MCQ Questions for Class 11 Commerce Applied Mathematics Relations Quiz 13

Let

$g(x)-f(x)=1$ . If

$f(x)+f(1-x)=2\forall x\in R$ , then

$g(x)$ of symmetrical about?

0%
The origin
0%
The line $x=\dfrac{1}{2}$
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The point $(1, 0)$
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The point $\left(\dfrac{1}{2}, 0\right)$

Let

$S=\{1, 2, 3, 4, 5\}$ and let A

$=S\times S$ . Defined the relation on the R on A as follows (a, b)R(c, d) if an only if ad

$=$ bc. Then R is?

0%
Reflexive only
0%
Symmetric only
0%
Transitive only
0%
An equivalence

A is a set having

$6$ distinct elements. The number of distinct function from A to A which are not bijections is?

0%
$6!-6$
0%
$6^6-6$
0%
$6^6-6!$
0%
$6!$

Explanation

Total

$-$ bijections

$6^6-6!$ .

N is the set of natural numbers. The relation R is defined on the N

$\times$ N as follows

$_{a\cdot b}R_{c\cdot d}\Leftrightarrow a+d=b+c$ . Then, R is?

0%
Reflexive only
0%
Symmetric only
0%
Transition only
0%
An equivalence

The relation

$R$ defined on the set

$A=\left\{ 1,2,3,4,5 \right\}$ by

$R=\left\{ \left( a,b \right) :\left| { a }^{ 2 }-{ b }^{ 2 } \right| <16 \right\}$ , is not given by

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$\left\{ \left( 1,1 \right) ,\left( 2,1 \right) ,\left( 3,1 \right) ,\left( 4,1 \right) ,\left( 2,3 \right) \right\}$
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$\left\{ \left( 2,2 \right) ,\left( 3,2 \right) ,\left( 4,2 \right) ,\left( 2,4 \right) \right\}$
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$\left\{ \left( 3,3 \right) ,\left( 4,3 \right) ,\left( 5,4 \right) ,\left( 3,4 \right) \right\}$
0%
none of these

Explanation

Given, the relation

$R$ defined on the set

$A=\left\{ 1,2,3,4,5 \right\}$ by

$R=\left\{ \left( a,b \right) :\left| { a }^{ 2 }-{ b }^{ 2 } \right| <16 \right\}$ .

Then

$R$ can be written as,

$R=\{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(3,4),(4,1),(4,2),$

$(4,3),(4,4),(4,5),(5,4),(5,5)\}$ .

Let

$A=\left\{ 2,3,4,5,....,17,18 \right\}$ . Let

$\simeq$ be the equivalence relation on

$A\times A$ , cartesian product of

$A$ with itself, defined by

$(a,b)\simeq (c,d)$ , iff

$ad=bc$ . The the number of ordered pairs of the equivalence class of

$(3,2)$ is

0%
$4$
0%
$5$
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$6$
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$7$

Explanation

Let

$\left(3,2\right)\cong\left(x,y\right)$

$\Rightarrow\,3y=2x$

This is possible in the cases:

$x=3,\,y=2$

$x=6,\,y=4$

$x=9,\,y=6$

$x=12,\,y=3$

$x=15,\,y=10$

$x=18,\,y=12$

Hence total pairs are

$6$ .

The maximum number of equivalence relations on the set

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

0%
$1$
0%
$2$
0%
$3$
0%
$5$

Explanation

${R}_{1}=\left\{\left(1,1\right),\left(2,2\right),\left(3,3\right)\right\}$

${R}_{2}=\left\{\left(1,1\right),\left(2,2\right),\left(3,3\right),\left(1,2\right),\left(2,1\right)\right\}$

${R}_{3}=\left\{\left(1,1\right),\left(2,2\right),\left(3,3\right),\left(1,3\right),\left(3,1\right)\right\}$

${R}_{4}=\left\{\left(1,1\right),\left(2,2\right),\left(3,3\right),\left(2,3\right),\left(3,2\right)\right\}$

${R}_{5}=\left\{\left(1,1\right),\left(2,2\right),\left(3,3\right),\left(1,2\right),\left(2,1\right),\left(1,3\right),\left(3,1\right),\left(2,3\right),\left(3,2\right)\right\}$

These are the

$5$ relations on

$A$ which are equivalence.

For real number

$x$ and

$y$ , define

$xRy$ iff

$x-y+\sqrt{2}$ 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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none of these

Explanation

For each value of x ∈ R,

$x − x + \sqrt 2$ that is

$\sqrt 2$ is an irrational number.

It is reflexive.

Let

$x = \sqrt 2$ and

$y =2$ then

$x − y + \sqrt 2 = 2 \sqrt 2 – 2$ which is irrational but when

$y = \sqrt 2$ and

$x = 2$ ,

$x − y + \sqrt 2$ is not irrational.

It is not symmetric.

Let

$x − y + \sqrt 2$ is irrational &

$y − z + \sqrt 2$ is irrational then in above case let

$x = 1; y = \sqrt 2 × 2$ &

$z = \sqrt 2$

Hence

$x − z + \sqrt 2$ is not irrational, so, the relation is not transitive.

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

$R=\left\{ \left( a,b \right) ,\left( b,a \right) ,\left( a,a \right) \right\}$ on

$A$ is

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

Explanation

Given if

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

$R=\left\{ \left( a,b \right) ,\left( b,a \right) ,\left( a,a \right) \right\}$ .

Since

$(b,b)\not \in R, (c,c)\not \in R$ so this relation is not reflexive.

The relation is symmetric as if

$(a,b)\in R\Rightarrow (b,a)\in R$ for all

$a,b\in A$ .

The relation is not transitive also as

$(b,a)\in R,(a,b)\in R$ but

$(b,b)\not \in R$ .

So the relation is only symmetric.

The number of ordered pairs (a, b) of positive integers such that

$\dfrac{2a - 1}{b}$ and

$\dfrac{2b - 1}{a}$ are both integers is

0%
$1$
0%
$2$
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$3$
0%
more than $3$

Explanation

The number of ordered pairs

$\left(a, b\right)$ of positive integers where

$a,b\in{I}^{+}$

$\Rightarrow\left(\dfrac{2a-1}{b},\dfrac{2b-1}{a}\right)\in{I}^{+}$

Let

$\dfrac{2a-1}{b}=1$

$\Rightarrow\,2a-1=b$

and

$\Rightarrow\,\dfrac{2b-1}{a}=\dfrac{2\left(2a-1\right)-1}{a}$

$=\dfrac{4a-2-1}{a}$

$=\dfrac{4a-3}{a}$

$\therefore \dfrac{2b-1}{a}=4-\dfrac{3}{a}$

For integer,

$a=1,3$ we get

$b=2a-1=2-1=1$ for

$a=1$

$b=2a-1=2\times 3-1=6-1=5$

$\therefore\,b=1,5$

So total sets are

$\left(1,1\right),\left(3,5\right),\left(5,3\right)$

Hence there are totally

$3$ sets.

$A=\{2, 4, 5\}, B=\{7, 8, 9\}$ then

$n(A\times B)$ is equal to

0%
$6$
0%
$9$
0%
$3$
0%
$0$

Let Z be the set of all integers and let R be a relation on Z defined by

$a$ R

$b\Leftrightarrow (a-b)$ is divisible by

$3$ . Then, R is?

0%
Reflexive and symmetric but not transitive
0%
Reflexive and transitive but not symmetric
0%
Symmetric and transitive but not reflexive
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An equivalence relation

Explanation

Let R={(a,b):a,b∈Z and (a−b) is divisible by 3}.

Show that R is an equivalence relation on Z.

Given:

R={(a,b):a,b∈Z and (a−b) is divisible by 3}.

$R = (a,b)$

$(a-b)$ is divisible by

$3$

Reflexive

$(a,a) \Rightarrow (a-a)$ is divisible by

$3$

Symmetric

$(a, b)\Leftrightarrow (b, a)$

$(a-b)$ is divisible by

$3$

$(b-a)$ is divisible by

$3$

Transitive

$(a,b), (b,c)\Leftrightarrow (a,c)$

$(a-b)$ is divisible by

$3$

$(b-c)$ is divisible by

$3$

$(a-c)$ is divisible by

$3$

R is an equivalent relation on Z

Let S be the set of all triangles in a plane and let R be a relation on S defined by

$\Delta_1S\Delta_2\Leftrightarrow \Delta_1\equiv \Delta_2$ . Then, R is?

0%
Reflexive and symmetric but not transitive
0%
Reflexive and transitive but not symmetric
0%
Symmetric and transitive but not reflexive
0%
An equivalence relation

Let R be a relation on

$N\times N$ , defined by

$(a, b)$ R

$(c, d)\Leftrightarrow a+d=b+c$ . Then, R is?

0%
Reflexive and symmetric but not transitive
0%
Reflexive and transitive but not symmetric
0%
Symmetric and transitive but not reflexive
0%
An equivalence relation

Let A be the set of all points in a plane and let O be the origin. Let

$R=\{(P, Q):OP=OQ\}$ . Then, R is?

0%
Reflexive and symmetric but not transitive
0%
Reflexive and transitive but not symmetric
0%
Symmetric and transitive but not reflexive
0%
An equivalence relation

Let us define a relation

$R$ in

$R$ as

$aRb$ if

$a \ge b$ . Then

$R$ is

0%
an equivalence relation
0%
reflexive, transitive but not symmetric
0%
symmetric, transitive but
0%
neither transitive nor reflexive but symmetric

Explanation

Given that

$aRb$ is

$a \ge b$

$\Rightarrow aRa \Rightarrow a\ge a$ which is true
let

$aRb, a\ge b$ , theb

$b \ge a$ which is not true

$R$ is not symmetric.
But

$aRb$ and

$bRc$

$\Rightarrow a\ge b$ and

$b\ge c$

$\Rightarrow a\ge c$
Hence,

$R$ is transitive.

Every relation which is symmetric and transitive is also reflexive.

0%
True
0%
False

Explanation

False
Let

$R$ be a relation defined by

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

$A=\left\{1,2,3\right\}$
It is clear that

$(3,3)\in R$ . So, it is not reflexive.

An integer

$m$ is said to be related to another integer,

$n$ if

$m$ is a integral multiple of

$n$ . This relation in

$Z$ is reflexive, symmetric and transitive.

0%
True
0%
False

$a$ relation

$R$ on the set

$\left\{1,2,3\right\}$ be defined by

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

$R$ is

0%
reflexive
0%
transitive
0%
symmetric
0%
none of these

Explanation

$R$ on the set

$\left\{1,2,3\right\}$ be defined by

$R=\left\{(1,2)\right\}$
It is clear that

$R$ is transitive.

a homogeneous relation R over a set X is transitive if for all elements a,b,c in X , whenever R relates a to b and b to c, then R also relates a to c.

Let

$R=\left\{(3,1),(1,3),(3,3)\right\}$ be a relation defined on the set

$A=\left\{1,2,3\right\}$ . Then

$R$ is symmetric, transitive but not reflexive.

0%
True
0%
False

Explanation

False
Given that,

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

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

$(1,1)\in R$
So,

$R$ is not reflexive

$(3,1)\in R, (1,3)\in R$
Hence,

$R$ is symmetric.
Since

$(3,1)\in R , (1,3)\in R$
But

$(1,1)\in R$
Hence,

$R$ is not transitive.

Consider the set

$A=\left\{1,2,3\right\}$ and the relation

$R=\left\{(1,2),(1,3)\right\} . R$ is a transitive relation.

0%
True
0%
False

Explanation

Here,

$A={1,2,3}$

$R={(1,1),(2,2),(3,3),(1,2),(2,3)}$
As

$R$ contains

$(1,1),(2,2),(3,3),$ it is reflexive.
R contains

$(1,2),(2,3)$ but does not contain

$(2,1),(3,2)$ , so it is not symmetric.
R contains

$(1,2),(2,3)$ but does not contain

$(1,3),$ so it is not transitive.

The relation

$R$ on the set

$A=\left\{1,2,3\right\}$ defined as

$R\left\{(1,1),(1,2),(2,1),(3,3)\right\}$ is reflexive, symmetric and transitive.

0%
True
0%
False

Explanation

False
Given that,

$R\left\{(1,1),(1,2),(2,1),(3,3)\right\}$

$(2,2)\in R$
So,

$R$ is not reflexive.

Let

$R$ be a relation from

$A$ to

$A$ defined by

$R=\left\{ (a, b): a, b \in N\ and\ a=b^2 \right\}$
Is the following true?

$(a, b) \in R$ , implies

$(b,a ) \in R$
Justify your answer.

0%
True
0%
False

Explanation

Given

$R=\left\{ (a, b):a, b \in N\ and\ a=b^2 \right\}=\left\{ (1, 1), (2, 4), (3, 9), (4, 16 ), ...\right\}$

$(a, a) \in R$ implies

$(b,a )\in R$ is not true, because

$(2, 4)\in R$ but

$(4, 2) \notin R$

Let

$R$ be a relation from

$A$ to

$A$ defined by

$R=\left\{ (a, b): a, b \in N\ and\ a=b^2 \right\}$
Is the following true?

$(a, a) \in R$ for all

$a \in A$
Justify your answer.

0%
True
0%
False

Explanation

Given

$R=\left\{ (a, b):a, b \in N\ and\ a=b^2 \right\}=\left\{ (1, 1), (2, 4), (3, 9), (4, 16 ), ...\right\}$

$(a, a) \in R$ for all

$a \in N$ is not true because

$(2, 2)\notin R$ .

Let

$R$ be a relation from

$A$ to

$A$ defined by

$R=\left\{ (a, b): a, b \in N\ and\ a=b^2 \right\}$
Is the following true?

$(a, b)\in R, (b, c)\in R$ , implies

$(a, c) \in R$ .
Justify your answer.

0%
True
0%
False

Explanation

Given

$R=\left\{ (a, b):a, b \in N\ and\ a=b^2 \right\}=\left\{ (1, 1), (2, 4), (3, 9), (4, 16 ), ...\right\}$

$(a, a) \in R, (b, c) \in R$ implies

$(a,c )\in R$ is not true, because

$(2, 4)$ and

$(4, 16) \in R$ but

$(2, 16)\notin R$ .

If A = { a , b , c , d} and B = { p , q ,r ,s} then relation from A and B is

0%
{(a ,p) , ( b, r), (c , r)}
0%
{(a , p), (p, q) , (c, r) , ( s , d)}
0%
{{ b , a) , (q , b ) , (c , r) }
0%
{ ( c,s) , ( d , s ,) ( r, a) , ( q , b )}

Explanation

Because

${ ( a , p), ( b ,r) ( c ,r)} \subseteq A \times B$

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

$B=\left \{ 4,5,6 \right \}$ then which of the following sets are relation from

$A$ to

$B$
(i)

$\displaystyle R_{1}=\left \{ (4,2) (2,6)(5,1)(2,4)\right \}$
(ii)

$\displaystyle R_{2}=\left \{ (1,4) (1,5)(3,6)(2,6) (3,4)\right \}$
(iii)

$\displaystyle R_{3}=\left \{ (1,5) (2,4)(3,6)\right \}$
(iv)

$\displaystyle R_{4}=\left \{ (1,4) (1,5)(1,6)\right \}$

0%
$\displaystyle R_{1},R_{2},R_{3}$
0%
$\displaystyle R_{1},R_{3},R_{4}$
0%
$\displaystyle R_{2},R_{3},R_{4}$
0%
$\displaystyle R_{1},R_{2},R_{3},R_{4}$

Explanation

A relation from

$A$ to

$B$ will include elements from

$A\times B$
Elements of

$A\times B$ are

$\{(1,4), (1,5) ,(1,6), (2,4), (2,5), (2,6), (3,4), (3,5), (3,6)\}$
Now the relation

$R_{1}$ consists of elements form both

$A\times B$ and

$B\times A$ .
Thus

$R_{1}$ is not a relation from

$A$ to

$B.$

Let 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)}. Choose the correct answer.

0%
R is reflexive and symmetric but not transitive
0%
R is reflexive and transitive but not symmetric
0%
R is symmetric and transitive but not reflexive
0%
R is an equivalence relation

Let A = {1, 2, 3}. Then number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is

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

If two sets

$A$ and

$B$ have

$99$ elements in common, then the number of elements common to each of the sets

$A \times B$ and

$B \times A$ are

0%
$2^{99}$
0%
$99^{2}$
0%
$100$
0%
$18$

Explanation

To find

$A \times B$ we take one element from set

$A$ and one from set

$B$ .
Given that

$99$ elements are common to both set

$A$ and set

$B$ .
Suppose these common elements are

$N_1, N_2, N_3,...N_{99}$ .
Select an ordered pair for

$A \times B$ such that both are selected out of these common elements. Examples:

$\left(N_1,N_2 \right),\; \left(N_3,N_5 \right)$
All these will also be elements of

$B \times A$ . Hence number of elements common to

$A \times B$ and

$B \times A$ is

$99 \times 99 = 99^2$ ( first element in ordered pair can be selected in 99 ways; second element can also be selected in 99 ways)

$n\left [ \left ( A\times B \right )\cap \left ( B\times A \right ) \right ]=n\left [ \left ( A\cap B \right )\cap \left ( B\cap A \right ) \right ]=(99)(99)=99^{2}$

CBSE Questions for Class 11 Commerce Applied Mathematics Relations Quiz 13 - 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 13 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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