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CBSE Questions for Class 11 Commerce Applied Mathematics Relations Quiz 13 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Relations
Quiz 13
Let $$g(x)-f(x)=1$$. If $$f(x)+f(1-x)=2\forall x\in R$$, then $$g(x)$$ of symmetrical about?
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The origin
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The line $$x=\dfrac{1}{2}$$
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The point $$(1, 0)$$
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The point $$\left(\dfrac{1}{2}, 0\right)$$
Let $$S=\{1, 2, 3, 4, 5\}$$ and let A$$=S\times S$$. Defined the relation on the R on A as follows (a, b)R(c, d) if an only if ad$$=$$bc. Then R is?
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Reflexive only
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Symmetric only
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Transitive only
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An equivalence
A is a set having $$6$$ distinct elements. The number of distinct function from A to A which are not bijections is?
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$$6!-6$$
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$$6^6-6$$
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$$6^6-6!$$
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$$6!$$
Explanation
Total$$-$$ bijections
$$6^6-6!$$.
N is the set of natural numbers. The relation R is defined on the N$$\times$$N as follows $$_{a\cdot b}R_{c\cdot d}\Leftrightarrow a+d=b+c$$. Then, R is?
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Reflexive only
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Symmetric only
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Transition only
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An equivalence
The relation $$R$$ defined on the set $$A=\left\{ 1,2,3,4,5 \right\} $$ by $$R=\left\{ \left( a,b \right) :\left| { a }^{ 2 }-{ b }^{ 2 } \right| <16 \right\} $$, is not given by
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$$\left\{ \left( 1,1 \right) ,\left( 2,1 \right) ,\left( 3,1 \right) ,\left( 4,1 \right) ,\left( 2,3 \right) \right\} $$
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$$\left\{ \left( 2,2 \right) ,\left( 3,2 \right) ,\left( 4,2 \right) ,\left( 2,4 \right) \right\} $$
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$$\left\{ \left( 3,3 \right) ,\left( 4,3 \right) ,\left( 5,4 \right) ,\left( 3,4 \right) \right\} $$
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none of these
Explanation
Given,
the relation $$R$$ defined on the set $$A=\left\{ 1,2,3,4,5 \right\} $$ by $$R=\left\{ \left( a,b \right) :\left| { a }^{ 2 }-{ b }^{ 2 } \right| <16 \right\} $$.
Then $$R$$ can be written as,
$$R=\{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(3,4),(4,1),(4,2),$$ $$(4,3),(4,4),(4,5),(5,4),(5,5)\}$$.
Let $$A=\left\{ 2,3,4,5,....,17,18 \right\} $$. Let $$\simeq $$ be the equivalence relation on $$A\times A$$, cartesian product of $$A$$ with itself, defined by $$(a,b)\simeq (c,d)$$, iff $$ad=bc$$. The the number of ordered pairs of the equivalence class of $$(3,2)$$ is
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$$4$$
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$$5$$
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$$6$$
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$$7$$
Explanation
Let $$\left(3,2\right)\cong\left(x,y\right)$$
$$\Rightarrow\,3y=2x$$
This is possible in the cases:
$$x=3,\,y=2$$
$$x=6,\,y=4$$
$$x=9,\,y=6$$
$$x=12,\,y=3$$
$$x=15,\,y=10$$
$$x=18,\,y=12$$
Hence total pairs are $$6$$.
The maximum number of equivalence relations on the set $$A=\left\{ 1,2,3 \right\} $$ is
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$$1$$
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$$2$$
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$$3$$
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$$5$$
Explanation
$${R}_{1}=\left\{\left(1,1\right),\left(2,2\right),\left(3,3\right)\right\}$$
$${R}_{2}=\left\{\left(1,1\right),\left(2,2\right),\left(3,3\right),\left(1,2\right),\left(2,1\right)\right\}$$
$${R}_{3}=\left\{\left(1,1\right),\left(2,2\right),\left(3,3\right),\left(1,3\right),\left(3,1\right)\right\}$$
$${R}_{4}=\left\{\left(1,1\right),\left(2,2\right),\left(3,3\right),\left(2,3\right),\left(3,2\right)\right\}$$
$${R}_{5}=\left\{\left(1,1\right),\left(2,2\right),\left(3,3\right),\left(1,2\right),\left(2,1\right),\left(1,3\right),\left(3,1\right),\left(2,3\right),\left(3,2\right)\right\}$$
These are the $$5$$ relations on $$A$$ which are equivalence.
For real number $$x$$ and $$y$$, define $$xRy$$ iff $$x-y+\sqrt{2}$$ is an irrational number. Then the relation $$R$$ is
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reflexive
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symmetric
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transitive
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none of these
Explanation
For each value of x ∈ R, $$x − x + \sqrt 2$$ that is $$\sqrt 2$$ is an irrational number.
It is reflexive.
Let $$x = \sqrt 2$$ and $$y =2$$ then $$x − y + \sqrt 2 = 2 \sqrt 2 – 2$$ which is irrational but when $$y = \sqrt 2$$ and $$x = 2$$, $$x − y + \sqrt 2$$ is not irrational.
It is not symmetric.
Let $$x − y + \sqrt 2$$ is irrational & $$y − z + \sqrt 2$$ is irrational then in above case let $$x = 1; y = \sqrt 2 × 2$$ & $$z = \sqrt 2$$
Hence $$x − z + \sqrt 2$$ is not irrational, so, the relation is not transitive.
If $$A=\left\{ a,b,c,d \right\} $$, then a relation $$R=\left\{ \left( a,b \right) ,\left( b,a \right) ,\left( a,a \right) \right\} $$ on $$A$$ is
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symmetric and transitive only
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reflexive and transitive only
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symmetric only
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transitive only
Explanation
Given if
$$A=\left\{ a,b,c,d \right\} $$, then a relation $$R=\left\{ \left( a,b \right) ,\left( b,a \right) ,\left( a,a \right) \right\} $$.
Since $$(b,b)\not \in R, (c,c)\not \in R$$ so this relation is not reflexive.
The relation is symmetric as if $$(a,b)\in R\Rightarrow (b,a)\in R$$ for all $$a,b\in A$$.
The relation is not transitive also as $$(b,a)\in R,(a,b)\in R$$ but $$(b,b)\not \in R$$.
So the relation is only symmetric.
The number of ordered pairs (a, b) of positive integers such that $$\dfrac{2a - 1}{b}$$ and $$\dfrac{2b - 1}{a}$$ are both integers is
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$$1$$
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$$2$$
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$$3$$
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more than $$3$$
Explanation
The number of ordered pairs $$\left(a, b\right)$$ of positive integers where $$a,b\in{I}^{+}$$
$$\Rightarrow\left(\dfrac{2a-1}{b},\dfrac{2b-1}{a}\right)\in{I}^{+}$$
Let $$\dfrac{2a-1}{b}=1$$
$$\Rightarrow\,2a-1=b$$
and
$$\Rightarrow\,\dfrac{2b-1}{a}=\dfrac{2\left(2a-1\right)-1}{a}$$
$$=\dfrac{4a-2-1}{a}$$
$$=\dfrac{4a-3}{a}$$
$$\therefore \dfrac{2b-1}{a}=4-\dfrac{3}{a}$$
For integer,$$a=1,3$$ we get
$$b=2a-1=2-1=1$$ for $$a=1$$
$$b=2a-1=2\times 3-1=6-1=5$$
$$\therefore\,b=1,5$$
So total sets are $$\left(1,1\right),\left(3,5\right),\left(5,3\right)$$
Hence there are totally $$3$$ sets.
If $$A=\{2, 4, 5\}, B=\{7, 8, 9\}$$ 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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$$0$$
Let Z be the set of all integers and let R be a relation on Z defined by $$a$$ R $$b\Leftrightarrow (a-b)$$ is divisible by $$3$$. Then, R is?
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Reflexive and symmetric but not transitive
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Reflexive and transitive but not symmetric
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Symmetric and transitive but not reflexive
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An equivalence relation
Explanation
Let R={(a,b):a,b∈Z and (a−b) is divisible by 3}.
Show that R is an equivalence relation on Z.
Given:
R={(a,b):a,b∈Z and (a−b) is divisible by 3}.
$$R = (a,b)$$
$$(a-b)$$ is divisible by $$3$$
Reflexive
$$(a,a) \Rightarrow (a-a)$$ is divisible by $$3$$
Symmetric
$$(a, b)\Leftrightarrow (b, a)$$
$$(a-b)$$ is divisible by $$3$$
$$(b-a)$$ is divisible by $$3$$
Transitive
$$(a,b), (b,c)\Leftrightarrow (a,c)$$
$$(a-b)$$ is divisible by $$3$$
$$(b-c)$$ is divisible by $$3$$
$$(a-c)$$ is divisible by $$3$$
R is an equivalent relation on Z
Let S be the set of all triangles in a plane and let R be a relation on S defined by $$\Delta_1S\Delta_2\Leftrightarrow \Delta_1\equiv \Delta_2$$. Then, R is?
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Reflexive and symmetric but not transitive
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Reflexive and transitive but not symmetric
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Symmetric and transitive but not reflexive
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An equivalence relation
Let R be a relation on $$N\times N$$, defined by $$(a, b)$$ R $$(c, d)\Leftrightarrow a+d=b+c$$. Then, R is?
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Reflexive and symmetric but not transitive
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Reflexive and transitive but not symmetric
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Symmetric and transitive but not reflexive
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An equivalence relation
Let A be the set of all points in a plane and let O be the origin. Let $$R=\{(P, Q):OP=OQ\}$$. Then, R is?
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Reflexive and symmetric but not transitive
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Reflexive and transitive but not symmetric
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Symmetric and transitive but not reflexive
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An equivalence relation
Let us define a relation $$R$$ in $$R$$ as $$aRb$$ if $$a \ge b$$. Then $$R$$ is
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an equivalence relation
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reflexive, transitive but not symmetric
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symmetric, transitive but
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neither transitive nor reflexive but symmetric
Explanation
Given that $$aRb$$ is $$a \ge b$$
$$\Rightarrow aRa \Rightarrow a\ge a$$ which is true
let $$aRb, a\ge b$$, theb $$b \ge a$$ which is not true $$R$$ is not symmetric.
But $$aRb$$ and $$bRc$$
$$\Rightarrow a\ge b$$ and $$b\ge c$$
$$\Rightarrow a\ge c$$
Hence, $$R$$ is transitive.
Every relation which is symmetric and transitive is also reflexive.
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True
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False
Explanation
False
Let $$R$$ be a relation defined by
$$R=\left\{(1,2),(2,1),(1,1),(2,2)\right\}$$ on the set $$A=\left\{1,2,3\right\}$$
It is clear that $$(3,3)\in R$$. So, it is not reflexive.
An integer $$m$$ is said to be related to another integer, $$n$$ if $$m$$ is a integral multiple of $$n$$. This relation in $$Z$$ is reflexive, symmetric and transitive.
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True
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False
Explanation
False
The given relation is reflexive and transitive but not symmetric.
If $$a$$ relation $$R$$ on the set $$\left\{1,2,3\right\}$$ be defined by $$R=\left\{(1,2)\right\}$$, then $$R$$ is
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reflexive
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transitive
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symmetric
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none of these
Explanation
$$R$$ on the set $$\left\{1,2,3\right\}$$ be defined by $$R=\left\{(1,2)\right\}$$
It is clear that $$R$$ is transitive.
a homogeneous relation R over a set X is transitive if for all elements a,b,c in X , whenever R relates a to b and b to c, then R also relates a to c.
Let $$R=\left\{(3,1),(1,3),(3,3)\right\}$$ be a relation defined on the set $$A=\left\{1,2,3\right\}$$. Then $$R$$ is symmetric, transitive but not reflexive.
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True
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False
Explanation
False
Given that, $$R=\left\{(3,1),(1,3),(3,3)\right\}$$ be defined on the set $$A=\left\{1,2,3\right\}$$.
$$(1,1)\in R$$
So, $$R$$ is not reflexive $$(3,1)\in R, (1,3)\in R$$
Hence, $$R$$ is symmetric.
Since $$(3,1)\in R , (1,3)\in R$$
But $$(1,1)\in R$$
Hence, $$R$$ is not transitive.
Consider the set $$A=\left\{1,2,3\right\}$$ and the relation $$R=\left\{(1,2),(1,3)\right\} . R$$ is a transitive relation.
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True
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False
Explanation
Here, $$A={1,2,3}$$
$$R={(1,1),(2,2),(3,3),(1,2),(2,3)}$$
As $$R$$ contains $$(1,1),(2,2),(3,3),$$ it is reflexive.
R contains $$(1,2),(2,3)$$ but does not contain $$(2,1),(3,2)$$, so it is not symmetric.
R contains $$(1,2),(2,3)$$ but does not contain $$(1,3),$$ so it is not transitive.
The relation $$R$$ on the set $$A=\left\{1,2,3\right\}$$ defined as $$R\left\{(1,1),(1,2),(2,1),(3,3)\right\}$$ is reflexive, symmetric and transitive.
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True
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False
Explanation
False
Given that, $$R\left\{(1,1),(1,2),(2,1),(3,3)\right\}$$
$$(2,2)\in R$$
So, $$R$$ is not reflexive.
Let $$R$$ be a relation from $$A$$ to $$A$$ defined by $$R=\left\{ (a, b): a, b \in N\ and\ a=b^2 \right\}$$
Is the following true?
$$(a, b) \in R$$, implies $$(b,a ) \in R$$
Justify your answer.
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True
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False
Explanation
Given $$R=\left\{ (a, b):a, b \in N\ and\ a=b^2 \right\}=\left\{ (1, 1), (2, 4), (3, 9), (4, 16 ), ...\right\}$$
$$(a, a) \in R$$ implies $$(b,a )\in R$$ is not true, because $$(2, 4)\in R$$ but $$(4, 2) \notin R$$
Let $$R$$ be a relation from $$A$$ to $$A$$ defined by $$R=\left\{ (a, b): a, b \in N\ and\ a=b^2 \right\}$$
Is the following true?
$$(a, a) \in R$$ for all $$a \in A$$
Justify your answer.
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True
0%
False
Explanation
Given $$R=\left\{ (a, b):a, b \in N\ and\ a=b^2 \right\}=\left\{ (1, 1), (2, 4), (3, 9), (4, 16 ), ...\right\}$$
$$(a, a) \in R$$ for all $$a \in N$$ is not true because $$(2, 2)\notin R$$.
Let $$R$$ be a relation from $$A$$ to $$A$$ defined by $$R=\left\{ (a, b): a, b \in N\ and\ a=b^2 \right\}$$
Is the following true?
$$(a, b)\in R, (b, c)\in R$$, implies $$(a, c) \in R$$.
Justify your answer.
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True
0%
False
Explanation
Given $$R=\left\{ (a, b):a, b \in N\ and\ a=b^2 \right\}=\left\{ (1, 1), (2, 4), (3, 9), (4, 16 ), ...\right\}$$
$$(a, a) \in R, (b, c) \in R$$ implies $$(a,c )\in R$$ is not true, because $$(2, 4)$$ and $$(4, 16) \in R$$ but $$(2, 16)\notin R$$.
If A = { a , b , c , d} and B = { p , q ,r ,s} then relation from A and B is
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{(a ,p) , ( b, r), (c , r)}
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{(a , p), (p, q) , (c, r) , ( s , d)}
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{{ b , a) , (q , b ) , (c , r) }
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{ ( c,s) , ( d , s ,) ( r, a) , ( q , b )}
Explanation
Because $$ { ( a , p), ( b ,r) ( c ,r)} \subseteq A \times B $$
If $$A=\left \{ 1,2,3 \right \} $$ and $$B=\left \{ 4,5,6 \right \}$$ then which of the following sets are relation from $$A$$ to $$B$$
(i) $$\displaystyle R_{1}=\left \{ (4,2) (2,6)(5,1)(2,4)\right \}$$
(ii) $$\displaystyle R_{2}=\left \{ (1,4) (1,5)(3,6)(2,6) (3,4)\right \}$$
(iii) $$\displaystyle R_{3}=\left \{ (1,5) (2,4)(3,6)\right \}$$
(iv) $$\displaystyle R_{4}=\left \{ (1,4) (1,5)(1,6)\right \}$$
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$$\displaystyle R_{1},R_{2},R_{3}$$
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$$\displaystyle R_{1},R_{3},R_{4}$$
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$$\displaystyle R_{2},R_{3},R_{4}$$
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$$\displaystyle R_{1},R_{2},R_{3},R_{4}$$
Explanation
A relation from $$A$$ to $$B$$ will include elements from $$A\times B$$
Elements of $$A\times B$$ are
$$\{(1,4), (1,5) ,(1,6), (2,4), (2,5), (2,6), (3,4), (3,5), (3,6)\}$$
Now the relation $$R_{1}$$ consists of elements form both $$A\times B$$ and $$B\times A$$.
Thus $$R_{1}$$ is not a relation from $$A$$ to $$B.$$
Let R be the relation in the set {1, 2, 3, 4} given by:
R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}. 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
Let A = {1, 2, 3}. Then number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is
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1
0%
2
0%
3
0%
4
If two sets $$A$$ and $$B$$ have $$ 99$$ elements in common, then the number of elements common to each of the sets $$A \times B$$ and $$B \times A$$ are
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$$2^{99}$$
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$$99^{2}$$
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$$100$$
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$$18$$
Explanation
To find $$A \times B$$ we take one element from set $$A$$ and one from set $$B$$.
Given that $$99$$ elements are common to both set $$A$$ and set $$B$$.
Suppose these common elements are $$N_1, N_2, N_3,...N_{99}$$.
Select an ordered pair for $$A \times B$$ such that both are selected out of these common elements. Examples: $$\left(N_1,N_2 \right),\; \left(N_3,N_5 \right)$$
All these will also be elements of $$B \times A$$. Hence number of elements common to $$A \times B$$ and $$B \times A$$ is $$99 \times 99 = 99^2$$ ( first element in ordered pair can be selected in 99 ways; second element can also be selected in 99 ways)
$$n\left [ \left ( A\times B \right )\cap \left ( B\times A \right ) \right ]=n\left [ \left ( A\cap B \right )\cap \left ( B\cap A \right ) \right ]=(99)(99)=99^{2}$$
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