chapter 1 - Relations Page/Excercise 1.2

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Also we need to find the set of all elements related to 1.

Since the relation is given by, R={(a,b):a=b}, and 1 is an element of A,

R={(1,1):1=1}

Thus, the set of all elements related to 1 is 1.

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

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 29