MCQ Questions for Class 11 Commerce Applied Mathematics Relations Quiz 15

$R$ is relation is "greater then or equal" from

$A=\left\{1,2,3,4\right\}$ to

$B=\left\{4,5,6\right\}$ , then

$R^{-1}=$

0%
$\left\{(4,4)\right\}$
0%
$\phi$
0%
$A\ x\ B$
0%
$R$

Which of the following is always true

0%
$(p\Longrightarrow q)\equiv \sim q\Longrightarrow \sim p$
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$\sim (p\Longrightarrow q)\equiv p\ \wedge \sim q$
0%
$\sim (p\ \vee q)\equiv \vee\ p\ \vee \sim q$
0%
$\sim (p\ \vee q)\equiv \sim\ p\ \wedge \sim q$

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,

$(p\Rightarrow q)\cong \sim q\Rightarrow \sim p$

$A={a,b,c,d,e}$ , then the number of equivalence relations on

$A$ is

0%
$26$
0%
$52$
0%
$54$
0%
$63$

The relation

$R=\left\{(1, 1), (2, 2), (3, 3)\right\}$ on the set

$\left\{1, 2, 3\right\}$ is

0%
symmetric only
0%
reflexive only
0%
an equivalence relation
0%
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?

0%
$0$
0%
$1009$
0%
$2018$
0%
$2019$

Let xyz =105 where x,y,z,then number of ordered triplets (x,y,z) satisfying the given equation

0%
15
0%
27
0%
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 :

0%
$\dfrac { 1 }{ 2 }$
0%
9
0%
$\dfrac { 1 }{ 11 }$
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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

0%
symmetric only
0%
reflexive only
0%
transitive only
0%
an equivalence relation

The total number of equivalence relations defined in the set

$S={a,b,c}$ is

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$5$
0%
$3!$
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$2^{3}$
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$3^{3}$
0%
$3^{2}$

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)

0%
Reflexive only
0%
Symmetric only
0%
Transitive only
0%
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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$3$
0%
$4$
0%
$7$
0%
$6$

The relation

$\bot$ is

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Reflective
0%
Symmetric
0%
Transitive
0%
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

0%
$5$
0%
$3$
0%
$1$
0%
$0$

$\sim ( p \leftrightarrow \sim q )$ is a tautology.

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True
0%
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

0%
5
0%
7
0%
9
0%
11

Explanation

$\begin{array}{l} \text { solution: } R=(1,2),(2,3),(3,4) \\ \text { To make it an equivalence relation, we need } \\ \text { to make it Reflexive, symmetric and transitive } \\ \text { at the same time. } \\ \text { Elements needed to make it Reflexive are } \\ (1,1),(2,2),(3,3),(4,4) \end{array}$

$\begin{array}{l} \text { Elements related to make it symmetric are } \\ (2,1),(3,2),(4,3) \\ \text { Elements needed to make it transitive are } \\ (1,3) \text { , }(2,4) \\ \text { so total number of terms required are } 9 \\ \text { Answer: option (c) } \end{array}$

$k\ \epsilon\ R^+$ and the middle term of

$(\dfrac{k}{2} + 2)^8$ is

$1120$ , then value of k is

0%
$3$
0%
$2$
0%
$1$
0%
$4$

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}
0%
{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

0%
Reflexive
0%
Symmetric
0%
Transitive
0%
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

0%
Reflexive
0%
Symmetric
0%
Transitive
0%
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

$R=\left\{ \left( 1,3 \right) ,\left( 2,5 \right) ,\left( 3,3 \right) \right\} .\quad Then\quad { R }^{ 1 }\quad is$

0%
$\left\{ \left( 3,3 \right) ,\left( 3,1 \right) ,\left( 5,2 \right) \right\}$
0%
$\left\{ \left( 1,2 \right) ,\left( 2,5 \right) ,\left( 3,3 \right) \right\}$
0%
$\left\{ \left( 1,3 \right) ,\left( 5,2 \right) \right\}$
0%
None of these

If n(A)=5 then number of relation on 'A' is:

0%
${ 2 }^{ 5 }$
0%
${ 2 }^{ 25 }$
0%
${ 5 }^{ 2 }$
0%
none

Consider the set

$A = \{ ( 1,2,3 ) \}$ and the relation

$R = \{ ( 1,2 ) , ( 1,3 ) \}$ then

$R$ is

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Reflexive
0%
Transitive
0%
Symmetric
0%
Equivalence

$A=\{ a,b,c\}$ then relation

$R=\left\{ \left( b,c \right) \right\}$ on A is

0%
reflexive only
0%
symmetric only
0%
transitive only
0%
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

0%
Reflexive
0%
Symmetric
0%
Transitive
0%
Equivalence

The number of equivalence relations that can be defined on a set {a,b,c}, is

0%
3
0%
5
0%
7
0%
8

Consider the set

$A=(1, 2, 3)$ and the R be the smallest equivalence relation on R then R is

0%
${(1, 1), (2, 2), (3,3)}$
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${(1,1), (2 ,2), (3, 3), (1, 2), (2, 1)}$
0%
${(1, 1,(2 ,2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2)}$
0%
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

0%
symmetric only
0%
reflexive only
0%
transitive only
0%
an equivalence relation

$A$ is a finite set containing n distinct elements, then the number of relations on A is equal to

0%
$2^{n}$
0%
$2^{n^{2}}$
0%
$2^{2}$
0%
none of these

If R is an equivalence relation on a set A, then

$R^1$ is

0%
reflexive only
0%
symmetric but not transitive
0%
equivalence
0%
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

0%
symmetric only
0%
reflexive only
0%
transitive only
0%
an equivalence relation

CBSE Questions for Class 11 Commerce Applied Mathematics Relations Quiz 15 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Relations Quiz 15 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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