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CBSE Questions for Class 11 Commerce Applied Mathematics Relations Quiz 6 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Relations
Quiz 6
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Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
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Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
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Assertion is incorrect but Reason is correct
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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.
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
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$$\displaystyle \left \{ (4,3),(2,4) ,(5,0) \right \}$$
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$$\displaystyle \left \{ (3,4),(4,2) ,(5,0) \right \}$$
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$$\displaystyle \left \{ (4,3),(2,4) ,(0,5) \right \}$$
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$$\displaystyle \left \{ (4,3),(4,2) ,(5,0) \right \}$$
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 $$A=\{0,1,2,3,4,5\}$$
So $$R$$ becomes $$\{(3,4),(4,2),(5,0)\}$$
Hence $$R^{-1}$$ is $$\{(4,3),(2,4),(0,5)\}$$
If $$\displaystyle R=\{ (x,y):x, y \in Z ,x^{2}+y^{2}\leq 4 \}$$ is a relation in $$Z$$ then domain $$D$$ is
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$$\displaystyle \left \{ -2,-1,0,1,2\right \}$$
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$$\displaystyle \left \{ -2,-1,0 \right \}$$
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$$\displaystyle \left \{ 0,1,2\right \}$$
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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 $$-2,-1,0,1,2.$$
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.
$$\displaystyle R_{1}= \left \{ \left ( a,1 \right ), \left ( a,3 \right ) \right \}$$
$$\displaystyle R_{2}= \left \{ \left ( a,1 \right ), \left ( c,2 \right ), \left ( d,1 \right ) \right \}$$
$$\displaystyle R_{3}= \left \{ \left ( a,1 \right ), \left ( b,2 \right ), \left ( 3,c \right ) \right \}.$$
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$$R_{1}$$ $$R_{2}$$ are relations but $$R_{3}$$ is not a relation.
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$$R_{1}$$ $$R_{3}$$ are relations but $$R_{2}$$ is not a relation.
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All are relations
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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
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$$5$$,$$8$$
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$$8$$,$$3$$
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$$-1$$,$$5$$
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$$7$$,$$1$$
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
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True
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False
Explanation
By rule of ordered pairs this statement is true.
Ordered pairs (a, 3) and (5, x) are equal ,the values of $$a$$ and $$x$$ are
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$$2$$ and $$4$$
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$$3$$ and $$6$$
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$$5$$ and $$3$$
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$$1$$ 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 ,$$ a = 5 , x = 6 $$
If $$(x, y) = (3, 5)$$ ; then values of $$x$$ and $$y $$ are
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3 and 5
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4 and 7
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-1 and 17
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2 and 4
Given $$M = (0, 1, 2)$$ and $$N = (1, 2, 3)$$. Find
$$(N - M) \times (N \cap M)$$
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{(3, 1), (3, 2)}
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{(3, -1), (3, -2)}
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{(-3, -1), (-3, 2)}
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{(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, $$ (N - M) \times (N \cap M) =\{ (3,1), (3,2) \}$$.
If $$A = \{5, 7\}, B= \{7, 9\}$$ and $$C = \{7, 9, 11\},$$ find
$$A \times (B \cup C)$$
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$$ \{(5, 7), (5, 9), (5, 11), (7, 7), (7, 9), (7, 11) \}$$
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$$ \{(5, 5), (5, 9), (5, 11), (7, 7), (7, 9), (7, 11) \}$$
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$$ \{(5, 7), (9, 9), (5, 11), (7, 7), (7, 9), (7, 11) \}$$
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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, $$ A \times (B \cup C )= $$ { $$ (5,7), (5,9), (5,11), (7,7),(7,9), (7,11) $$ }
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
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$$\displaystyle \left \{ \left ( 1,3 \right ),\left ( 1,5 \right ),\left ( 2,3 \right ),\left ( 2,5 \right ),\left ( 3,5 \right ),\left ( 4,5 \right ) \right \}$$
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$$\displaystyle \left \{ \left ( 3,1 \right ),\left ( 5,1 \right )\left ( 3,2 \right ),\left ( 5,2 \right ) ,\left ( 5,3 \right ),\left ( 5,4 \right )\right \}$$
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$$\displaystyle \left \{ \left ( 3,3 \right ),\left ( 3,5 \right ) ,\left ( 5,3 \right ),\left ( 5,5 \right )\right \}$$
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$$\displaystyle \left \{ \left ( 3,3 \right ),\left ( 3,4 \right ) ,\left ( 4,5 \right )\right \}$$
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
$$\displaystyle A\times \left ( B\cap C \right )=.........$$
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$$\displaystyle A\times \left ( B\cap C \right )=\left \{ \left ( 2, 4 \right ),\left ( 3, 5 \right ) \right \}$$
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$$\displaystyle A\times \left ( B\cap C \right )=\left \{ \left ( 2, 5 \right ),\left ( 3, 5 \right ) \right \}$$
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$$\displaystyle A\times \left ( B\cap C \right )=\left \{ \left ( 2, 5 \right ),\left ( 4, 5 \right ) \right \}$$
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$$\displaystyle A\times \left ( B\cap C \right )=\left \{ \left ( 2, 3 \right ),\left ( 3, 5 \right ) \right \}$$
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 ?
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True
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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
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$$6$$
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$$36$$
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$$15$$
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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
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$$\displaystyle R=A\cup B$$
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$$\displaystyle R=A\cap B$$
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$$\displaystyle R\subseteq A\times B$$
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$$\displaystyle R\subseteq B\times A$$
Explanation
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 $$B$$.
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
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$$\displaystyle \left \{ \left ( 2, 2 \right ),\left ( 3, 4 \right ),\left ( 4, 2 \right ),\left ( 5, 4 \right ) \right \}$$
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$$\displaystyle \left \{ \left ( 2, 3 \right ),\left ( 4, 3 \right ),\left ( 4, 5 \right ) \right \}$$
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$$\displaystyle \left \{ \left ( 2, 4 \right ),\left ( 3, 4 \right ),\left ( 4, 4 \right ),\left ( 4, 5 \right ) \right \}$$
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$$\displaystyle \left \{ \left ( 4, 2 \right ),\left ( 4, 3 \right ),\left ( 4, 4 \right ),\left ( 4, 5 \right ) \right \}$$
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
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$$18$$
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$$6$$
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$$4$$
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$$0$$
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 $$\ n\left [ \left ( A
\times B \right )\cap \left ( B \times A \right ) \right ]=n\left [
\left ( A \cap B \right )\times \left ( B \cap A \right ) \right ]$$
$$\displaystyle \Rightarrow \ n\left [ \left ( A
\times B \right )\cap \left ( B \times A \right ) \right ]=2 \times 2=4$$
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
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$$\displaystyle \left \{ \left ( a, d \right ),\left ( a, e \right ),\left ( a, c \right ) \right \}$$
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$$\displaystyle \left \{ \left ( a, d \right ),\left ( b, d \right ),\left ( c, d \right ) \right \}$$
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$$\displaystyle \left \{ \left ( d, a \right ),\left ( d, b \right ),\left ( d, c \right ) \right \}$$
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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.
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)$$\displaystyle F= \left \{ \left ( x,y \right ) \because y= x+2, x \in X, y \in Y \right \}$$
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It is clearly a one-one onto mapping i.e. a bijection. It is also a relation.
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It is clearly a many-one onto mapping. It is also a relation.
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It is clearly a one-one but not onto mapping. It is also a relation.
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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.
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
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$$\displaystyle \left \{ \left ( 3, 1 \right ), \left ( 3,2 \right ), \left ( 3, 3 \right ), \left ( 3, 8 \right ) \right \}$$
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$$\displaystyle \left \{ \left ( 1, 3 \right ), \left ( 2,3 \right ), \left ( 3, 3 \right ), \left ( 8, 3\right ) \right \}$$
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$$\displaystyle \left \{ \left ( 1, 2 \right ), \left ( 2,2 \right ), \left ( 3, 3 \right ), \left ( 8, 8\right ) \right \}$$
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$$\displaystyle \left \{ \left ( 8, 3 \right ), \left ( 8,2 \right ), \left ( 8, 1 \right ), \left ( 8, 8\right ) \right \}$$
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:
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$$mn$$
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$$2^{mn}$$
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$$m^{n}$$
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$$n^{m}$$
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
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$$m\ge n$$
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$$m\le n$$
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$$m=n$$
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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 $$m\ge n$$.
If $$A=\{a,b,c,d\}, B=\{p,q,r,s\}$$, then which of the following are relations from $$A$$ to $$B$$?
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$$\displaystyle R_{1}= \left \{ \left ( a,p \right ), \left ( b,r \right ), \left ( c,s \right ) \right \}$$
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$$\displaystyle R_{2}= \left \{ \left ( q,b \right ), \left ( c,s \right ), \left ( d,r \right ) \right \}$$
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$$\displaystyle R_{3}= \left \{ \left ( a,p \right ), \left ( a,q \right ), \left ( d,p \right ), \left ( c,r \right ), \left ( b,r \right ) \right \}$$
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$$\displaystyle R_{4}= \left \{ \left ( a,p \right ), \left ( q,a \right ), \left ( b,s \right ), \left ( s,b \right ) \right \}$$
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.
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True
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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.
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$$0$$
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Domain of $$\displaystyle R= \left \{ 0, \pm 3 \right \}= $$ domain of $$\displaystyle R^{-1}.$$
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Domain of $$\displaystyle R= \left \{ 0, \pm 3, \pm 4 \right \}= $$ domain of $$\displaystyle R^{-1}.$$
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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 $${ R }^{ -1 }$$
If $$A=\left\{ 2,3 \right\} $$ and $$B=\left\{ 1,2,3,4 \right\} $$, then which of the following is not a subset of $$A\times B$$
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$$\left\{ (2,3),(2,4),(3,3),(3,4) \right\} $$
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$$\left\{ (2,2),(3,1),(3,4),(2,3) \right\} $$
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$$\left\{ (2,1),(3,2) \right\} $$
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$$\left\{ (1,2),(2,3) \right\} $$
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 $$R$$
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is reflexive
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is symmetric
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is transitive
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possess all the above three properties
Explanation
A relation R is said to be an equivalence relation if it is
(i)Reflexive
(ii)Symmetric
(iii)Transitive
Which one of the following relations on $$R$$ is equivalence redlation-
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$$x{R}_{1}y\Leftrightarrow \left| x \right| =\left| y \right| $$
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$$x{R}_{2}y\Leftrightarrow \left| x \right| > \left| y \right| $$
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$$x{R}_{3}y\Leftrightarrow x|y$$
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$$x{R}_{4}y\Leftrightarrow x< y$$
Explanation
Because $$\left| x \right| =\left| x \right| $$ (Reflexive)
$$\left| x \right| =\left| y \right| \Rightarrow \left| y \right| =\left| x \right| $$ (Symmetric)
$$\left| x \right| =\left| y \right| $$ and $$\left| y \right| =\left| z \right| \Rightarrow \left| x \right| =\left| z \right| $$ (Transitive)
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
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$${2}^{5}$$
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$${2}^{10}-1$$
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$${2}^{12}-1$$
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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 $$2^{12}-1$$
If $$A=\left\{ 2,4,5 \right\} , B=\left\{ 7,8,9 \right\} $$ then $$n(A\times B)$$ is equal to-
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$$6$$
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$$9$$
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$$3$$
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$$0$$
Explanation
Given $$n(A)=3,$$ and $$n(B) =3$$
Hence $$n(A\times B) = 3\times 3=9$$
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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
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