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 A = {1, 2, 3} and B = {5, 6, 7, 8, 9} and let f(x) = {(1, 8), (2, 7), (3, 6)} then f is

A
bijective

B
Injective

C
not a function

D
surjective

Solution

Here, f(1) = 8, f(2) = 7, f(3) = 6.

Since, different points of domain have the different f-image in the range, therefore f is a one-one function.
Question 2
1 / -0

The law a + b = b + a is called

A
Distributive law

B
Associative law

C
Commutative law

D
Closure law

Solution

For a binary operation ‘+’, a + b = b + a, is a commutative law on a and b.
Question 3
1 / -0

For which one of these mappings the function f(x) = x ² will be one-one?

A
N→N

B
Z→Z

C
R→R

D
None of these

Solution

Let x₁ and x₂ ∈ Domain such that f(x₁) = f(x₂)

i.e x₁ ² = x₂ ²

⇒ x₁ =± x₂

If the domain is N then x₁ = x₂ (which is true)
Question 4
1 / -0

Let A = {1,2,3,4,5,6,7}. P={1,2}, Q = {3, 7}. Write the elements of the set R so that P, Q and R form a partition that results in equivalence relation.

A
{0}

B
{ }

C
{1,2,3,4,5,6,7}

D
{4,5,6}

Solution

With the option (i) the sets P , Q and R will be pair wise disjoint and their Union would result in A
Question 5
1 / -0

A function f: A→B is said to be invertible, if there exists a function g: B→A such that

A
g o f = I_A and f o g = I_B

B
g o f = I_B and f o g = I_A

C
g o f = I_B and f o g = I_B

D
g o f = I_A and f o g = I_A

Solution

A function f: A→B is said to be invertible, if there exists a function g: B→A

such that :

g o f = I_A and f o g = I_B
Question 6
1 / -0

Let R be the relation on the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3,3), (3,2)}. then R is

A
R is reflexive and symmetric but not transitive.

B
R is symmetric and transitive but not reflexive.

C
R is an equivalence relation.

D
R is reflexive and transitive but not symmetric.

Solution

(i) All the ordered pairs (1, 1), (2,2), (3,3) (4,4) are in R, i.e the relation is reflexive.

(ii) (1, 2) ^∈ R but (2,1) ^∉ R so R is not symmetric

(ii) For pairs (a, b) and (b, c) ^∈ R, (a, c) ^∈ R . i.e for (1, 2), (2, 2) ^∈ R we have (1,1) ^∈ R so R is transitive
Question 7
1 / -0

If A = N x N and * be any binary operation on A defined by

(a, b) * (c, d) = (a + c, b + d), then the binary operation is

A
Commutative and associative

B
Commutative but not associative

C
Associative

D
Commutative

Solution

(a, b) * (c, d) = (a + c, b + d) and

(c, d) * (a, b) = (c + a, d + b) = (a + c, b + d) = (a, b) * (c, d)

So, * is commutative.

(a, b) * [(c, d) * (e, f)] = (a, b) * (c + e, d + f) = (a + c + e, b + d + f) and

[(a, b) * (c, d)] * (e, f) = (a + c, b + d) * (e, f) = (a + c + e, b + d + f).

So, * is associative also.

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Relations and Functions Test - 4

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Relations and Functions Test - 4

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Question 1

bijective

Injective

not a function

surjective

Solution

Question 2

Distributive law

Associative law

Commutative law

Closure law

Solution

Question 3

N→N

Z→Z

R→R

None of these

Solution

Question 4

{0}

{ }

{1,2,3,4,5,6,7}

{4,5,6}

Solution

Question 5

g o f = IA and f o g = IB

g o f = IB and f o g = IA

g o f = IB and f o g = IB

g o f = IA and f o g = IA

Solution

Question 6

R is reflexive and symmetric but not transitive.

R is symmetric and transitive but not reflexive.

R is an equivalence relation.

R is reflexive and transitive but not symmetric.

Solution

Question 7

Commutative and associative

Commutative but not associative

Associative

Commutative

Solution

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g o f = I_A and f o g = I_B

g o f = I_B and f o g = I_A

g o f = I_B and f o g = I_B

g o f = I_A and f o g = I_A