Relations Test 16

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Relations Test 16

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Weekly Quiz Competition

Question 1
1 / -0

Given $$\displaystyle A=\left \{ 2, 3 \right \},B=\left \{ 4, 5 \right \},C=\left \{ 5, 6 \right \},$$ find $$\displaystyle A\times \left ( B\cap C \right )=.........$$

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

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

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

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

Solution

$$\displaystyle A=\left \{ 2, 3 \right \},B=\left \{ 4, 5 \right \},C=\left \{ 5, 6 \right \}$$
$$\displaystyle B\cap C=\left\{5\right\}$$
Now, $$\displaystyle A\times \left ( B\cap C \right )=\left \{ \left ( 2, 5 \right ),\left ( 3, 5 \right ) \right \}$$
Hence, option B.
Question 2
1 / -0

$$A$$ 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

A
$$6$$

B
$$36$$

C
$$15$$

D
none of these

Solution

$$\displaystyle \left ( a\times b \right )$$ 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.
Question 3
1 / -0

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

A
$$\displaystyle R=A\cup B$$

B
$$\displaystyle R=A\cap B$$

C
$$\displaystyle R\subseteq A\times B$$

D
$$\displaystyle R\subseteq B\times A$$

Solution

If $$R$$ 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 $$B$$.
Question 4
1 / -0

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

A
$$18$$

B
$$6$$

C
$$4$$

D
$$0$$

Solution

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$$
Question 5
1 / -0

If $$\displaystyle A=\left \{ a, b, c \right \},B=\left \{ c, d, e \right \},C=\left \{ a, d, f \right \}$$, then $$A\times \left ( B\cup C \right )$$ is

A
$$\displaystyle \left \{ \left ( a, d \right ),\left ( a, e \right ),\left ( a, c \right ) \right \}$$

B
$$\displaystyle \left \{ \left ( a, d \right ),\left ( b, d \right ),\left ( c, d \right ) \right \}$$

C
$$\displaystyle \left \{ \left ( d, a \right ),\left ( d, b \right ),\left ( d, c \right ) \right \}$$

D
none of these

Solution

$$\displaystyle A\times \left ( B\cup C \right )=\left \{ a, b, c \right\}\times \left \{ a, c, d, e, f \right \}$$
The above set will consist of 15 ordered pairs and not 3.
Hence option 'D' is correct choice.
Question 6
1 / -0

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

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

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

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

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

Solution

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 \}$$
Question 7
1 / -0

If $$\displaystyle X= \left \{ 1,2,3,4,5 \right \}, Y= \left \{ 1,3,5,7,9 \right \}$$ determine which of the following sets are mappings, relations or neither from A to B:
(i)$$\displaystyle F= \left \{ \left ( x,y \right ) \because y= x+2, x \in X, y \in Y \right \}$$

A
It is clearly a one-one onto mapping i.e. a bijection. It is also a relation.

B
It is clearly a many-one onto mapping. It is also a relation.

C
It is clearly a one-one but not onto mapping. It is also a relation.

D
It is not a mapping but a relation

Solution

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.
Question 8
1 / -0

If $$\displaystyle A=\left \{ 1, 2, 3 \right \}$$ and $$B=\left \{ 3, 8 \right \},$$ then $$\displaystyle \left ( A\cup B \right )\times \left ( A\cap B \right )$$ is

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

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

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

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

Solution

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 \}$$
Question 9
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
$$2^{mn}$$

C
$$m^{n}$$

D
$$n^{m}$$

Solution

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}$$
Question 10
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Let $$R$$ be a reflexive on a finite set $$A$$ having $$n$$ elements, and let there be $$m$$ ordered pairs in $$R$$. Then

A
$$m\ge n$$

B
$$m\le n$$

C
$$m=n$$

D
None of these

Solution

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 $$m\ge n$$.

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Relations Test 16

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Relations Test 16

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Question 1

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

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

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

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

Solution

Question 2

$$6$$

$$36$$

$$15$$

none of these

Solution

Question 3

$$\displaystyle R=A\cup B$$

$$\displaystyle R=A\cap B$$

$$\displaystyle R\subseteq A\times B$$

$$\displaystyle R\subseteq B\times A$$

Solution

Question 4

$$18$$

$$6$$

$$4$$

$$0$$

Solution

Question 5

$$\displaystyle \left \{ \left ( a, d \right ),\left ( a, e \right ),\left ( a, c \right ) \right \}$$

$$\displaystyle \left \{ \left ( a, d \right ),\left ( b, d \right ),\left ( c, d \right ) \right \}$$

$$\displaystyle \left \{ \left ( d, a \right ),\left ( d, b \right ),\left ( d, c \right ) \right \}$$

none of these

Solution

Question 6

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

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

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

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

Solution

Question 7

It is clearly a one-one onto mapping i.e. a bijection. It is also a relation.

It is clearly a many-one onto mapping. It is also a relation.

It is clearly a one-one but not onto mapping. It is also a relation.

It is not a mapping but a relation

Solution

Question 8

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

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

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

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

Solution

Question 9

$$mn$$

$$2^{mn}$$

$$m^{n}$$

$$n^{m}$$

Solution

Question 10

$$m\ge n$$

$$m\le n$$

$$m=n$$

None of these

Solution

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