Relations Test 14

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Relations Test 14

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Weekly Quiz Competition

Question 1
1 / -0

If $$\displaystyle :n(A)= m, $$ then number of relations in $$A$$ are

A
$$\displaystyle 2^{m}$$

B
$$\displaystyle 2^{m}-2 $$

C
$$\displaystyle 2^{m^{2}} $$

D
None of these

Solution

Given $$\displaystyle :n(A)= m $$ Relation in a set $$A$$ is subset of

$$\displaystyle A\times A $$ For Number of subset of $$A$$ We have

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

\therefore$$ Number of subset of $$\displaystyle A\times A = 2^{m^{2}}$$

every subset of $$\displaystyle A\times A $$ is a relation

$$\displaystyle \therefore$$ Number of relations in A is $$\displaystyle

2^{m^{2}}= 2^{m\times m}$$
Question 2
1 / -0

In order that a relation $$R$$ defined on a non-empty set $$A$$ is an equivalence relation.
It is sufficient, if $$R$$

A
is reflexive

B
is symmetric

C
is transitive

D
possesses all the above three properties.

Solution

An equivalent relation needs to be symmetric,transitive and reflexive.
Question 3
1 / -0

Let $$ A= \{ 1,2,3,.......50\} $$ and $$B=\{2,4,6.......100\}$$ .The number of elements $$\left ( x, y \right )\in A\times B$$ such that $$x+y=50$$

A
$$24$$

B
$$25$$

C
$$50$$

D
$$75$$

Solution

The elements will be
$$(2,48),(48,2)$$
$$(4,46), (46,4)$$
:
:
$$(2n,50-2n), (50-2n,2n)$$
Now we have
$$2,4,6,8...$$ upto $$48$$
This forms an A.P
The number terms is $$24$$.
Question 4
1 / -0

Let $$\displaystyle A= \left \{ a,b,c \right \} $$ and $$\displaystyle B= \left \{ 4,5 \right \} $$ Consider a relation $$R$$ defined from set $$A$$ to set $$B$$ then $$R$$ is subset of

A
$$A$$

B
$$B$$

C
$$\displaystyle A\times B $$

D
$$\displaystyle B\times A $$

Solution

If A and B are two non empty sets, the relation from set $$A$$ to set $$B$$ is denoted as $$\displaystyle A\times B $$
Question 5
1 / -0

If $$A=\{2, 4\}$$ and $$B=\{3, 4, 5\}$$ then $$(A\cap B)\times (A\cup B)$$ is

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

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

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

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

Solution

$$A\cap B=\{4\}$$
$$A\cup B=\{2,3,4,5\}$$
Hence
$$(A\cap B)\times(A\cup B)$$
$$=\{(4,2),(4,3),(4,4),(4,5)\}$$
Question 6
1 / -0

If $$R$$ is an anti symmetric relation in $$\displaystyle A$$ such that $$(a,b),(b,a)\:\in\: R$$ then

A
$$\displaystyle a\leq b$$

B
$$\displaystyle a= b$$

C
$$\displaystyle a\geq b$$

D
None of these

Solution

For $$(a,b), (b,a)$$ belongs to the same relation which is anti symmetric that is possible only if the relation is reflexive i.e $$a=b$$
Question 7
1 / -0

$$R$$ is a relation from $$\displaystyle \left \{ 11,12,13 \right \}$$ to $$\displaystyle \left \{ 8,10,12 \right \}$$ defined by $$y=x-3$$ then the $$R^{-1}$$ .

A
$$\displaystyle \left \{ (11,8),(13,10) \right \}$$

B
$$\displaystyle \left \{ (8,11),(10,13) \right \}$$

C
$$\displaystyle \left \{ (8,11),(9,12),(10,13) \right \}$$

D
$$None\ of\ these$$

Solution

$$x=\{11,12,13\}, y=\{8,10,12\}$$
Now, $$y=x-3$$
So, the relation $$R$$ becomes $$\{(11,8),(13,10)\}$$
So, $$R^{-1}$$ becomes $$\{(8,11),(10,13)\}$$
Question 8
1 / -0

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

A
$$mn$$

B
$$\displaystyle 2^{mn}$$

C
$$\displaystyle m^{n}$$

D
$$\displaystyle n^{m}$$

Solution

Consider an element $$\displaystyle a\in A,$$ 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 mapping is
$$\displaystyle n\times n\times n\times ...m \mbox{times}=n^{m}.$$
Question 9
1 / -0

If $$\displaystyle A=\left \{ 0,1,2,3,4,5 \right \}$$ and relation $$R$$ defined by $$a R b$$ such that $$2a+b=10$$ then $$ R^{-1}$$ equals

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

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

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

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

Solution

$$2a+b=10$$, $$2a$$ 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 $$A=\{0,1,2,3,4,5\}$$
So $$R$$ becomes $$\{(3,4),(4,2),(5,0)\}$$
Hence $$R^{-1}$$ is $$\{(4,3),(2,4),(0,5)\}$$
Question 10
1 / -0

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

A
$$\displaystyle \left \{ -2,-1,0,1,2\right \}$$

B
$$\displaystyle \left \{ -2,-1,0 \right \}$$

C
$$\displaystyle \left \{ 0,1,2\right \}$$

D
None of these

Solution

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

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Re-Attempt Weekly Quiz Competition

Relations Test 14

Result

Relations Test 14

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SHARING IS CARING

Question 1

$$\displaystyle 2^{m}$$

$$\displaystyle 2^{m}-2 $$

$$\displaystyle 2^{m^{2}} $$

None of these

Solution

Question 2

is reflexive

is symmetric

is transitive

possesses all the above three properties.

Solution

Question 3

$$24$$

$$25$$

$$50$$

$$75$$

Solution

Question 4

$$A$$

$$B$$

$$\displaystyle A\times B $$

$$\displaystyle B\times A $$

Solution

Question 5

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

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

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

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

Solution

Question 6

$$\displaystyle a\leq b$$

$$\displaystyle a= b$$

$$\displaystyle a\geq b$$

None of these

Solution

Question 7

$$\displaystyle \left \{ (11,8),(13,10) \right \}$$

$$\displaystyle \left \{ (8,11),(10,13) \right \}$$

$$\displaystyle \left \{ (8,11),(9,12),(10,13) \right \}$$

$$None\ of\ these$$

Solution

Question 8

$$mn$$

$$\displaystyle 2^{mn}$$

$$\displaystyle m^{n}$$

$$\displaystyle n^{m}$$

Solution

Question 9

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

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

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

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

Solution

Question 10

$$\displaystyle \left \{ -2,-1,0,1,2\right \}$$

$$\displaystyle \left \{ -2,-1,0 \right \}$$

$$\displaystyle \left \{ 0,1,2\right \}$$

None of these

Solution

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