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CBSE Questions for Class 11 Commerce Applied Mathematics Relations Quiz 8 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Relations
Quiz 8
If $$\displaystyle n\left ( P\times Q \right )=0$$ then n(P) can possibly be
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0
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10
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20
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Any value
Explanation
$$ n(P) $$ can be of any value as we are not sure of $$ n(Q) $$
Hence, $$ n(P) $$ can take any of the given values.
Let R be a relation from N to N defined by $$R=\{(a, b) :a, b\in N$$ and $$a=b^2\}$$. Is the following statement true for $$(a, b)\in R\Rightarrow (b, a)\in R ?$$.
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True
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False
Explanation
No, because $$(4, 2)\in R$$ but $$(2, 4)\not{\in}R$$.
If a R b, b R c and a R c, and a,b,c,$$\displaystyle \epsilon $$ A, then the relation R on the set A is said to be a/an_____ relation
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reflexive
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symmetric
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transitive
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equivalence
Explanation
If in a relation, an element $$ a $$ is related to $$ b $$, and $$ b $$ is related to a third element $$ c $$, the relation $$ R $$ is said to be transitive if $$ a $$ is also related to $$ c $$
Let $$R$$ be the relation in the set $$\left\{1, 2, 3, 4\right\}$$ given by $$R=\left\{(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)\right\}$$.
Choose the correct answer.
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$$R$$ is reflexive and symmetric but not transitive.
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$$R$$ is reflexive and transitive but not symmetric.
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$$R$$ is symmetric and transitive but not reflexive.
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$$R$$ is an equivalence relation.
Explanation
$$R=\left\{(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)\right\}$$
It is seen that $$(a, a)\in R$$, for every $$a\in \left\{1, 2, 3, 4\right\}$$.
$$\therefore $$ $$R$$ is reflexive.
It is seen that $$(1, 2)\in R$$, but $$(2, 1)\notin R$$.
$$\therefore$$ $$R$$ is not symmetric.
Also, it is observed that $$(a, b), (b, c)\in R\Rightarrow (a, c)\in R$$ for all $$a, b, c \in \left\{1, 2, 3, 4\right\}$$
$$\therefore$$ $$R$$ is transitive.
Hence,$$R$$ is reflexive and transitive but not symmetric.
The correct answer is B.
If $$A=\left \{1, 2,3\right \}$$ and $$B=\left \{3,8\right \}$$, then $$(A\cup B)\times (A\cap B)$$ is equal to
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$$\left \{(8,3), (8,2), (8,1), (8,8)\right \}$$
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$$\left \{(1,2), (2,2), (3,3), (8,8)\right \}$$
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$$\left \{(3,1), (3,2), (3,3), (3,8)\right \}$$
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$$\left \{(1,3), (2,3), (3,3), (8,3)\right \}$$
Explanation
Given,
$$A=\left \{1, 2,3\right \}$$ and $$B=\left \{3,8\right \}$$,
Therefore, $$A\cup B=\left \{1,2,3\right \}\cup \left \{3,8\right \}=\left \{1,2,3,8\right \}$$
and $$A\cap B=\left \{1,2,3\right \}\cap \left \{3,8\right \}=\left \{3\right \}$$
$$\therefore (A\cup B)\times (A\cap B)=\left \{1,2,3,8\right \}\times \left \{3\right \}$$
$$=\left \{(1,3), (2,3), (3,3), (8,3)\right \}$$
If $$R$$ is a relation from a set $$A$$ to the set $$B$$ and $$S$$ is a relation from $$B$$ to $$C,$$ then the relation $$SoR$$
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is from $$C$$ to $$A$$
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does not exist
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is from $$A$$ to $$C$$
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None of these
Explanation
Since $$R\subseteq A\times B$$ and $$S\subseteq B\times C$$, we have
So $$R\subseteq A\times C$$.
$$\therefore$$ So R is a relation from $$A$$ to $$C.$$
$$A = \left \{1, 2, 3, 4\right \}$$ and $$B = \left \{a, b, c\right \}$$. The relations from $$A$$ to $$B$$ is
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$$\left \{(1, 2), (1, 3), (2, 3), (2, 4), (3, 4), (3, 1)\right \}$$
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$$\left \{(a, b), (a, c), (b, a), (b, c), (c, a)\right \}$$
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$$\{(1,a),(1,b)(1,c),(2,a),(2,b),(2,c),(3,a),(3,b),(3,c),(4,a),(4,b),(4,c)\}$$
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$$\left \{(a, 1), (1, 3), (b, 2), (c, 3), (b, 3), (b, 4)\right \}$$
Explanation
$$A=\{1,2,3,4\}$$
$$B=\{a,b,c\}$$
Possible Relation $$R:A\to B$$
$$=\{(1,a),(1,b)(1,c),(2,a),(2,b),(2,c),(3,a),(3,b),(3,c),(4,a),(4,b),(4,c)\}$$
Option C contains
Relation $$R:A\to B$$
$$A =\left \{a, b, c\right \}$$ and $$B = \left \{5, 7, 9\right \}$$. The relation from $$B$$ to $$A$$ is
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$$\left \{(a, 5), (a, 7), (b, 7), (c, 9)\right \}$$
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$$\left \{(5, 7), (9, 9), (7, 5)\right \}$$
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$$\left \{(5, a), (5, b), (5, c)\right \}$$
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$$\left \{(5, b), (7, c), (7, a), (9, b)\right \}$$
Explanation
$$A=\{a,b,c\}$$
$$B=\{5,7,9\}$$
$$R:B \longrightarrow A$$
$$R=\{(5,a),(5,b),(5,c),(7,a),(7,b),(7,c),(9,a),(9,b),(9,c)\}$$
Option C and D contain relations in R.
What is the first component of an ordered pair $$(1, -1)$$?
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$$1$$
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$$-1$$
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$$2$$
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$$0$$
Explanation
In an ordered pair $$(x,y)$$, the first component is $$x$$ and the second component is $$y$$.
Therefore, in
an ordered pair $$(1,-1)$$, the first component is $$1$$.
$$R$$ is a relation defined in $$R\times T$$ by $$(a,b) R (c,d)$$ iff $$a-c$$ is an integer and $$b=d$$. The relation $$R$$ is
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an identity relation
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an universal relation
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an equivalence relation
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None of these
Explanation
We have $$R=\left \{((a,b), (c,d)):a-c\in Z\ and \ b=d; a,b,c,d\in R\right \}$$.
Let $$(a,b)\in R\times R$$
$$\therefore (a,b)R(a,b)$$, because $$a-a=0\in Z$$ and $$b=b$$
$$\therefore$$ R is reflexive.
Let $$(a,b) R (c,d).\Rightarrow a-c\in Z$$ and $$b=d$$
$$\Rightarrow c-a\in Z$$ and $$d=b$$
$$\Rightarrow (c,d) R (a,b)\Rightarrow R$$ is symmetric.
Let $$(a,b)R(c,d)$$ and $$(c,d)R(e,f)$$.
$$\Rightarrow a-c\in Z, b=d, c-e\in Z, d=f$$
$$\Rightarrow (a-c)+(c-e)\in z, b=f$$
$$\Rightarrow a-e\in z, b=f\Rightarrow (a,b)R(e,f)$$
$$\Rightarrow R$$ is transitive.
$$\therefore R$$ is an equivalence relation.
$$\therefore$$ the correct answer is (c).
Let a relation $$R$$ be defined by $$R=\left \{(4,5), (1,4), (4,6), (7,6), (3,7)\right \}$$. The relation $$R^{-1}\circ R$$ is given by
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$$\left \{(1,1), (4,4), (7,4), (4,7), (7,7)\right \}$$
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$$\left \{(1,1), (4,4), (4,7), (7,4), (7,7),(3,3)\right \}$$
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$$\left \{(1,5), (1,6), (3,6)\right \}$$
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None of these
Explanation
We have $$R=\left \{(4,5), (1,4), (4,6), (7,6), (3,7)\right \}$$.
$$\therefore R^{-1}=\left \{(5,4), (4,1), (6,4), (6,7), (7,3)\right \}$$
$$(4,4)\in R^{-1}\circ R$$ because $$(4,5)\in R$$ and $$(5,4)\in R^{-1}$$
$$(1,1)\in R^{-1}\circ R$$ because $$(1,4)\in R$$ and $$(4,1)\in R^{-1}$$
$$(4,4)\in R^{-1}\circ R$$ because $$(4,6)\in R$$ and $$(6,4)\in R^{-1}$$
$$(4,7)\in R^{-1}\circ R$$ because $$(4,6)\in R$$ and $$(6,7)\in R^{-1}$$
$$(7,4)\in R^{-1}\circ R$$ because $$(7,6)\in R$$ and $$(6,4)\in R^{-1}$$
$$(7,7)\in R^{-1}\circ R$$ because $$(7,6)\in R$$ and $$(6,7)\in R^{-1}$$
$$(3,3)\in R^{-1}\circ R$$ because $$(3,7)\in R$$ and $$(7,3)\in R^{-1}$$
$$\therefore R^{-1}\circ R=\left \{(4,4), (1,1), (4,7), (7,4), (7,7), (3,3)\right \}$$.
$$\therefore$$ The correct answer is $$B$$.
Given $$(a - 2, b + 3) = (6, 8)$$, are equal ordered pair. Find the value of $$a$$ and $$b$$.
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$$a = 8$$ and $$b = 5$$
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$$a = 8$$ and $$b = 3$$
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$$a = 5$$ and $$b = 5$$
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$$a = 8$$ and $$b = 6$$
Explanation
By equality of ordered pairs, we have
$$(a - 2, b + 3) = (6, 8)$$
On equating we get
$$a - 2 = 6$$
$$a = 8$$
$$b + 3 = 8$$
$$b = 5$$
So, the value of$$ a = 8 $$ and $$ b = 5.$$
What is the second component of an ordered pair $$(3, -0.2)$$?
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$$3$$
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$$0.2$$
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$$1$$
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$$-0.2$$
Explanation
In an ordered pair $$(x,y)$$, the first component is $$x$$ and the second component is $$y$$.
Therefore, in
an ordered pair $$(3,-0.2)$$, the second component is $$-0.2$$.
What is the Cartesian product of $$A = \left \{1, 2\right \}$$ and $$B = \left \{a, b\right \}$$?
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$$\left \{(1, a), (1, b), (2, a), (b, b)\right \}$$
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$$\left \{(1, 1), (2, 2), (a, a), (b, b)\right \}$$
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$$\left \{(1, a), (2, a), (1, b), (2, b)\right \}$$
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$$\left \{(1, 1), (a, a), (2, a), (1, b)\right \}$$
Explanation
If $$A $$ and $$B$$ are two non empty sets, then the Cartesian product $$A \times B$$ is set of all ordered pairs $$(a,b)$$ such that $$a\in A$$ and $$b\in B$$.
Given $$A =\{1,2\}$$ and $$B = \{a,b\}$$
Hence $$A\times B = \{(1,a),(1,b),(2,a),(2,b)\}$$
The minimum number of elements that must be added to the relation $$R =\{(1,2)(2,3)\}$$ on the set of natural numbers so that it is an equivalence is
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$$4$$
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$$7$$
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$$6$$
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$$5$$
Explanation
For $$R$$ to be equivalence relation, it should be reflexive, symmetric and transitive.
Now, for $$R$$ reflexive:
$$\Rightarrow (x, x) \in R$$ for all $$x \in \{1, 2, 3\}$$
$$\Rightarrow (1,1), (2,2), (3,3) \in R$$
For $$R$$ is symmetric:
$$\Rightarrow (1,2), (2,3) \in R$$
$$\Rightarrow (2,1), (3,2) \in R$$
For $$R$$ is transitive:
$$\Rightarrow (1,2), (2,3) \in R$$
$$\Rightarrow (1,3) \in R$$
Also $$(3,1) \in R$$ as $$R$$ is symmetric.
So, the total number of elements is $$9$$.
Hence, minimum $$7$$ elements must be added.
The relation $$R$$ defined on the set $$A=\left \{1,2,3,4,5\right \}$$ by $$R=\left \{(x,y):|x^2-y^2| < 16\right \}$$ is given by
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$$\left \{(1,1), (2,1), (3,1), (4,1), (2,3)\right \}$$
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$$\left \{(2,2), (3,2), (4,2), (2,4)\right \}$$
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$$\left \{(3,3), (4,3), (5,4), (3,4)\right \}$$
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None of these
Explanation
We have $$R=\left \{(x,y):|x^2-y^2| < 16\right \}$$.
Take $$x=1,$$ we have
$$ |x^2-y^2| < 16\Rightarrow |1-y^2| < 16$$
$$\Rightarrow |y^2-1| < 16\Rightarrow y=1,2,3,4$$.
Take $$x=2,$$ we have
$$ |x^2-y^2| < 16 \Rightarrow |4-y^2| < 16$$
$$\Rightarrow |y^2-4| < 16\Rightarrow y=1,2,3,4$$.
Take $$x=3,$$ we have
$$ |x^2-y^2| < 16\Rightarrow |9-y^2| < 16$$
$$\Rightarrow |y^2-9| < 16\Rightarrow y=1,2,3,4$$.
Take $$x=4,$$ we have
$$ |x^2-y^2| < 16\Rightarrow |16-y^2| < 16$$
$$\Rightarrow |y^2-16| < 16\Rightarrow y=1,2,3,4,5$$.
Take $$x=5,$$
$$|x^2-y^2| < 16 \Rightarrow |25-y^2| < 16$$
$$\Rightarrow |y^2-25|<16 \Rightarrow y = 4,5$$
$$\therefore R=\left \{(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), \\(3,3), (3,4), (4,1), (4,2), (4,3), (4,4), (4,5), (5,4), (5,5)\right \}$$
$$\therefore$$ The correct answer is $$D$$.
$$(x, y)$$ and $$(p, q)$$ are two ordered pairs. Find the values of $$p$$ and $$y$$, if $$(4y + 5, 3p - 1) = (25, p + 1)$$
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$$p = 0, y = 5$$
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$$p = 1, y = 5$$
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$$p = 0, y = 1$$
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$$p = 1, y = 1$$
Explanation
Given $$(x,y)=(p,q)$$
$$(4y + 5, 3p - 1) = (25, p + 1)$$
On equating we get
$$4y + 5 = 25$$
$$4y = 25 - 5$$
$$4y = 20$$
$$y = 5$$
$$3p - 1 = p + 1$$
$$3p - p = 1 + 1$$
$$2p = 2$$
$$p = 1$$
So, the value of$$ p = 1, y = 5$$
$$(x, y)$$ and $$(p, q)$$ are two ordered pairs. Find the values of $$x$$ and $$p$$, if $$(3x - 1, 9) = (11, p + 2)$$
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$$x = 4, p = 9$$
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$$x = 6, p = 7$$
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$$x = 4, p = 5$$
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$$x = 4, p = 7$$
Explanation
Given$$(x,y)=(p,q)$$
$$(3x - 1, 9) = (11, p + 2)$$
By equating
$$3x - 1 = 11$$
$$3x = 12$$
$$x = 4$$
$$9 = p + 2$$
$$p = 7$$
So, the value of $$x = 4, p = 7.$$
If $$A \times B = \{(3, a), (3, -1), (3, 0), (5, a), (5, -1), (5, 0)\}$$, find $$A$$.
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$$\{a, 5\}$$
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$$\{a, -1\}$$
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$$\{0, 5\}$$
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$$\{3, 5\}$$
What is the relation for the following diagram?
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$$R =\{(2, 5), (2, 6), (3, 7)\}$$
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$$R = \{(1, 5), (2, 6), (3, 7)\}$$
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$$R =\{(1, 5), (2, 6), (1, 7)\}$$
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$$R = \{(1, 5), (2, 6), (2, 7)\}$$
Explanation
$$R:A\to B$$
$$A\times B = \{(1,5),(2,6),(3,7)\}$$
Which of the following do(es) not belong to $$A \times B$$ for the sets $$A = \{1, 2\}$$ and $$B =\{0, 2\}$$?
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$$R = \{(1, 0), (2, 2)\}$$
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$$R = \{(1, 1), (2, 1)\}$$
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$$R = \{(1, 0), (1, 2)\}$$
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$$R = \{(1, 2), (2, 2)\}$$
Explanation
A relation for the sets $$A$$ and $$B$$ from $$A$$ to $$B$$ will be a subset of the $$A\times B.$$
Now the ordered pair belonging to $$A\times B$$ are
$$\{1,2\}\times\{0,2\}$$ $$=\{(1,0),(1,2),(2,0),(2,2)\}$$
Hence, $$\{(1,1)\}$$ and $$\{(2,1)\}$$ do not belong to $$A\times B$$.
If $$A = \{2, 3\}$$ and $$B = \{1, 2\}$$, find $$A \times B$$.
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$$\{(2, 1), (2, 2), (3, 1), (3, 2)\}$$
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$$\{(2, 1), (2, 1), (3, 1), (3, 2)\}$$
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$$\{(2, 1), (2, 2), (2, 1), (3, 2)\}$$
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$$\{2, 1), (2, 2), (3, 1), (2, 2)\}$$
If $$A \times B =$$ $${(2, 4), (2, a), (2, 5), (1, 4), (1, a), (1, 5)}$$, find $$B$$.
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$$\{4, 2, 5\}$$
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$$\{4, a, 5\}$$
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$$\{4, 1, 5\}$$
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$$\{2, a, 5\}$$
What is the relation for the set $$ A =\{-1, 0, 3\}$$ and $$B =\{1, 2, 3\}$$ ?
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$$R =\{(-1, 1), (1, 2), (3, 3)\}$$
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$$R = \{(-1, 1), (0, 2), (3, 3)\}$$
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$$R = \{(-1, 1), (2, 2), (3, 3)\}$$
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$$R = \{(-1, 1), (0, 2), (-1, 3)\}$$
Explanation
$$A\times B=\{(-1,1),(-1,2),(-1,3),(0,1),(0,2),(0,3),(3,1),(3,2),(3,3)\} $$
R is subset of $$A\times B$$
So option B is ans.
Ordered pairs $$(x, y)$$ and $$(3, 6)$$ are equal if $$x = 3$$ and $$y = ?$$
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$$3$$
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$$6$$
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$$-6$$
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$$-3$$
Explanation
Given
$$(x,y)= (3,6)$$
$$x=3$$
$$y=6$$
The value of $$x=3$$ and $$y=6$$.
Determine whether each of the following relations are reflexive, symmetric and transitive.
Relation R in the set A = {1, 2, 3, 4, 5, 6} as R = {(x, y): y is divisible by x}
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R is reflexive
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R is symmetric
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R is transitive
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None of these
Let $$S=\{(a, b, c) \in N \times N\times N:a+b+c=21, a \le b \le c\}$$ and $$T=\{(a, b, c)\in N\times N\times N: a, b, c\, are\, in\,AP\}$$, where $$N$$ is the set of all natural numbers. Then, the number of elements in the set $$S\cap T$$ is
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$$6$$
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$$7$$
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$$13$$
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$$14$$
Explanation
We have, $$a+b+c=21$$ and $$\dfrac{a+c}{2}=b$$
$$\Rightarrow a+c+\dfrac{a+c}{2}=21$$
$$\Rightarrow a+c=14\Rightarrow \dfrac{a+c}{2}=7$$
$$\Rightarrow b=7$$
So, a can take values from $$1$$ to $$6$$ and
$$c$$ can have values from $$8$$ to $$13$$
$$a=b=c=7$$
$$[\because a\le b \le c]$$
So, there are $$7$$ such triplets.
$$x^2=xy$$ is a relation which is:
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Symmetric
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Reflexive
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Transitive
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All of these
Explanation
$$xRx \Rightarrow x^2=x.x$$ Hence reflexive
$$xRy \Rightarrow x^2=xy. \Rightarrow x=y$$ $$ yRx \Rightarrow y^2=xy \Rightarrow y=x$$ Hence symmetric
$$xRy $$ and $$yRz$$ $$\Rightarrow x^2=xy\Rightarrow x=y. $$ Also $$y^2=yz\Rightarrow y=z$$
Since $$x = z, x.x=z.x \Rightarrow x^2=xz \Rightarrow xRz$$. Hence transitive.
Let $$A = \left \{1, 2, 3\right \}$$. Then number of relations containing $$(1, 2)$$ and $$(1, 3)$$ which are reflexive and symmetric but not transitive is
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$$1$$
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$$2$$
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$$3$$
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$$4$$
Explanation
Total possible pairs$$=\{ (1,1),(1,2),(1,2),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)\} $$
Reflexive means $$(a,a)$$ should be in relation.So $$(1,1),(2,2),(3,3)$$ should be in a relation.
Symmetric means if $$(a,b)$$ is in relation,then $$(b,a)$$ should be in relation.
So,since $$(1,2)$$ is in relation,$$(2,1)$$ should be in relation and since $$(1,3)$$ should be in relation.
Transitive means if $$(a,b)$$ is in relation,and $$(b,c)$$ is in relation,then $$(a,c)$$ is in relation.
If we add $$(2,3)$$,we need to add $$(3,2)$$ for symmetric,but it could become transitive then
Relation $$R1=\{ (1,1),(1,2),(1,2),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)\} $$
So,there is only $$1$$ possible relation.
The relation $$R=\{(1, 1), (2, 2), (3, 3)\}$$ on the set $$\{1, 2, 3\}$$ is
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Symmetric only
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Reflexive only
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An equivalence relation
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transitive only
Explanation
Given Relation
$$R = \{(1,1),(2,2), (3,3)\}$$
Reflexive: If a relation has $$\{(a,b)\}$$ as its element, then it should also have $$\{(a,a), (b,b)\}$$ as its elements too.
Symmetric:
If a relation has $$(a,b)$$ as its element, then it should also have $$\{(b,a)\}$$ as its element too.
Transitive:
If a relation has $$\{(a,b), (b,c)\}$$ as its elements, then it should also have $$\{(a,c)\}$$ as its element too.
Now, the given relation satisfies all these three properties.
Therefore, its an equivalence relation.
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