Relations and Functions Test

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Relations and Functions Test - 4

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Weekly Quiz Competition

Question 1
1 / -0

Let R be a relation defined on the set of N natural numbers as

R = {(x, y): y is a factor of x, x, y∈ N} then,

A
(2, 4) ∈ R

B
(4, 2) ∈ R

C
(3, 9) ∈ R

D
(9, 5) ∈ R

Solution

R = {(x, y): y is a factor of x, x, y ∈ N}

Now if y is a factor of x then y divides x exactly. Since 2 divides 4 so (4, 2) ∈ R
Question 2
1 / -0

If A = {1, 2, 3 }, B = {4, 5, 6 }, which of the following is not a relation from A to B

A
R₃ = {(1, 4), (1, 5), (3, 6), (2, 6), (3, 4)}

B
R₄ = {(4, 2), (2, 6), (5, 1), (2, 4)}

C
R₂= {(1, 5), (2, 4), (3, 6)}

D
R₁ = {(1, 4), (1, 5),(1, 6)}

Solution

R₄is a set of ordered pairs where each first element does not come from A.
Question 3
1 / -0

For a constant function f(x)=k where k is a real number ; {k} is

A
the range of f

B
the co-domain of f

C
a superset of the range

D
the domain

Solution

Range of a function is the set of all images of the elements of the domain of the function. In this case each x from the domain is mapped with the number k. So {k} becomes the range of the function.
Question 4
1 / -0

Given A = {1, 2} and B = {5, 6, 7} then A x B =

A
{(1, 5), (1, 6), (1, 7), (2, 5), (2, 6), (2, 7)}

B
{ 1, 6), (1, 7), (2, 5), (2, 6), (2, 7)}

C
{(5, 1), (5, 2), (6, 1), (6, 2), (7, 1), (7, 2)}

D
Φ

Solution

A x B = {(1, 5), (1, 6), (1, 7), (2, 5), (2, 6), (2, 7)}
Question 5
1 / -0

For the relation R= {(1,2),(2,3),(4,5),(3,6)}

The domain is

A
{1,2,3,4}

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

C
{1,3,5,6}

D
{2,3,5,6}

Solution

The domain of a relation is the set of all first entries in the ordered pairs of the relation.
Question 6
1 / -0

If f(x) = x² - x + 1, g(x) = 7x - 3, be two real functions then (f - g)(8) is

A
64

B
57

C
4

D
53

Solution

For functions f : X ^→ R and g : X ^→ R, we have

(f - g)(x) = f(x) - g(x), x ^∈ X

Hence,

(f - g)(8) = f(8) - g(8) = 4
Question 7
1 / -0

If n(A) = p and n(B) = q, then how many relations are there from A to B.

A
(pq)²

B
3^pq

C
2^pq

D
pq

Solution

The number of relations from a set A with m elements to a set B with n elements is given by 2^mn

So given n(A) = p and n(B) = q

Then 2^pq relations are defined from A to B.
Question 8
1 / -0

If A = {1, 2, 3, 4, 5}, let relation R be defined on A as R={(x, y): y = x+3, x, y∈ A } Then R is

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

B
{(1, 2), (2, 3), (3, 4)}

C
{(1, 4), (2, 5)}

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

Solution

A = {1, 2, 3, 4, 5}. A relation R is defined as

R={(x, y): y = x+3, x, y∈ A }

1 R 4 and 2 R 5 and 3R 6 but 6∉ A.

So R = {(1, 4), (2, 5)}
Question 9
1 / -0

If f(x) = x² – x + 1; g(x) = 7x – 3, be two real functions then (f + g)(3) is

A
25

B
3

C
18

D
7

Solution

For functions f : X ^→ R and g : X ^→ R, we have

(f + g)(x) = f(x) + g(x), x ∈ X

Hence,

(f + g)(3) = f(3) + g(3) = 25

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Relations and Functions Test - 4

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Relations and Functions Test - 4

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Question 1

(2, 4) ∈ R

(4, 2) ∈ R

(3, 9) ∈ R

(9, 5) ∈ R

Solution

Question 2

R3 = {(1, 4), (1, 5), (3, 6), (2, 6), (3, 4)}

R4 = {(4, 2), (2, 6), (5, 1), (2, 4)}

R2 = {(1, 5), (2, 4), (3, 6)}

R1 = {(1, 4), (1, 5),(1, 6)}

Solution

Question 3

the range of f

the co-domain of f

a superset of the range

the domain

Solution

Question 4

{(1, 5), (1, 6), (1, 7), (2, 5), (2, 6), (2, 7)}

{ 1, 6), (1, 7), (2, 5), (2, 6), (2, 7)}

{(5, 1), (5, 2), (6, 1), (6, 2), (7, 1), (7, 2)}

Φ

Solution

Question 5

{1,2,3,4}

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

{1,3,5,6}

{2,3,5,6}

Solution

Question 6

64

57

4

53

Solution

Question 7

(pq)2

3pq

2pq

pq

Solution

Question 8

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

{(1, 2), (2, 3), (3, 4)}

{(1, 4), (2, 5)}

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

Solution

Question 9

25

3

18

7

Solution

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R₃ = {(1, 4), (1, 5), (3, 6), (2, 6), (3, 4)}

R₄ = {(4, 2), (2, 6), (5, 1), (2, 4)}

R₂= {(1, 5), (2, 4), (3, 6)}

R₁ = {(1, 4), (1, 5),(1, 6)}

(pq)²

3^pq

2^pq