Relations and Functions Test

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Relations and Functions Test - 7

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Weekly Quiz Competition

Question 1
1 / -0

Let A = {1, 2, 3}, then the domain of the relation R = {(1, 1), (2, 3), (2, 1)} defined on A is

A
{1, 2, 3}

B
{1, 3}

C
{1, 2}

D
None of these.

Solution

Since the domain is represented by the x- coordinate of the ordered pair ( x , y ).Therefore, domain of the given relation is { 1 , 2 }.
Question 2
1 / -0

Let A = {a, b, c}, then the range of the relation R= {(a, b), (a, c), (b, c)} defined on A is

A
{b, c}

B
{a, b, c}

C
{c}

D
{a, b}

Solution

Since the range is represented by the y- coordinate of the ordered pair ( x , y ). Therefore, range of the given relation is { b , c }.
Question 3
1 / -0

Number of relations that can be defined on the set A = {a, b, c, d} is

A
24

B
4⁴

C
16

D
2¹⁶

Solution

No. of elements in the set A = 4 . Therefore , the no. of elements in A × A = 4 × 4 = 16 As, the no. of relations in A × A = no. of subsets of A × A = 2¹⁶ .
Question 4
1 / -0

Let A = {1, 2, 3, 4, 5, 6}. Which of the following partitions of A correspond to an equivalence relation on A?

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

B
{1, 2, }, {3, 4}, {2, 3, 5, 6}

C
{1, 2,3 }, {4, 5, 6}

D
{1, 3}, {2, 4, 5}, {6}

Solution

Conditions for the partition sub-sets to be an equivalence relation:

(i) The partition sub-sets must be disjoint i.e.there is no common elements between them

(ii) Their union must be equal to the main set (super-set)

Here, the set A={1,2,3,4,5,6},the partition sub-sets {1,3},{2,4,5},{6} are pairwise disjoint and their union i.e. {1,3} U {2,4,5} U {6} = {1,2,3,4,5,6} = A, which is the condition for the partition sub-sets to be an equivalence relation of the set A.
Question 5
1 / -0

A relation R on a non – empty set A is an equivalence relation if it is

A
Reflexive, antisymmetric, transitive

B
symmetric and transitive

C
reflexive

D
reflexive, symmetric and transitive

Solution

By definition of Equivalence Relation, a relation is said to be equivalence if it is reflexive,symmetric and transitive.
Question 6
1 / -0

Let R={ (x,y) : x²+y²= 1 and x, y ∈∈ R } be a relation in R. The relation R is

A
symmetric

B
anti – symmetric

C
reflexive

D
transitive

Solution

A relation R on a non empty set A is said to be symmetric if xRy ⇔⇔ yRx, for all x,y ∈∈ R. Clearly, x²+ y²= 1 is same as y²+ x²= 1 for all x,y ∈∈R. Therefore, R is symmetric.
Question 7
1 / -0

The void relation (a subset of A x A ) on a non empty set A is :

A
Reflexive

B
Only transitive

C
Only symmetric

D
Transitive and symmetric

Solution

The relation { }⊂ A x A on A is surely not reflexive.However, neither symmetry nor transitivity is contradicted. So { } is a transitive and symmetric relation on A.
Question 8
1 / -0

If A = { 1, 2, 3}, then the relation R = {(1, 2), (2, 3), (1, 3)} in A is _

A
reflexive only

B
transitive only

C
symmetric only

D
symmetric and transitive only

Solution

A relation R on a non-empty set A is said to be transitive if xRy and yRz , for all x ∈ R. Here, (1, 2) and (2, 3) belongs to R implies that (1, 3) belongs to R

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Relations and Functions Test - 7

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Relations and Functions Test - 7

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Question 1

{1, 2, 3}

{1, 3}

{1, 2}

None of these.

Solution

Question 2

{b, c}

{a, b, c}

{c}

{a, b}

Solution

Question 3

24

44

16

216

Solution

Question 4

{1, 2, 3}, {3, 4, 5, 6}.

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

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

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

Solution

Question 5

Reflexive, antisymmetric, transitive

symmetric and transitive

reflexive

reflexive, symmetric and transitive

Solution

Question 6

symmetric

anti – symmetric

reflexive

transitive

Solution

Question 7

Reflexive

Only transitive

Only symmetric

Transitive and symmetric

Solution

Question 8

reflexive only

transitive only

symmetric only

symmetric and transitive only

Solution

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