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CBSE Questions for Class 11 Commerce Applied Mathematics Relations Quiz 15 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Relations
Quiz 15
If $$R$$ is relation is "greater then or equal" from $$A=\left\{1,2,3,4\right\}$$ to $$B=\left\{4,5,6\right\}$$, then $$R^{-1}=$$
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$$\left\{(4,4)\right\}$$
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$$\phi$$
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$$A\ x\ B$$
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$$R$$
Which of the following is always true
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$$(p\Longrightarrow q)\equiv \sim q\Longrightarrow \sim p$$
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$$\sim (p\Longrightarrow q)\equiv p\ \wedge \sim q$$
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$$\sim (p\ \vee q)\equiv \vee\ p\ \vee \sim q$$
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$$\sim (p\ \vee q)\equiv \sim\ p\ \wedge \sim q$$
Explanation
Truth table:
$$p$$
$$q$$
$$p\Rightarrow q$$
$$\sim q$$
$$\sim p$$
$$\sim q\Rightarrow \sim p$$
$$T$$
$$T$$
$$T$$
$$F$$
$$F$$
$$T$$
$$T$$
$$F$$
$$F$$
$$T$$
$$F$$
$$F$$
$$F$$
$$T$$
$$T$$
$$F$$
$$T$$
$$T$$
$$F$$
$$F$$
$$T$$
$$T$$
$$T$$
$$T$$
So, $$(p\Rightarrow q)\cong \sim q\Rightarrow \sim p$$
If $$A={a,b,c,d,e}$$, then the number of equivalence relations on $$A$$ is
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$$26$$
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$$52$$
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$$54$$
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$$63$$
The relation $$R=\left\{(1, 1), (2, 2), (3, 3)\right\}$$ on the set $$\left\{1, 2, 3\right\}$$ is
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symmetric only
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reflexive only
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an equivalence relation
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transistive only
The number of ordered pairs (x, y) of natural numbers satisfy the equation $$x^2+y^2+2xy-2018x-2018y-2019^o=0$$ is?
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$$0$$
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$$1009$$
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$$2018$$
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$$2019$$
Let xyz =105 where x,y,z,then number of ordered triplets (x,y,z) satisfying the given equation
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0%
15
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27
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6
0%
33
Let $$f:R\rightarrow R$$ be defined as
$$f(x)={ x }^{ 3 }+{ 2x }^{ 2 }+4x+sin\left( \dfrac { \pi x }{ 2 } \right) $$ and $$g(x)$$be the inverse function of f(x), then $${ g }^{ ' }(8)$$is equal to :
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$$\dfrac { 1 }{ 2 } $$
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9
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$$\dfrac { 1 }{ 11 } $$
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11
Let $$N$$ be the set of all positive integers and $$\phi$$ be a relation on $$N\times \ N$$ defined by $$(a,b)\times (c,d)$$. If $$ad(b+c)=bc(a+d)$$ then $$\phi$$ is
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symmetric only
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reflexive only
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transitive only
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an equivalence relation
The total number of equivalence relations defined in the set $$S={a,b,c}$$ is
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$$5$$
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$$3!$$
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$$2^{3}$$
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$$3^{3}$$
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$$3^{2}$$
Let R be a relation over set N$$\times N$$defined by (a,b)R (c,d) if a + d = b + c then R is ...... ( Here N is the set of all natural numbers)
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Reflexive only
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Symmetric only
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Transitive only
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Equivalence relation
The minimum number of elements that must be added to the relation $$R=\{(1, 2), (2, 3)\}$$ on the set $$\{1, 2, 3\}$$ so that it is an equivalence relation.
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$$3$$
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$$4$$
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$$7$$
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$$6$$
The relation $$\bot $$ is
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Reflective
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Symmetric
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Transitive
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Equivalence
A relation $$R_1$$ is defined set $$A={1,2,3}$$ such that $$R_1\equiv {(1,1),(2,2),(2,3),(3,2)}$$, then minimum number of elements required in $$R_1$$ so that $$R_1$$ becomes R which in an equivalence relation is
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$$5$$
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$$3$$
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$$1$$
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$$0$$
$$\sim ( p \leftrightarrow \sim q )$$ is a tautology.
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True
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False
$$R={(1, 2), (2,3),(3 4)}$$ be a relation on the set of natural numbers. Then the last number of elements that must be included inn R to get a new relation S where S is an equivalence relation, is
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5
0%
7
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9
0%
11
Explanation
$$ \begin{array}{l} \text { solution: } R=(1,2),(2,3),(3,4) \\ \text { To make it an equivalence relation, we need } \\ \text { to make it Reflexive, symmetric and transitive } \\ \text { at the same time. } \\ \text { Elements needed to make it Reflexive are } \\ (1,1),(2,2),(3,3),(4,4) \end{array} $$
$$ \begin{array}{l} \text { Elements related to make it symmetric are } \\ (2,1),(3,2),(4,3) \\ \text { Elements needed to make it transitive are } \\ (1,3) \text { , }(2,4) \\ \text { so total number of terms required are } 9 \\ \text { Answer: option (c) } \end{array} $$
If $$k\ \epsilon\ R^+$$ and the middle term of $$(\dfrac{k}{2} + 2)^8$$ is $$1120$$, then value of k is
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$$3$$
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$$2$$
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$$1$$
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$$4$$
Explanation
$$K\epsilon { R }^{ + }$$
The general term, $${ \left( r+1 \right) }^{ th }$$ term of the expansion
$${ \left( \dfrac { K }{ 2 } +2 \right) }^{ 8 }$$ is $${ T }_{ r+1 }={ 8 }_{ Cr }{ \left( \dfrac { K }{ 2 } \right) }^{ 8-r }{ 2 }^{ r }$$
In the above expansion there are $$9$$ terms, hence the middle term will be the $$5th$$ term $$(r=4)$$
$$\therefore$$ $${ T }_{ 5 }={ 8 }_{ { C }_{ 4 } }{ \left( \dfrac { K }{ 2 } \right) }^{ 8-4 }{ 2 }^{ 4 }$$
$$=\dfrac { 8! }{ 4!4! } \dfrac { { K }^{ 4 } }{ { 2 }^{ 4 } } .{ 2 }^{ 4 }$$
$$=\dfrac { 8\times 7\times 6\times 5 }{ 4\times 3\times 2 } { K }^{ 4 }$$
$$=70{ K }^{ 4 }$$
Given,
$${ 70K }^{ 4 }=1120$$
$$\Rightarrow { K }^{ 4 }=112/7$$
$$\Rightarrow { K }^{ 4 }=16$$
$$\Rightarrow { K }^{ 4 }={ 2 }^{ 4 }$$
$$\therefore$$ $$K=2$$
If a relation R is defined on the set Z of integers as follows : (a,b) $$\epsilon \quad R\Leftrightarrow { a }^{ 2 }+{ b }^{ 2 }=25,$$ , Then domain (R)=
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{3,4,5}
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{0,3,4,5}
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{0,$$\pm $$3, $$\pm $$4, $$\pm $$5}
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{3,4}
Let A={1,2,3} and R={(1,1),(2,2),(1,2),(2,1),(1,3)}, then R is
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Reflexive
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Symmetric
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Transitive
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None of these
Let $$R_1$$ and $$R_2$$ be equivalence relations on a set A, the $$R_1 \cup R_2$$ may or may not be
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Reflexive
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Symmetric
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Transitive
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Cannot say anything
Let $$A=\left\{ 1,2,3 \right\} ,\quad B=\left\{ 1,3,5 \right\} .$$ If relation R from A to B is given by $$R=\left\{ \left( 1,3 \right) ,\left( 2,5 \right) ,\left( 3,3 \right) \right\} .\quad Then\quad { R }^{ 1 }\quad is$$
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$$\left\{ \left( 3,3 \right) ,\left( 3,1 \right) ,\left( 5,2 \right) \right\} $$
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$$\left\{ \left( 1,2 \right) ,\left( 2,5 \right) ,\left( 3,3 \right) \right\} $$
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$$\left\{ \left( 1,3 \right) ,\left( 5,2 \right) \right\} $$
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None of these
If n(A)=5 then number of relation on 'A' is:
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$${ 2 }^{ 5 }$$
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$${ 2 }^{ 25 }$$
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$${ 5 }^{ 2 }$$
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none
Consider the set $$A = \{ ( 1,2,3 ) \}$$ and the relation $$R = \{ ( 1,2 ) , ( 1,3 ) \}$$ then $$R$$ is
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Reflexive
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Transitive
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Symmetric
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Equivalence
If $$A=\{ a,b,c\}$$ then relation $$R=\left\{ \left( b,c \right) \right\}$$ on A is
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reflexive only
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symmetric only
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transitive only
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reflexive and transitive only
For real numbers x and y, we write $$xRy\Leftrightarrow { x }^{ 2 }-{ y }^{ 2 }+\sqrt { 3 } $$ is an irrational number, then the relation R is
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Reflexive
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Symmetric
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Transitive
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Equivalence
The number of equivalence relations that can be defined on a set {a,b,c}, is
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3
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5
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7
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8
Consider the set $$A=(1, 2, 3)$$ and the R be the smallest equivalence relation on R then R is
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$${(1, 1), (2, 2), (3,3)}$$
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$${(1,1), (2 ,2), (3, 3), (1, 2), (2, 1)}$$
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$${(1, 1,(2 ,2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2)}$$
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None of these
The relation $$R=\left\{ \left( 1,1 \right) ,\left( 2,2 \right) ,\left( 3,3 \right) \right\} $$ on the set $$\left\{ 1,2,3 \right\} $$ is
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symmetric only
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reflexive only
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transitive only
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an equivalence relation
If $ A $ is a finite set containing n distinct elements, then the number of relations on A is equal to
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$$ 2^{n} $$
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$$ 2^{n^{2}} $$
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$$ 2^{2} $$
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none of these
If R is an equivalence relation on a set A, then $$R^1$$ is
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reflexive only
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symmetric but not transitive
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equivalence
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transitive
Let $$N$$ denotes the set of all natural numbers and $$R$$ be the relation on $$N \times N$$ defined
by $$( a , b ) R ( c , d )$$ iff $$a d ( b + c ) = b c ( a + d ) ,$$ then $$R$$ is
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symmetric only
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reflexive only
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transitive only
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an equivalence relation
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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
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