MCQ Questions for Class 11 Commerce Applied Mathematics Relations Quiz 3

Let

$R$ be a relation such that

$R = \{(1,4), (3, 7), (4, 5), (4, 6), (7, 6) \}$ then

$(R^{-1} oR)^{-1} =$

0%
$\{(1, 1), (3, 3), (4, 4), (7, 7), (4, 7), (7, 4), (4, 3)\}$
0%
$\{(1, 1), (3, 3), (4, 4), (7, 7), (4, 7), (7, 4) \}$
0%
$\{(1, 1), (3, 3), (4, 4) \}$
0%
$\phi$

$A$ and

$B$ are two sets, then

$A \times B = B \times A$ if

0%
$A \subset B$
0%
$B \subset A$
0%
$A = B$
0%
None of these

Let

$R$ be the relation on

$Z$ defined by

$R = \{(a, b): a, b \in z, a - b$ is an integer

$\}$ . Find the domain and Range of

$R$ .

0%
$z, z$
0%
$z^+, z$
0%
$z, z^-$
0%
None of these

Explanation

Given:

$R=\{ (a, b) : a, b, \in z, a-b \text{ is an integer}\}$

As difference of integers are also integers so,

Domain of

$R = z$

Range of

$R = z$ , as

$a-b$ spans the whole integer values.

Let

$R$ be a relation on the set

$N$ given by

$R=\left\{ \left( a,b \right) :a=b-2,b>6 \right\}$ . Then

0%
$(2,4)\in R$
0%
$(3,8)\in R$
0%
$(6,8)\in R$
0%
$(8,7)\in R$

Explanation

$a=b-2,b>6$

Above relation also imply

$a>6-2\rightarrow a>4$

Among the options

$C$ is correct answer as

$6>4$ and aslo

$6=8-2$ .

Which of the following is not an equivalence relation on

$Z$ ?

0%
$aRb\Leftrightarrow a+b$ is an even integer
0%
$aRb\Leftrightarrow a-b$ is an even integer
0%
$aRb\Leftrightarrow a< b$
0%
$aRb\Leftrightarrow a= b$

Explanation

Among the Options

$Opt:[C]$

$aRb\Leftrightarrow a< b$

As,

$a<b$ doesn't imply

$b<a$ ,which make the relation asymmetric and thus it is not an equivalence relation .

A relation

$R$ is defined from

$\left\{ 2,3,4,5 \right\}$ to

$\left\{ 3,6,7,10 \right\}$ by:

$xRy\Leftrightarrow x$ is relatively prime to

$y$ . Then, domain of

$R$ is

0%
$\left\{ 2,3,5 \right\}$
0%
$\left\{ 3,5 \right\}$
0%
$\left\{ 2,3,4 \right\}$
0%
$\left\{ 2,3,4,5 \right\}$

Explanation

Let's first write R as a set of ordered pairs,

$R=\{(2,3),(2,7),(3,7),(3,10),(4,3),(4,7),(5,3),(5,6),(5,7)\}$

So, Domain of R is

$\{2,3,4,5\}$

The relation

$R$ in

$N\times N$ such that

$(a,b)R(c,d)\Leftrightarrow a+d=b+c$ is

0%
reflexive but not symmetric
0%
reflexive and transitive but not symmetric
0%
an equivalence relation
0%
none of these

Explanation

If (a,b)R(c,d) then (c,d)R(a,b) as if a+d=b+c then b+d=a+c .

Thus relation R is symmetric .

For some a,b in N , (a,b)R(a,b) as a+b=b+a.

Thus every element of

$N\times N$ is related to itself,

So relation R is also reflexive .

For some a,b,c,d,e,f in N, if (a,b)R(c,d) and (c,d)R(e,f) then (a,b)R(e,f),

As if a+d=b+c...1 and c+f=d+e….2 then a+f=b+e [since eq.1- eq.2 gives a-e=b-f

$\rightarrow$ a+f=b+e ]

So, R is also symmetric relation .

Thus R is a equivalence relation.So, C is correct answer .

Let

$R$ be the relation over the set of all straight lines in a plane such that

${l}_{1}$

$R$

${l}_{2}\Leftrightarrow {l}_{1}\bot {l}_{2}$ . Then,

$R$ is

0%
symmetric
0%
reflexive
0%
transitive
0%
an equivalence relation

Explanation

For some lines

$l_1,l_2$ , We know that

$l_1\perp l_2 \leftrightarrow l_2\perp l_1$

So, given relation is symmetric .

For some lines

$l_1,l_2,l_3$ , We know that if

$l_1\perp l_2 \ and \ l_2\perp l_3$ , then

$l_1\parallel l_3$

So, given relation is not transitive .

For some line

$l$ , We know that it is never possible that

$l\perp l$ .

So, given relation is not reflexive .

Thus

$A$ is correct answer .

$A=\left\{ a,b,c \right\}$ , then the relation

$R=\left\{ \left( b,c \right) \right\}$ on

$A$ is

0%
reflexive only
0%
symmetric only
0%
transitive only
0%
reflexive and transitive only

Let

$A=\left\{ 1,2,3 \right\}$ . Then, the number of equivalence relations containing

$(1,2)$ over set A is

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

Relation to be equivalent firstly must contain (1,1),(2,2),(3,3) to make it reflexive relation over given set .

Also since (1,2) is to be contained in set (2,1) must also be there to make relation symmetric .

So, one option of relation R is

$R=\{(1,1)(2,2)(3,3)(1,2)(2,1)\}$

Now if we add (1,3) we have to also add (3,1) to keep relation symmetric .

So, second option of relation R is

$R=\{(1,1)(2,2)(3,3)(1,2)(2,1)(1,3)(3,1)\}$
And, third option of relation R is

$R=\{(1,1)(2,2)(3,3)(1,2)(2,1)(1,3)(3,1)(2,3)(3,2)\}$

Thus, there can be three equivalence relations containing (1,2).

Let

$A=\left\{ 1,2,3 \right\}$ and

$R=\left\{ \left( 1,2 \right) ,\left( 2,3 \right) ,\left( 1,3 \right) \right\}$ be a relation on

$A$ . Then

$R$ is

0%
neither reflexive nor transitive
0%
neither symmetric nor transitive
0%
transitive
0%
none of these

Explanation

The given relation is not reflexive as

$(1,1),(2,2),(3,3)\not \in R$ .

Again the relation is not symmetric as

$(1,2)\in R$ but

$(2,1)\not \in R$ .

But the relation is transitive as

$(1,2)\in R,(2,3)\in R\Rightarrow (1,2)\in R$ for all

$1,2,3\in A$ .

$A=\left\{ 1,2,3 \right\} , B=\left\{ 1,4,6,9 \right\}$ and

$R$ is a relation from

$A$ to

$B$ defined by

$x$ is greater than

$y$ . The range of

$R$ is

0%
$\left\{ 1,4,6,9 \right\}$
0%
$\left\{ 4,6,9 \right\}$
0%
$\left\{ 1 \right\}$
0%
none of these

Explanation

First Let's write R as set of ordered pairs,

$R=\{(2,1),(3,1)\}$

We can simply see from set that range of

$R = \{1\}$

A relation

$\phi$ from

$C$ to

$R$ is defined by

$x\phi y\Leftrightarrow \left| x \right| =y$ . Which one is correct?

0%
$(2+3i)\phi 13$
0%
$3\phi (-3)$
0%
$(1+i)\phi 2$
0%
$i\phi 1$

Explanation

Given, a relation

$\phi$ from

$C$ to

$R$ is defined by

$x\phi y\Leftrightarrow \left| x \right| =y$ .

Now,

$|(2+3i)|=\sqrt{2^2+3^2}=\sqrt{13}$ .

$(2+3i)$ and

$13$ are not related.

Again, since

$y>0$ this gives

$3$ not related to

$-3$ .

$|1+i|=\sqrt{1+1}=\sqrt 2$

$1+i$ and

$2$ are not related.

Also

$|i|=1$ so

$i\phi 1$ holds.

Let

$R$ be a relation on

$N$ defined by

$x+2y=8$ . The domain of

$R$ is

0%
$\left\{ 2,4,8 \right\}$
0%
$\left\{ 2,4,6,8 \right\}$
0%
$\left\{ 2,4,6 \right\}$
0%
$\left\{ 1,2,3,4 \right\}$

Explanation

$x+2y=8$

$x=8-2y$

Since y and x both must be natural,

Possible values of y are 1,2,3 which corresponds to possible value of x to be 6,4,2 .

Thus range of R =

$\{2,4,6\}$

$A=\left\{ 1,2,3 \right\}$ , then a relation

$R=\left\{ \left( 2,3 \right) \right\}$ on

$A$ is

0%
symmetric and transitive only
0%
symmetric only
0%
transitive only
0%
none of these

Explanation

A binary relation R over Set X is transitive if for all elements a,b,c in X , if a is related to b and b is related to c, then a is related to c.

A binary relation R over Set X is symmetric iff

$\forall a,b \in X (aRb \leftrightarrow bRa)$

But Given relation is not symmetric as it contains (2,3) but do not contain (3,2).

Since Given relation R contains only one pair (2,3), it is transitive .

Let

$A=\left\{ 1,2,3 \right\}$ and consider the relation

$R=\left\{ \left( 1,1 \right) ,\left( 2,2 \right) \left( 3,3 \right) ,\left( 1,2 \right) ,\left( 2,3 \right) ,\left( 1,3 \right) \right\}$ , then

$R$ is

0%
reflexive but not symmetric
0%
reflexive but not transitive
0%
symmetric and transitive
0%
neither symmetric nor transitive

Explanation

Relation

$R$ is reflexive relation over set

$A$ as every element of

$A$ is related to itself in

$R$ .

Relation

$R$ is not symmetric as

$(1,2)$ is in

$R$ but

$(2,1)$ is not in

$R$ .

Relation

$R$ is also transitive as for

$a,b,c$ in

$A$ , if

$(a,b)$ is in

$R$ and

$(b,c)$ is in

$R$ , then

$(a,c)$ is also in

$R$ .

Thus

$A$ is correct answer .

$n(A) = 4$ and

$n(B) = 5$ , then

$n(A \times B) =$

0%
$20$
0%
$25$
0%
$4$
0%
$15$

Explanation

Given,

$n\left( A \right) =4$ and

$n\left( B \right) =5$

$\Rightarrow n\left( A\times B \right) =n\left( A \right) \cdot n\left( B \right)$

$\Rightarrow n(A\times B)=4\cdot 5=20$

So, answer is

$20$ .

In the set

$Z$ of all integers, which of the following relation

$R$ is an equivalence relation?

0%
$xRy:$ if $x\le y$
0%
$xRy:$ if $x-= y$
0%
$xRy:$ if $x- y$ is an even integer
0%
$xRy:$ if $x\equiv y$ (mod 3)

Explanation

For option C, xRy:if x-y is an even integer .

Let a be any number in Z,

Then, a-a=0 is an even integer .

So, Every number is related to itself .

Thus the relation is reflexive .

Let a,b be two numbers in Z,

Also let

$a-b=k$ be an even integer (which implies aRb)

Then,

$b-a=-k$ is also an even integer (which implies bRa).

Thus, It is evident

$aRb\leftrightarrow bRA$

Thus the relation is also symmetric .

Let a,b,c be three numbers in Z,

Also let aRb and bRc,

then

$a-b=2k_1$ and

$b-c=2k_2$ for

$k_1,k_2$ belonging to Z

which gives

$a-b+b-c=2k_1+2k_2\Rightarrow a-c=2(k_1+k_2)$

which implies

$aRc$ .

Thus relation is also transitive .

Thus we finally arrive at a conclusion that relation is Equivalence relation .

Let

$L$ denote the set of all straight lines in a plane, Let a relation

$R$ be defined by

$lRm$ , iff

$l$ is perpendicular to

$m$ for all

$l \in L$ . Then,

$R$ is

0%
reflexive
0%
symmetric
0%
transitive
0%
none of these

Explanation

We know that if line l is perpendicular to line m, line m must also be perpendicular to line l,

So,

$lRm \Leftrightarrow mRl$ and thus Relation R is symmetric .

Let

$T$ be the set of all triangles in the Euclidean plane, and let a relation

$R$ on

$T$ be defined as

$aRb$ , if

$a$ is congruent to

$b$ for all

$a,b\in T$ . Then,

$R$ is

0%
reflexive but not symmetric
0%
transitive but not symmetric
0%
equivalence
0%
none of these

Explanation

Let there be three triangles a,b,c in Euclidean plane such that aRb and bRc.

We know that if triangle a is congruent to triangle b and triangle b is congruent to triangle c, triangle a must also be congruent to triangle c .

So, aRc and thus relation R is transitive .

Also we know that any triangle (say t) in Euclidean plane is always congruent to itself .

So, tRt and thus relation R is reflexive .

Let there be two triangles n and m in Euclidean plane.

We know that if n is congruent to m, m must also be congruent to n.

So,

$nRm \Leftrightarrow mRn$ and thus relation R is symmetric .

Thus, finally we arrive at conclusion that realation R is equivalence relation .

Let

$R$ be a relation on the set

$N$ of natural numbers defined by

$nRm$ , iff

$n$ divides

$m$ . Then,

$R$ is

0%
Reflexive and symmetric
0%
Transitive and symmetric
0%
Equivalence
0%
Reflexive, transitive but not symmetric

Explanation

Let there be a natural number

$n$ ,

We know that

$n$ divides

$n$ , which implies

$nRn$ .

So, Every natural number is related to itself in relation

$R$ .

Thus relation

$R$ is reflexive .

Let there be three natural numbers

$a,b,c$ and let

$aRb,bRc$

$aRb$ implies

$a$ divides

$b$ and

$bRc$ implies

$b$ divides

$c$ , which combinedly implies that

$a$ divides

$c$ i.e.

$aRc$ .

So, Relation R is also transitive .

Let there be two natural numbers

$a,b$ and let

$aRb$ ,

$aRb$ implies

$a$ divides

$b$ but it can't be assured that

$b$ necessarily divides

$a$ .

For ex,

$2R4$ as 2 divides 4 but 4 does not divide 2 .

Thus Relation R is not symmetric .

The relation

$R=\left\{ \left( 1,1 \right) ,\left( 2,2 \right) \left( 3,3 \right) \right\}$ on the set

$A=\left\{ 1,2,3 \right\}$ is

0%
symmetric only
0%
reflexive only
0%
an equivalence relation
0%
transitive only

Which one of the following relations on Z is equivalence relation?

0%
$xR_1 y\Leftrightarrow |x| = |y|$
0%
$xR_2 y\Leftrightarrow x \geq y$
0%
$xR_3 y\Leftrightarrow \dfrac{x}{y}$
0%
$xR_4 y\Leftrightarrow x < y$

Explanation

A relation is equivalence if it is symmetric, transitive and reflexive
Consider,

$xR_1y\Leftrightarrow \left | x \right |=\left | y \right |$

For reflexive:

$xR_1x\Leftrightarrow \left | x \right |=\left | x \right |$ , clearly it satisfies

$R_{1}$

Thus,

$R_{1}$ is reflexive.

For symmetric:

$xR_{1}y\Leftrightarrow \left | x \right |=\left | y \right |$

$\implies yR_{1}x\Rightarrow \left | y \right |=\left | x \right |$ as equality communicative

So,

$R_{1}$ is symmetric.

For transitive:
Let

$(x,y)$ and

$(y,z)$ satisfy

$R_{1}$

$\implies xR_{1}y\Leftrightarrow \left | x \right |=\left | y \right |$ and

$yR_{1}z\Leftrightarrow \left | y \right |=\left | z \right |$

$\implies |x|=|y|$ and

$|y |=|z|$

$\implies |x |=|z|$

$\therefore xR_{1}z\Leftrightarrow \left | x \right |=\left | z \right |$

$\therefore \left ( x,z \right )$ satisfies

$R_{1}$

Thus,

$R_1$ is a transitive relation.
Hence,

$R_{1}$ is equivalent.

Also, relations

$R_{2},\ R_{3}$ and

$R_{4}$ are not symmeteric on

$Z,$ hence not an equivalence relations.

Let

$R$ be a reflexive relation on a finite set

$A$ having

$n$ elements, and let there be

$m$ ordered pairs in

$R,$ then:

0%
$m\geq n$
0%
$m\leq n$
0%
$m=n$
0%
$m< n$

Explanation

By the definition for a relation with

$n$ elements to be reflexive relation it must have at least

$n$ ordered pairs.
it has

$m$ ordered pairs therefore

$\mathrm{m}\geq \mathrm{n}$ .

Total number of equivalence relations defined in the set

$S =\{a,b,c\}$ is

0%
$5$
0%
$3$ !
0%
$2^{3}$
0%
$3^{3}$

Explanation

The smallest equivalence relation is the identity relation

$R_1=\{(a,a),(b,b),(c,c)\}$

Then two ordered pairs of two distinct elements can be added to give three more equivalence relations.

$R_2=\{(a,a),(b,b),( c,c),(a,b),(b,a)\}$

Similarly

$R_3$ and

$R_4$ can be made by taking

$(b,c),(c,b)$ and

$(a,c),(c,a)$ respectively.

Finally the largest equivalence relation i.e., the universal relation

$R_5=\{(a,a),(b, b),(c, c),(a, b),(b, a),(a, c),(c,a ),(b, c),(c, b) \}$

Hence, total

$5$ equivalence relations can be created.

If relation

$R$ is defined by

$\mathrm{R}=\{(\mathrm{x},\ \mathrm{y}):2\mathrm{x}^{2}+3\mathrm{y}^{2}\leq 6\}$ , then the domain of

$\mathrm{R}$ is

0%
$[-3,3]$
0%
$[-\sqrt{3},\sqrt{3}]$
0%
$[-\sqrt{2},\sqrt{2}]$
0%
$[-2,2]$

Explanation

Domain

$=$ value of

$\mathrm{x}$ in

$f=2\mathrm{x}^{2}+3\mathrm{y}^{2}\leq 6$ then

$2\mathrm{x}^{2}+3\mathrm{y}^{2}\leq 6$ =

$\displaystyle \frac{\mathrm{x}^{2}}{3}+\frac{\mathrm{y}^{2}}{2}\leq 1$
Since it is an equation of ellipse then value of

$x$

$\Rightarrow \mathrm{x}\in[-\sqrt{3},\sqrt{3}]$

$\displaystyle A=\left \{ 2, 3, 5 \right \}, B=\left \{ 2, 5, 6 \right \}$ then

$\left ( A-B \right )\times \left ( A\cap B \right )$ is

0%
$\displaystyle \left \{ \left ( 3, 2 \right ), \left ( 3, 3 \right ), \left ( 3, 5 \right )\right \}$
0%
$\displaystyle \left \{ \left ( 3, 2 \right ), \left ( 3, 5 \right ), \left ( 3, 6 \right )\right \}$
0%
$\displaystyle \left \{ \left ( 3, 2 \right ), \left ( 3, 5 \right )\right \}$
0%
None of these

Explanation

$\displaystyle A-B=\left \{ 3 \right \},A\cap B=\left ( 2, 5 \right ).$

$\displaystyle \therefore \left ( A-B \right )\times \left ( A\cap B \right )=\left \{ \left ( 3, 2 \right ),\left ( 3, 5 \right ) \right \}.$

The domain and range of relation

$R=\{(x,y) | x, y \in N$ ,

$x+2y=5\}$ is?

0%
$\{1,3\}, \{2,1\}$
0%
$\{2,1\}, \{3,2\}$
0%
$\{1,3\}, \{1,1\}$
0%
$\{1,2\}, \{1,3\}$

Explanation

The possible values of

$x$ which satisfies the given relation is the domain of the relation

The possible values of

$y$ which satisfies the given relation is the range of the relation

Given that

$x,y$ are natural numbers and the relation is

$x+2y=5$

For

$x=1$ , the value of

$y$ is

$2$

For

$x=3$ , the value of

$y$ is

$1$

$x\geq 5$ , then

$y$ is negative which does not belong to naturals.

$x=2,4$ then

$y$ becomes rational number which is not in naturals.

Thus, only

$1$ and

$3$ gives

$y$ as natural numbers

$2$ and

$1$ .

Therefore, the domain is

$\{1,3\}$ and the range is

$\{2,1\}$

Let

$A$ be a non-empty set such that

$A \times A$ has

$9$ elements among which are found

$(-1, 0)$ and

$(0, 1)$ , then

0%
$A=\left\{ -1,0 \right\}$
0%
$A=\left\{ 0,1 \right\}$
0%
$A=\left\{ -1,0,1 \right\}$
0%
$A=\left\{ -1,1 \right\}$

Explanation

Given

$A\times A$ has

$9$ elements

$\Rightarrow n(A)=3$
Elements of

$A\times A$ are

$(-1,0)$ &

$(0,1)$ .

$\therefore$

$A$ needs to contain

$-1, 0$ and

$1$ .

$\therefore$ Since

$n(A)=3 \Rightarrow A=\left \{ -1,0,1 \right \}$

Given the relation

$R =\{(1,2) (2,3)\}$ on the set

$A=\{1,2,3\}$ , the minimum number of ordered pairs which when added to

$R$ to make it an equivalence relation is

0%
$5$
0%
$6$
0%
$7$
0%
$8$

Explanation

$R$ is reflexive if it contains

$(1, 1), (2, 2), (3, 3)$ .

$\because (1, 2)\in R$ and

$(2,3)\in R$

$\therefore R$ is symmetric if

$(2, 1), (3, 2) \in R.$

$(1, 2), (2,3)\in R$

Then,

$R$ is transitive is

$(1,3)\in R$

Also,

$(3,1)\in R$ for symmetric property.

Now,

$R=\{(1,1),(2,2),(3,3),(2,1),(3,2),(2,3),(1,2),(1,3),(3,1)\}$
Thus,

$R$ becomes an equivalence relation by adding

$(1,1),(2,2),(3,3),(2,1),(3,2),(1,3),(3,1)$

Hence, the total number of ordered pairs is

$7$ .

CBSE Questions for Class 11 Commerce Applied Mathematics Relations Quiz 3 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Relations Quiz 3 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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