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

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

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