Note : We don't have to list out all these things. In order to understand, we have listed out.

R1 ∈ (2, 1) ∉ A x B. Hence R1 is not a relation.

(ii) R2 = {(–1,1)}

Solution :

(-1, 1) is a element of R2, but in Cartesian product, we have (1, -1).

(1, -1) ≠ (-1, 1). Hence R2 is not a relation.

(iii) R3 = {(2, –1), (7, 7), (1, 3)}

Solution :

R3 is a subset of A x B, Hence it is a relation from A to B.

(iv) R4= {(7,–1), (0,3), (3,3), (0,7)}

Solution :

The element (7,–1) ∈ A x B

But (0,3), (3,3), (0,7) ∉ A x B

Hence it is not a relation.

(2) Let A = {1, 2, 3, 4,..., 45} and R be the relation defined as “is square of ” on A. Write R as a subset of A x A. Also, find the domain and range of R.

Solution :

A = {1, 2, 3, 4,..., 45}

Let "A" be a set which contains some of elements of A

A = {1, 2, 3, 4, 5, 6}

Now, we have to write the other set of elements from A. The elements in this set denotes the square value of previous set.

Square values of set A = {1, 4, 9, 16, 25, 36}

A x A = {(1, 1)(1,4)(1, 9)(1, 16) (1, 25) (1, 36)..................}

Domain = {1, 2, 3, 4, 5, 6}

Range = {1, 4, 9, 16, 25, 36}

(3) A Relation R is given by the set {(x, y) /y = x + 3, x ∈ {0, 1, 2, 3, 4, 5}}. Determine its domain and range.

Solution :

Given that :

y = x + 3

y = x + 3

x = 0

y = 0 + 3

y = 3

y = x + 3

x = 1

y = 1 + 3

y = 4

y = x + 3

x = 2

y = 2 + 3

y = 5

y = x + 3

x = 3

y = 3 + 3

y = 6

y = x + 3

x = 4

y = 4 + 3

y = 7

y = x + 3

x = 5

y = 5 + 3

y = 8

R = {(0, 3) (1, 4) (2, 5) (3, 6) (4, 7) (5, 8)}

Domain of R = {0, 1, 2, 3, 4, 5}

Range of R = {3, 4, 5, 6, 7, 8}

(4) Represent each of the given relations by (a) an arrow diagram, (b) a graph and (c) a set in roster form, wherever possible.

(i) {(x, y)| x = 2y, x ∈ {2, 3, 4, 5}, y ∈ {1, 2, 3, 4} Solution

Solution :

Given that :

x = 2y

y = x/2

if x = 2

y = 2/2

y = 1

f x = 3

y = 3/2

if x = 4

y = 4/2

y = 2

if x = 5

y = 5/2

Hence the required relation is {(2, 1) (3, 3/2) (4, 2) (5, 5/2)}

(i) Arrow diagram

(ii) A graph

(iii) in roster form

= {(2, 1) (4, 2)}

(ii) {(x, y) | y = x + 3, x, y are natural numbers < 10} Solution

Solution :

Given that :

y = x + 3

if x = 1

y = 4

if x = 2

y = 5

if x = 3

y = 6

if x = 4

y = 7

if x = 5

y = 8

if x = 6

y = 9

R = {(1, 4) (2, 5) (3, 6) (4, 7) (5, 8) (6, 9)}

An arrow diagram :

(ii) A graph

In Roster form :

= {(1, 4) (2, 5) (3, 6) (4, 7) (5, 8) (6, 9)}

(5) A company has four categories of employees given by Assistants (A), Clerks (C), Managers (M) and an Executive Officer (E). The company provide ₹10,000, ₹25,000, ₹50,000 and ₹1,00,000 as salaries to the people who work in the categories A, C, M and E respectively. If A1, A2, A3, A4and A5 were Assistants; C1, C2, C3, C4 were Clerks; M1, M2, M3 were managers and E1, E2 were Executive officers and if the relation R is defined by xRy, where x is the salary given to person y, express the relation R through an ordered pair and an arrow diagram.