If $$A = \{1, 2, 3\}, B = \{3, 4\}$$ find (A $$\times$$ B) $$\cup$$ (B $$\times$$ A)

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

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

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

If $$A = \{a, b\}, B=\{1, 2, 3\}$$, find B $$\times$$ A

$$B$$ $$\times$$ $$A$$$$ = \{(1, a), (2, a), (3, a), (1, b) (2, b), (3, b)\}$$

$$B$$ $$\times$$ $$A$$$$ = \{ (2, a), (3, a), (1, b) (2, b), (3, b)\}$$

$$B$$ $$\times$$ $$A$$$$ = \{(1, a), (2, a), (3, a), (1, b) (2, b)\}$$

MCQ Questions for Class 11 Commerce Applied Mathematics Relations Quiz 2

Let the number of elements of the sets

$A$ and

$B$ be

$p$ and

$q$ respectively. Then, the number of relations from the set

$A$ to the set

$B$ is

0%
${ 2 }^{ p+q }$
0%
${ 2 }^{ pq }$
0%
$p+q$
0%
$pq$

Explanation

Given, number of elements of

$A$ and

$B$ are

$p$ and

$q$ respectively.

$\therefore$ The number of relations from the set

$A$ to the set

$B$ is

${2}^{pq}$ .

Find the second component of an ordered pair

$(2, -3)$

0%
$2$
0%
$3$
0%
$0$
0%
$-3$

Explanation

In an ordered pair

$(x,y)$ , the first component is

$x$ and the second component is

$y$ .

Therefore, in an ordered pair

$(2,-3)$ , the second component is

$-3$ .

A ______ maps elements of one set to another set.

0%
order
0%
set
0%
relation
0%
function

Explanation

A relation map elements of one set to another set

$R:A\to B$

Elements in A is mapped to elements in set B.

The ______ product of two sets is the set of all possible ordered pairs whose first component is a member of the first set and whose second component is a member of the second set.

0%
cartesian
0%
coordinate
0%
simple
0%
discrete

Explanation

The cartesian product of two sets is the set of all possible ordered pairs whose first component is a member of the first set and whose second component is a member of the second set.
Example:

$A = \{1, 2\} \quad B = \{2\}$
cartesian product,

$A \times B = {(1,2), (2, 2)}$

Identify the first component of an ordered pair

$(2, 1)$ .

0%
$1$
0%
$2$
0%
$-1$
0%
$0$

Explanation

In an ordered pair

$(x,y)$ , the first component is

$x$ and the second component is

$y$ .

Therefore, in an ordered pair

$(2,1)$ , the first component is

$2$ .

Cartesian product of sets

$A$ and

$B$ is denoted by _______.

0%
$A \times B$
0%
$B \times A$
0%
$A \times A$
0%
$B \times B$

Explanation

Cartesian product of Set

$A$ and

$B$ is denoted by

$A\times B$ .

What is the relation for the statement "A is taller than B"?

0%
is taller than
0%
A is taller
0%
B is taller
0%
is less than

Identify the first component of an ordered pair

$(0, -1)$ .

0%
$0$
0%
$-1$
0%
$2$
0%
$1$

Explanation

In an ordered pair

$(x,y)$ , the first component is

$x$ and the second component is

$y$ .

Therefore, in an ordered pair

$(0,-1)$ , the first component is

$0$ .

$A = \{1, 2, 3\}, B = \{3, 4\}$
find (A

$\times$ B)

$\cup$ (B

$\times$ A)

0%
$\{(1, 3), (2, 3), (3, 3), (1, 4), (2, 4), (3, 4), (3, 1), (4, 1), (3, 2), (4, 2), (4, 3)\}$
0%
$\{(2, 3), (3, 3), (1, 4), (2, 4), (3, 4), (3, 1), (4, 1), (3, 2), (4, 2), (4, 3)\}$
0%
$\{(1, 3), (2, 3), (3, 3), (1, 4), (2, 4), (3, 4), (3, 1), (4, 1), (4, 2), (4, 3)\}$
0%
None of these

Explanation

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

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

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

If A= {0, 1} and B ={1, 0}, then what is A x B equal to ?

0%
{(0, 1), (1, 0)}
0%
{(0, 0), (1, 1)}
0%
{(0, 1), (1, 0), (1, I)}
0%
A X A

Explanation

$\left\{ { 0,1 } \right\} \times \left\{ 1,0 \right\} ={ \left\{ (0,1),(0,0),(1,1),(1,0) \right\} }$

$\left\{ { 0,1 } \right\} \times \left\{ 0,1 \right\} ={ \left\{ (0,0),(0,1),(1,0),(1,1) \right\} }$

So,

$A\times B=A\times A$

Hence, D is correct.

$A = \{a, b\}, B=\{1, 2, 3\}$ , find B

$\times$ A

0%
$B$ $\times$ $A$ $= \{(1, a), (2, a), (3, a), (1, b) (2, b), (3, b)\}$
0%
$B$ $\times$ $A$ $= \{ (2, a), (3, a), (1, b) (2, b), (3, b)\}$
0%
$B$ $\times$ $A$ $= \{(1, a), (2, a), (3, a), (1, b) (2, b)\}$
0%
None of these

Let R be the set of real numbers and the mapping

$f:R\rightarrow R$ and

$g:R\rightarrow R$ be defined by

$f(x)=5-x^2$ and

$g(x)=3\lambda-4$ , then the value of

$(fog)(-1)$ is

0%
$-44$
0%
$-54$
0%
$-32$
0%
$-64$

Explanation

Given,

$f(x)=5-x^2,\,g(x)=3x-4$
Now,

$(fog)(-1)=f\begin{Bmatrix}g(-1)\end{Bmatrix}$

$=f(-7)\;[\because\;g(-1)=3(-1)-4=-7]$

$=5-(-7)^2$

$=-44$

Let

$A = \left \{x, y, z\right \}$ and

$B = \left \{p, q, r, s\right \}$ . What is the number of distinct relations from

$B$ to

$A$ ?

0%
$4096$
0%
$4094$
0%
$128$
0%
$126$

Explanation

Since, No. of elements in set A

$=3$
No. of elements in set B

$=4$

$\therefore$ No. of elements in B

$\times$ A

$=3\times 4=12$

$\therefore$ No. of distinct relations from B to A

$= 2^{12}=4096$
Option A is correct.

Let R be the relation in the set N given by = {(a, b): a = b - 2, b > 6}. Choose the correct answer

0%
$(2, 4) \epsilon R$
0%
$(3, 8) \epsilon R$
0%
$(6, 8) \epsilon R$
0%
$(8, 7) \epsilon R$

Let

$A=\left\{ u,v,w,z \right\} ;B=\left\{ 3,5 \right\}$ , then the number of relations from

$A$ to

$B$ is

0%
$256$
0%
$1024$
0%
$512$
0%
$64$

Explanation

Given

$A=\left\{ u,v,w,z \right\} ;B=\left\{ 3,5 \right\}$
Here, number of element in set

$A$ is

$n(A)=4$
And number of element in set

$B$ is

$n(B)=2$

$\therefore$ Number of relations from

$A$ to

$B$

$={ 2 }^{ n(A).n(B) }={ 2 }^{ 4\times 2 }={ 2 }^{ 8 }=256$

Let A be a finite set containing n distinct elements. The number of relations that can be defined on A is.

0%
$2^n$
0%
$n^2$
0%
$2n^2$
0%
$2n$

On the set

$N$ of all natural numbers define the relation

$R$ by

$a R b$ if and only if the G.C.D. of

$a$ and

$b$ is

$2$ . Then

$R$ is:

0%
Reflexive, but not symmetric
0%
Symmetric only
0%
Reflexive and transitive
0%
Reflexive, symmetric and transitive

Explanation

$aRa$ only if GCD of a and a is 2 which is clearly not the case.

Hence not reflexive.

$aRb \Rightarrow bRa$ if GCD of a and b is 2 then GCD of b and a is also 2.

The relation is symmetric.

Transitiveness is not necessary,i.e, if and b have GCD 2 and b and c also have GCD 2 a and c need not have GCD 2.

The relation is symmetric only.

Consider two sets

$A=\{a, b, c\}, B=\{e, f\}$ . If maximum numbers of total relations from A to B; symmetric relation from A to A and from B to B are

$l, m, n$ respectively, then the value of

$2l+m-n$ is

0%
$212$
0%
$184$
0%
$240$
0%
$64$

Explanation

We have

$n(A)=3$ and

$n(B)=2$ .

Now cartesian product of

$A$ and

$B$ can contain maximum

$3\times 2=6$ elements.

Since relation is the possible subsets of cartesian product.

So maximum number of relations from

$A$ to

$B$ be

$(l)=2^{6}=64$ .

Now maximum possible symmetric relation from

$A$ to

$A$ be

$(m)=2^{\cfrac{3(3+1)}{2}}=2^6=64$ and that of from

$B$ to

$B$ be

$(m)=2^{\cfrac{2(2+1)}{2}}=2^3=8$ . [Using formula for number of symmetric relation]

Now

$2l+m-n=128+64-8=184$ .

The number of reflexive relation in set A = {a, b, c} is equal to

0%
$2^9$
0%
$2^4$
0%
$2^7$
0%
$2^6$

Explanation

$\therefore$ Total number of relation in set A

$2 ^{3 \times 3} = 2^9$

and maximum number of cartesian product = 9 out of which 3 ordered pair is necessary for reflexive.

So, for remaining 6 ordered pair

Number of ordered pair required.

$=^6 C _0\, +\, ^6C_1 \,+ \, ^6C_2 +......^6C_6$ =

$2^6$

If the relation is defined on

$R-\left\{ 0 \right\}$ by

$\left( x,y \right) \in S\Leftrightarrow xy>0$ , then

$S$ is ________

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

Explanation

$s\Leftrightarrow xy$

for (x,x) on relation

$s\Leftrightarrow xy$

and

$x^2>0$

so,

$S$ is symmetric

$xy>0$

thus

$xy>0$

so,

$S$ is reflexive.

Now, if

$xy>0$

and

$yz>0$

so,

$xz>0$

Thus,

$S$ is transitive.

So, relation is equivilance.

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

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

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

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

Explanation

From given we get,

$A\cap B=\{ 4\} ,A\cup B=\{ 2,3,4,5\}$

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

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

Ans.

$(d)$ .

Let

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

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

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

Explanation

Ans.

$(c)$ .
Common elements will be

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

Let A={ 1, 2, 3, 4} and R= {( 2, 2), (3, 3), (4, 4), (1, 2)} be a relation on A. Then R is

0%
Reflexive
0%
Symmetric
0%
Transitive
0%
None of these

Explanation

$\neq$ R as every element is not related to itself.

$\neq$ S as (1, 2)

$\in$ R but (2, 1)

$\notin$ R
T (1,2),

$\in$ R, (2, 2)

$\in$ R

$\Rightarrow$ (1, 2)o (2, 2) = (1, 2)

$\in$ R

If A = (a, b, c, d), B= (p, q, r, s). then which of the following are relations from A to B? Give reasons for your answer:

0%
$R_1$ = {( a, p), (b, r), (c, s)}
0%
$R_2$ = { ( q, b), (c, s), (d, r)}
0%
$R_3$ = {(a, p), (a, q), (d, p), (c, r), (b, r)}
0%
$R_4$ = {(a, p), (a, q), (b, s), (s, b)}

Explanation

(A) .(C)
If R is a relation from A to B, then R

$\subset (A\times B)$ which consists of all ordered pairs of the type. (x, y) such that

$x\in A, y \in B$ .
Clearly,

$R_1$ and

$R_3$ are such sets of

$A\times B$ . The elements (q, b), of

$R_2 \in (B\times A)$ and not

$A\times B$ and the elements (q, b), and (s, b), of

$R_4$ also not belong to

$A\times B$ and as such

$R_2$ and

$R_4$ are not relations from A to B.

Given the relation R= {(1,2), (2,3) } on the set {1, 2, 3}, the minimum number of ordered pairs which when added to R make it an equivalence relation is

0%
5
0%
6
0%
7
0%
8

The union of two equivalence relations is:

0%
always an equivalence relation
0%
reflexive and symmetric but needn't be transitive
0%
reflexive and transitive but need not be symmetric
0%
None of these

Let R be a relation from a set A to a set B then

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

Explanation

Let R be a relation from a set A to a set B then

$R\subseteq A\times B$

$n(A) = 2, n(B) = m$ and the number of relation from

$A$ to

$B$ is

$64$ , then the value of

$m$ is

0%
$6$
0%
$3$
0%
$16$
0%
$8$

Explanation

The number of relations between sets can be calculated using

$2^{mn}$ where m and n represent

the number of members in each set.

Given that

$n(A) = 2$ and

$n(B) = m$

So,

$2^{2m} = 64$

$2^{2m} = 2^{6}$

On comparing powers of 2 we get,

$2m = 6$

$m = 3$

1126604_1070745_ans_9c11a5191a434e25b07455e0bbb975cb.jpg

$A$ and

$B$ are two sets containing four and two elements, respectively. Then the number of subsets of the set

$A\times B$ each having at least three elements is

0%
$219$
0%
$256$
0%
$275$
0%
$510$

Explanation

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

$B=\left\{x,y\right\}$

$A\times B=2\times 4=8$ Elements.

Total number of subsets of

$A\times B$ having

$3$ or more elements

$=2^8=256$

$\Rightarrow 256-(1$ null set

$+$

$8$ singleton set

$+$

$^8e_2$ having

$2$ elements

$)$

$=256-1-8-28$

$=219$

Find x and y, if

$(x+3,5)=(6,2x+y)$ .

0%
x=3, y=-1
0%
x=6, y=2
0%
x=2,y=3
0%
none of these

Explanation

By the definition of equality of ordered pairs

$(x+3,5)=(6,2x+y)$

$x+3=6$ and

$5=2x+y$

$x=3$ and

$5=2x+y$

$x=3$ and

$5=6+y$

$x=3$ and

$y=-1$

CBSE Questions for Class 11 Commerce Applied Mathematics Relations Quiz 2 - 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 2 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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