Relations and Functions Test

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Relations and Functions Test - 6

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Weekly Quiz Competition

Question 1
1 / -0.25

If A is a finite set containing n distinct elements, then the number of relations on A is equal to

A
2n²

B
2n

C
n²

D
None of these

Solution
Question 2
1 / -0.25

R is a relation from { 11, 12, 13} to {8, 10, 12} defined by y = x –3. The relation R⁻¹

A
{(8, 11), (9, 12), (10, 13)}

B
{(11, 8), (13, 10)}

C
{(8, 11), (10, 13)}

D
None of these

Solution

R is a relation from {11, 12, 13} to {8, 10, 12} defined by y = x –3. The relation is given by x = y + 3,from{8, 10, 12} to {11, 12, 13} ⇒ relation = {(8,11),(10,13)}.
Question 3
1 / -0.25

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

A
Symmetric and transitive only

B
Symmetric only

C
Transitive only

D
Not transitive

Solution

A = {1,2,3}
B = {(2,3)} is not reflexive or symmetric on A but it is transitive
∵if (a,b) exists but (b,c) does not exist then (a,c) does not need to exist and the relation is still transitive.
Question 4
1 / -0.25

The range of the function f(x) = [sin x] is

A
[1, 1].

B
(–1, 1)

C
{–1, 0, 1}

D
{–1, 1}

Solution

The only possible integral values of sin x are {-1 ,0, 1}.
Question 5
1 / -0.25

The period of the function f(x) = sin² x + tan x is

A
2 π

B
3 π

C
π

D
None of these

Solution

We have the function f(x) = sin² x + tan x, therefore, f(π+x)
= {sin² (π+x)+tan(π+x)}
= {sin² x+tanx} = f(x).
This implies that period of the function f(x) is π.
Question 6
1 / -0.25

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}

B
{1, 3}

C
{1, 2, 3}

D
None of these.

Solution

Since the domain is represented by the x- co ordinate of the ordered pair (x , y).Therefore, domain of the given relation is {1, 2}.
Question 7
1 / -0.25

Given the relation R = {(1, 2), (2, 3)} on eht set {1, 2, 3}, the minimum number of ordered pairs which when added to R make it an equivalence

A
5

B
6

C
8

D
7

Solution

To make the relation an equivalence relation , the following ordered pairs are required (1,1),(2,2),(3,3)(2,1)(3,2)(1,3),(3,1)
Question 8
1 / -0.25

Let R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} be a relations on the set A = {1, 2, 3, 4}. the relation R is

A
Reflexive

B
Transitive

C
Not symmetric

D
A function

Solution

Given, R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} be a relation on the set A = {1, 2, 3, 4}.
(a) Since, (1, 1), (2, 2), (3, 3), (4, 4) ∉R. So, R is not reflexive.
(b) Since, (1, 3) ∈R and (3, 1) ∈R but (1, 1) ∉R. So, R is not transitive.
(c) Since, (2, 3) ∈R but (3,2) ∉R. So, B is not symmetric.
(d) Since, (2, 4) ∈R and (2, 3) ∈R. So, R is not a function.
Question 9
1 / -0.25
The relation R in the set Z of integers given by R = {(a, b): 2 divides a –b}(a, b) where a, b ∈Z is
A
An empty relation

B
Non commutative relation

C
A universal relation

D
An equivalence relation

Solution
Question 10
1 / -0.25

The range of the function f(x) = x –[x] is

A
[0, 1)

B
(0, 1)

C
[0, 1]

D
None of these.

Solution

Since [x] ≤x, therefore , x –[x] ≥0. Also, x –[x] <1,∴0 ⩽x −[x]<1. Therefore ,Rf = [0,1).
Question 11
1 / -0.25

Let A contain n distinct numbers. How many bijections from A to A can be defined?

A

B
n²

C
N

D
None of these.

Solution

A bijection from A to A is infact an arrangement of its n elements, taken all at a time, which can be done in ways. Hence, the number of bijections from A to A is
Question 12
1 / -0.25

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 13
1 / -0.25

A relation R in a set A is called universal relation, if

A
No element of A is related to any element of A

B
Every element of A is related to one element of A

C
Each element of A is related to every element of A

D
One element of A is related to all elements of A

Solution

The relation R = A x A is called Universal relation.
Question 14
1 / -0.25
Let R be the relation defined in the set A = {1, 2, 3, 4, 5, 6, 7} by R = {(a, b) : both a and b are either odd or even}. Then R is
A
An equivalence relation

B
Non commutative relation

C
A universal relation

D
An empty relation

Solution

Consider any a ,b , c ∈A .

1) Since both a and a must be either even or odd, so (a , a) ∈R ⇒R is reflexive.

2) Let (a ,b) ∈R ⇒ both a and b must be either even or odd, ⇒ both b and a must be either even or odd, ⇒ (b ,a) ∈R. Thus , (a ,b) ∈R ⇒ (b ,a) ∈R ⇒R is symmetric.

3) Let (a ,b) ∈R and (b ,c) ∈R ⇒ both a and b must be either even or odd, also ,both b and c must be either even or odd, ⇒ all elements a, b and c must be either even or odd, ⇒ (a ,c) ∈R . Thus, (a ,b) ∈R ⇒ (b,c) ∈R ⇒ (a ,c) ∈R ⇒R is transitive.
Question 15
1 / -0.25

If f(x) = x –x² , then f(a + 1) –f(a –1) , a ∈ R is :

A
2 –4a

B
2a + 4

C
4 –2a

D
None of these

Solution

f(a+1)−f(a −1)
=(a+1)−(a+1)2 −{(a −1)−(a −1)2} = 2 −4a(a −1)2} = 2 −4a

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Relations and Functions Test - 6

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Relations and Functions Test - 6

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Question 1

2n2

2n

n2

None of these

Solution

Question 2

{(8, 11), (9, 12), (10, 13)}

{(11, 8), (13, 10)}

{(8, 11), (10, 13)}

None of these

Solution

Question 3

Symmetric and transitive only

Symmetric only

Transitive only

Not transitive

Solution

Question 4

[1, 1].

(–1, 1)

{–1, 0, 1}

{–1, 1}

Solution

Question 5

2 π

3 π

π

None of these

Solution

Question 6

{1, 2}

{1, 3}

{1, 2, 3}

None of these.

Solution

Question 7

5

6

8

7

Solution

Question 8

Reflexive

Transitive

Not symmetric

A function

Solution

Question 9

An empty relation

Non commutative relation

A universal relation

An equivalence relation

Solution

Question 10

[0, 1)

(0, 1)

[0, 1]

None of these.

Solution

Question 11

n2

N

None of these.

Solution

Question 12

24

44

16

216

Solution

Question 13

2n²

n²

n²

2⁴

4⁴

2¹⁶