Test: Introduction To Relations - JEE MCQ

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 = 

(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 + m²n = nm² + n²m
⇒ mn² + m²n = mn² + m²n
⇒ 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.

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.

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}