A $$\times$$ (B - C) =

$$(A \times B) - (A \times C)$$

$$(A \times C) - (A \times B)$$

$$(A \times B) \bigcup (A \times C)$$

$$(A \times B) \bigcap (A \times C)$$

MCQ Questions for Class 11 Commerce Applied Mathematics Relations Quiz 4

The number of relations from

A = {1, 2, 3}

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

B = {4, 6, 8, 10}

$B = \left \{ 4,6,8,10 \right \}$ is

0%
$4^{3}$
0%
$2^{7}$
0%
$2^{12}$
0%
$3^{4}$

Explanation

n (B) = 4

$n\left ( B \right )=4$

n (A) = 3

$n\left ( A \right )=3$
Therefore there exists

2^{3 \times 4}

$2^{3 \times 4}$ relations from

A

$A$ to

B

$B$
i.e

2^{12}

$2^{12}$ relations

The range of the function

f (x) = \frac{sin (π [x])}{x^{2} + 1}

$f(x)=\dfrac{\sin(\pi [x])}{x^{2}+1}$ (where

[.]

$[.]$ denotes greatest integer function) is

0%
$\{0\}$
0%
$R$
0%
$(0,1)$
0%
$(0,\infty )$

Explanation

f (x) = \frac{sin (π [x])}{x^{2} + 1}

$f(x)=\dfrac {\sin(\pi[x])}{x^2+1}$

$\because$

$[x]$ always gives integer value and

$\sin n\pi =0$

$\therefore f(x)=0$

Which of the following are not equivalence relations on

$I$ ?

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

Explanation

Option A,

$aRb$ if

$a+b$ is an even integer.

Let

$a, b,c \in I$
Clearly,

$a+a =2a$ is an even integer.
Hence,

$aRa$ for all

$a\in I$

Now, let

$aRb$

$\Rightarrow a+b$ is an even integer.
or

$b+a$ is an even integer.

$\Rightarrow bRa$

Next, let

$aRb, bRc$

$\Rightarrow a+b$ in an even integer and

$b+c$ is an even integer

$\Rightarrow a$ and

$b$ both are even or both are odd . Also,

$b$ and

$c$ both are even or both are odd.
Hence, if

$a$ and

$b$ are even .So

$c$ is also even . Hence,

$a+c$ is even.
And if

$a$ and

$b$ are odd.So,

$c$ is also odd . Again

$a+c$ is even.
Hence,

$aRc$

Hence, option A is equivalence relation.

Option B,

$aRb$ if

$a-b$ is an even integer.

Let

$a, b,c \in I$
Clearly,

$a-a =0$ is an even integer.
Hence,

$aRa$ for all

$a\in I$

Now, let

$aRb$

$\Rightarrow a-b$ is an even integer.
or

$-(a-b)$ is also an even integer.

$\Rightarrow \ bRa$

Next, let

$aRb, bRc$

$\Rightarrow a-b$ in an even integer and

$b-c$ is an even integer

$a+c=a-b+b-c=$ even+even
Hence,

$a+c$ is even.
Hence,

$aRc$

Hence, option

$B$ is equivalence relation.

Option C

$aRb$ if

$a<b$

Let

$a, b,c \in I$
Clearly,symmetry does not hold here
Now, let

$aRb$

$\Rightarrow a<b$
But it does not implies

$b<a$

Hence, option C is not an equivalence relation.

Option D,

$aRb$ if

$a=b$

Let

$a, b,c \in I$
Clearly,

$a=a$
Hence,

$aRa$ for all

$a\in I$

Now, let

$aRb$

$\Rightarrow a=b$
or

$b=a$ is an even integer.

$\Rightarrow bRa$

Next, let

$aRb, bRc$

$\Rightarrow a=b$ and

$b=c$

$\Rightarrow a=b=c$
Hence,

$aRc$

Hence, option D is equivalence relation.

Which one of the following is an elementary symmetric function of

$x_{1},x_{2},x_{3},x_{4}$ .

0%
$x_{1}x_{2}x_{3}+x_{2}x_{3}x_{4}$
0%
$x_{1}x_{2}+x_{2}x_{3}+x_{3}x_{1}$
0%
$x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}$
0%
$x_{1}x_{2}+x_{1}x_{3}+x_{1}x_{4}+x_{2}x_{3}+x_{2}x_{4}+x_{3}x_{4}$

Explanation

Numbers of elements of is

$\{x_{1},x_{2},x_{3},x_{4}\}$ .

$\therefore n(\mathrm{A})=4$

$\therefore$ Number of elementary functions is given by

$^{4}\textrm{c}_{2}=6$
Hence, option 'D' is correct.

The solution of

$8x\equiv 6(mod \ 14)$ is

0%
$\{8, 6\}$
0%
$\{6,14\}$
0%
$\{6,13\}$
0%
$\{8,14,6\}$

Explanation

Since,

$8\mathrm{x}\equiv 6(mod \ 14)$
i.e.,

$8\mathrm{x}-6=14\mathrm{k}$ for

$\mathrm{k}\in \mathrm{I}$ .

$\Rightarrow 8x = 14k+6$

$\Rightarrow 4x = 7k+3$
The values

$6$ and

$13$ satisfy this equation (when

$k =3$ and

$k=7$ ),

while

$8, 14$ and

$16$ do not.

If relation R

$=\left \{ (x, x+2) : x \in N, 1 \leq x <4 \right \}$ then R is

0%
$\{(1, 3), (2, 4), (3, 5)\}$
0%
$\{(2, 3), (4,2), (5, 3)\}$
0%
$\{(1, 3), (2, 4), (3, 5), (4, 6)\}$
0%
none of these

Explanation

We have

$x\in N, 1\le x<4$

$\Rightarrow x =1,2,3$
Hence

$R =\{(1,3),(2,4),(3,5)\}$

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

0%
$4$
0%
$7$
0%
$6$
0%
$5$

Explanation

for equivalence relation R

$(a,a)\in R$
if

$(a,b)\in R$
then

$(b,a)\in R$
if

$(a,b)\in R$

$(b,c)\in R$
then

$(a,c)\in R$

$R={(1,2), (2,3), (1,1), (2,2), (3,3), (2,1), (3,2), (1,3), (3,1)}$
No of added ordered pairs

$=9-2=7$

$R$ is the relation 'less than' from

$A=\{1, 2, 3, 4, 5\}$ to

$B=\{1, 4\}$ , the set of ordered pairs corresponding to

$R$ , then the inverse of

$R$ is

0%
$\{(3, 1), (3, 2), (3, 3)\}$
0%
$\{(4, 1), (4, 2), (4, 3)\}$
0%
$\{(4, 3), (4, 4), (4, 5)\}$
0%
$\{(1, 3), (2, 4), (3, 5)\}$

Explanation

$\because R$ is the relation 'less than' from

$A$ to

$B$ .

$R=\{(1,4), (2,4), (3,4)\}$

$R^{-1}=\{(4,1), (4,2), (4, 3)\}$

$A =\{1,2,3\}$ ,

$B=\{1,4,6,9\}$ and

$R$ is a relation from

$A$ to

$B$ defined by '

$x$ is greater than

$y$ '. The range of

$R$ is

0%
$\{1, 4, 6, 9\}$
0%
$\{4, 6, 9\}$
0%
$\{1\}$
0%
$\{4, 6\}$

Explanation

$R=\left \{ ( x,y \right ): x\in A, y\in B$ and

$x> y\}$

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

Therefore,

$R=\left \{ 1 \right \}$

Let

$R = \{(2,3),(3, 4)\}$ be relation defined on the set of natural numbers. The minimum number of ordered pairs required to be added in

$R$ so that enlarged relation becomes an equivalence relation is

0%
$3$
0%
$5$
0%
$7$
0%
$9$

Explanation

We know that a relation is an equivalence relation if it is reflexive, symmetric and transitive.

For reflexive relation

$(a, a)\in R$

For symmetric: If

$(a,b)\in R,$ then

$(b,a)\in R$

For transitive relation:

Let

$(a,b)\in R$ and

$(b,c)\in R,$ then

$(a,c)\in R$

$\therefore$ Required

$R= \{(2,3), (3,4), (2,2), (3,3), (2,4), (4,2), (3,2), (4,3), (4,4)\}$

So minimum number of ordered pairs to be added

$=7$ .

$A$ is the set of even natural numbers less than

$8$ and

$B$ is the set of prime numbers less than

$7$ , then the number of relations from

$A$ to

$B$ is

0%
$2^{9}$
0%
$9^{2}$
0%
$3^{2}$
0%
$2^{9}-1$

Explanation

$A = \{2, 4, 6\}, B = \{2,3,5\}$

$n (A\times B) = 3\times 3 = 9$
Number of relations from

$A$ to

$B$ is

$2^9$

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

$R$ is a relation from

$A$ to

$B$ defined by '

$x$ is greater than

$y$ '. Then 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

Here

$R$ is a relation

$A$ to

$B$ defined by '

$x$ is greater then

$y$ '

$R=\left\{ (2,1);(3,1) \right\}$ .Hence, range of

$R=\left\{ 1 \right\}$

$X =\{1, 2,3,4,5\}$ and

$Y =\{1,3,5,7,9\}$ , determine which of the following sets represent a relation and also a mapping?

0%
$R_{1}= \{(x,y)$ : $y=x+2, x \in Y,y \in Y\}$
0%
$R_{2}=\{(1,1), (1,3), (3,5), (4,7), (5,9)\}$
0%
$R_{3}=\{(1,1), (2,3), (3,5), (3,7), (5,7)\}$
0%
$R_{4}=\{(1,3), (2,5), (4,7), (5,9), (3,1)\}$

Explanation

Here we will check all options one by one:

Option A:

$R_1 = \{(x,y): y= x+2, x\in X, y\in Y\}$

$\implies \mathrm{R}_{1}=\{(1,3),(2, 4),(3,5),(4,6),(5,7)\}$

Since

$4$ amd

$6$ are the images of

$2$ and

$4$ respectively but

$4$ and

$6$ do not belong to

$Y$

$\therefore (2, 4),(4,6)\not\in X \times Y$

Hence

$R_{1}$ is not a relation as well as not a mapping.

Option B:

$R_{2}$ : It is a relation but not a mapping because the element

$1$ has two different images.

Option C:

$R_{2}$ : It is a relation but not a mapping because the element

$3$ has two different images.

Option D:

$R_{4}$ : It is both a relation and a mapping because every element in

$X$ is mapped to the elements in

$Y$ . Also, every element of

$X$ has a one and only one image in

$Y$ and every element in

$Y$ has its pre-image in

$X$ . Hence, it is also one-one and onto mapping and hence it is a bijection.

If A

$=$ {

$x : x^{2}-3x+2= 0$ }, and

$R$ is a universal relation on

$A$ , then

$R$ is

0%
$\{(1,1),(2, 2)\}$
0%
$\{(1,1)\}$
0%
$\phi$
0%
$\{(1,1),(1, 2)(2,1),(2,2)\}$

Explanation

Consider,

$x^2-3x+2=0$

$\implies x^2-2x-x+2=0$

$\implies (x-2)(x-1)=0$

$\implies x=1,2$

$\therefore A=\{1,2\}$
Also R is universal relation on set A, then every element of set A is related every other element of A

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

$A=\{b,c,d\}$ and

$B=\{x,y\}$ . Find which of the following are elements of

$A \times A$ .

0%
$\{b,b\}$
0%
$\{b,c\}$
0%
$\{b,d\}$
0%
All of the above

Explanation

$A\times A \Rightarrow$ the first element will be from

$A$ and the second element will also be from

$A$ .

$A \times A = \left\{\{b,b\}, \{b,c\}, \{b,d\},\{c,b\},\{c,c\},\{c,d\},\{d,b\},\{d,c\},\{d,d\}\right\}$

Thus, all the options

$A,B$ and

$C$ are the elements of

$A \times A$

n(A)=m and n(B)=n ; then

0%
n(A)+n(B)=n(A+B)
0%
n(A)-n(B)=n(A+B)
0%
A $\times$ B=mn
0%
n(A) $\times$ n(B =n(A $\times$ B)

Suppose

$S=\{1,2\}$ and

$T=\{a,b\}$ then

$T \times S$

0%
$(a,1),(a,2),(b,1),(b,2)$
0%
$(1,a),(2,b),(b,1),(b,2)$
0%
$(a,1),(a,2),(1,b),(2,b)$
0%
None of the above

Explanation

Given :

$S=\{ 1,2\} ,T=\{ a,b\}$

$T\times S$ is the set of ordered pair of elements of

$T$ and

$S$ respectively

Then,

$T\times S=\{a,b\}\times \{1,2\}$

$T\times S=\{ (a,1),(a,2),(b,1),(b,2)\}$

Given

$A=\{b,c,d\}$ and

$B=\{x,y\}$ : find element of

$A\times B$ .

0%
$\{b,x\}$
0%
$\{b,y\}$
0%
$\{c,x\}$
0%
All of the above

Explanation

Given

$A=\{b,c,d\}$ and

$B=\{x,y\}$ , then

$A\times B=\{(b,x),(c,x),(d,x),(b,y),(c,y),(d,y)\}$

Hence, all the elements given in options are elements of

$A\times B$ .

Given M={0,1,2} and N={1,2,3}, then (M

$\cup$ N)

$\times$ (M-N) contains

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

Explanation

$M=\{0,1,2\}$

$N=\{1,2,3\}$

$M\cup N=\{0,1,2,3\}$

$M-N=\{0\}$

$(M\cup N)\times (M-N)$

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

$\left (A \cap B \right ) \times C$

0%
$\left (A \times B \right ) \cap \left (B \times C \right )$
0%
$\left (A \times C \right ) \cap \left (B \times C \right )$
0%
$\left (A \times B \right ) \cup \left (B \times C \right )$
0%
$\left (A \times B \right ) \cup \left (A \times C \right )$

$M={0,1,2}$ and

$N={1,2,3}$ : find (N-M)

$\times$ (N

$\cap$ M)

0%
$\{3,1\}$
0%
$\{3,2\}$
0%
$\{3,3\}$
0%
None of the above

Explanation

For two sets, A and B the set difference of set B from set A is the set of all element in A but not in B.
Example:

$A=\{a,b,c,d\}$

$B=\{c,d,e\}$

$A-B=\{a,b\}$

$M=\{0,1,2\}$

$N=\{1,2,3\}$

$N-M=\{3\}$

$N \cap M=\{1,2\}$

$(N-M)\times (N\cap M)$

$=\{3,1\},\{3,2\}$

Given

$M=\{5,6,7\}$ and

$N=\{6,8,10\}$ find element of

$(M\cup N)\times N$

0%
$\{5,6\}$
0%
$\{5,8\}$
0%
$\{5,10\}$
0%
All of the above

Explanation

For two non-empty sets

$A$ and

$B$ , the Cartesian product

$A\times B$ is the set of all ordered pair of elements from

$A$ and

$B$ .

$A \times B = \{(x, y) : x \in A, y \in B\}$

If

$A= \{a,b\}$ and

$B= \{x, y\},$
then

$A\times B = \{(a, x); (a, y); (b, x); (b, y)\}$

here,

$A=M\cup N$ and

$B=N$

$A=M\cup N$ ={

$5,6,7,8,10$ }

$B=$ {

$6,8,10$ }

$A\times B = (M\cup N)\times N=(A,6),(A,8),(A,10)$

hence A,B,C all are correct so answer is D

$n(A)=4$ and

$n(B) =5$ :

$n(A \times B)=$

0%
$20$
0%
$10$
0%
$30$
0%
None of the above

Explanation

$n(A)=m$ ,and

$n(B)=n$ ,then

$n(A\times B)=mn$

$n(A\times B)=5.4=20$

n (A

$\times$ B) =

0%
n(A) x n(B)
0%
n(A $\bigcap$ B)
0%
n(A $\bigcup$ B)
0%
all of these

Which one of the statement is false ?

0%
$\phi \times A = \phi$
0%
A $\times$ B = B $\times$ A
0%
A $\times$ B = {(x $\times$ y) : x A and y B}
0%
$R^{-1}$ = {(y, x) : (x, y) R}

$\times$ (B - C) =

0%
$(A \times B) - (A \times C)$
0%
$(A \times C) - (A \times B)$
0%
$(A \times B) \bigcup (A \times C)$
0%
$(A \times B) \bigcap (A \times C)$

R is a relation in A and (a, b)

$\notin$ r, implies (b, a)

$\notin$ R then R is said to be ____ relation

0%
symmetric
0%
contradictory
0%
skew symmetric
0%
none of these

Explanation

$R$ is a relation in

$A$ and

$(a,b)$ .

$\Rightarrow (b,a)\notin R$

Hence

$'R'$ is a skew symmetric relation.

Answer:-

$C$

The minimum number of elements that must be added to the relation

$R=\left \{ (1, 2),(2, 3) \right \}$ on the set of natural numbers, so that it is an equivalence is:

0%
$4$
0%
$7$
0%
$6$
0%
$5$

Explanation

Since

$R=\left \{ (1, 2),(2, 3) \right \}$
i.e.,

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

$B=\left \{ 2, 3 \right \}$
Now, if

$R_{r}$ is the reflexive relation, such that

$R_{1}=\left \{ (1, 2),(2, 3),(1, 1),(2, 2),(3, 3) \right \}$ has

$5$ elements.
Now, if R' is both symmetric and reflexive relation, then

$R_{2}=\left \{ (1, 2),(2, 3),(1, 1),(2, 2),(2, 1),(3, 2),(3, 3) \right \}$ has

$7$ elements
Again, if

$R_{3}$ is reflexive, symmetric and transitive all together, than

$R_{3}=\begin{Bmatrix} (1, 2) & (2, 3) & (1, 1) \\ (2, 2) & (2, 1) & (3, 2) \\ (3, 3) & (1, 3) & (3, 1)\end{Bmatrix}$
has

$9$ elements. Starting from

$2$ elements, therefore, the minimum number of elements to be added is

$7$ .

Let

$A=R-\left\{3\right\},B=R-\left\{1\right\}$ and

$f:A \rightarrow B$ defined by

$f(x)\displaystyle =\frac{x-2}{x-3}$ Is

$f$ bijective ?
If yes enter 1 else enter 0

0%
True
0%
False

Explanation

As monotonic and range

$=$ Codomain

$\Rightarrow$ Bijective

If A

$=$ {1, 2}, B

$=$ {3, 4}, then A

$\times$ B

$=$

0%
{(1, 3), (1, 4), (2, 3), (2, 4)}
0%
{(1, 1), (2, 2), (3, 3), (4, 4)}
0%
{(4, 1), (3, 1), (4, 2), (3, 2)}
0%
All the above

Explanation

${\textbf{Step -1: Define the Cartesian product.}}$

${\text{Cartesian product: If }}A{\text{ and }}B{\text{ are two non empty sets, then }}$

${\text{Cartesian product }}A \times B{\text{ is set of all ordered pairs }}\left( {a,b} \right)$ ${\text{such that }}a \in A{\text{ and }}b \in B.$

${\textbf{Step -2: Find the Cartesian product of given sets.}}$

${\text{We have given,}}$

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

${\text{So,}}$ $A \times B = \left\{ {\left( {1,3} \right),\left( {1,4} \right),\left( {2,3} \right),\left( {2,4} \right)} \right\}$

${\textbf{Hence, option A. }}\left\{ \mathbf{\left( {1,3} \right),\left( {1,4} \right),\left( {2,3} \right),\left( {2,4} \right)} \right\}$ ${\textbf{is correct answer.}}$

CBSE Questions for Class 11 Commerce Applied Mathematics Relations Quiz 4 - 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 4 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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