If $$A = \{1, 2 \}$$ and $$B = \{3, 4\}$$ then find $$A \times B$$

$$A \times B = \{(1,3),(1,4),(2,3),(2,4)\}$$

$$A \times B = \{(1,3),(1,2),(2,3),(2,4)\}$$

$$A \times B = \{(1,3),(1,4),(2,1),(2,4)\}$$

$$A \times B = \{(1,3),(1,4),(2,3),(2,1)\}$$

MCQ Questions for Class 11 Commerce Applied Mathematics Relations Quiz 7

Let

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

$Y=\left\{ 1,2,3,4 \right\}$ . Which of the following is a relation from

$X$ to

$Y$ .

0%
${R}_{1}=\left\{ (x,y)| y=2+x, x\in X, y\in Y \right\}$
0%
${R}_{2}=\left\{ (1,1),(2,1),(3,3),(4,3),(5,5) \right\}$
0%
${R}_{3}=\left\{ (1,1),(1,3),(3,5),(3,7),(5,7) \right\}$
0%
${R}_{4}=\left\{ (1,3),(2,5),(2,4),(7,9) \right\}$

Explanation

Given

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

$Y=\left\{ 1,2,3,4 \right\}$

$X\times Y=\{ (1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)$

$(3,1),(3,2),(3,3),(3,4),(4,1),(4,2),(4,3),(4,4)\}$

Now,

$R_1=\{(1,3),(2,2)\}$

Since,

$R_1\subset X\times Y$

Hence,

$R_1$ is a relation from

$X$ to

$Y$

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

Since,

$(5,5) \in R_2$ but

$\notin X\times Y$

So,

$R_2$ is not a relation from

$X$ to

$Y$

${R}_{3}=\left\{ (1,1),(1,3),(3,5),(3,7),(5,7) \right\}$

Since,

$(3,5),(3,7),(5,7) \in R_3$ but

$\notin X\times Y$

So,

$R_3$ is not a relation from

$X$ to

$Y$

${R}_{4}=\left\{ (1,3),(2,5),(2,4),(7,9) \right\}$

Since,

$(2,5),(7,9) \in R_4$ but

$\notin X\times Y$

So,

$R_4$ is not a relation from

$X$ to

$Y$

Let

$L$ be the set of all straight lines in the Euclidean plane. Two lines

${l}_{1}$ and

${l}_{2}$ are said to be related by the relation

$R$ if

${l}_{1}$ is parallel to

${l}_{2}$ . Then the relation

$R$ is-

0%
Reflexive
0%
Symmetric
0%
Transitive
0%
Equivalence

Explanation

$l_1 R l_1 \Rightarrow l_1$ is parallel to itself. Reflexive

$l_1 Rl_2\Rightarrow l_2Rl_1$ which is true. Symmetric

$l_1Rl_2$ and

$l_2Rl_3 \Rightarrow l_1Rl_3$ which is true. Transitive. Hence all options are correct.

Let

$g\left( x \right)=1+x-\left[ x \right]$ and

$f\left( x \right)=\begin{cases} \begin{matrix} -1 & x<0 \end{matrix} \\ \begin{matrix} 0 & x=0 \end{matrix} \\ \begin{matrix} 1 & x>0 \end{matrix} \end{cases}$ . Then for all

$x,f\left[ g\left( x \right) \right]$ is equal to

0%
$x$
0%
$1$
0%
$f(x)$
0%
$g(x)$

Explanation

We have,

$g\left( x \right)=\begin{cases} 1+n-n=1,x=n\in Z \\ 1+n+k-n=1+k,x=n+k \end{cases}$
where

$n\in Z,0<k<1$
Now

$f\left[ g\left( x \right) \right] =\begin{cases} \begin{matrix} -1 & g\left( x \right)<0 \end{matrix} \\ \begin{matrix} 0 & g\left( x \right)=0 \end{matrix} \\ \begin{matrix} 1 & g\left( x \right)>1 \end{matrix} \end{cases}$
Clearly,

$g(x)>0$ for all

$x$ . So

$f\left[ g\left( x \right) \right] =1$ , for all

$x$ .

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

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

$R$ is 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 than

$y$ .

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

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

from the given statement

$N$ denotes the natural number and

$W$ denotes the whole number, so which statement in the following is correct

0%
N=W
0%
N $\subset$ W
0%
W $\subset$ N
0%
N $\cong$ W

Explanation

Natural numbers are naturally occurring counts of an object

$1, 2, 3, 4,…$ forming an infinite list in which the first number is

$1$ and every next number is equal to the preceding number

$+1.$

Whole number include the set of all Natural numbers and also

$0.$

Let

$N$ and

$W$ elements of natural and whole numbers.

$N=\{1,2,3,4,....\infty\}$

$W=\{0,1,2,3,4,...\infty\}$

So, here we can say

$N$ is subset of

$W.$

$\therefore$

$N\subset W$ is correct statement.

Which one of the following relations on

$R$ is equivalence relation

0%
$x \space R_1 \space y \Leftrightarrow |x| = |y|$
0%
$x \space R_2 \space y \Leftrightarrow x \ge y$
0%
$x \space R_3 \space y \Leftrightarrow x | y$
0%
$x \space R_4 \space y \Leftrightarrow x < y$

Explanation

Let's take option A,

$x \space R_1 \space y \Leftrightarrow |x| = |y|$
Now,

$|x|=|x|$

$\Rightarrow xR_{1}x$
Hence,

$R_1$ is reflexive.

Now, let

$xR_1 y$

$\Rightarrow |x|=|y|$
or

$|y|=|x|$

$\Rightarrow yR_{1}x$
Hence,

$R_1$ is symmetric

Now, let

$xR_1 y, yR_1 z$

$\Rightarrow |x|=|y|, |y|=|z|$

$\Rightarrow |x|=|z|$

$\Rightarrow xR_{z}x$
Hence,

$R_1$ is transitive.
Hence,

$R_1$ is an equivalence relation.

Clearly,

$R_2$ is not an equivalence relation as

$2 \ge 3$ does not implies

$3\ge 2$
Also,

$R_3$ is not an equivalence relation as

$x$ divides

$y$ , does not implies

$y$ divides

$x$
Also

$R_4$ is not an equivalence relation because

$2 <3$ but

$3 >2$

Given the relation

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

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

$R$ make it an equivalence relation is

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

Explanation

$R$ is reflexive if it contains

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

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

$\therefore R$ is symmetric if

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

Now,

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

$R$ will be transitive if

$(3,1); (1,3)\in R$ . Thus,

$R$ becomes and

equivalence relation by adding

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

the total number of ordered pairs is

$7$ .

$A$ 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

0%
$2^5$
0%
$2^{10} - 1$
0%
$2^{12} - 1$
0%
$none\ of\ these$

Explanation

`if

$A$ and

$B$ are two sets which contains 3 and 4 elements respectively.

Then no of elements in relation from

$A$ to

$B$ is

$2^(3\times 4)$

means

$2^{12}$

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

$B = \left\{1,2\right\}$ , then

$A \times B$ is equal to

0%
$\left\{(2,1), (2,2), (3,1), (3,2)\right\}$
0%
$\left\{(1,2), (1,3), (2,2), (2,3)\right\}$
0%
$\left\{(2,1), (3,2)\right\}$
0%
$\left\{(1,2), (2,3)\right\}$

Explanation

$A$ and

$B$ are any two non-empty sets.
then

$A\times B$

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

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

$B = \left\{1,2\right\}$

$A \times B$

$=\left\{(2,1), (2,2), (3,1), (3,2)\right\}$
Hence, option A.

Let

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

$Y = \left\{1,3,5,7,9\right\}$ . Which of the following is relations from

$X$ to

$Y$

0%
$R_1 = \left\{(x,y) | y = 2x+1, x \in X, y \in Y\right\}$
0%
$R_2 = \left\{(1,1),(2,1),(3,3),(4,3),(5,5)\right\}$
0%
$R_3 = \left\{(1,1),(1,3),(3,5),(3,7),(5,7)\right\}$
0%
$R_4 = \left\{(1,3),(2,5), (2,4), (7,9)\right\}$

Explanation

We have

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

$Y = \left\{1,3,5,7,9\right\}$ .

$X \times Y=\{(1,1),(1,3),(1,5),(1,7),(1,9), (2,1),(2,3),(2,5),$

$(2,7),(2,9), (3,1),(3,3),(3,5),(3,7),(3,9),(4,1),(4,3),(4,5),(4,7),(4,9)\}$

Let's take option A,

$R_1 = \left\{(x,y) | y = 2x+1, x \in X, y \in Y\right\}$

$R_{1}= \{(1,3),(2,5),(3,7), (4,9)\}$
Since,

$R_1 \subseteq X\times Y$
So,

$R_1$ is a relation from

$X$ to

$Y$

Clearly,

$R_2$ is not a relation from

$X$ to

$Y$ as

$(5,5) \notin X\times Y$
Similarly,

$R_3$ is not a relation from

$X$ to

$Y$ as

$(5,7) \notin X\times Y$
In the same manner,

$R_4$ is also not a relation from

$X$ to

$Y$ as

$(7,9) \notin X\times Y$

$R$ is a relation from a finite set

$A$ having

$m$ elements to a finite set

$B$ having

$n$ elements, then the number of relations from

$A$ to

$B$ is:

0%
$2^{mn}$
0%
$2^{mn} - 1$
0%
$2mn$
0%
$m^n$

Explanation

$R$ is a relation from a finite set

$A$ having

$m$ elements to a finite set

$B$ having

$n$ elements.
Number of elements in

$A\times B$

$=mn$
No of relations

$=$ number of subsets of

$A\times B$

$=2^{mn}$
Hence, option A is correct.

The relation

$R$ is defined in

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

$a\ R$

$b$ if

$|a^2 - b^2| \le 5$ . Which of the following is false?

0%
$R = \left\{(1,1), (2,2), (3,3), (2,1), (1,2), (2,3), (3,2)\right\}$
0%
$R^{-1} = R$
0%
Domain of $R = \left\{1,2,3\right\}$
0%
Range of $R = \left\{5\right\}$

Explanation

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

$R^{-1} = R$
Domain

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

$=\{1,2,3\}$
Hence option 'D' is correct choice.

$R = \{(x, y):3x + 2y = 15 \text{ and }x\,, y\,\in \, N\}$ , the range of the relation R is .........

0%
$\{1, 2\}$
0%
$\{1, 2,..., 5\}$
0%
$\{1, 2,..., 7\}$
0%
$\{3, 6\}$

Explanation

Since x and y belong to natural numbers we need to find positive integral solutions for

$3x+2y=15$

$2y=15-3x$

$\Rightarrow y =\dfrac{15-3x}{2}$

For

$x=0,2,4$

$y$ will become fractional which is not possible

For

$x\geq 5$ y becomes 0 or negative which again is not possible.

Hence for

$x=1,3$ y= 6,3$$ respectively.

Hence the range is

$\{3,6\}$

Let

$n(A) = n$ . Then the number of all relations on

$A$ is

0%
$\displaystyle 2^{n}$
0%
$\displaystyle 2^{\left ( n \right )!}$
0%
$\displaystyle 2^{n^{2}}$
0%
none

Explanation

For any set

$A$ such that

$n(A)=n$
then number of all relations on

$A$ is

${ 2 }^{ n^{2} }$

As the total number of Relations that can be defined from a set

$A$ to

$B$ is the number of possible subsets of

$A \times B$ . If

$n(A) = p$ and

$n(B) = q$ then

$n(A \times B) = pq$ and the number of subsets of

$A \times B$ =

$2^{pq}$ .

$R$ is an equivalence relation in a set

$A$ , then

$R^{-1}$ is

0%
Reflexive but not symmetric
0%
Symmetric but not transitive
0%
An equivalence relation
0%
None of these

Explanation

Let R be a relation on A i.e.

$R\subseteq A\times A$

$R=\{(a,b)|a,b \in A \}$
Also, given

$R$ is equivalence relation,

Now, let

$R^{-1}=\{(b,a) | (a,b) \in R\}$
We will check whether

$R^{-1}$ is reflexive, symmetric, transitive or an equivalence relation.

Reflexive:
Since,

$R$ is reflexive

$\Rightarrow (a,a)\in R$

$\Rightarrow (a,a) \in R^{-1}$ (by def of

$R^{-1}$ )
Hence,

$R^{-1}$ is reflexive.

Symmetric: Let

$(b,a) \in R^{-1}$

$\Rightarrow (a,b) \in R$ (by def of

$R^{-1}$ )

$\Rightarrow (b,a) \in R$ (Since, R is symmetric)

$\Rightarrow (a,b) \in R^{-1}$ (by def of

$R^{-1}$ )
Hence,

$R^{-1}$ is symmetric.

Transitive : Let

$(b,a), (a,c) \in R^{-1}$

$\Rightarrow (a,b), (c,a) \in R$ (by def of

$R^{-1}$ )
or

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

$\Rightarrow (c,b) \in R$ (since, R is transitive.)

$\Rightarrow (b,c) \in R^{-1}$
Hence,

$R^{-1}$ is transitive.
Hence,

$R^{-1}$ is an equivalence relation.

$\displaystyle A=\left\{ 2,4,5 \right\} ,B=\left\{ 7,8,9 \right\}$ then

$\displaystyle n\left( A \times B \right)$ is equal to

0%
$6$
0%
$9$
0%
$3$
0%
$0$

Explanation

$\displaystyle A=\left\{ 2,4,5 \right\} ,B=\left\{ 7,8,9 \right\}$

$\Rightarrow n(A)=3$ and

$n(B)=3$

$\therefore n(A\times B)=n(A)n(B)=9$

Hence, option B.

$(3p+q,p-q)=(p-q,3p+q)$ , then:

0%
$p=q=0$
0%
$p=q$
0%
$p=2q$
0%
$p+q=0$

Explanation

As the ordered pairs are equal,

$3p + q = p - q$

$=> 2p = -2q$

$=> 2p + 2q = 0$

$=> p + q = 0$

Let

$A = \left\{p,q,r\right\}$ . Which of the following is an equivalence relation in

$A$ ?

0%
$R_1 = \left\{(p,q), (q,r), (p,r), (p,p)\right\}$
0%
$R_2 = \left\{(r,q), (r,p), (r,r), (q,q)\right\}$
0%
$R_3 = \left\{(p,p), (q,q), (r,r), (p,q)\right\}$
0%
None of these

Explanation

Given

$A=\{p,q,r\}$

$R_1 = \left\{(p,q), (q,r), (p,r), (p,p)\right\}$

$R_1$ is not reflexive as it does not have

$(q,q),(r,r)$ . So,

$R_1$ is not an equivalence relation on A.

$R_2 = \left\{(r,q), (r,p), (r,r), (q,q)\right\}$

$R_2$ is not reflexive as it does not have

$(p,p)$ . So,

$R_2$ is not an equivalence relation on A.

$R_3 = \left\{(p,p), (q,q), (r,r), (p,q)\right\}$

$R_3$ is reflexive but not symmetric as

$(p,q)$ is there ,but not

$(q,p)$

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

$B = \{3, 4\}$ then find

$A \times B$

0%
$A \times B = \{(1,3),(1,2),(2,3),(2,4)\}$
0%
$A \times B = \{(1,3),(1,4),(2,3),(2,4)\}$
0%
$A \times B = \{(1,3),(1,4),(2,1),(2,4)\}$
0%
$A \times B = \{(1,3),(1,4),(2,3),(2,1)\}$

$\displaystyle R_{n}=\left \{ x:\frac{-1}{n}< x< \frac{1}{n} \right \}$ then

$\displaystyle R_{5}\cup R_{15}=$ ________

0%
$\displaystyle R_{5}$
0%
$\displaystyle R_{15}$
0%
$\displaystyle R_{3}$
0%
$\displaystyle R_{20}$

Explanation

$R_5 =$ {

$x:-\frac {1}{5} < x < \frac {1}{5}$ }
And

$R_{15} =$ {

$x:-\frac {1}{15} < x < \frac {1}{15}$ }

But,

$\frac {1}{15} < x < \frac {1}{15}$ is within the limits of

$\frac {1}{5} < x < \frac {1}{5}$

Hence,

$R_5 \cup R_{15} = \frac {1}{5} < x < \frac {1}{5} = R_5$

a R b if "a and b are animals in different zoological parks" then R is

0%
only reflexive
0%
only symmetric
0%
only transitive
0%
equivalence

R is a relation on set A then

$\displaystyle \left [ \left ( R^{-1} \right )^{-1} \right ]^{-1}$ is_______

0%
R
0%
$\displaystyle R^{-1}$
0%
$\displaystyle A\times A$
0%
None of these

Explanation

We know that Inverse of inverse of a Relation is the original Relation itself.

So,

$\left[ \left( { R }^{ -1 } \right) ^{ -1 } \right] ^{ -1 } = \left[ R \right] ^{ -1 }$

A relation which satisfies reflexive symmetric and transitive is ________ relation

0%
an identity
0%
a constant
0%
an equivalence
0%
None of these

The domain of the function f(x) =

$\displaystyle \frac{\left | x \right |-2}{\left | x \right |-3}$ is ________

0%
R
0%
R - {2, 3}
0%
R - {2, -2}
0%
R - {-3, 3}

Explanation

The function has a real value only when denominator of the fraction is not equal to zero.

So,

$\left| x \right| - 3 \neq = 0$

$=> \left| x \right| \neq = 3$

$=> x \neq = 3 ; -3$

So, the domain of the function is

$R -$ {

$-3,3$ }

$\displaystyle R=R^{-1}$ then the relation R is ________

0%
reflexive
0%
symmetric
0%
anti-symmetric
0%
transitive

Write the properties that the relation "is greter that" satisfies in the set of all positive integers

0%
Reflexive
0%
Symmetric
0%
Antisymmetric
0%
Transitive

Explanation

If in a relation, an element

$a$ is related to

$b$ , and

$b$ is related to a third element

$c$ , the relation

$R$ is said to be transitive if

$a$ is also related to

$c$

Clearly, when

$a > b$ and

$b > c$ then

$a$

$> c$
Hence, the relation is transitive.

$\displaystyle n\left ( A\times B \right )=36$ then n(A) can possibly be____

0%
$7$
0%
$8$
0%
$9$
0%
$10$

Explanation

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

Hence

$n(A)$ must be a factor of

$36$ . only possible answer is

$B:9$

The domain of the relation R =

$\displaystyle \left \{ \left ( x,y \right ):x,y\epsilon N \ and\ x+y\leq 3 \right \}$ is____

0%
{1,2,3}
0%
{1,2}
0%
{...-1,0,1,2,3}
0%
None of these

Explanation

The only natrual numbers, less than equal to

$3$ are

$1, 2, 3$

From these, only the sum of

$1 ,2$ gives

$3$

So the domain of he relation is {

$1,2$ }

The relation 'is a sister of' in the set of human beings is____

0%
only transitive
0%
only symmetric
0%
equivalent
0%
None of these

Explanation

A relation is said to be transitive when an element

$a$ is related to

$b$ and element

$b$ is related to

$c$ , then

$c$ is also related to

$a$

We know that if

$a$ is the sister of

$b$ and

$b$ is sister of

$c$ , then

$c$ is also a sister of

$a$

Thus, the relation " is a sister of " is a transitive relation.

The relation 'is a factor of' on the set of natural numbers is not___________

0%
reflective
0%
symmetric
0%
anti symmetric
0%
transitive

Explanation

A relation is said to be symmetric if an element a is related to b then b is also related to a.

Suppose factor of a number say

$10 = 1,2,5,10$
We can notice that factors of

$10$ are all these numbers, but the reverse is not true for each factor. That is

$10$ is not a factor of

$1$ or

$2$ or

$5$ .

So, this relation is not symmetric.

CBSE Questions for Class 11 Commerce Applied Mathematics Relations Quiz 7 - 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 7 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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