Relations and Functions Test

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Relations and Functions Test - 9

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Weekly Quiz Competition

Question 1
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Let R be a relation on set A of triangles in a plane.

R = {(T₁ , T₂ ) : T₁ , T₂ element of A and T₁ is congruent to T₂ } Then the relation R is______

A
Equivalence relation

B
Transitive

C
Symmetric

D
Reflexive

Solution

It is equivalence relation...
as a triangle is congruent to itself that means (T₁ , T₁ ) exist in relation which implies it is reflexive.
And also if T₁ is congruent to T₂ then T₂ is also congruent to T₁ as simple that means (T₁ ,T₂ ) and (T₂ ,T₁ ) both belongs to relation ...which implies it is symmetric.
And is T1 is congruent to T₂ and T₂ is congruent to T₃ , then T₁ is also congruent to T₃ , congruency rule that means (T₁ ,T₂ ), (T₂ ,T₃ ) and (T₁ ,T₃ ) belongs to relation which implies it transitive.
this relation is reflexive, symmetric, and transitive, hence it is equivalence relation.
Question 2
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Which one of the following relations on set of real numbers is an equivalence relation?

A
a R b ⇔ a ≥b

B
a R b ⇔|a| = |b|

C
a R b ⇔ a > b

D
a R b ⇔ a < b

Solution

If |a| = |b | then |b| = |a| so this is symmetric as well as reflexive and if |a| = |b| and |b| = |c| then |c| = |a| then it is transitive as well so it is an equivalence relation.
Question 3
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Let C = {(a, b): a² + b² = 1; a, b ∈ R} a relation on R, set of real numbers. Then C is

A
Equivalence relation

B
Reflexive

C
Transitive

D
Symmetric

Solution

Correct Answer :- D

Explanation:- Check for reflexive

Consider (a,a)

∴ a²+a²

=1 which is not always true.

If a=2

∴ 2²+2²

=1⇒4+4=1 which is false.

∴ R is not reflexive ---- ( 1 )

Check for symmetric

aRb⇒a²+b²=1

bRa⇒b²+a² =1

Both the equation are the same and therefore will always be true.

∴ R is symmetric ---- ( 2 )

Check for transitive

aRb⇒a²+b²=1

bRc⇒b²+c²=1

∴ a²+c²=1 will not always be true.

Let a=−1,b=0 and c=1

∴ (−1)²+0²=1,

= 0²+1²

=1 are true.

But (−1)²+1²

=1 is false.

∴ R is not transitive ---- ( 3 )
Question 4
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Let A = {1, 2, 3, 4} and B = {x, y, z}. Then R = {(1, x), (2, z), (1, y), (3, x)} is

A
Relation from B to A

B
Is not a relation

C
Relation from A to B

D
Relation from B to B

Solution

Let a set of A = (1, 2, 3, 4) and set B (x, y, z) so. set A of all elements in set B then the relation of A to B
Question 5
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Let R be a relation on N, set of natural numbers such that m R n ⇔ m divides n. Then R is

A
Reflexive and symmetric

B
Neither reflexive nor transitive

C
Reflexive and transitive

D
Symmetric and transitive

Solution

Let there be a natural number n,
We know that n divides n, which implies nRn.
So, Every natural number is related to itself in relation R.
Thus relation R is reflexive .

Let there be three natural numbers a,b,c and let aRb,bRc
aRb implies a divides b and bRc implies b divides c, which combinedly implies that a divides c i.e. aRc.
So, Relation R is also transitive .

Let there be two natural numbers a,b and let aRb,
aRb implies a divides b but it can 't be assured that b necessarily divides a.
For ex, 2R4 as 2 divides 4 but 4 does not divide 2 .
Thus Relation R is not symmetric .
Question 6
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If R be a relation “less than” from set A = {1, 2, 3, 4} to B = {1, 3, 5}, i.e. (a, b) ∈ R if a < b, if (b, a) ∈ R^-1elements in R^-1 are

A
{(3, 3), (3, 5), (5, 3), (5, 5)}

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

C
{(3, 3), (3, 4), (4, 5)}

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

Solution

A = {1, 2, 3, 4}
B = {1, 3, 5}
(a, b) ∈ R if a < b
(b, a) ∈ R^{- 1}
R ^{- 1} will have all (b, a) pairs where b < a, for all b ∈ B, a ∈ A
R ^{- 1} = {(3, 1), (3, 2), (5, 1), (5, 2), (5, 3), (5, 4)}
Question 7
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Let R be a relation on a finite set A having n elements. Then, the number of relations on A is

A
n x n

B
2ⁿ

C
n²

D

Solution

Number of relations on A =

Step-by-step explanation:

If there are n elements in set A then the total number of ordered pairs in the set A × A = n²

In other words A × A will have n² elements.

We also know that if a set has N elements then the number of subsets of A are 2ⁿ

Therefore, for A × A there can as many relations as the number of subsets of A × A
Question 8
1 / -0.25

Let R be a relation on N (set of natural numbers) such that (m, n) R (p, q) mq(n + p) = np(m + q). Then, R is

A
An Equivalence Relation

B
Only Reflexive

C
Transitive and reflexive.

D
Only Transitive

Solution

(m, n) R (p, q) <=>mq(n + p) = np(m + q)
For all m,n,p,q €N
Reflexive:
(m, n) R (m, n) <=>mn(n + m) = nm(m + n)
⇒mn² + m² n = nm² + n² m
⇒mn² + m² n = mn² + m² n
⇒LHS = RHS
So, (m, n) R (m, n) exists.
Hence, it is Reflexive
Symmetric:
Let (m, n) R (p, q) exists
mq(n + p) = np(m + q) --- (eqn1)
(p, q) R (m, n) <=>pn(q + m) = qm (p + n)
⇒np(m + q) = mq(n + p)
⇒mq(n + p) = np(m + q)
This equation is true by (eqn1).
So, (p, q) R (m, n) exists
Hence, it is not symmetric.
Transitive:
Let (m, n) R (p, q) and (p, q) R (r, s) exists.
Therefore,
mq(n + p) = np(m + q) --- (eqn1)
ps(q + r) = qr (p + s) --- (eqn2)
We cannot obtain ms(n+r) = nr(m+s) using eqn1 and eqn2.
So, ms(n + r) ≠nr(m + s)
Therefore, (m, n) R (r, s) doesn ’t exist.
Hence, it is transitive.
Question 9
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A situation in which significant power is distributed among three or more states is known as what?

A
Bipolarity

B
A balance of power

C
A state of nature

D
Multipolarity

Solution

Definition of Multipolar system: A multipolar system is a system in which power is distributed at least among 3 significant poles concentrating wealth and/or military capabilities and able to block or disrupt major political arrangements threatening their major interests.
Question 10
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Let R be an equivalence relation on Z, the set of integers.

R = {(a, b): a,b ∈Z and a –b is a multiple of 3} The Equivalence class of [1] is

A
{-7, -4, 2, 5, 8}

B
{-4, -1, 2, 5, 8}

C
{-6, -1, 3, 5, 9}

D
{…..-5, -2, 1, 4, 7}

Solution

Correct Answer :- d

Explanation : R = (a,b) : 3 divides (a-b)

⇒(a−b) is a multiple of 3.

To find equivalence class 1, put b=1

So, (a−0) is a multiple of 3

⇒ a is a multiple of 3

So, In set z of integers, all the multiple

of 3 will come in equivalence

class {1}

Hence, equivalence class {1} = {3x+1}

{-5,-2,1,4,7}

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Relations and Functions Test - 9

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Question 1

Equivalence relation

Transitive

Symmetric

Reflexive

Solution

Question 2

a R b ⇔ a ≥b

a R b ⇔|a| = |b|

a R b ⇔ a > b

a R b ⇔ a < b

Solution

Question 3

Equivalence relation

Reflexive

Transitive

Symmetric

Solution

Question 4

Relation from B to A

Is not a relation

Relation from A to B

Relation from B to B

Solution

Question 5

Reflexive and symmetric

Neither reflexive nor transitive

Reflexive and transitive

Symmetric and transitive

Solution

Question 6

{(3, 3), (3, 5), (5, 3), (5, 5)}

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

{(3, 3), (3, 4), (4, 5)}

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

Solution

Question 7

n x n

2n

n2

Solution

Question 8

An Equivalence Relation

Only Reflexive

Transitive and reflexive.

Only Transitive

Solution

Question 9

Bipolarity

A balance of power

A state of nature

Multipolarity

Solution

Question 10

{-7, -4, 2, 5, 8}

{-4, -1, 2, 5, 8}

{-6, -1, 3, 5, 9}

{…..-5, -2, 1, 4, 7}

Solution

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