0

Cartesian product of sets - class-XI

Description: cartesian product of sets
Number of Questions: 74
Created by:
Tags: maths mathematics and statistics sets and relations sets, relations and functions relations and functions relations sets
Attempted 0/71 Correct 0 Score 0

If $A$ and $B$ are two sets containing four and two elements, respectively. Then the number of subsets of the set $A\times B$ each having at least three elements is

  1. $219$

  2. $256$

  3. $275$

  4. $510$


Correct Option: A
Explanation:

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

$B=\left{x,y\right}$
$A\times B=2\times 4=8$ Elements.
Total number of subsets of $A\times B$ having $3$ or more elements $=2^8=256$
$\Rightarrow 256-(1$ null set $+$ $8$ singleton set $+$ $^8e _2$having $2$ elements$)$
$=256-1-8-28$
$=219$

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

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

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

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

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


Correct Option: B
Explanation:

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

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

  1. $(0, 0)$

  2. $(2, 2)$

  3. $(0, 2)$

  4. $(2, 0)$


Correct Option: D
Explanation:

Ordered pair for co-ordinate of a point in $XY$ plane is written as $(x, y).$

So, option D is correct.

What is general representation of ordered pair for two variables $a$ and $b$?

  1. $a, b$

  2. $(a, b)$

  3. $(a), b$

  4. $a, (b)$


Correct Option: B
Explanation:

An ordered pair is written in the form $(x-$coordinate, $y-$coordinate$)$.

If $a$ and $b$ are two variables and $(a, b)=(b, a)$, then 

  1. $a = 0$

  2. $b = 0$

  3. $a = b$

  4. $\displaystyle a\pm b$


Correct Option: C
Explanation:

 $(a, b) = (b, a)$ only if $a = b.$.

So, option C is correct.

If $x$ and $y$ co-ordinate of a point is $(3, 10)$, then $y$ co-ordinate is

  1. $0$

  2. $3$

  3. $10$

  4. None of the above


Correct Option: C
Explanation:

$y$ coordinate will be second entry in ordered pair.

So, option C is correct.

If $x$ and $y$ coordinate of a point is $(3, 10)$, then the $x$ co-ordinate is

  1. $0$

  2. $3$

  3. $10$

  4. None of the above


Correct Option: B
Explanation:

Here, $x$ co-ordinate will be the first entry in ordered pair.

So, option B is correct.

If ordered pair $(a, b)$ is given as $(-2, 0)$, then $a =$

  1. $-2$

  2. $0$

  3. $2$

  4. None of the above


Correct Option: A
Explanation:

$a$ will be first entry in ordered pair $(a, b).$

So, option A is correct.

If $y$ is second entry and $x$ is first entry then its ordered pair will be ............

  1. $(x, y)$

  2. $(y, y)$

  3. $(y, x)$

  4. None of the above


Correct Option: A
Explanation:

If $x$ is first entry and $y$ is second, then its ordered pair will be $(x, y)$.


So, option A is correct.

Identify the first component of an ordered pair $(2, 1)$.

  1. $1$

  2. $2$

  3. $-1$

  4. $0$


Correct Option: B
Explanation:

In an ordered pair $(x,y)$, the first component is $x$ and the second component is $y$.

Therefore, in an ordered pair $(2,1)$, the first component is $2$.

Cartesian product of sets $A$ and $B$ is denoted by _______.

  1. $A \times B$

  2. $B \times A$

  3. $A \times A$

  4. $B \times B$


Correct Option: A
Explanation:

Cartesian product of Set $A$ and $B$ is denoted by $A\times B$.

Identify the first component of an ordered pair $(0, -1) $.

  1. $0$

  2. $-1$

  3. $2$

  4. $1$


Correct Option: A
Explanation:

In an ordered pair $(x,y)$, the first component is $x$ and the second component is $y$.

Therefore, in an ordered pair $(0,-1)$, the first component is $0$.

Find the second component of an ordered pair $(2, -3)$

  1. $2$

  2. $3$

  3. $0$

  4. $-3$


Correct Option: D
Explanation:

In an ordered pair $(x,y)$, the first component is $x$ and the second component is $y$.

Therefore, in an ordered pair $(2,-3)$, the second component is $-3$.

The ______ product of two sets is the set of all possible ordered pairs whose first component is a member of the first set and whose second component is a member of the second set.

  1. cartesian

  2. coordinate

  3. simple

  4. discrete


Correct Option: A
Explanation:

The cartesian product of two sets is the set of all possible ordered pairs whose first component is a member of the first set and whose second component is a member of the second set. 
Example:$ A = {1, 2} \quad B = {2}$
cartesian product, $A \times B = {(1,2), (2, 2)}$

If $A = {a, b}, B={1, 2, 3}$, find B $\times$ A

  1. $B$ $\times$ $A$$ = {(1, a), (2, a), (3, a), (1, b) (2, b), (3, b)}$

  2. $B$ $\times$ $A$$ = { (2, a), (3, a), (1, b) (2, b), (3, b)}$

  3. $B$ $\times$ $A$$ = {(1, a), (2, a), (3, a), (1, b) (2, b)}$

  4. None of these


Correct Option: A
Explanation:

To find B × A multiply each element of B with that of A & form an ordered pair.

 i.e. ordered pairs are (1,a); (2,a); (3,a); (1,b); (2,b); (3,b)
Therefore B × A = {  (1,a), (2,a), (3,a), (1,b), (2,b), (3,b)}

If A= {0, 1} and B ={1, 0}, then what is A x B equal to ?

  1. {(0, 1), (1, 0)}

  2. {(0, 0), (1, 1)}

  3. {(0, 1), (1, 0), (1, I)}

  4. A X A


Correct Option: D
Explanation:

$\left{ { 0,1 } \right} \times \left{ 1,0 \right} ={ \left{ (0,1),(0,0),(1,1),(1,0) \right}  }$

$\left{ { 0,1 } \right} \times \left{ 0,1 \right} ={ \left{ (0,0),(0,1),(1,0),(1,1) \right}  }$

So, $A\times B=A\times A$
Hence, D is correct.

If $A = {2, 3, 5}$ and $B = {5, 7}$, find the set with highest number of elements:

  1. $A \times B$

  2. $ B \times A$

  3. $A \times A$

  4. $B \times B$


Correct Option: C
Explanation:
$A=\left \{ 2,3,5 \right \}$

$B=\left \{ 5,7 \right \}$

$A\times B=\left \{ (2,5),(2,7),(3,5),(3,7),(5,5),(5,7) \right \}$

$B\times A=\left \{ (5,2),(5,3),(5,5),(7,2),(7,3),(7,5) \right \}$

$A\times A=\left \{ (2,2),(2,3),(2,5),(3,2),(3,3),(3,5),(5,2),(5,3),(5,5) \right \}$

$B\times B=\left \{ (5,5),(5,7),(7,5),(7,7) \right \}$

$\therefore A\times A$ has the highest number of elements

For two sets $A$ and $B$, $A\times B=B\times A$.

  1. True

  2. False


Correct Option: B
Explanation:

The given statement is false.

Example:-
Let us consider $A={1,2}$ and $B={3,4}$.

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

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

Form (1) and (2) it's evident that $A\times B\ne B\times A$.

Let A and B be sets containing 2 and 4 elements respecetively. The number of subsets $A \times B$ having 3 or more elements is 

  1. $219$

  2. $211$

  3. $256$

  4. $220$


Correct Option: A
Explanation:
Let $A=\left\{x,y\right\}$
$B=\left\{a,bc,d\right\}$
$A\times B$ has $2\times 4=8$ elements
Total substance of $A\times B={2}^{8}=256$
$\therefore\,$Total number of subsets of $A\times B$ having $3$ or more elements
$=256-\left(1\,null \,set+8\,single\,ton\,set-^{8}C _{2}\,having\,2\,elements\right)$
$=256-1-8-\dfrac{8!}{6!2!}$
$=256-1-8-\dfrac{8\times 7\times 6!}{6!2!}$
$=256-1-8-28=219$

If $A$ and $B$ are independent event such that $P(A \cap B')=\dfrac {3}{25}$ and $P(A' \cap B)=\dfrac {8}{25}$, then $P(A)=$

  1. $1/5$

  2. $3/8$

  3. $2/5$

  4. $4/5$


Correct Option: C
Explanation:
$A$ & $B$ are independent
$P(A\cap B)=\dfrac {3}{25}$ and $P(A'\cap B)=\dfrac {8}{25}\quad P(A)=(?)$
$\rightarrow \ P(A)+P(B)=1---(i)$ ($A$& $B$ are independent )
$\rightarrow \ P(A\cap B')=P(A)-P(A\cap B)$
$P(A)-P(A\cap B)=\dfrac {3}{25}----(ii)$
$\rightarrow \ P(A' \cap B)=P(B)-P(A\cap B)$
$P(B)-P(A\cap B)=\dfrac {8}{25}-----(iii)$
$\rightarrow \ $ solving equation $(ii)$ and $(iii)$
$\dfrac {\,\,\, P\left( A \right) -P\left( A\cap B \right) =\dfrac { 3 }{ 25 } \\\,\,\, P\left( B \right) -P\left( A\cap B \right) =\dfrac { 8 }{ 25 } \\ -\quad \,\,\,\,\,\,\,+\quad \quad\quad\quad- }{ P\left( A \right) -P\left( B \right) =\dfrac { 3 }{ 25 } -\dfrac { 8 }{ 25 } \\ P\left( A \right) -P\left( B \right) =\dfrac { 3-8 }{ 25 }  } $
                           $=\dfrac {-5}{25}$
$\rightarrow \ P(A)-P(B)=\dfrac {-1}{5}$
$P(A)-(1-P(A))=\dfrac {-1}{5}$
$P(A)-I+P(A)=\dfrac {-1}{5}$
$\therefore \ 2P(A)=\dfrac {-1}{5}+1$
$\therefore \ 2P(A)=\dfrac {4}{5}$
$\therefore \ P(A)=\dfrac {4}{5\times 2}$
$\therefore \ P(A)=\dfrac {2}{5}$


The number of ordered pairs (x, y) of natural numbers satisfy the equation $x^2+y^2+2xy-2018x-2018y-2019^o=0$ is?

  1. $0$

  2. $1009$

  3. $2018$

  4. $2019$


Correct Option: A

The number of ordered triplet $\left(x,y,z\right),x,y,z$ are positive integers satisfying $xyz=105$

  1. $15$

  2. $27$

  3. $54$

  4. $35$


Correct Option: A

If $\int\dfrac{2\cos x-\sin x+\lambda}{\cos x-\sin x-2}dx=A In\left|\cos x+\sin x-2\right|+Bx+C$. Then the ordered triplet $\left(A,B,\lambda\right)$, is 

  1. $\left(\dfrac{1}{2},\dfrac{3}{2},-1\right)$

  2. $\left(\dfrac{3}{2},\dfrac{1}{2},-1\right)$

  3. $\left(\dfrac{1}{2},-1, \dfrac{3}{2}\right)$

  4. $\left(\dfrac{3}{2},-1, \dfrac{1}{2}\right)$


Correct Option: A

If $A={1, 2, 3}$ and $B={3, 8}$, then $(A\cup B)\times (A\cap B)$ is

  1. ${(3, 1), (3, 2), (3, 3), (3, 8)}$

  2. ${(1, 3), (2, 3), (3, 3), (8, 3)}$

  3. ${(1, 2), (2, 2), (3, 3), (8, 8)}$

  4. ${(8, 3), (8, 2), (8, 1), (8, 8)}$


Correct Option: B
Explanation:

$A\cup B={1,2,3,8}$
$A\cap B={3}$
$\therefore (A\cup B)\times (A\cap B)$
$={(1,3),(2,3),(3,3),(8,3)}$

Let $ A= { 1,2,3,.......50} $ and $B={2,4,6.......100}$ .The number of elements $\left ( x, y \right )\in A\times B$ such that $x+y=50$

  1. $24$

  2. $25$

  3. $50$

  4. $75$


Correct Option: A
Explanation:

The elements will be
$(2,48),(48,2)$
$(4,46), (46,4)$
:
:
$(2n,50-2n), (50-2n,2n)$
Now we have
$2,4,6,8...$ upto $48$
This forms an A.P
The number terms is $24$.

If the cardinality of a set $A$ is $4$ and that of a set $B$ is $3$, then what is the cardinality of the set $A\Delta B$.

  1. $1$

  2. $5$

  3. $7$

  4. $Cannot\ be\ determined$


Correct Option: D
Explanation:

$n(A\Delta B)=n(A)+n(B)-n(A\cap B)$

Here we don`t know the $n(A\cap B)$

So the answer cannot be Determined.


If $(x, y) = (3, 5)$ ; then values of $x$  and $y $ are 

  1. 3 and 5

  2. 4 and 7

  3. -1 and 17

  4. 2 and 4


Correct Option: A
Explanation:

x has to be 3 and y has to be 5.

If $n(A) = 4$ and $n(B) = 5$, then $n(A \times  B) = $

  1. $20$

  2. $25$

  3. $4$

  4. $15$


Correct Option: A
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 $20$.

State True or False
Let $A = \{1, 2\}$ and $B = \{2, 3, 4\}$, then A $\times$ B = B $\times$ A ?
  1. True

  2. False


Correct Option: B
Explanation:

Product of two sets is the set of ordered pairs formed by mapping every element from the first set to every element of the second set.

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

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

Clearly, $ A \times B \neq B \times A $

If $A=\left{ 2,4,5 \right} , B=\left{ 7,8,9 \right} $ then $n(A\times B)$ is equal to-

  1. $6$

  2. $9$

  3. $3$

  4. $0$


Correct Option: B
Explanation:

Given $n(A)=3,$  and $n(B) =3$
Hence $n(A\times B) = 3\times 3=9$

If $A = \left{2,3\right}$ and $B = \left{1,2\right}$, then $A \times B$ is equal to 

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

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

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

  4. $\left{(1,2), (2,3)\right}$


Correct Option: A
Explanation:

If $A$ 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.

If $\displaystyle A=\left{ 2,4,5 \right} ,B=\left{ 7,8,9 \right} $ then $\displaystyle n\left( A \times B \right) $ is equal to

  1. $6$

  2. $9$

  3. $3$

  4. $0$


Correct Option: B
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.

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

  1. $7$

  2. $8$

  3. $9$

  4. $10$


Correct Option: C
Explanation:

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


Hence $n(A)$ must be a factor of $36$. only possible answer is $B:9$

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

  1. $p=q=0$

  2. $p=q$

  3. $p=2q$

  4. $p+q=0$


Correct Option: D
Explanation:

As the ordered pairs are equal,
$ 3p + q = p - q $
$  => 2p = -2q $
$ => 2p + 2q = 0 $
$ => p + q = 0 $


If $\displaystyle n\left ( P\times Q \right )=0$  then n(P) can possibly be

  1. 0

  2. 10

  3. 20

  4. Any value


Correct Option: D
Explanation:

$ n(P) $ can be of any value as we are not sure of $ n(Q) $
Hence, $ n(P) $ can take any of the given values.

If $A=\left {1, 2,3\right }$ and $B=\left {3,8\right }$, then $(A\cup B)\times (A\cap B)$ is equal to

  1. $\left {(8,3), (8,2), (8,1), (8,8)\right }$

  2. $\left {(1,2), (2,2), (3,3), (8,8)\right }$

  3. $\left {(3,1), (3,2), (3,3), (3,8)\right }$

  4. $\left {(1,3), (2,3), (3,3), (8,3)\right }$


Correct Option: D
Explanation:

Given, $A=\left {1, 2,3\right }$ and $B=\left {3,8\right }$,
Therefore, $A\cup B=\left {1,2,3\right }\cup \left {3,8\right }=\left {1,2,3,8\right }$
and $A\cap B=\left {1,2,3\right }\cap \left {3,8\right }=\left {3\right }$
$\therefore (A\cup B)\times (A\cap B)=\left {1,2,3,8\right }\times \left {3\right }$
$=\left {(1,3), (2,3), (3,3), (8,3)\right }$

What is the Cartesian product of $A = \left {1, 2\right }$ and $B = \left {a, b\right }$?

  1. $\left {(1, a), (1, b), (2, a), (b, b)\right }$

  2. $\left {(1, 1), (2, 2), (a, a), (b, b)\right }$

  3. $\left {(1, a), (2, a), (1, b), (2, b)\right }$

  4. $\left {(1, 1), (a, a), (2, a), (1, b)\right }$


Correct Option: C
Explanation:

If $A $ and $B$ are two non empty sets, then the Cartesian product $A \times B$ is set of all ordered pairs $(a,b)$ such that $a\in A$ and $b\in B$.


Given $A ={1,2}$ and $B = {a,b}$

Hence $A\times B = {(1,a),(1,b),(2,a),(2,b)}$ 

Let a relation $R$ be defined by $R=\left {(4,5), (1,4), (4,6), (7,6), (3,7)\right }$. The relation $R^{-1}\circ R$ is given by

  1. $\left {(1,1), (4,4), (7,4), (4,7), (7,7)\right }$

  2. $\left {(1,1), (4,4), (4,7), (7,4), (7,7),(3,3)\right }$

  3. $\left {(1,5), (1,6), (3,6)\right }$

  4. None of these


Correct Option: B
Explanation:
We have $R=\left \{(4,5), (1,4), (4,6), (7,6), (3,7)\right \}$.
$\therefore R^{-1}=\left \{(5,4), (4,1), (6,4), (6,7), (7,3)\right \}$
$(4,4)\in R^{-1}\circ R$ because $(4,5)\in R$ and $(5,4)\in R^{-1}$
$(1,1)\in R^{-1}\circ R$ because $(1,4)\in R$ and $(4,1)\in R^{-1}$
$(4,4)\in R^{-1}\circ R$ because $(4,6)\in R$ and $(6,4)\in R^{-1}$
$(4,7)\in R^{-1}\circ R$ because $(4,6)\in R$ and $(6,7)\in R^{-1}$
$(7,4)\in R^{-1}\circ R$ because $(7,6)\in R$ and $(6,4)\in R^{-1}$
$(7,7)\in R^{-1}\circ R$ because $(7,6)\in R$ and $(6,7)\in R^{-1}$
$(3,3)\in R^{-1}\circ R$ because $(3,7)\in R$ and $(7,3)\in R^{-1}$
$\therefore R^{-1}\circ R=\left \{(4,4), (1,1), (4,7), (7,4), (7,7), (3,3)\right \}$.
$\therefore$ The correct answer is $B$.

Given $(a - 2, b + 3) = (6, 8)$, are equal ordered pair. Find the value of $a$ and $b$.

  1. $a = 8$ and $b = 5$

  2. $a = 8$ and $b = 3$

  3. $a = 5$ and $b = 5$

  4. $a = 8$ and $b = 6$


Correct Option: A
Explanation:

By equality of ordered pairs, we have
$(a - 2, b + 3) = (6, 8)$
On equating we get
$a - 2 = 6$
$a = 8$
$b + 3 = 8$
$b = 5$
So, the value of$ a = 8 $ and $ b = 5.$

What is the second component of an ordered pair $(3, -0.2)$?

  1. $3$

  2. $0.2$

  3. $1$

  4. $-0.2$


Correct Option: D
Explanation:

In an ordered pair $(x,y)$, the first component is $x$ and the second component is $y$.

Therefore, in an ordered pair $(3,-0.2)$, the second component is $-0.2$.

What is the first component of an ordered pair $(1, -1)$?

  1. $1$

  2. $-1$

  3. $2$

  4. $0$


Correct Option: A
Explanation:

In an ordered pair $(x,y)$, the first component is $x$ and the second component is $y$.

Therefore, in an ordered pair $(1,-1)$, the first component is $1$.

Ordered pairs $(x, y)$ and $(-1, -1)$ are equal if $y = -1$ and $x =$ _____

  1. $1$

  2. $-1$

  3. $0$

  4. $2$


Correct Option: B
Explanation:

Given , $(x-y)=(-1,1)$

$x=-1, y=-1$
The value of $x=-1$.

Ordered pairs $(x, y)$ and $(3, 6)$ are equal if $x = 3$ and $y = ?$

  1. $3$

  2. $6$

  3. $-6$

  4. $-3$


Correct Option: B
Explanation:

Given 

$(x,y)= (3,6)$
$x=3$
$y=6$
The value of $x=3$ and $y=6$.

If $A \times B = {(3, a), (3, -1), (3, 0), (5, a), (5, -1), (5, 0)}$, find $A$.

  1. ${a, 5}$

  2. ${a, -1}$

  3. ${0, 5}$

  4. ${3, 5}$


Correct Option: D
Explanation:

$A\times B = {(3,a),(3,-1),(3,0),(5,a),(5,-1),(5,0)}$

$A={3,5}$
as we know A is set of all entries in ordered  pair $A\times B$.

$(x, y)$ and $(p, q)$ are two ordered pairs. Find the values of $x$ and $p$, if $(3x - 1, 9) = (11, p + 2)$

  1. $x = 4, p = 9$

  2. $x = 6, p = 7$

  3. $x = 4, p = 5$

  4. $x = 4, p = 7$


Correct Option: D
Explanation:

Given$(x,y)=(p,q)$
$(3x - 1, 9) = (11, p + 2)$
By equating 
$3x - 1 = 11$
$3x = 12$
$x = 4$
$9 = p + 2$
$p = 7$
So, the value of $x = 4, p = 7.$

 $(x, y)$ and $(p, q)$ are two ordered pairs. Find the values of $p$ and $y$, if $(4y + 5, 3p - 1) = (25, p + 1)$

  1. $p = 0, y = 5$

  2. $p = 1, y = 5$

  3. $p = 0, y = 1$

  4. $p = 1, y = 1$


Correct Option: B
Explanation:

Given $(x,y)=(p,q)$
$(4y + 5, 3p - 1) = (25, p + 1)$
On equating we get
$4y + 5 = 25$
$4y = 25 - 5$
$4y = 20$
$y = 5$
$3p - 1 = p + 1$
$3p - p = 1 + 1$
$2p = 2$
$p = 1$
So, the value of$ p = 1, y = 5$

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

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

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

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

  4. ${2, 1), (2, 2), (3, 1), (2, 2)}$


Correct Option: A
Explanation:

$A= {2,3}$

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

If $A \times B =$ ${(2, 4), (2, a), (2, 5), (1, 4), (1, a), (1, 5)}$, find $B$.

  1. ${4, 2, 5}$

  2. ${4, a, 5}$

  3. ${4, 1, 5}$

  4. ${2, a, 5}$


Correct Option: B
Explanation:

$ A\times B = {(2,4),(2,a),(2,5),(1,4),(1,a),(1,5)}$

$B={4,a,5}$
as we know B is a set of all second entries in ordered pair $A\times B$.

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 $B\times A$?

  1. $n$

  2. $n^2$

  3. $2n$

  4. Zero


Correct Option: B
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?

  1. $6$

  2. $8$

  3. $10$

  4. $12$


Correct Option: B
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$

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$?

  1. $6$

  2. $7$

  3. $12$

  4. $64$


Correct Option: C
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 $3\times 4=12$

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) $

  1. $8$

  2. $6$

  3. $2$

  4. $1$


Correct Option: C
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.

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?

  1. $0$

  2. $9$

  3. $15$

  4. $20$


Correct Option: D
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 two sets $A$ and $B$ are having $39$ elements in common, then the number of elements common to each of the sets $A\times B$ and $B\times A$ are

  1. ${ 2 }^{ 39 }$

  2. ${ 39 }^{ 2 }$

  3. $78$

  4. $351$


Correct Option: B
Explanation:

If set $A$ and set $B$ have $39$ common elements, then the number of common elements in set $A\times B$ and set $B\times A\,=39^2$

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

  1. $\quad \left( \cfrac { 1 }{ 2 } ,-\cfrac { 1 }{ 2 } ,1 \right) $

  2. $\left( 3,1,0 \right) $

  3. $\left( 1,-1,2 \right) $

  4. $\left( 3,-\cfrac { 5 }{ 2 } ,2 \right) $


Correct Option: A
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.

Suppose $S=\{1,2\}$  and $T=\{a,b\}$  then $T \times S$
  1. $(a,1),(a,2),(b,1),(b,2)$
    B×A={(a,1),(a,2),(b,1),(b,2)}

  2. $(1,a),(2,b),(b,1),(b,2)$

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

  4. None of the above


Correct Option: A
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  $A\times B$ .

  1. ${b,x}$

  2. ${b,y}$

  3. ${c,x}$

  4. All of the above


Correct Option: D
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 $A\times B$.

$M={0,1,2}$ and $N={1,2,3}$: find (N-M) $\times$(N $\cap$M)

  1. ${3,1}$

  2. ${3,2}$

  3. ${3,3}$

  4. None of the above


Correct Option: A,B
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={0,1,2} and N={1,2,3}, then (M $\cup$ N) $\times$(M-N) contains

  1. ${0,0}$

  2. ${1,0}$

  3. ${2,0}$

  4. ${3,0}$


Correct Option: A,B,C,D
Explanation:
$M=\{0,1,2\}$
$N=\{1,2,3\}$
$M\cup N=\{0,1,2,3\}$
$M-N=\{0\}$
$(M\cup N)\times (M-N)$ $=\{0,0\},\{1,0\},\{2,0\},\{3,0\}$

If $A={b,c,d}$ and $B={x,y}$. Find which of the following are elements of $A \times A$.

  1. ${b,b}$

  2. ${b,c}$

  3. ${b,d}$

  4. All of the above


Correct Option: D
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)=4 $ and  $n(B) =5$: $n(A \times B)=$

  1. $20$

  2. $10$

  3. $30$

  4. None of the above


Correct Option: A
Explanation:

If $n(A)=m$,and $n(B)=n$,then $n(A\times B)=mn$

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

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

  1. n(A)+n(B)=n(A+B)

  2. n(A)-n(B)=n(A+B)

  3. A$\times$B=mn

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


Correct Option: D
Explanation:

C is not even logical while in the first 2 cases:


If there are elements common in A and B L.H.S of the first statement will be greater than R.H.S.

Clearly option B is also untrue as the combination of A and B cannot have lesser number of elements than A alone.

The last option has to be true as cartesian product of 2 sets gives a matrix having n(A) and n(B) as columns and rows.whose product will give the number of elements in the matrix.

n (A $\times$ B) =

  1. n(A) x n(B)

  2. n(A $\bigcap$ B)

  3. n(A $\bigcup$ B)

  4. all of these


Correct Option: A

$\left (A \cap B  \right ) \times C$

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

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

  3. $\left (A \times B \right ) \cup \left (B \times C \right )$

  4. $\left (A \times B \right ) \cup \left (A \times C \right )$


Correct Option: B

$\left (A \cap B  \right ) \times C$


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

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

  3. $\left (A \times B \right ) \cup \left (B \times C \right )$

  4. $\left (A \times B \right ) \cup \left (A \times C \right )$


Correct Option: A

Which one of the statement is false ?

  1. $\phi \times A = \phi$

  2. A $\times$ B = B $\times$ A

  3. A $\times$ B = {(x $\times$ y) : x A and y B}

  4. $R^{-1}$ = {(y, x) : (x, y) R}


Correct Option: B

A $\times$ (B - C) =

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

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

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

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


Correct Option: A

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

  1. {(1, 3), (1, 4), (2, 3), (2, 4)}

  2. {(1, 1), (2, 2), (3, 3), (4, 4)}

  3. {(4, 1), (3, 1), (4, 2), (3, 2)}

  4. All the above


Correct Option: A
Explanation:

Given, $A={1,2}$ and $B={3,4}$.

Then,
$A\times B$
$={(1,3),(1,4),(2,3),(2,4)}$. [ Using (1) and (2)]

If ,$(x-1, y+2)= (7, 5)$ then values of $x$ and $y$ are

  1. $5$,$8$

  2. $8$,$3$

  3. $-1$,$5$

  4. $7$,$1$


Correct Option: B
Explanation:

as the given ordered pairs are equal,

$x-1=7$  and  $y+2=5$
$\therefore x=8$ and $y=3$

Ordered pairs (a, 3) and (5, x) are equal ,the values of $a$ and $x$ are

  1. $2$ and $4$

  2. $3$ and $6$

  3. $5$ and $3$

  4. $1$ and $-1$


Correct Option: C
Explanation:

If two ordered pairs are equal, then both x - coordinates are equal and both y-coordinates are equal.

Since  $ (a, 3)  =  (5,x)  $ then ,$ a = 5 , x = 6 $

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

  1. $3$

  2. $4$

  3. $5$

  4. $6$


Correct Option: B
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.

- Hide questions