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CBSE Questions for Class 11 Commerce Applied Mathematics Relations Quiz 1 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Relations
Quiz 1
What is general representation of ordered pair for two variables $$a$$ and $$b$$?
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$$a, b$$
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$$(a, b)$$
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$$(a), b$$
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$$a, (b)$$
Explanation
An ordered pair is written in the form $$(x-$$coordinate, $$y-$$coordinate$$)$$.
Ordered pairs $$(x, y)$$ and $$(-1, -1)$$ are equal if $$y = -1$$ and $$x =$$ _____
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$$1$$
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$$-1$$
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$$0$$
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$$2$$
Explanation
Given , $$(x-y)=(-1,1)$$
$$x=-1, y=-1$$
The value of $$x=-1$$.
Is the following statement is true for $$(a, a)\in R$$ $$\forall a\in N?$$
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True
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False
Explanation
$$(2,2) \in N$$
But Not $$\in R$$
as
$$2^2 \neq 2$$
So given statement is False.
If R={$$(x,y)/3x+2y=15$$ and x,y $$\displaystyle \epsilon $$ N}, the range of the relation R is________
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$$\{1,2\}$$
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$$\{1,2,..,5\}$$
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$$\{1,2,.,7\}$$
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$$\{3,6\}$$
Explanation
Given, $$ 3x + 2y = 15 $$
$$ => y = \dfrac {15-3x}{2} $$
As both $$ x , y $$ are natural numbers, we should find $$ x $$ such that $$ 15-3x $$ is divisible by $$ 2 $$
We can see that when $$ x =1$$ or $$3 $$, we get $$ y = 6$$ or $$3 $$
Hence, the range is $$ {3,6} $$
If $$x$$ co-ordinate of a point is $$2$$ and $$y$$ co-ordinate is $$0$$, then ordered pair for its coordinate on $$XY$$ plane is
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$$(0, 0)$$
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$$(2, 2)$$
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$$(0, 2)$$
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$$(2, 0)$$
Explanation
Ordered pair for co-ordinate of a point in $$XY$$ plane is written as $$(x, y).$$
So, option D is correct.
Let $$R$$ be the relation in the set $$N$$ given by $$R=\left\{(a, b): a=b-2, b>6\right\}$$. Choose the correct answer.
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$$(2, 4)\in R$$
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$$(3, 8)\in R$$
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$$(6, 8)\in R$$
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$$(8, 7)\in R$$
Explanation
$$R=\left\{(a, b):a=b-2, b>6\right\}$$
Now, since $$b>6, (2, 4)\notin R$$
Also, as $$3\neq 8-2, (3, 8)\notin R$$
And, as $$8\neq 7-2$$
$$\therefore (8, 7)\notin R$$
Now, consider $$(6, 8)$$.
We have $$8 > 6$$ and also, $$6=8-2$$.
$$\therefore (6, 8)\in R$$
Hence the correct answer is C.
If ordered pair $$(a, b)$$ is given as $$(-2, 0)$$, then $$a =$$
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$$-2$$
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$$0$$
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$$2$$
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None of the above
Explanation
$$a$$ will be first entry in ordered pair $$(a, b).$$
So, option A is correct.
If $$x$$ and $$y$$ coordinate of a point is $$(3, 10)$$, then the $$x$$ co-ordinate is
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$$0$$
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$$3$$
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$$10$$
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None of the above
Explanation
Here, $$x$$ co-ordinate will be the first entry in ordered pair.
So, option B is correct.
What are the ways of representing a relation?
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Set builder form
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Roster form
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Arrow diagram
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All of these
Explanation
Relation can be represented in set builder form, roster form and Arrow diagram.
If $$x$$ and $$y$$ co-ordinate of a point is $$(3, 10)$$, then $$y$$ co-ordinate is
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$$0$$
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$$3$$
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$$10$$
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None of the above
Explanation
$$y$$ coordinate will be second entry in ordered pair.
So, option C is correct.
The first component of all ordered pairs is called
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Range
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Domain
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Function
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None of these
Explanation
The first components of all order pair is called Domain.
The second components of all ordered pair is called Range.
If $$a$$ and $$b$$ are two variables and $$(a, b)=(b, a)$$, then
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$$a = 0$$
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$$b = 0$$
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$$a = b$$
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$$\displaystyle a\pm b$$
Explanation
$$(a, b) = (b, a)$$ only if $$a = b.$$.
So, option C is correct.
The second component of all ordered pairs of a relation is
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Range
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Domain
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mapping
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none of these
Explanation
The second components of all ordered pairs of a relation is Range.
Which of the ordered pair satisfies the relation $$x<y$$?
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$$(1,2)$$
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$$(2,1)$$
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$$(1,1)$$
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None of these
Explanation
$$(A) (1,2)$$
$$x<y$$
$$1<2$$
True , $$x<y$$
$$(B) (2,1)$$
$$2>1$$
$$(C) (1,1)$$
$$x=y$$
If $$y$$ is second entry and $$x$$ is first entry then its ordered pair will be ............
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$$(x, y)$$
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$$(y, y)$$
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$$(y, x)$$
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None of the above
Explanation
If $$x$$ is first entry and $$y$$ is second, then its ordered pair will be $$(x, y)$$.
So, option A is correct.
x varies directly as y and inversely as the square of z. When y = 4 and z is 14 x =If y = 16 and z = 7 what is x?
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180
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160
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280
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200
Explanation
Given
x varies directly as y and inversely as the square of z. When y = 4 and z is 14 x = 10. If y = 16 and z = 7
Then
$$x=k\frac{y}{z^{2}}$$
$$\Rightarrow 10=k\times \frac{4}{14\times14}$$
$$\Rightarrow k=490$$
Thus
$$x=490\times \frac{16}{7\times 7}=160$$
Let N denote the set of all natural numbers and R a relation on $$N\times N$$. Which of the following is an equivalence relation?
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$$(a, b) R (c, d) $$ if $$ad(b+c)=bc(a+d)$$
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$$(a, b) R (c, d) $$ if $$a+d=b+c$$
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$$(a, b) R (c, d) $$ if $$ad=bc$$
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$$(a, b) R (c, d) $$ if $$a-d=b-c$$
Explanation
$$(a,b)R(c,d)$$ if $$ad(b+c)=bc(a+d)$$ defined on $$N\times N$$ is reflexive, symmetric and transitive. So, it is equivalence relation $$(a,b)R(c,d)$$ if $$a+d=b+c$$ and $$(a,b)R(c,d)$$ if $$ad=bc$$ is also equivalence relation.
$$A$$ 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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$$2^{12}$$
Explanation
Given:
$$n(A) = 3 = m$$
$$n(B) = 4 = n$$
Number of relation
$$= { 2 }^{ mn }$$
$$= { 2 }^{ 3\times 4 }$$
$$= { 2 }^{ 12 }$$.
$$N$$ is the set of positive integers. The relation $$R$$ is defined on N x N as follows: $$(a,b) R (c,d)\Longleftrightarrow ad=bc$$ Prove that
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$$R$$ is an equivalence relation.
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$$R$$ is symmetric relation.
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$$R$$ is transitive relation.
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$$R$$ is not an equivalence relation.
Explanation
Reflexive: let $$(a,b)$$ be an ordered pair of positive integer. to show R is reflexive we must show $$\left( \left( a,b \right) ,\left( a,b \right) \right) \in R$$. Multiplication of integer is commutative, so $$ab=ba$$. Thus $$\left( \left( a,b \right) ,\left( a,b \right) \right) \in R$$
Symmetric: Let $$(a,b)$$ and $$(c,d)$$ be an ordered pair of positive integer such that $$\left( a,b \right) R\left( c,d \right) $$. Then $$ad=bc$$. This equation is equivalent to $$cb=da$$, so $$\left( c,d \right) R\left( a,b \right) $$. This shows R is symmetric
Transitive: Let $$(a,b),(c,d)$$ and $$(e,f)$$ be ordered pairs of positive integer such that $$\left( a,b \right) R\left( c,d \right) $$ and $$\left( c,d \right) R\left( e,f \right) $$. Then $$ad=bc$$ and $$cf=de$$.
Thus $$adf=bcf$$ and $$bcf=bde$$, which implies $$adf=bde$$.
Since $$d\neq 0$$, we can cancel it from both sides of this equation to get $$af=be$$.
This shows $$\left( a,b \right) R\left( e,f \right) $$ and R is transitive
Let $$R$$ be a relation from a set $$A$$ to a set $$B$$, then:
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$$R=A\cup B$$
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$$R=A\cap B$$
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$$R\subseteq A\times B$$
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$$R\subseteq B\times A$$
Explanation
$$R : $$$$A\rightarrow B$$
then $$R$$ is a subset $$A\times B$$
$$\therefore$$ $$R\quad \subseteq \quad A\times B$$
Let $$x$$ be a real number $$\left [ x \right ]$$ denotes the greatest integer function, and $$\left \{ x \right \}$$ denotes the fractional part and $$(x)$$ denotes the least integer function,then solve the following.
$$\left [ 2x \right ]-2x=\left [ x+1 \right ]$$
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$$\left \{ -1,\displaystyle-\frac{1}{2} \right \}$$
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$$\left \{ -1,\displaystyle \frac{1}{2} \right \}$$
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$$\left \{ 1,\displaystyle-\frac{1}{2} \right \}$$
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$$\left \{ 1,\displaystyle \frac{1}{2} \right \}$$
Explanation
$$\left [ 2x \right ]-2x=\left [ x+1 \right ]$$
$$\Rightarrow \left \{2x\right\}=\left [ x \right ]+1$$ (i)
But range of fractional part is $$[0,1)$$
$$\Rightarrow 0\leq [x]+1 < 1$$
$$\Rightarrow -1 \leq [x] < 0$$
$$ \Rightarrow -1 \leq x < 0$$
$$\Rightarrow [x] =-1$$
Thus (i) becomes $$\{2x\} =0 $$ it means that fraction part is zero for that,
$$\therefore x = -\cfrac{1}{2}, -1$$ in $$[-1,0)$$
If $$A = \{x, y\}$$ and $$B = \{3, 4, 5, 7, 9\}$$ and $$C = \{4, 5, 6, 7\}$$, find $$\displaystyle A\times (B\cap C)$$
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$$\displaystyle \left \{ (x,4),(x,5),(x,7),(y,4),(y,5),(y,7) \right \}$$
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$$\displaystyle \left \{ (x,3),(x,5),(x,7),(y,4),(y,5),(y,7) \right \}$$
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$$\displaystyle \left \{ (x,4),(x,5),(x,7),(y,4),(y,5),(y,9) \right \}$$
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none of the above
Explanation
$$\displaystyle B\cap C=\left \{ 3,4,5,7,9 \right \}\cap \left \{ 4,5,6,7 \right \}=\left \{ 4,5,7 \right \}$$
$$\displaystyle \therefore A\times (B\cap C)=\left \{ x,y \right \}\times \left \{ 4,5,7 \right \}$$
=$$\displaystyle \left \{ (x,4),(x,5),(x,7),(y,4),(y,5),(y,7) \right \}$$
$$\displaystyle x^{2} = xy$$ is a relation (defined on set R) which
is
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Symmetric
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Reflexive
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Transitive
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None of these
Explanation
Given , R:$$x^{2}=xy$$
Symmetric: If $$ (x,x) \in $$ R
$$x^{2}=x \times x$$
$$x^{2}=x^{2},(x,x) \in$$ R
So the relation is symmetric
Which one of the following relations on R (set of real numbers) is an equivalence relation
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$$\displaystyle aR_{1}b\Leftrightarrow \left | a \right |= \left | b \right |$$
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$$\displaystyle aR_{2}b\Leftrightarrow a\geq b $$
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$$\displaystyle aR_{3}b\Leftrightarrow a$$ divides $$b $$
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$$\displaystyle aR_{4}b\Leftrightarrow a< b $$
Explanation
$$a R_1 b \iff |a| = |b| $$
$$a=b$$
$$R_1 $$is a reflexive symmetric and transitive.( as we know $$a=b$$)
$$\therefore (a,a) $$ is present.
$$\therefore (a,b) (b,a)$$ will also be present.
So it is an equivalence relation .
Suppose y is equal to the sum of two quantities of which one varies directly as x and the other inversely as x If y = 6 when x = 4 and $$\displaystyle y=\frac{10}{3}$$ when x = 3 then what is the relation between x and y?
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$$\displaystyle y=x+\frac{4}{x}$$
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$$\displaystyle y+2x=\frac{4}{x}$$
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$$\displaystyle y=2x+\frac{8}{x}$$
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$$\displaystyle y=2x-\frac{8}{z}$$
Explanation
(A)If$$y=X+\frac{4}{x}\Rightarrow 4+\frac{4}{4}=3 \neq 6=6$$
(B)$$y+2x=\frac{4}{x}\Rightarrow 6+2(4)=\frac{4}{4}\Rightarrow 6+8\neq 1$$
(C)$$y=2x+\frac{8}{x}\Rightarrow 6=2(4)+\frac{8}{4}\Rightarrow 6\neq 8+2$$
(D)$$y=2x-\frac{8}{x}\Rightarrow 6=2(4)-\frac{8}{4}=8-2=6$$ If x=4 and y=6
$$y=2x-\frac{8}{x}\Rightarrow \frac{10}{3}=2\times 3-\frac{8}{3}\Rightarrow \frac{10}{3}=\frac{10}{3}$$ If x=3 and y=$$y=\frac{10}{3}$$
Then option (D) $$Y=2x-\frac{8}{x}$$
If X is brother of the son of Y's son. How is X related to Y?
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Son
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Brother
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Cousin
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Grandson
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Uncle
Explanation
Son of Y's Son- Grandson, Brother of Y's Grandson- Y's Grandson
Option D is correct.
If $$\displaystyle A\prime $$ is symmetric to A and $$\displaystyle B\prime $$ is symmetric to B with respect to a line of symmetry, then
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$$\displaystyle AB>A'B'$$
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$$\displaystyle AB=A'B'$$
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$$\displaystyle AB\neq A'B'$$
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None of these
If $$A=\left \{x:x^2-3x+2=0\right \}$$ and $$B=\left \{x:x^2+4x-5=0\right \}$$ then the value of A-B is
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$$\left \{1, 2\right \}$$
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$$\left \{2\right \}$$
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$$\left \{1\right \}$$
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$$\left \{5, 2\right \}$$
Explanation
$$A=\left \{x:x^2-3x+2=0\right \}\Rightarrow A=\left \{1, 2\right \}$$
$$B=\left \{x:x^2+4x-5=0\right \}\Rightarrow A=\left \{1, -5\right \}$$
$$\therefore A-B=\left \{2\right \}$$
The number of reflexive relations of a set with four elements is equal to
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2$$^{16}$$
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2$$^{12}$$
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2$$^{8}$$
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2$$^{4}$$
Explanation
Total number of reflexive relations in a set with n elements = 2$$^n$$ Therefore, total number of reflexive relations set with 4 elements = 2$$^4$$
Let $$A = \left \{1, 2, 3\right \}$$. Then number of equivalence relations containing $$(1, 2)$$ is
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$$1$$
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$$2$$
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$$3$$
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$$4$$
Explanation
Total possible pair=$$\{ (1,1),(1,2),(1,2),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)\} $$
Reflexive means $$(a,a)$$ should be in relation.
So,$$(1,1),(2,2),(3,3)$$ should be in 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 also be in relation.
Transitive means if $$(a,b)$$ is in relation and $$(b,c)$$ is in relation.
So if $$(1,2)$$ is in relation and $$(2,1)$$ is in relation,then $$(1,1)$$ should be in relation.
Relative $$R1=\{ (1,2),(1,1),(2,2),(3,3),(2,1)\} $$
So,smallest relation is $$R1=\{(1,2),(2,1),(1,1),(2,2),(3,3)\}$$
If we add $$(2,3)$$ ,then we have to add $$(3,2)$$ also,as it is symmetric but as $$(1,2)\&(2,3)$$ are there,we need to add $$(1,2)$$ also,as it is transitive.
As we are adding $$(1,3)$$ we should add $$(3,1)$$ also,as it is symmetric.
Relation $$R2=\{(1,2),(2,1),(1,1),(2,2),(3,3),(2,3),(1,3),(3,1)\}$$
Hence only two possible relation are there which are equivalence.
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