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Test: Introduction To Relations


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20 Questions MCQ Test Mathematics (Maths) Class 12 | Test: Introduction To Relations

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Test: Introduction To Relations - Question 1

Let R = {(P, Q) : OP = OQ , O being the origin} be an equivalence relation on A. The equivalence class [(1, 2)] is

Detailed Solution for Test: Introduction To Relations - Question 1

Correct Answer : a

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

=> (1,2) {(1)2 + (4)2 = 5}

=> {1 + 4 = 5}

=> {5 = 5}

Test: Introduction To Relations - Question 2

Let a relation T on the set R of real numbers be T = {(a, b) : 1 + ab < 0, a, ∈ R}. Then from among the ordered pairs (1, 1), (1, 2), (1, -2), (2, 2), the only pair that belongs to T is________.​

Detailed Solution for Test: Introduction To Relations - Question 2

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

Test: Introduction To Relations - Question 3

For real number x and y, we write  xRy ⇔ x-y + √2 is an irrational number. Then the relation R is:​

Detailed Solution for Test: Introduction To Relations - Question 3

xRy => 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 xRx 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, xRy => x - y + √2 is irrational    ............1
and yRz => 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

Test: Introduction To Relations - Question 4

Let R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} be a relation on the set A = {1, 2, 3, 4}. The relation R is

Detailed Solution for Test: Introduction To Relations - Question 4

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.

Test: Introduction To Relations - Question 5

If A = {1, 3, 5, 7} and we define a relation R = {(a, b), a, b ∈ A: |a - b| = 8}. Then the number of elements in the relation R is

Detailed Solution for Test: Introduction To Relations - Question 5

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

Test: Introduction To Relations - Question 6

If A = {1, 3, 5, 7} and define a relation, such that R = {(a, b) a, b ∈ A : |a + b| = 8}. Then how many elements are there in the relation R

Detailed Solution for Test: Introduction To Relations - Question 6

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

Test: Introduction To Relations - Question 7

In the set N x N, the relation R is defined by (a, b) R (c, d) ⇔  a + d = b + c. Then R is

Detailed Solution for Test: Introduction To Relations - Question 7

(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

Test: Introduction To Relations - Question 8

If A = {1, 2, 3, 4} and B = {1, 3, 5} and R is a relation from A to B defined by(a, b) ∈ element of R ⇔ a < b. Then, R = ?

Detailed Solution for Test: Introduction To Relations - Question 8

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)}

Test: Introduction To Relations - Question 9

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​

Detailed Solution for Test: Introduction To Relations - Question 9

 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.

Test: Introduction To Relations - Question 10

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.​

Detailed Solution for Test: Introduction To Relations - Question 10

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

Test: Introduction To Relations - Question 11

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______​

Detailed Solution for Test: Introduction To Relations - Question 11

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.

Test: Introduction To Relations - Question 12

Which one of the following relations on set of real numbers is an equivalence relation?

Detailed Solution for Test: Introduction To Relations - Question 12

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.

Test: Introduction To Relations - Question 13

Let C = {(a, b): a2 + b2 = 1; a, b ∈ R} a relation on R, set of real numbers. Then C is​

Detailed Solution for Test: Introduction To Relations - Question 13

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 )

Test: Introduction To Relations - Question 14

Let A = {1, 2, 3, 4} and B = {x, y, z}. Then R = {(1, x), (2, z), (1, y), (3, x)} is​

Detailed Solution for Test: Introduction To Relations - Question 14

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

Test: Introduction To Relations - Question 15

Let R be a relation on N, set of natural numbers such that m R n ⇔ m divides n. Then R is​

Detailed Solution for Test: Introduction To Relations - Question 15

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 .

Test: Introduction To Relations - Question 16

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​

Detailed Solution for Test: Introduction To Relations - Question 16

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)}

Test: Introduction To Relations - Question 17

Let R be a relation on a finite set A having n elements. Then, the number of relations on A is​

Detailed Solution for Test: Introduction To Relations - Question 17

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 = 

Test: Introduction To Relations - Question 18

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​

Detailed Solution for Test: Introduction To Relations - Question 18

(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.

Test: Introduction To Relations - Question 19

A situation in which significant power is distributed among three or more states is known as what?

Detailed Solution for Test: Introduction To Relations - Question 19

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.

Test: Introduction To Relations - Question 20

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​

Detailed Solution for Test: Introduction To Relations - Question 20

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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