Relations Test 15

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Relations Test 15

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Weekly Quiz Competition

Question 1
1 / -0

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

A
$$R_{1}$$ $$R_{2}$$ are relations but $$R_{3}$$ is not a relation.

B
$$R_{1}$$ $$R_{3}$$ are relations but $$R_{2}$$ is not a relation.

C
All are relations

D
none of these

Solution

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 $$B\times A.$$
Question 2
1 / -0

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

A
$$5$$,$$8$$

B
$$8$$,$$3$$

C
$$-1$$,$$5$$

D
$$7$$,$$1$$

Solution

as the given ordered pairs are equal,
$$x-1=7$$ and $$y+2=5$$
$$\therefore x=8$$ and $$y=3$$
Question 3
1 / -0

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

A
$$2$$ and $$4$$

B
$$3$$ and $$6$$

C
$$5$$ and $$3$$

D
$$1$$ and $$-1$$

Solution

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

Since $$ (a, 3) = (5,x) $$ then ,$$ a = 5 , x = 6 $$
Question 4
1 / -0

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

A
3 and 5

B
4 and 7

C
-1 and 17

D
2 and 4

Solution
Question 5
1 / -0

Given $$M = (0, 1, 2)$$ and $$N = (1, 2, 3)$$. Find $$(N - M) \times (N \cap M)$$

A
{(3, 1), (3, 2)}

B
{(3, -1), (3, -2)}

C
{(-3, -1), (-3, 2)}

D
{(3, -1), (-3, -2)}

Solution

$$ N - M $$ 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, $$ (N - M) \times (N \cap M) =\{ (3,1), (3,2) \}$$.
Question 6
1 / -0

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

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

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

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

D
none of these

Solution

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, $$ A \times (B \cup C )= $$ { $$ (5,7), (5,9), (5,11), (7,7),(7,9), (7,11) $$ }
Question 7
1 / -0

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

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

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

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

D
none of these

Solution

Product of two sets is found by forming ordered pairs by multiplying every element of the first set with the second set.
So, $$ A \times B = $$ { $$ (5,7), (5,9), (7,7),(7,9) $$ }
And $$ A \times C = $$ { $$ (5,7), (5,9), (5,11), (7,7),(7,9), (7,11) $$ }
Union of two sets has all the elements of both the sets.
So, $$ (A \times B) \cup ( A \times C) = $$ { $$ (5,7), (5,9), (5,11), (7,7),(7,9), (7,11) $$ }
Question 8
1 / -0

If $$n(A) = 4$$ and $$n(B) = 5$$, then $$n(A \times B) = $$

A
$$20$$

B
$$25$$

C
$$4$$

D
$$15$$

Solution

Given, $$n\left( A \right) =4$$ and $$n\left( B \right) =5$$
$$ \Rightarrow n\left( A\times B \right) =n\left( A \right) \cdot n\left( B \right) $$
$$\Rightarrow n(A\times B)=4\cdot 5=20$$
So, answer is $$20$$.
Question 9
1 / -0

If $$\displaystyle A=\left \{ 2, 3, 5 \right \}, B=\left \{ 2, 5, 6 \right \}$$ then $$\left ( A-B \right )\times \left ( A\cap B \right )$$ is

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

B
$$\displaystyle \left \{ \left ( 3, 2 \right ), \left ( 3, 5 \right ), \left ( 3, 6 \right )\right \}$$

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

D
None of these

Solution

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

$$\displaystyle \therefore \left ( A-B \right )\times \left ( A\cap B \right )=\left \{ \left ( 3, 2 \right ),\left ( 3, 5 \right ) \right \}.$$
Question 10
1 / -0

If $$R$$ 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

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

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

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

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

Solution

$$\displaystyle R;A\in B$$ 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.

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Relations Test 15

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Relations Test 15

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Question 1

$$R_{1}$$ $$R_{2}$$ are relations but $$R_{3}$$ is not a relation.

$$R_{1}$$ $$R_{3}$$ are relations but $$R_{2}$$ is not a relation.

All are relations

none of these

Solution

Question 2

$$5$$,$$8$$

$$8$$,$$3$$

$$-1$$,$$5$$

$$7$$,$$1$$

Solution

Question 3

$$2$$ and $$4$$

$$3$$ and $$6$$

$$5$$ and $$3$$

$$1$$ and $$-1$$

Solution

Question 4

3 and 5

4 and 7

-1 and 17

2 and 4

Solution

Question 5

{(3, 1), (3, 2)}

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

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

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

Solution

Question 6

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

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

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

none of these

Solution

Question 7

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

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

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

none of these

Solution

Question 8

$$20$$

$$25$$

$$4$$

$$15$$

Solution

Question 9

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

$$\displaystyle \left \{ \left ( 3, 2 \right ), \left ( 3, 5 \right ), \left ( 3, 6 \right )\right \}$$

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

None of these

Solution

Question 10

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

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

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

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

Solution

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