If $$A = \{5, 7\}, B= \{7, 9\}$$ and $$C = \{7, 9, 11\},$$ find $$A \times (B \cup C)$$

$$ \{(5, 5), (5, 9), (5, 11), (7, 7), (7, 9), (7, 11) \}$$

$$ \{(5, 7), (9, 9), (5, 11), (7, 7), (7, 9), (7, 11) \}$$

MCQ Questions for Class 11 Commerce Applied Mathematics Relations Quiz 6

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is incorrect but Reason is correct
0%
Both Assertion and Reason are incorrect

Explanation

Here,

$\displaystyle A=\left \{ 2,4,6 \right \}$

$B=\left \{ 2,3,5 \right \}$

$\displaystyle \therefore n(A)=3,\ n(B)=3$

$A\times B=\{(2,2), (2,3), (2,5),(4,2),(4,3),(4,5), (6,2),(6,3),(6,5)\}$

$n(A\times B)=9$

So, the number of subsets of

$A \times B$ is

$2^9$

Since the number of relations from

$A$ to

$B$ is the number of subsets of

$A \times B$ .

So, the number of relation from

$A$ to

$B$ is

$2^9$

Hence, the assertion is false.

Now, number of subsets of

$\displaystyle A \times B=2^{n(A)\times n(B)}$

$\displaystyle =2^{n(A)\times n(B)}=2^{3 \times 3}=2^{9}$

$\displaystyle \Rightarrow$ Assertion

$(A)$ is false but Reason

$(R)$ is true.

$\displaystyle A=\left \{ 0,1,2,3,4,5 \right \}$ and relation

$R$ defined by

$a R b$ such that

$2a+b=10$ then

$R^{-1}$ equals

0%
$\displaystyle \left \{ (4,3),(2,4) ,(5,0) \right \}$
0%
$\displaystyle \left \{ (3,4),(4,2) ,(5,0) \right \}$
0%
$\displaystyle \left \{ (4,3),(2,4) ,(0,5) \right \}$
0%
$\displaystyle \left \{ (4,3),(4,2) ,(5,0) \right \}$

Explanation

$2a+b=10$ ,

$2a$ is always even and

$10$ is even.So,

$b$ should also be even.
Hence the solution set is

$\{(0,10),(1,8),(2,6),(3,4),(4,2),(5,0)\}$
But

$A=\{0,1,2,3,4,5\}$

$R$ becomes

$\{(3,4),(4,2),(5,0)\}$

Hence

$R^{-1}$ is

$\{(4,3),(2,4),(0,5)\}$

$\displaystyle R=\{ (x,y):x, y \in Z ,x^{2}+y^{2}\leq 4 \}$ is a relation in

$Z$ then domain

$D$ is

0%
$\displaystyle \left \{ -2,-1,0,1,2\right \}$
0%
$\displaystyle \left \{ -2,-1,0 \right \}$
0%
$\displaystyle \left \{ 0,1,2\right \}$
0%
None of these

Explanation

Relation

$Z$ consists of all the integral points inside and on the circle of radius

$2$ and center

$(0,0)$ .

For all the points

$(x,y)$ which lie inside and on the circle

$x\epsilon [-2,2]$ and

$y\epsilon [-2,2]$ . Hence the domain is the integer values of

$x$ i.e

$-2,-1,0,1,2.$

$\displaystyle A= \left \{ a,b,c,d \right \}, B= \left \{ 1,2,3 \right \}$ find whether or not the following sets of ordered pairs are relations from

$A$ to

$B$ or not.

$\displaystyle R_{1}= \left \{ \left ( a,1 \right ), \left ( a,3 \right ) \right \}$

$\displaystyle R_{2}= \left \{ \left ( a,1 \right ), \left ( c,2 \right ), \left ( d,1 \right ) \right \}$

$\displaystyle R_{3}= \left \{ \left ( a,1 \right ), \left ( b,2 \right ), \left ( 3,c \right ) \right \}.$

0%
$R_{1}$ $R_{2}$ are relations but $R_{3}$ is not a relation.
0%
$R_{1}$ $R_{3}$ are relations but $R_{2}$ is not a relation.
0%
All are relations
0%
none of these

Explanation

Given,

$A= \left \{ a,b,c,d \right \}, B= \left \{ 1,2,3 \right \}$ and

$R_{1}= \left \{ \left ( a,1 \right ), \left ( a,3 \right ) \right \}$

$R_{2}= \left \{ \left ( a,1 \right ), \left ( c,2 \right ), \left ( d,1 \right ) \right \}$

$R_{3}= \left \{ \left ( a,1 \right ), \left ( b,2 \right ), \left ( 3,c \right ) \right \}.$

We know that every relation from

$A$ to

$B$ will be a subset of

$A \times B$ .
Thus, both

$R_{1}$ and

$R_{2}$ are subsets of

$A\times B$ and hence are relations but

$R_{3}$ is not a relation because

$R_{3}$ is not a subset of

$A\times B$ because the element

$(3,c) \notin (A\times B).$ It belongs to

$B\times A.$

If ,

$(x-1, y+2)= (7, 5)$ then values of

$x$ and

$y$ are

0%
$5$ , $8$
0%
$8$ , $3$
0%
$-1$ , $5$
0%
$7$ , $1$

Explanation

as the given ordered pairs are equal,

$x-1=7$ and

$y+2=5$

$\therefore x=8$ and

$y=3$

State whether the following statement is True or False.

If (x, y) = (3, 5) ; then x= 3 and y = 5

0%
True
0%
False

Ordered pairs (a, 3) and (5, x) are equal ,the values of

$a$ and

$x$ are

0%
$2$ and $4$
0%
$3$ and $6$
0%
$5$ and $3$
0%
$1$ and $-1$

Explanation

If two ordered pairs are equal, then both x - coordinates are equal and both y-coordinates are equal.

Since

$(a, 3) = (5,x)$ then ,

$a = 5 , x = 6$

$(x, y) = (3, 5)$ ; then values of

$x$ and

$y$ are

0%
3 and 5
0%
4 and 7
0%
-1 and 17
0%
2 and 4

Given

$M = (0, 1, 2)$ and

$N = (1, 2, 3)$ . Find

$(N - M) \times (N \cap M)$

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

Explanation

$N - M$ means subtracting the common elements of

$M$ and

$N$ from

$N$ .
So,

$N - M =\{3\}$

Intersection of two sets has the elements which are common in both the sets.
So,

$N \cap M =\{1,2\}$
Cartesian product of two sets is found by forming ordered pairs, where

$x$ -coordinate is from first set and

$y$ -coordinate is from second set.
So,

$(N - M) \times (N \cap M) =\{ (3,1), (3,2) \}$ .

$A = \{5, 7\}, B= \{7, 9\}$ and

$C = \{7, 9, 11\},$ find

$A \times (B \cup C)$

0%
$\{(5, 7), (5, 9), (5, 11), (7, 7), (7, 9), (7, 11) \}$
0%
$\{(5, 5), (5, 9), (5, 11), (7, 7), (7, 9), (7, 11) \}$
0%
$\{(5, 7), (9, 9), (5, 11), (7, 7), (7, 9), (7, 11) \}$
0%
none of these

Explanation

Given,

$A = \{5, 7\}, B= \{7, 9\}$ and

$C = \{7, 9, 11\},$

Union of two sets has all the elements of both the sets.
So,

$B \cup C =$ {

$7,9, 11$ }
Product of two sets is found by forming ordered pairs by multiplying every element of the first set with the second set.
So,

$A \times (B \cup C )=$ {

$(5,7), (5,9), (5,11), (7,7),(7,9), (7,11)$ }

$R$ be a relation defined from

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

$\displaystyle B=\left \{ 1,3,5 \right \},i.e.\left ( a,b \right )\in R$ iff

$a<b$ then

$\displaystyle R o R^{-1}$ is

0%
$\displaystyle \left \{ \left ( 1,3 \right ),\left ( 1,5 \right ),\left ( 2,3 \right ),\left ( 2,5 \right ),\left ( 3,5 \right ),\left ( 4,5 \right ) \right \}$
0%
$\displaystyle \left \{ \left ( 3,1 \right ),\left ( 5,1 \right )\left ( 3,2 \right ),\left ( 5,2 \right ) ,\left ( 5,3 \right ),\left ( 5,4 \right )\right \}$
0%
$\displaystyle \left \{ \left ( 3,3 \right ),\left ( 3,5 \right ) ,\left ( 5,3 \right ),\left ( 5,5 \right )\right \}$
0%
$\displaystyle \left \{ \left ( 3,3 \right ),\left ( 3,4 \right ) ,\left ( 4,5 \right )\right \}$

Explanation

$\displaystyle R;A\in B$ under given condition

$a<b$ is given by,

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

$\displaystyle R^{-1}=\left \{ \left ( 3,1 \right ),\left ( 5,1 \right ),\left ( 3,2 \right ) ,\left ( 5,2 \right ),\left ( 5,3 \right )\left ( 5,4 \right )\right \}$

$\displaystyle R o R^{-1}:$ For composing

$\displaystyle R o R^{-1}$ we will pick up an element of

$\displaystyle R^{-1}$ first and then of

$R$

$\displaystyle \left ( 3,1 \right )\in R^{-1}\left ( 1,3 \right )\in R\rightarrow \left ( 3,3 \right )\in R o R^{-1}$

$\displaystyle \therefore R o R^{-1}=\left \{ \left ( 3,3 \right ),\left ( 3,5 \right ),\left ( 5,3 \right ),\left ( 5,5 \right ) \right \}$ only.

Given

$\displaystyle A=\left \{ 2, 3 \right \},B=\left \{ 4, 5 \right \},C=\left \{ 5, 6 \right \},$ find

$\displaystyle A\times \left ( B\cap C \right )=.........$

0%
$\displaystyle A\times \left ( B\cap C \right )=\left \{ \left ( 2, 4 \right ),\left ( 3, 5 \right ) \right \}$
0%
$\displaystyle A\times \left ( B\cap C \right )=\left \{ \left ( 2, 5 \right ),\left ( 3, 5 \right ) \right \}$
0%
$\displaystyle A\times \left ( B\cap C \right )=\left \{ \left ( 2, 5 \right ),\left ( 4, 5 \right ) \right \}$
0%
$\displaystyle A\times \left ( B\cap C \right )=\left \{ \left ( 2, 3 \right ),\left ( 3, 5 \right ) \right \}$

Explanation

$\displaystyle A=\left \{ 2, 3 \right \},B=\left \{ 4, 5 \right \},C=\left \{ 5, 6 \right \}$

$\displaystyle B\cap C=\left\{5\right\}$
Now,

$\displaystyle A\times \left ( B\cap C \right )=\left \{ \left ( 2, 5 \right ),\left ( 3, 5 \right ) \right \}$
Hence, option B.

State True or False

Let

$A = \{1, 2\}$ and

$B = \{2, 3, 4\}$ , then A

$\times$ B = B

$\times$ A ?

0%
True
0%
False

Explanation

Product of two sets is the set of ordered pairs formed by mapping every element from the first set to every element of the second set.

So,

$A \times B =$ {

$(1,2), (1,3), (1,4), (2,2), (2,3), (2,4)$ }

And

$B \times A =$ {

$(2,1), (2,2), (3,1), (3,2), (4,1), (4,2)$ }

Clearly,

$A \times B \neq B \times A$

$A$ and

$B$ are two sets having

$3$ and

$5$ elements respectively and having

$2$ elements in common. Then the number of elements in

$\displaystyle A\times B$ is

0%
$6$
0%
$36$
0%
$15$
0%
none of these

Explanation

$\displaystyle \left ( a\times b \right )$ is different from

$\displaystyle \left ( b\times a \right )$
Thue if

$\displaystyle A= \left \{ a,b,c \right \}, B=\left \{ a, b, d, e, f \right \}$
then

$A \times B$ will contain

$3\times 5 =15$ elements.

Let

$R$ be a relation from a set

$A$ to a set

$B$ ,then

0%
$\displaystyle R=A\cup B$
0%
$\displaystyle R=A\cap B$
0%
$\displaystyle R\subseteq A\times B$
0%
$\displaystyle R\subseteq B\times A$

Explanation

$R$ is a relation from

$A$ to

$B$ then

$R$ is the subset of

$A \times B$ because it contains all the possible mappings from

$A$ to

$B$ .

$\displaystyle A=\left \{ 2, 4 \right \}$ and

$B\left \{ 3, 4, 5 \right \},$ then

$\displaystyle \left ( A\cap B \right )\times \left ( A\cup B \right )$ is

0%
$\displaystyle \left \{ \left ( 2, 2 \right ),\left ( 3, 4 \right ),\left ( 4, 2 \right ),\left ( 5, 4 \right ) \right \}$
0%
$\displaystyle \left \{ \left ( 2, 3 \right ),\left ( 4, 3 \right ),\left ( 4, 5 \right ) \right \}$
0%
$\displaystyle \left \{ \left ( 2, 4 \right ),\left ( 3, 4 \right ),\left ( 4, 4 \right ),\left ( 4, 5 \right ) \right \}$
0%
$\displaystyle \left \{ \left ( 4, 2 \right ),\left ( 4, 3 \right ),\left ( 4, 4 \right ),\left ( 4, 5 \right ) \right \}$

Explanation

Given

$\displaystyle A=\left \{ 2, 4 \right \}$ and

$B=\left \{ 3, 4, 5 \right \}$

$\therefore \displaystyle A\cap B=\left \{ 4 \right \}$ and

$\left ( A\cup B \right )=\left \{ 2, 3, 4, 5 \right \}$

$\displaystyle \therefore \left ( A\cap B \right )\times \left ( A\cup B \right )$

$\displaystyle =\left \{ \left ( 4, 2 \right ), \left ( 4, 3 \right ),\left ( 4, 4 \right ),\left ( 4, 5 \right ) \right \}$

Let

$\displaystyle A=\left \{ 1, 2, 3, 4, 5 \right \}, B=\left \{ 2, 3, 6, 7 \right \}.$ Then the number of elements in

$\displaystyle \left ( A\times B \right )\cap \left ( B\times A \right )$ is

0%
$18$
0%
$6$
0%
$4$
0%
$0$

Explanation

Given

$\displaystyle A=\left \{ 1, 2, 3, 4, 5 \right \}, B=\left \{ 2, 3, 6, 7 \right \}.$

$\therefore \displaystyle n\left ( A \cap B \right )=2$

and we know

$\ n\left [ \left ( A \times B \right )\cap \left ( B \times A \right ) \right ]=n\left [ \left ( A \cap B \right )\times \left ( B \cap A \right ) \right ]$

$\displaystyle \Rightarrow \ n\left [ \left ( A \times B \right )\cap \left ( B \times A \right ) \right ]=2 \times 2=4$

$\displaystyle A=\left \{ a, b, c \right \},B=\left \{ c, d, e \right \},C=\left \{ a, d, f \right \}$ , then

$A\times \left ( B\cup C \right )$ is

0%
$\displaystyle \left \{ \left ( a, d \right ),\left ( a, e \right ),\left ( a, c \right ) \right \}$
0%
$\displaystyle \left \{ \left ( a, d \right ),\left ( b, d \right ),\left ( c, d \right ) \right \}$
0%
$\displaystyle \left \{ \left ( d, a \right ),\left ( d, b \right ),\left ( d, c \right ) \right \}$
0%
none of these

Explanation

$\displaystyle A\times \left ( B\cup C \right )=\left \{ a, b, c \right\}\times \left \{ a, c, d, e, f \right \}$
The above set will consist of 15 ordered pairs and not 3.
Hence option 'D' is correct choice.

$\displaystyle X= \left \{ 1,2,3,4,5 \right \}, Y= \left \{ 1,3,5,7,9 \right \}$ determine which of the following sets are mappings, relations or neither from A to B:

(i)

$\displaystyle F= \left \{ \left ( x,y \right ) \because y= x+2, x \in X, y \in Y \right \}$

0%
It is clearly a one-one onto mapping i.e. a bijection. It is also a relation.
0%
It is clearly a many-one onto mapping. It is also a relation.
0%
It is clearly a one-one but not onto mapping. It is also a relation.
0%
It is not a mapping but a relation

Explanation

Given def of F as

$\displaystyle F= \left \{ \left ( x,y \right ) \because y= x+2, x \epsilon X, y \epsilon Y \right \}$
where

$\displaystyle X= \left \{ 1,2,3,4,5 \right \}, Y= \left \{ 1,3,5,7,9 \right \}$

So, we get

$\left \{ \left ( 1,3 \right ), \left ( 2,4 \right ), \left ( 3,5 \right ), \left ( 4,6 \right ), \left ( 5,7 \right ) \right \}$
Since,

$\displaystyle y \in Y$ and in the above 4,6 do not belong to Y.

$\displaystyle \therefore$ we exclude the ordered pair (2,4) and (4,6)

$\displaystyle \therefore \displaystyle F=\left \{ \left ( 1,3 \right ), \left ( 3,5 \right ), \left ( 5,7 \right ) \right \}$
F is not a mapping because the elements 2,4

$\displaystyle \epsilon X$ do not have any image.

$\displaystyle F\subset X\times Y$ and hence F is a relation from X to Y.

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

$B=\left \{ 3, 8 \right \},$ then

$\displaystyle \left ( A\cup B \right )\times \left ( A\cap B \right )$ is

0%
$\displaystyle \left \{ \left ( 3, 1 \right ), \left ( 3,2 \right ), \left ( 3, 3 \right ), \left ( 3, 8 \right ) \right \}$
0%
$\displaystyle \left \{ \left ( 1, 3 \right ), \left ( 2,3 \right ), \left ( 3, 3 \right ), \left ( 8, 3\right ) \right \}$
0%
$\displaystyle \left \{ \left ( 1, 2 \right ), \left ( 2,2 \right ), \left ( 3, 3 \right ), \left ( 8, 8\right ) \right \}$
0%
$\displaystyle \left \{ \left ( 8, 3 \right ), \left ( 8,2 \right ), \left ( 8, 1 \right ), \left ( 8, 8\right ) \right \}$

Explanation

Given

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

$B=\left \{ 3, 8 \right \},$

$\therefore \displaystyle \left ( A\cup B \right )=\left \{ 1, 2, 3, 8 \right \}$ and

$\displaystyle \left ( A\cap B \right )=\left \{ 3 \right \}$

$\displaystyle \therefore \left ( A \cup B \right )\times \left \{ A\cap B \right \}$

$\displaystyle =\left \{ \left ( 1, 3 \right ),\left ( 2, 3 \right ),\left ( 3, 3 \right ),\left ( 8, 3 \right ) \right \}$

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:

0%
$mn$
0%
$2^{mn}$
0%
$m^{n}$
0%
$n^{m}$

Explanation

Consider an element

$a\in$ 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 mappings is

$\displaystyle n\times n\times n\times$ ...m times

$=$

$n^{m}$

Let

$R$ be a reflexive on a finite set

$A$ having

$n$ elements, and let there be

$m$ ordered pairs in

$R$ . Then

0%
$m\ge n$
0%
$m\le n$
0%
$m=n$
0%
None of these

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

$m\ge n$ .

$A=\{a,b,c,d\}, B=\{p,q,r,s\}$ , then which of the following are relations from

$A$ to

$B$ ?

0%
$\displaystyle R_{1}= \left \{ \left ( a,p \right ), \left ( b,r \right ), \left ( c,s \right ) \right \}$
0%
$\displaystyle R_{2}= \left \{ \left ( q,b \right ), \left ( c,s \right ), \left ( d,r \right ) \right \}$
0%
$\displaystyle R_{3}= \left \{ \left ( a,p \right ), \left ( a,q \right ), \left ( d,p \right ), \left ( c,r \right ), \left ( b,r \right ) \right \}$
0%
$\displaystyle R_{4}= \left \{ \left ( a,p \right ), \left ( q,a \right ), \left ( b,s \right ), \left ( s,b \right ) \right \}$

Explanation

From then definiation of Relation

Let

$R$ be

$a$ relation from

$A$ to

$B$

i.e.,

$R\subseteq A\times B,$ then

Domain of {

$a:a\in A,\left( a,b \right) \in R$ for some

$b\in B$ }

Range of

$R=$ {

$b:b\in B,\left( a,b \right) \in R$ for some

$a\in A$ }.

Therefore,

(A)

$\displaystyle R,=\left\{ \left( a,p \right) ,\left( b,s \right) ,\left( c,s \right) \right\}$ is a relation as

$\left\{ a,b,c \right\} \subseteq A$ and

$\left\{ p,r,s \right\} \subseteq B.$

(B)

${ R }_{ 2 }=\left\{ \left( q,b \right) ,\left( c,s \right) ,\left( d,r \right) \right\}$ is not a relation as

$q$ is not from set

$A.$

(C)

${ R }_{ 3 }=\left\{ \left( a,p \right) ,\left( a,q \right) ,\left( d,p \right) ,\left( c,r \right) ,\left( b,r \right) \right\}$ is a relation as

$\displaystyle \left\{ a,d,c,b \right\} \subseteq A$ and

$\displaystyle \left\{ p,q,r \right\} \subseteq B.$

(D)

${ R }_{ 4 }=\left\{ \left( a,p \right) ,\left( q,a \right) ,\left( b,s \right) ,\left( s,b \right) \right\}$ is not a relation as

$q$ and

$s$ does not belong to set

$A.$

N is the set of positive integers and

$\displaystyle \sim$ be a relation on

$\displaystyle N\times N\:defined\:\left ( a,b \right )\sim \left ( c,d \right )$ iff ad=bc.
Check the relation for being an equivalence relation.

0%
True
0%
False

Explanation

$\left ( a,b \right )\sim \left ( c,d \right )$ iff

$ad=bc$

Reflexivity:
We can write

$ab=ab$

$\Rightarrow (a,b) \sim (a,b)$ (by given def)
Hence,

$\sim$ is reflexive.

Symmetry:
Let

$(a,b) \sim (c,d)$

$\Rightarrow ad=bc$

$\Rightarrow da=cb$ (Product of real numbers is commutative)

$or cb=da$

$\Rightarrow (c,d)\sim (a,b)$
Hence,

$\sim$ is symmetric.

Transitivity:
Let

$(a,b)\sim (c,d) ; (c,d)\sim (e,f)$

$\Rightarrow ad=bc ; cf=de$

$\Rightarrow af=be$ (Since

$\dfrac{d}{c}=\dfrac{f}{e}$ )

$\Rightarrow (a,b) \sim (e,f)$
Hence,

$\sim$ is transitive.

Hence,

$\sim$ is an equivalence relation.

A relation

$R$ is defined on the set

$Z$ of integers as follows: R=

$(x,y)$

$\displaystyle \in {R}:x^{2}+y^{2}= 25$ . Express

$R$ and

$\displaystyle R^{-1}$ as the sets of ordered pairs and hence find their respective domains.

0%
$0$
0%
Domain of $\displaystyle R= \left \{ 0, \pm 3 \right \}=$ domain of $\displaystyle R^{-1}.$
0%
Domain of $\displaystyle R= \left \{ 0, \pm 3, \pm 4 \right \}=$ domain of $\displaystyle R^{-1}.$
0%
Domain of $\displaystyle R= \left \{ 0, \pm 3, \pm 4, \pm 5 \right \}=$ domain of $\displaystyle R^{-1}.$

Explanation

$R=\left\{ \left( 0,5 \right) ,\left( 0,-5 \right) ,\left( 3,4 \right) ,\left( -3,4 \right) ,\left( 3,-4 \right) ,\left( -3,-4 \right) ,\\ \left( 4,3 \right) ,\left( -4,3 \right) ,\left( 4,-3 \right) ,\left( -4,3 \right) ,\left( 5,0 \right) ,\left( -5,0 \right) \right\} \\ { R }^{- 1 }=\left\{ \left( 5,0 \right) ,\left( -5,0 \right) ,\left( 4,3 \right) ,\left( 4,-3 \right) ,\left( -4,3 \right) ,\left( -4,-3 \right) ,\\ \left( 3,4 \right) ,\left( 3,-4 \right) ,\left( -3,4 \right) ,\left( 3,-4 \right) ,\left( 0,5 \right) ,\left( 0,-5 \right) \right\}$
Therefore domain of

$R\equiv \left\{ 0,3,-3,-4,4,-5,-5 \right\} =$ domain of

${ R }^{ -1 }$

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

$B=\left\{ 1,2,3,4 \right\}$ , then which of the following is not a subset of

$A\times B$

0%
$\left\{ (2,3),(2,4),(3,3),(3,4) \right\}$
0%
$\left\{ (2,2),(3,1),(3,4),(2,3) \right\}$
0%
$\left\{ (2,1),(3,2) \right\}$
0%
$\left\{ (1,2),(2,3) \right\}$

Explanation

Given

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

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

$\therefore A\times B =\left\{(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(3,4)\right\}$

$\therefore$ options A,B,C are the subsets of the

$A \times B$ except option D.

In order that a relation

$R$ defined in a non-empty set

$A$ is an equivalence relation, it is sufficient that

$R$

0%
is reflexive
0%
is symmetric
0%
is transitive
0%
possess all the above three properties

Which one of the following relations on

$R$ is equivalence redlation-

0%
$x{R}_{1}y\Leftrightarrow \left| x \right| =\left| y \right|$
0%
$x{R}_{2}y\Leftrightarrow \left| x \right| > \left| y \right|$
0%
$x{R}_{3}y\Leftrightarrow x|y$
0%
$x{R}_{4}y\Leftrightarrow x< y$

Explanation

Because

$\left| x \right| =\left| x \right|$ (Reflexive)

$\left| x \right| =\left| y \right|$ and

Hence option (A) is equivalence.

None of the relations given in option (b),(c)and(d) are equivalence relations because none of them are symmetric.

$A$ and

$B$ are two sets having

$3$ and

$4$ elements respectively and having

$2$ elements in common. The number of relations which can be defined from

$A$ to

$B$ is

0%
${2}^{5}$
0%
${2}^{10}-1$
0%
${2}^{12}-1$
0%
None of these

Explanation

Given

$n(A)=3$ ,

$n(B)=4$

Now, number of elements (ordered pairs) in

$A\times B$ is

$3 \times 4=12$

Number of subsets of

$A\times B$ is

$2^{12}$ including empty set.

Since, every subset of

$A\times B$ is a relation from

$A$ to

$B.$

Hence, number of relations from

$A$ to

$B$ is

$2^{12}-1$

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

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

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

Explanation

Given

$n(A)=3,$ and

$n(B) =3$
Hence

$n(A\times B) = 3\times 3=9$

CBSE Questions for Class 11 Commerce Applied Mathematics Relations Quiz 6 - 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 6 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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