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CBSE Questions for Class 11 Commerce Applied Mathematics Relations Quiz 5 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Relations
Quiz 5
If $$ABC\sim PQR$$, then AB:PQ =
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$$AC:PB$$
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$$AC:PR$$
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$$AB:PR$$
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$$AC:RQ$$
Let $$A=\left \{ 1, 2, 3 \right \}$$. Then number of equivalence relations containing (1, 2) is:
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1
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2
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3
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4
Explanation
Observe that 1 is related to 2. So, we have two possible cases.
Case 1: When 1 is not related to 3, then the relation. $$R_{1}=\left \{ (1, 1),(1, 2),(2, 1),(2, 2),(3, 3) \right \}$$ is the only equivalence
relation containing $$(1, 2)$$.
Case 2: When 1 is related to 3, then $$A \times A$$ = $$\{(1,1),(2,2),(3,3),(1,2),(2,1),(1,3),(3,1),(2,3),(3,2)\}$$ is the only equivalence relation containing $$ (1,2) $$
$$\therefore $$ There are two required equivalence relations.
Let A and B be finite sets containing m and n elements respectively. The number of relations that can be defined from A and B is:
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mn
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$$2^{mn}$$
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$$2^{m+n}$$
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$$n^{m}$$
Explanation
Here, $$O(A)=m$$ and $$O(B)=n$$.
Hence $$O(A×B)=mn$$
Since every subset of $$A×B$$ is a relation from $$A$$ to $$B$$, therefore, number of relations from A to B is equal to the number of the subsets of $$A×B$$, i.e., $$2^{m{n}}$$
Which of the following are not equivalence relations on I?
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a R b if $$a+b$$ is an even integer
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a R b if $$a-b$$ is an even integer
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a R b if $$a< b$$
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a R b if $$a= b$$
Explanation
For Option $$[A]$$
aRb if $$a+b$$ is an even integer
$$Reflexive:$$ $$a+a=2a\Rightarrow$$ is $$even$$ $$number$$
$$Symmetric$$ $$a+b=b+a\Rightarrow$$ is $$even$$ $$number$$
$$Transitive$$ $$a+b=2k;b+c=2p$$ then $$a+c=2k+2p-2b=2(k+p-b)\Rightarrow$$ is an $$even$$ $$number$$
$$\therefore$$ this is a equivalence relation
For option $$[B]$$
aRb if $$a-b$$ is an even number
$$Reflexive:$$
$$a-a=0\Rightarrow$$
is $$even$$ $$number$$
$$Symmetric$$ $$a-b=2k;b-a=-2k\Rightarrow$$
is $$even$$ $$number$$
$$Transitive$$ $$a-b=2k;b-c=2p$$
$$a-b+b-c=a-c=2k+2p=2(k+p)$$
is $$even$$ $$number$$
$$\therefore$$ this is a equivalence relation
For option $$[C]$$
$$Reflexive:$$ $$a\not <a\Rightarrow$$
$$not$$ $$Reflexive:$$
$$Symmetric:$$ $$a<b$$ but $$b\not< a$$ $$\Rightarrow$$ $$not$$ $$Symmetric$$
$$Transitive$$ $$a<b;b<c\Rightarrow a<c$$ $$\Rightarrow$$
$$Transitive$$
$$\therefore$$ not a Equivalence relation
For option $$[D]$$
aRb if a=b $$\Rightarrow$$ $$Reflexive,Symmetric,Transitive$$
$$\therefore$$ this is a equivalence relation
If $$X=\left \{ 1, 2, 3, 4, 5 \right \}$$ and $$Y=\left \{ 1, 3, 5, 7, 9 \right \}$$, determine which of the following sets represent a relation and also a mapping.
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$$R_{1}=\left \{ (x, y):y=x+2, x\in X, y\in Y \right \}$$
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$$R_{2}=\left \{ (1, 1),(2, 1),(3, 3),(4, 3),(5, 5) \right \}$$
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$$R_{3}=\left \{ (1, 1),(1, 3),(3, 5),(3, 7),(5, 7) \right \}$$
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$$R_{4}=\left \{ (1, 3),(2, 5),(4, 7),(5, 9),(3, 1) \right \}$$
Explanation
$$R_1={(1,3),(2,4),(3,5),(4,6),(5,7)}$$
Since $$4$$ and $$6$$ do not belong to $$Y$$
$$(2,4),(4,6)∉R_1$$
$$R_1={(1,3),(3,5),(5,7)}⊂A×B$$
Hence $$R_1$$ is a relation but not a mapping as the elements $$2$$ and $$4$$ do not have any image.
$$R_2$$ : It is certainly a mapping and since every mapping is a relation, it is a relation as well.
$$R_3$$: It is a relation being a subset of $$A×B$$ but the elements $$1$$ and $$3$$ do not have a unique image and hence it is not mapping.
$$R_4$$ : It is both a mapping and a relation. Each element in A has a unique image. It is also one-one and onto mapping and hence a bijection
Let R be a reflexive relation on a finite set A having n elements, and let there be m ordered pairs in R. Then:
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$$m\geq n$$
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$$m\leq n$$
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$$m=n$$
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$$m=-n$$
Explanation
The set consists of n elements and for relation to be reflexive it must have at least n ordered pairs. It has m ordered pairs therefore $$ m≥n$$.
Let A be a finite set containing n distinct elements. The number of relations that can be defined on A is:
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$$2^{n}$$
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$$n^{2}$$
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$$2^{n^{2}}$$
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2n
Explanation
Since $$A$$ contains $$n$$ distinct elements, therefore, $$A×A$$ contains $$n×n=n^2$$ distinct elements. since every subset of $$A×A$$ is a relation on $$A$$., therefore, number of relations on $$A$$ is equal to the order of the power set of $$A×A$$, is equal to the order of the power set of $$A×A$$, i.e., $$2^{n^{2}}$$
Let $$A=\left \{ 1, 2, 3 \right \}$$. Which of the following is not an equivalence relation on A?
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$$\left \{ (1, 1),(2, 2),(3, 3) \right \}$$
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$$\left \{ (1, 1),(2, 2),(3, 3),(1, 2),(2, 1) \right \}$$
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$$\left \{ (1, 1),(2, 2),(3, 3),(2, 3),(3, 2) \right \}$$
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$$\left \{ (1, 1),(2, 1) \right \}$$
Explanation
every relation in options follow the equivalence property.
but in fourth option this option only contains two elements $$[[{1,1}],[{2,1}]]$$
this option doesn't satisfy reflexive i.e., doesn't contain all $$(a,a)$$ type $$(2,2),(3,3)$$
Not symmetric i.e., doesn't contain this type of elements $$(1,2)$$ missing
and as well transitive.
and all others options satisfy these relations.
"Every reflexive relation is an anti-symmetric relation", Is it true?
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True
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False
The relation "congruence modulo $$m$$" is:
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reflexive only
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symmetric only
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transitive only
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an equivalence relation
Explanation
If $$R$$ be the relation, then
$$x R y$$
$$\Leftrightarrow x-y$$ is divisible by $$m$$.
$$x R x$$ because $$x-x=0$$ is divisible by $$m$$.
So, $$R$$ is reflexive.
If $$xRy\Rightarrow x-y$$ is divisible by $$m$$
$$y-x=-(x-y)$$ is divisible by $$m$$
$$\Rightarrow yRx$$
So, $$R$$ is symmetric
Let $$x R y$$ and $$y R z$$
$$\Rightarrow x-y=k_{1}m, y-z=k_{2}m$$
$$\therefore x-z=x-y+y-z=k_{1}m+k_{2}m=(k_{1}+k_{2})m$$.
So, $$R$$ is transitive.
As R is reflexive, symmetric and transitive, it is an equivalence relation.
Let R be the relation in the set N given by $$=\left \{ (a, b):a=b-2, b >6 \right \}$$. Choose the correct answer.
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$$(2, 4)\in R$$
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$$(3, 8)\in R$$
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$$(6, 8)\in R$$
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$$(8, 7)\in R$$
Explanation
$$(a,b)∈R$$ only if $$a=b−2$$ and $$b>6$$
$$(6,8)∈R$$ as $$6=8−2$$ and $$8 >6$$
Hence $$(c)$$ is the correct alternative
Find the number of relations from $$\left \{ m, o, t, h, e, r \right \}$$ to $$\left \{ c, h, i, l, d \right \}$$
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$$30$$
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$$30^{2}$$
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$$2^{30}$$
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$$60$$
Explanation
$$n \{m.o,t,b,e,r\} = 6 = m$$ (let)
$$n \{c,h,i,l,d\} = 5 = n$$ (let)
$$\therefore $$ numbers of Relations
$$= { 2 }^{ mn }$$
$$= { 2 }^{ 6\times 5 }$$
$$= { 2 }^{ 30 }$$
Which of the following statements is true?
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The x-axis is a vertical line
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The y-axis is a horizontal line
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The scale on both axes must be the same in a Cartesian plane
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The point of intersection between the x-axis and the y-axis is called the origin
Explanation
Since, Origin is the point of intersection of $$x$$ and $$y,$$ we can say that option $$D$$ is correct.
If $$R$$ is a relation from a set $$A$$ to a set $$B$$ and $$S$$ is a relation from $$B$$ to a set $$C$$, then the relation $$SOR$$
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is from $$A$$ to $$C$$
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is from $$C$$ to $$A$$
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does not exist
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none of these
Explanation
Domain of $$SOR$$ is the domain of $$R$$ and the range is the range of $$S$$. Hence the mapping is from set $$A$$ to set $$C$$.
Let $$A=\left \{ 1,2,3 \right \}$$. The total number of distinct relations that can be defined over $$A$$ is:
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$$2^{9}$$
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$$6$$
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$$8$$
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None of the above
Explanation
Total number of distinct binary relations over the set $$ A$$ will be
$$2^{n^2}$$
$$=2^{3^{2}}$$
$$=2^{9}$$
$$=512$$
Let $$X=\{1, 2, 3, 4, 5 \} $$ and $$Y=\{1, 3, 5, 7, 9\}$$ . Which of the following is/are relations from $$X$$ to $$Y$$?
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$$R_1=\{(x, y):y=2+x, x\in X,y\in Y\}$$
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$$R_2=\{(1, 1), (2, 1), (3, 3), (4, 3), (5, 5)\}$$
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$$R_3=\{(1, 1), (1, 3), (3, 5), (3, 7), (5, 7)\}$$
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$$R_4=\{(1, 3), (2, 5), (2, 4), (7, 9)\}$$
Explanation
$$R_1=\{(x, y):y=2+x, x\in X,y\in Y\}$$
$$x=1 \rightarrow y =1+2=3$$
$$x=2 \rightarrow y =2+2=4$$
$$x=3 \rightarrow y =3+2=5$$
$$x=4 \rightarrow y =4+2=6$$
$$x=5 \rightarrow y =5+2=7$$
$$R=\{(1,3),(2,4),(3,5),(4,6),(5,7)\}$$
Here in$$(4,6)$$ ,$$6$$ does not belong to either $$X or Y$$
So $$R_1 $$ is not a relation between $$X$$ and $$Y$$
In $$R_{4}$$, since it contains an element $$(7,9)$$ which relates set Y to set Y, while rest elements relate set X to Y.
$$\therefore R_4$$ is
not a relation between
$$X$$ and $$Y$$
Let $$R=\{(1, 3), (2, 2), (3, 2)\}$$ and $$S=\{(2, 1), (3, 2), (2, 3)\}$$ be two relations on set $$A=\{1, 2, 3\}$$.Then $$R$$o$$S$$ is equal
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$$\{(2, 3), (3, 2), (2, 2)\}$$
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$$\{(1, 3), (2, 2), (3, 2), (2 ,1), (2, 3)\}$$
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$$\{(3, 2), (1, 3)\}$$
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$$\{(2, 3), (3, 2)\}$$
Explanation
If $$f\epsilon X\times Y$$ and $$g\epsilon Y\times Z$$
Then
$$gof\epsilon X\times Z$$ i.e
$$ (x,z)\epsilon gof$$
Applying the above concept, we gives us
$$RoS=\{(2,1)|(1,3),(3,2)|(2,2),(2,3)|(3,2)\}$$
$$=\{(2,3),(3,2),(2,2)\}$$
Give a relation R={(1,2), (2,3)} on the set of natural numbers, add a minimum number of ordered pairs.
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enlarged relation is symmetric
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enlarged relation is transitive
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enlarged relation is reflexive.
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enlarged relation is equivalence relation
Explanation
$$R=\{(1,2), (2,3)\}$$
A)To make the relation symmetric, we have to add (2,1) and (3,2) in R.
Enlarged symmetric relation $$R'=\{(1,2), (2,3),(2,1), (3,2)\}$$
B) To make enlarged relation transitive , we have to add (1,3), (1,1),(2,2) in R'.
Also, we need to add (3,1) to make it symmetric
So, the new enlarged relation is $$R''=\{(1,2), (2,3),(2,1), (3,2),(1,1),(2,2),(3,1)\}$$
C) For enlarged relation to be symmetric, we need to add (1,1) in R''
$$R''=\{(1,2), (2,3),(2,1), (3,2),(1,1),(2,2),(3,1),(1,1)\}$$
D) For enlarged relation to be equivalence,
If $$A=\{1, 2, 3\}$$ and $$B=\{3, 8\}$$, then $$(A\cup B)\times (A\cap B)$$ is
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$$\{(3, 1), (3, 2), (3, 3), (3, 8)\}$$
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$$\{(1, 3), (2, 3), (3, 3), (8, 3)\}$$
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$$\{(1, 2), (2, 2), (3, 3), (8, 8)\}$$
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$$\{(8, 3), (8, 2), (8, 1), (8, 8)\}$$
Explanation
$$A\cup B=\{1,2,3,8\}$$
$$A\cap B=\{3\}$$
$$\therefore (A\cup B)\times (A\cap B)$$
$$=\{(1,3),(2,3),(3,3),(8,3)\}$$
If $$\displaystyle :n(A)= m, $$ then number of relations in $$A$$ are
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$$\displaystyle 2^{m}$$
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$$\displaystyle 2^{m}-2 $$
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$$\displaystyle 2^{m^{2}} $$
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None of these
Explanation
Given $$\displaystyle :n(A)= m $$ Relation in a set $$A$$ is subset of
$$\displaystyle A\times A $$ For Number of subset of $$A$$ We have
$$\displaystyle n(A\times A) = n(A)\times n(A)= m^{2}$$
$$\displaystyle
\therefore$$ Number of subset of $$\displaystyle A\times A = 2^{m^{2}}$$
every subset of $$\displaystyle A\times A $$ is a relation
$$\displaystyle \therefore$$ Number of relations in A is $$\displaystyle
2^{m^{2}}= 2^{m\times m}$$
If $$\displaystyle X= \left \{ 1,2,3,4,5 \right \}$$ and $$\displaystyle Y= \left \{ 1,3,5,7,9 \right \}$$ then which of the following sets are relation from $$X$$ to $$Y$$
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$$\displaystyle R_{1}= \left \{ (x,a):a= x+2,x\in X,a\in Y \right \} $$
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$$\displaystyle R_{2}= \left \{ (1,1),(2,1),(3,3),(4,3),(5,5) \right \} $$
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$$\displaystyle R_{3}= \left \{ (1,1),(1,3),(3,5),(3,7),(5,7) \right \} $$
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$$\displaystyle R_{4}= \left \{ (1,3),(2,5),(4,7),(5,9),(3,1) \right \} $$
Explanation
$${ R }_{ 1 }=\left\{ \left( 1,3 \right) ,\left( 2,4 \right) ,\left( 3,5 \right) ,\left( 4,6 \right) ,\left( 5,7 \right) \right\} $$
since $$4$$ and $$6$$ do not belong to $$y$$.
$$\therefore \left( 2,4 \right) ,\left( 4,6 \right) \notin { R }_{ 1 }$$
$$\therefore { R }_{ 1 }=\left\{ \left( 1,3 \right) ,\left( 3,5 \right) ,\left( 5,7 \right) \right\} \subset X\times Y$$
Hence, $${ R }_{ 1 }$$ is a relation but not a mapping as the elements $$2$$ and $$4$$ do not have any image.
$${ R }_{ 2 }:$$ It is certainly a mapping and since every mapping is a relation, it is a relation as well.
$${ R }_{ 3 }:$$ It is a relation being a subset of $$X\times Y$$.
$${ R }_{ 4 }:$$ It is both mapping and a relation. Each element in $$X$$ has a unique image. It is also one-one and on-to mapping and hence a bijection.
In order that a relation $$R$$ defined on a non-empty set $$A$$ is an equivalence relation.
It is sufficient, if $$R$$
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is reflexive
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is symmetric
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is transitive
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possesses all the above three properties.
Explanation
An equivalent relation needs to be symmetric,transitive and reflexive.
Let $$ A= \{ 1,2,3,.......50\} $$ and $$B=\{2,4,6.......100\}$$ .The number of elements $$\left ( x, y \right )\in A\times B$$ such that $$x+y=50$$
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$$24$$
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$$25$$
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$$50$$
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$$75$$
Explanation
The elements will be
$$(2,48),(48,2)$$
$$(4,46), (46,4)$$
:
:
$$(2n,50-2n), (50-2n,2n)$$
Now we have
$$2,4,6,8...$$ upto $$48$$
This forms an A.P
The number terms is $$24$$.
Let $$\displaystyle A= \left \{ a,b,c \right \} $$ and $$\displaystyle B= \left \{ 4,5 \right \} $$ Consider a relation $$R$$ defined from set $$A$$ to set $$B$$ then $$R$$ is subset of
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$$A$$
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$$B$$
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$$\displaystyle A\times B $$
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$$\displaystyle B\times A $$
Explanation
If A and B are two non empty sets, the relation from set $$A$$ to set $$B$$ is denoted as $$\displaystyle A\times B $$
If $$A = \{5, 7\}, B= \{7, 9\}$$ and $$C = \{7, 9, 11\},$$ find
$$(A \times B) \cup (A \times C)$$
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$$\{(5, 7), (9, 9), (5, 11), (7, 7), (7, 9), (7,11)\}$$
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$$\{(5, 7), (5, 9), (5, 11), (7, 7), (7, 9), (7,11)\}$$
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$$\{(5, 5), (5, 9), (5, 11), (7, 7), (7, 9), (7,11)\}$$
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none of these
Explanation
Product of two sets is found by forming ordered pairs by multiplying every element of the first set with the second set.
So, $$ A \times B = $$ { $$ (5,7), (5,9), (7,7),(7,9) $$ }
And $$ A \times C = $$ { $$ (5,7), (5,9), (5,11), (7,7),(7,9), (7,11) $$ }
Union of two sets has all the elements of both the sets.
So, $$ (A \times B) \cup ( A \times C) = $$ { $$ (5,7), (5,9), (5,11), (7,7),(7,9), (7,11) $$ }
Let $$A=\left \{ 1,2,3 \right \}$$ and $$B=\left \{ a,b \right \}$$.Which of the following subsets of $$A\times B$$ is a mapping from $$A$$ to $$B$$
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$$\left \{ \left ( 1,a \right ),\left ( 3,b \right ),\left ( 2,a \right ),\left ( 2,b \right ) \right \}$$
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$$\left \{ \left ( 1,b \right ),\left ( 2,a \right ),\left ( 3,a \right ) \right \}$$
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$$\left \{ \left ( 1,a \right ),\left ( 2,b \right ) \right \}$$
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none of these
Explanation
$$A=\left\{ 1,2,3 \right\} \\ B=\left\{ a,b \right\} \\ A\times B=\left\{ \left( 1,a \right) ,\left( 2,a \right) ,\left( 3,a \right) ,\left( 1,b \right) ,\left( 2,b \right) ,\left( 3,b \right) \right\} .$$
$$ \left\{ \left( 1,a \right) ,\left( 3,b \right) ,\left( 2,a, \right) \left( 2,b \right) \right\} \subset A\times B$$
$$ \left\{ \left( 1,b \right) ,\left( 2,a \right) ,\left( 3,a \right) \right\} \subset A\times B$$
$$ \left\{ \left( 1,a \right) ,\left( 2,b \right) \right\} \subset A\times B$$
If $$A=\{2, 4\}$$ and $$B=\{3, 4, 5\}$$ then $$(A\cap B)\times (A\cup B)$$ is
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$$\{(2, 2), (3, 4), (4, 2), (5, 4)\}$$
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$$\{(2, 3), (4, 3), (4, 5)\}$$
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$$\{(2, 4), (3, 4), (4, 4), (4, 5)\}$$
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$$\{(4, 2), (4, 3), (4, 4), (4, 5)\}$$
Explanation
$$A\cap B=\{4\}$$
$$A\cup B=\{2,3,4,5\}$$
Hence
$$(A\cap B)\times(A\cup B)$$
$$=\{(4,2),(4,3),(4,4),(4,5)\}$$
If $$R$$ is an anti symmetric relation in $$\displaystyle A$$ such that $$(a,b),(b,a)\:\in\: R$$ then
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$$\displaystyle a\leq b$$
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$$\displaystyle a= b$$
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$$\displaystyle a\geq b$$
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None of these
Explanation
For $$(a,b), (b,a)$$ belongs to the same relation which is anti symmetric that is possible only if the relation is reflexive i.e $$a=b$$
$$R$$ is a relation from $$\displaystyle \left \{ 11,12,13 \right \}$$ to $$\displaystyle \left \{ 8,10,12 \right \}$$ defined by $$y=x-3$$ then the $$R^{-1}$$ .
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$$\displaystyle \left \{ (11,8),(13,10) \right \}$$
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$$\displaystyle \left \{ (8,11),(10,13) \right \}$$
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$$\displaystyle \left \{ (8,11),(9,12),(10,13) \right \}$$
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$$None\ of\ these$$
Explanation
$$x=\{11,12,13\}, y=\{8,10,12\}$$
Now, $$y=x-3$$
So, the relation $$R$$ becomes $$\{(11,8),(13,10)\}$$
So, $$R^{-1}$$ becomes $$\{(8,11),(10,13)\}$$
Let $$A$$ and $$B$$ be two finite sets having $$m$$ and $$n$$ elements respectively. Then the total number of mapping from $$A$$ to $$B$$ is
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$$mn$$
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$$\displaystyle 2^{mn}$$
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$$\displaystyle m^{n}$$
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$$\displaystyle n^{m}$$
Explanation
Consider an element $$\displaystyle a\in A,$$ it can be assigned to any of the $$n $$ elements of $$B$$ i.e. it has $$n$$ images. Similarly each of the $$m$$ elements of $$A$$ can have $$n$$ images in $$B.$$ Hence the number of mapping is
$$\displaystyle n\times n\times n\times ...m \mbox{times}=n^{m}.$$
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