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CBSE Questions for Class 11 Commerce Applied Mathematics Relations Quiz 9 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Relations
Quiz 9
Determine all ordered pairs that satisfy $$(x - y)^{2} + x^{2} = 25$$, where $$x$$ and $$y$$ are integers and $$x \geq 0$$. Find the number of different values of $$y$$ that occur
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$$3$$
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$$4$$
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$$5$$
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$$6$$
Explanation
$${ \left( x-y \right) }^{ 2 }+{ x }^{ 2 }=25$$
Now, $${ 3 }^{ 2 }+{ 4 }^{ 2 }={ 5 }^{ 2 }$$
$$9+16=25$$
$$\therefore $$There are 2 possibilities:
$$I.{ \left( x-y \right) }^{ 2 }=9$$ and $${ x }^{ 2 }=16$$
$$\therefore x=\pm 4$$ and $$x-y=\pm 3$$
$$\left( i \right) .x-y=3\Rightarrow \left( 4,1 \right) $$ and $$\left( -4,-7 \right) $$
$$\left( ii \right) .x-y=-3\Rightarrow \left( 4,7 \right) $$ and $$\left( -4,-1 \right) $$
$$II.{ \left( x-y \right) }^{ 2 }=16$$ and $${ x }^{ 2 }=9$$
$$\therefore x=\pm 3$$ and $$x-y=\pm 4$$
$$\left( i \right) .x-y=4\Rightarrow \left( 3,-1 \right) $$ and $$\left( -3,-7 \right) $$
$$\left( ii \right) .x-y=-4\Rightarrow \left( 3,7 \right) $$ and $$\left( -3,-1 \right) $$
$$\therefore $$ Different values of y are $$1,-1,7,-7$$
$$\therefore 4$$ different values of y occur.
Let $$R = \{(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)\}$$ be a relation on the set $$A = \{1, 2, 3, 4\}$$. The relation $$R$$ is
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A function
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Transitive
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Not symmetric
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Reflexive
Explanation
Let $$R = \{(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)\}$$ be a relation on the set $$A = \{1, 2, 3, 4\}$$, then
(a) Since $$(2, 4)$$ $$\in$$ $$R$$ and $$(2, 3)$$ $$\in$$ $$R$$, so $$R$$ is not a function.
(b) Since $$(1, 3)$$ $$\in$$ $$R$$ and $$(3, 1) $$ $$\notin$$ $$R$$ but $$(1, 1)$$ $$\notin$$ $$R$$, so $$R$$ is not transitive:
(c) Since $$(2, 3)$$ $$\in$$ $$R$$ but $$(3, 2)$$ $$\notin$$ $$'R$$, so $$R$$ is not symmetric.
(d) Since $$(1, 1)$$ $$\notin$$ $$R$$, so $$R$$ is not reflexive.
Hence, option C is correct.
For any two real numbers $$\theta$$ $$\phi $$, we define $$\theta R \phi $$ if and only if $$\sec ^{ 2 }{ \theta } -\tan ^{ 2 }{ \phi } =1$$. The relation $$R$$ is
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Reflexive but not transitive
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Symmetric but not reflexive
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Both reflexive and symmetric but not transitive
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An equivalence relation
Explanation
Given ,$$\theta$$ R$$\phi$$ where $$\theta$$ and $$\phi$$ are two real numbers.
$$\sec^{2} { \theta } -\tan ^{2} { \phi } =1$$
$$\theta=\phi$$
R is reflexive and symmetric.
R is not transitive since there are only two numbers.
The relation $$R$$ defined on set $$A = \left \{x :|x| < 3, x\epsilon I\right \}$$ by $$R = \left \{(x, y) : y = |x|\right \}$$ is
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$$\left \{(-2, 2), (-1, 1), (0, 0), (1, 1), (2, 2)\right \}$$
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$$\left \{(-2, 2), (-2, 2), (-1, 1), (0, 0), (1, -2), (2, -1), (2, -2)\right \}$$
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$$\left \{(0, 0), (1, 1), (2, 2)\right \}$$
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None of the above
Explanation
$$R$$ defined on set $$A = \left \{x :|x| < 3, x\epsilon I\right \}$$ by $$R = \left \{(x, y) : y = |x|\right \}$$
$$A = \left \{x :|x| < 3, x\epsilon I\right \}$$
$$=\left\{x:-3<x<3, x\epsilon I\right\}$$
$$\therefore$$
$$R = \left \{(x, y) : y = |x|\right \}$$
$$=\left\{(-2,2),(-1,1),(0,0),(1,1),(2,2)\right\}$$
Hence, option A is correct.
Let the number of elements of the sets $$A$$ and $$B$$ be $$p$$ and $$q$$ respectively. Then the number of relations from the set $$A$$ to the set $$B$$ is
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$${ 2 }^{ p+q }$$
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$${ 2 }^{ pq }$$
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$$p+q$$
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$$pq$$
Explanation
Given $$A$$ and $$B$$ are two sets with number of elements $$p$$ and $$q$$ respectively.
The cartesian product of $$A$$ and $$B = A \times B = \{(a,b): (a\in A)$$ and $$(b\in B)\}$$
Number of elements in $$A \times B = |A\times B| =|A|.|B| = pq$$
Any relation from $$A$$ to $$B$$ is a subset of $$A\times B$$.
Hence number of relations from $$A$$ to $$B$$ is the number of subsets of $$A\times B$$
$$= 2^{|A\times B|} = 2^{pq}$$
A relation $$\rho$$ on the set of real number $$R$$ is defined as follows:
$$x\rho y$$ if any only if $$xy > 0$$. Then which of the following is/are true?
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$$\rho$$ is reflexive and symmetric
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$$\rho$$ is symmetric but not reflexive
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$$\rho$$ is symmetric and transitive
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$$\rho$$ is an equivalence relation
Explanation
$$x\rho x : x^{2} > 0$$ not true for $$x = 0$$
$$x\rho y\Rightarrow y\rho x$$
$$x\rho y$$ so $$xy > 0, y\rho z$$ so $$yz > 0$$.
$$xzy^{2} > 0$$, Hence $$xz > 0$$ so $$x\rho z$$.
If $$N$$ denote the set of all natural numbers and $$R$$ be the relation on $$N\times N$$ defined by $$(a, b)R(c, d)$$. if $$ad(b + c) = bc (a + d)$$, then $$R$$ is
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Symmetric only
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Reflexive only
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Transitive only
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An equivalence relation
Explanation
For $$(a, b), (c, d)\epsilon N\times N$$
$$(a, b)R(c, d)$$
$$\Rightarrow ad(b + c) = bc(a + d)$$
Reflexive: $$ab(b + a) = ba(a + b), \forall ab\epsilon N$$
$$\therefore (a, b)R(a, b)$$
So, $$R$$ is reflexive,
Symmetric: $$ad(b + c) = bc(a + d)$$
$$\Rightarrow bc(a + d) = ad(b + c)$$
$$\Rightarrow cd(d + a) = da(c + b)$$
$$\Rightarrow (c, d)R(a, b)$$
So, $$R$$ is symmetric.
Transitive: For $$(a, b), (c, d), (e, f)\epsilon N\times N$$
Let $$(a, b)R(c, d),\ (c, d) R(e, f)$$
$$\therefore ad(b + c) = bc(a+ d)\ and\ cf(d + e) = de(c + f)$$
$$\Rightarrow adb + adc = bca + bcd .... (i)$$
and $$cfd + cfe = dec + def ... (ii)$$
On multiplying eq. (i) by $$ef$$ and eq. (ii) by $$ab$$ a
nd then adding, we have
$$adbef + adcef + cfdab + cfeab$$
$$= bcaef + bcdef + decab + defab$$
$$\Rightarrow adcf (b + e) = bcde (a + f)$$
$$\Rightarrow af (b + e) = be (a + f)$$
$$\Rightarrow (a, b)R(e, f)$$
So, $$R$$ is transitive.
Hence, $$R$$ is an equivalence relation.
If $$p - q > 0$$, which of the following is true?
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If q = 0, then p < 0
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If q < 0, then p < 0
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If p > 0, then q > 0
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If q = 0, then p > 0
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If q < 1, then p > 1
Explanation
Given $$p-q>0$$
If $$q=0$$ , then $$p-0>0$$ , which gives $$p>0$$
Therefore option $$D$$ is correct
Let $$R$$ be a relation defined on the set $$Z$$ of all integers and $$xRy$$ when $$x + 2y$$ is divisible by $$3$$. Then
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$$R$$ is not transitive
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$$R$$ is symmetric only
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$$R$$ is an equivalence relation
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$$R$$ is not an equivalence relation
Explanation
$$aRa$$ is positive as $$a+2a=3a$$ is divisible by $$3$$
Hence reflexive
If $$aRb$$ is positive
$$a+2b$$ is divisible by $$3$$
Hence $$3a+3b-(2a+b)$$ is divisible by $$3$$
$$\Rightarrow 2a+b$$ is divisible by $$3$$
ie
$$aRb$$ is positive
$$\Rightarrow bRa$$ is positive hence symmetric
$$aRb,bRc$$ is $$a+2b,b+2c$$
divisible by $$3$$
$$\Rightarrow a+2b+b+2c$$
divisible by $$3$$
$$\Rightarrow a+3b+2c$$
divisible by $$3$$
So $$a+2c$$
divisible by $$3$$ hence transitive
So it is an equivalence relation.
The relation R define on the set of natural numbers as {(a, b) : a differs from b by 3} is given.
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$$\{(1, 4), (2, 5), (3, 6), ........\}$$
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$$\{(4, 1), (5, 2), (6, 3), ........\}$$
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$$\{(1, 3), (2, 6), (3, 9), ........\}$$
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None of above
Explanation
Let $$R=\{(a,b): a, b \in N, a-b=3\}$$
$$=\{(n+3,n):n \in N\}$$
$$=\{(4,1), (5, 2), (6, 3).......\}$$
Hence, option B is correct.
A relation R in N is defined such that $$ xRy \Leftrightarrow x + 4y = 16 ,$$ then the range of R is
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{ 1 , 2 , 4}
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{ 1 , 3 , 4}
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{ 1 , 2 , 3}
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{ 2 , 3 , 4}
Rule from of relation {(1 ,2) , ( 2 , 5) , (3 , 10) , ( 4 , 17), ......} in N
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$$ {(x ,y) : x , y \in N = y = 2x + 1 } $$
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$$ {( x , y ) : x , y \in N , y = x^{2} + 1 } $$
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$$ {( x , y ) : x , y \in N , y = 3x - 1} $$
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$$ {( x , y ) : x , y \in N , y = x + 3} $$
The relation 'has the same father as' over the set of children is:
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Only reflective
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Only symmetric
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Only transitive
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An equivalence relation
Explanation
Let this relation be called as $$R$$.
Then, if $$xRx$$ then clearly $$xRx$$ for all $$x$$. Also $$xRy$$ it means $$yRx$$
If $$xRy, yRz$$ it means $$x$$ has same father as $$y$$ and $$y$$ has same father as $$z$$.
So we can say that, $$x$$ has same father as $$z$$
Thus, $$xRz$$.
Thus, the given relation is reflexive, symmetric and transitive.
Hence, the relation is an equivalence relation.
Option D is correct.
Let $$X$$ be the set of all persons living in a city. Persons $$x, y$$ in $$X$$ are said to be related as $$x < y$$ if $$y$$ is at least $$5$$ years older than $$x$$. Which one of the following is correct?
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The relation is an equivalence relation on $$X$$
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The relation is transitive but neither reflexive nor symmetric
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The relation is reflexive but neither nor symmetric
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The relation is symmetric but neither transitive nor reflexive
Explanation
Given, $$X$$ is the set of persons living ins a city.
Relation(R) = {(x,y): x,y $$\in$$ $$R$$ $$\&$$ x<y if y is atleast $$5$$ years older than x}
For Reflexive: $$(x,x)$$ should $$\in R$$ for all $$x \in X $$
now, $$x$$ cannot be be $$5$$ years older than himself. So the relation is not reflexive.
For Symmetric: If $$(x,y) \in R \Rightarrow (y,x) \in R$$
$$(x,y) \in R \Rightarrow y $$ is at least $$5$$ years older than $$x$$.
$$(y,x) \in R \Rightarrow x$$ is at least $$5$$ years older than $$y$$. This contradicts the above statement. Hence the relation is not symmetric
For Transitive: If $$(x,y) \in R$$ and $$(y,z) \in R \Rightarrow (x,z) \in R$$.
$$(x,y) \in R \Rightarrow y$$ is at least $$5$$ years older than $$x$$.
$$(y,z) \in R \Rightarrow z$$ is at least $$5$$ years older than $$y$$.
Then, $$(x,z) \in R \Rightarrow z $$ is at least $$5$$ years older than $$x$$.
Since, $$z $$ is at least $$10 $$ years older than $$x$$. The relation is transitive.
If A and B are two non-empty sets having n elements in common, then what is the number of common elements in the sets $$A\times B$$ and $$B\times A$$?
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$$n$$
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$$n^2$$
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$$2n$$
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Zero
Explanation
Say A has x elements and B has y elements in total.
Their cartesian product $$A\times B$$ will have $$x\times y$$ elements.
Hence if they have n elements in common. $$n^2$$ common elements are present in the products $$A\times B$$ and $$B\times A$$
Let $$A=\left\{ x\in W,the\quad set\quad of\quad whole\quad numbers\quad and\quad x<3 \right\} $$
$$B=\left\{ x\in N,the\quad set\quad of\quad natural\quad numbers\quad and\quad 2\le x<4 \right\} $$ and $$C=\left\{ 3,4 \right\} $$, then how many elements will $$\left( A\cup B \right) \times C$$ conatin?
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$$6$$
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$$8$$
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$$10$$
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$$12$$
Explanation
$$A=\{0,1,2\}, B=\{2,3\}$$ and $$C=\{3,4\}$$
$$A \cup B=\{0,1,2,3\}$$
No. of elements in $$(A\cup B)\times C$$
$$=$$ No. of elements in $$(A \cup B)\times $$ No. of elements in $$C$$
$$=4\times 2=8$$
If $$A = \left\{ 1,2 \right\}$$, $$B = \left\{ 2,3 \right\}$$ and $$ C = \left\{ 3,4 \right\}$$, then what is the cardinality of $$ \left( A\times B \right) \cap \left( A\times C \right) $$
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$$8$$
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$$6$$
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$$2$$
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$$1$$
Explanation
$$A = \left\{ 1,2 \right\}$$, $$B = \left\{ 2,3 \right\}$$ and $$ C = \left\{ 3,4 \right\}$$
Now, $$A\times B=\{ (1,2),(1,3),(2,2),(2,3)\}$$
$$A\times C=\{ (1,3),(1,4),(2,3),(2,4)\} $$ And
$$ (A\times B)\cap (A\times C)=\{ (1,3),(2,3)\} $$
So cardinality is $$2$$.
Hence, option C is correct.
Let $$A = \left\{ a,b,c,d \right\}$$ and $$ B=\left\{ x,y,z \right\}$$. What is the number of elements in $$ A\times B$$?
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$$6$$
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$$7$$
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$$12$$
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$$64$$
Explanation
Given sets are $$A=\{a,b,c,d\}$$ and $$\{x,y,z\}$$
So $$A\times B=\{(a,x),(a,y),(a,z),(b,x),(b,y),(b,z),(c,x),(c,y),(c,z),(d,x),(d,y),(d,z)\}$$
No. of elements in $$A\times B$$ is $$3\times 4=12$$
Let $$Z$$ be the set of integers and $$aRb$$, where $$a, b\epsilon Z$$ if an only if $$(a - b)$$ is divisible by $$5$$.
Consider the following statements:
$$1.$$ The relation $$R$$ partitions $$Z$$ into five equivalent classes.
$$2.$$ Any two equivalent classes are either equal or disjoint.
Which of the above statements is/are correct?
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$$1$$ only
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$$2$$ only
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Both $$1$$ and $$2$$
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Neither $$1$$ nor $$2$$
Explanation
A relation is defined on $$Z$$ such that $$aRb \Rightarrow (a-b)$$ is divisible by $$5$$,
For Reflexive: $$(a,a) \in R$$.
Since, $$(a-a) = 0$$ is divisible by $$5$$. Therefore, the realtion is reflexive.
For symmetric: If $$(a,b) \in R \Rightarrow (b,a) \in R$$.
$$(a,b) \in R \Rightarrow (a-b)$$ is divisible by $$5$$.
Now, $$(b-a) =-(a-b)$$ is also divisible by $$5$$. Therefore, $$(b,a) \in R$$
Hence, the relation is symmetric.
For Transitive: If $$(a,b) \in R $$ and $$ (b,a) \in R \Rightarrow (a,c) \in R$$.
$$(a,b) \in R \Rightarrow (a-b) $$ is divisible by $$5$$.
$$(b,c) \in R \Rightarrow (b-c)$$ is divisible by $$5$$.
Then $$(a-c)=(a-b+b-c)= (a-b)+(b-c) $$ is also divisible by $$5$$. Therefore, $$(a,c) \in R$$.
Hence, the relation is transitive.
There, the relation is equivalent.
Now, depending upon the remainder obtained when dividing $$(a-b)$$ by $$5$$ we can divide the set $$Z$$ iinto $$5$$ equivalent classes and they are disjoint i.e., there are no common elements between any two classes.
Therefore, the correct option is $$(C)$$.
If $$A$$ is a finite set having $$n$$ elements, then the number of relations which can be defined in $$A$$ is
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$${ 2 }^{ n }$$
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$${ n }^{ 2 }$$
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$${ 2 }^{ { n }^{ 2 } }$$
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$${ n }^{ n }$$
Explanation
A relation is simply a subset of cartesian product $$ A\times A$$.
If $$ {A\times A}= [(a_1,a_1), (a_1,a_2),.....(a_1,a_n), $$
$$ (a_2,a_1),(a_2,a_2),........(a_2,a_n) $$
$$ ...... $$
$$ (a_n,a_1), (a_n,a_2).........(a_n,a_n)]$$
We can select first element of ordered pair in $$n$$ ways and second element in $$n$$ ways.
So, clearly this set of ordered pairs contain $$n^2 $$ pairs.
Now, each of these $$n^{2}$$ ordered pairs can be present in the relation or can't be. So, there are $$2$$ possibilities for each of the $$n^{2}$$ ordered pairs.
Thus, the total no. of relations is $$2^{n^{2}}$$.
Hence, option C is correct.
A and B are two sets having $$3$$ elements in common. If $$n(A)=5, n(B)=4$$, then what is $$n(A\times B)$$ equal to?
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$$0$$
0%
$$9$$
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$$15$$
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$$20$$
Explanation
If $$n(A) =5$$ and $$n(B) = 4$$
For this type cases we know that the formula for the no. of elements in $$n(A\times B)$$ = $$5\times4 = 20$$
If A = { a ,b , c} , then numbers of possible non zero relations in A is
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$$ 511$$
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$$ 512 $$
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$$ 7 $$
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$$ 8 $$
Let R be a relation from A = {1, 2, 3, 4} to B = {1, 3, 5} such that
R = [(a, b) : a < b, where a $$\varepsilon$$ A and b $$\varepsilon$$ B].
What is RoR$$^{-1}$$ equal to?
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$${(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)}$$
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$${(3, 1), (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)}$$
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$${(3, 3), (3, 5), (5, 3), (5, 5)}$$
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$${(3, 3), (3, 4), (4, 5)}$$
Explanation
$$R$$ gives the set $$\{(1,3),(1,5),(2,3),(2,5),(3,5),(4,5)\}$$
$$R^{-1}$$ gives the set $$\{(3,1),(5,1),(3,2),(5,2),(5,3),(5,4),(3,3)\}$$
To compute $$RoR^{-1}$$, we pick 1 element from $$R^{-1}$$ and its corresponding relation from $$R$$.
eg: $$(3,1)\in R^{-1}$$ and $$(1,3)\in R\implies (3,3)\in RoR^{-1}$$
Similarly, computing for all such pairs, we have
$$RoR^{-1}=\{(3,3),(3,5),(5,3),(5,5)\}$$
Let $$ A = \{ 1,2,3,4 \} $$ and $$R$$ be a relation in $$A$$ given by $$ R = \{ (1,1) , (2,2) (3,3) , (4,4) , (1,2) , (3,1) , (1,3) \} $$ then $$R$$ is :
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Reflexive and transitive only
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Transitive and symmetric only
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An equivalence relation
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None of the above
Explanation
$$ A = \{ 1, 2, 3, 4\} $$
$$ R = \{ (1,1) , (2,2),(3,3),(4,4),(1,2),(3,1)(1,3) \} $$
clearly $$ (1,1) \epsilon R, (1,2) \epsilon (2,2) \epsilon R , (3,3) \epsilon R, (4,4) \epsilon R $$
$$ \Rightarrow R $$ is Reflective.
also $$ (1,1) \epsilon R, (1,2) \epsilon R \Rightarrow (1,2) R $$
and $$(1,2) \epsilon R, (1,3) \epsilon R \Rightarrow (1,3) \epsilon R \Rightarrow R $$ is transitive
But for $$ (1,2) \epsilon R (2,1) \notin R, \Rightarrow R $$ is not symmetric
$$ \Rightarrow R $$ is reflective and transitive only.
How many ordered pairs of (m, n) integers satisfy $$\displaystyle\frac{m}{12}=\frac{12}{n}$$?
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$$30$$
0%
$$15$$
0%
$$12$$
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$$10$$
Explanation
$$m \times n = 144$$
144 can be written in multiplication of two integers in the following ways:
$$1 \times 144 = 2 \times 72 = 3 \times 48 = 4 \times 36 = 6 \times 24 = 8 \times 18 = 9 \times 16 = 12 \times 12$$
For all the factors except $$(12, 12)$$ there would be 14 ordered pairs.
Adding one more, yields 15 possibilities
If $${ y }^{ 2 }={ x }^{ 2 }-x+1$$ and $$\quad { I }_{ n }=\int { \cfrac { { x }^{ n } }{ y } } dx$$ and $$A{ I }_{ 3 }+B{ I }_{ 2 }+C{ I }_{ 1 }={ x }^{ 2 }y$$ then ordered triplet $$A,B,C$$ is
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$$\quad \left( \cfrac { 1 }{ 2 } ,-\cfrac { 1 }{ 2 } ,1 \right) $$
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$$\left( 3,1,0 \right) $$
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$$\left( 1,-1,2 \right) $$
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$$\left( 3,-\cfrac { 5 }{ 2 } ,2 \right) $$
Explanation
We have given
$${y^2} = {x^2} - x + 1$$
$${I_n} =\displaystyle \int {\dfrac{{{x^n}}}{y}dx} $$
And,
$$A{I_3} + B{I_2} + C{I_1} = {x^2}y$$
Order triplet $$A,\, B,\,C$$ $$ = \left( {\dfrac{1}{2},\,\dfrac{{ - 1}}{2},1} \right)$$
Hence, the option $$(A)$$ is correct.
The number of ordered pairs $$(x, y)$$ of real numbers that satisfy the simultaneous equations.
$$x + y^{2} = x^{2} + y = 12$$ is.
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$$0$$
0%
$$1$$
0%
$$2$$
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$$4$$
Explanation
$$x + y^{2} = x^{2} + y = 12$$
$$x + y^{2} = x^{2} + y$$
$$x - y = x^{2} - y^{2} \Rightarrow x = y, x + y = 1$$
when $$x = y, x^{2} + x = 12$$
$$x^{2} + x - 12 = 0$$
$$x^{2} + 4x - 3x - 12 = 0$$
$$(x + 4)(x - 3) = 0$$
$$x = -4, 3$$
$$(3, 3),(-4, -4)$$
When $$y = 1 - x$$
$$x + (1 - x)^{2} = 12$$
$$x^{2} - x - 11 = 0$$
$$x = \dfrac {1\pm \sqrt {1 + 44}}{2}$$ for two value of $$x$$ there are two value of $$y$$.
So four pair.
$$R$$ is a relation on $$N$$ given by $$R = \left \{(x, y)|4x + 3y = 20\right \}$$. Which of the following doesnot belong to $$R$$?
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$$(-4, 12)$$
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$$(5, 0)$$
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$$(3, 4)$$
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$$(2, 4)$$
Explanation
$$R$$ is given by $$\{(x,y)|4x+3y=20\}$$
$$4 \times (-4)+3 \times 12=-16+36=20$$
$$4 \times 5+3 \times 0=20+0=20$$
$$4 \times 3+3 \times 4=12+12=24\neq 20$$
$$4 \times 2+3 \times 4=8+12=20$$
So, among the given options $$C$$ does not satisfy the given condition.
All except $$C$$, does not belong to the set $$R.$$
Hence, option C is correct.
If $$ n (A) = 5 $$ and $$ n (B) = 7 , $$ then the number of relations on $$ A \times B $$ is :
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$$ 2^{35} $$
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$$ 2^{49} $$
0%
$$ 2^{25} $$
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$$ 2^{70} $$
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$$ 2^{35 \times35 } $$
Explanation
Given, $$ n (A) = 5 $$ and $$ n (B) = 7 $$
Therefore, number of relations on
$$ A \times B = 2^{[ n(A) \times n(B) ]} $$
$$ = 2^{(5 \times 7)} = 2^{(35)} $$
The number of ordered pairs $$\left( m,n \right)$$, where $$ m, n \in \left\{ 1,2,3,\dots ,50 \right\}$$, such that $$ { 6 }^{ m }+{ 9 }^{ n }$$ is a multiple of $$5$$ is
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$$1250$$
0%
$$2500$$
0%
$$625$$
0%
$$500$$
Explanation
For a number to be divisible by $$5$$ the last digit shall be a multiple of $$5$$
Any power of 6 has its last digit(or unit's digit) equal to 6.
The power of $$9$$ of the form
$$4p+1$$ has it's last digit as 9
$$4p+2$$ has it's last digit as 1
$$4p+3$$ has it's last digit as 9
$$4p$$ has it's last digit as 1
Thus $$6^{m}+9^{n}$$ will be a multiple of $$5$$ only if $$m$$ is any number from the given set and $$n$$ is a number of the form $$4p+1$$ or $$4p+3$$.
Therefore the number of ways to select $$m$$ is $$\binom{50}{1}$$ and the number of ways to select $$n$$ from the given set is $$\binom{25}{1}$$
hence the total number of ordered pairs $$(m,n)$$ is $$\binom{50}{1} \times \binom{25}{1}=50 \times 25=1250$$
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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
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