Introduction To Relations - Free MCQ Practice Test with solutions, JEE

Correct Answer : a

Explanation :  A = {(x,y) : x² + y² = 5}

=> (1,2) {(1)² + (4)² = 5}

=> {1 + 4 = 5}

=> {5 = 5}

Since T is a set of real number for it's given ordered pairs we have a condition provided that is 1 + ab is less than zero. so in the given option c if we put order of a and b then it satisfy our given condition

_xR_y => x - y + √2 is an irrational number.
Let R is a binary relation on real numbers x and y.
Clearly, R is reflexive relation
As _xR_x iff  x – x +√2 = √2 ,which is an irrational number.
Here R is not symmteric if we take x =√2  and y =1 then x – y + √2 is an irrational
number but y – x + √2 = 1, which is not irrational number
Now, R is transitive iff for all (x, y) ∈ R and (y, z) ∈ R implies (x, z) ∈ R
But here R is not transitive as we take x = 1, y = 2√2, z=√2
Given, _xR_y => x - y + √2 is irrational    ............1
and _yR_z => y - z + √2 is irrational       ............2
Add equation 1 and 2, we get
(x - y + √2) + (y - z + √2)
= x - z + √2  = 1, which is not an irrational

Let R={(1,3),(4,2),(2,4),(2,3),(3,1)} be a relation on the set A={1,2,3,4}, then
(a) Since (2,3) ∈ R but (3,2) ∈/​ ′R, so R is not symmetric.
(b) Since (1,3) ∈ R and (3,1) ∈/​ R but (1,1) ∈/​ R, so R is not transitive:
(c) Since (1,1) ∈/​ R, so R is not reflexive.
(d) Since (2,4) ∈ R and (2,3) ∈ R, so R is not a function.

Clearly there is no pair in set A whose difference is 8 or -8.
so D is the correct option.

Number of relations would be 4 as.. 1 + 7,7 + 1, 3 + 5, 5 + 3 all are equal to 8

(a, b) R (c, d) <=> a + d = b + c
Reflexive:
(a, b) R (a, b) <=> a + b = b + a
This is true for all (a, b) € N x N
Hence, it is reflexive.
Symmetric:
Let (a, b) R (c, d) <=> a + d = b + c
a + d = b + c
c + b = d + a (same)
By the above equation,
(c, d) R (a, b)
Hence,
(c, d) R (a, b) <=> c + b = d + a
Hence, it is symmetric
Transitive:
Let (a, b) R (c, d) <=> a + d = b + c, --- eqn 1
(c, d) R (e, f) <=> c + f = d + e --- eqn 2
for all a, b, c, d, e, f € N
eqn 1 : a + d = b + c
⇒  a - b = c - d
eqn 2 : c + f = d + e
⇒ c - d = e - f
So, a-b = e-f
⇒ a + f = b + e
⇒ (a, b) R (e, f)
Hence, it is transitive.
This is an equivalence relation

A = {1, 2, 3, 4} 
B = {1, 3, 5}
(a, b) ∈ element of R ⇔ a < b for all a ∈ A, b ∈ B
(a, b) pairs satisfying the condition of R are:
(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)
So, 
R = {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)}

 R be the relation in the set {1, 2,3, 4] given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}
it is seen that (a, a) ∈ R for every a ∈ {1, 2, 3, 4}
so,R is reflexive.
it is seen that (a, b) = (b, a) ∈ R 
because, (1, 2)∈ R but (2, 1) ∉ R
so, R is not symmetric.
it is seen that (a, b), (b, c) ∈ R ⇒ (a, c) ∈ R for all a, b, c ∈ {1, 2, 3, 4}.
so, R is transitive.
Hence, R is reflexive and transitive but not symmetric.

P{1, 2} union Q {3, 7} union R{} = A {1, 2, 3, 4, 5, 6, 7}
therefore R = {4, 5, 6}

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.

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.

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 )

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

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 .

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)}

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

The number of subsets of A × A = 

Therefore the number of relations = 

A. The relation is an equivalence relation.

From the defining equality we get

m n (q - p) = p q (n - m).

Dividing both sides by m n p q (nonzero for natural numbers) gives

1/p - 1/q = 1/m - 1/n.

Thus the value 1/m - 1/n is invariant for related pairs. Equivalently, (m,n)R(p,q) exactly when this invariant is equal for the two pairs.

Reflexive: for any pair the invariant equals itself, so the relation holds for every pair; hence it is reflexive.

Symmetric: equality of the invariant is symmetric, so if (m,n)R(p,q) then (p,q)R(m,n); hence it is symmetric.

Transitive: equality of the invariant is transitive, so if (m,n)R(p,q) and (p,q)R(r,s) then (m,n)R(r,s); hence it is transitive.

Since the relation is reflexive, symmetric and transitive, it is an equivalence relation. Therefore the correct option is A.

Correct answer: D

Multipolarity is a system in which power is distributed among three or more significant states or centres of power.

In a multipolar arrangement, each major centre possesses substantial military, economic or political capability and can influence or check the actions of others.

By contrast, Bipolarity involves power concentrated in two dominant states, while a balance of power describes a situation or policy aimed at preventing any one state from becoming dominant.

Therefore, option D is correct.

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}