Relations and Functions Test

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Relations and Functions Test - 23

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Question 1
1 / -0

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

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

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

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

D
$\left (A \times B \right ) \cup \left (A \times C \right )$

Solution
Question 2
1 / -0

Given $M=\{5,6,7\}$ and $N=\{6,8,10\}$ find element of $(M\cup N)\times N$

A
$\{5,6\}$

B
$\{5,8\}$

C
$\{5,10\}$

D
All of the above

Solution

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
Question 3
1 / -0

If $f(x) \displaystyle = \frac{2^x+2^{-x}}{2}$ , then $f(x+y) \cdot f(x-y)=$ ____________

A
$\displaystyle \frac{1}{4} [f(2x)+f(2y)]$

B
$\displaystyle \frac{1}{2} [f(2x)+f(2y)]$

C
$\displaystyle \frac{1}{2} [f(2x)-f(2y)]$

D
$\displaystyle \frac{1}{4} [f(2x)-f(2y)]$

Solution

Given that $f(x)=\dfrac { { 2 }^{ x }+{ 2 }^{ -x } }{ 2 }$
$=\dfrac { { 2 }^{ x }+\frac { 1 }{ { 2 }^{ x } } }{ 2 }$

$=\dfrac { { 2 }^{ 2x }+1 }{ { 2 }^{ x+1 } }$

$\therefore f(x+y)=\dfrac { { 2 }^{ 2(x+y) }+1 }{ { 2 }^{ x+y+1 } }$

$=\dfrac { 1 }{ 2 } \left[ { 2 }^{ x+y }+{ 2 }^{ -(x+y) } \right]$

Similarly, $f(x-y)=\dfrac { 1 }{ 2 } \left[ { 2 }^{ x-y }+{ 2 }^{ -(x-y) } \right]$

$\therefore f(x+y).f(x-y)=\dfrac { 1 }{ 2 } \left[ { 2 }^{ x+y }+{ 2 }^{ -x-y } \right] .\dfrac { 1 }{ 2 } \left[ { 2 }^{ x-y }+{ 2 }^{ -x+y } \right]$

$=\dfrac { 1 }{ 4 } \left( { 2 }^{ 2x }+{ 2 }^{ 2y }+{ 2 }^{ -2y }+{ 2 }^{ -2x } \right)$

$=\dfrac { 1 }{ 4 } \left[ \left( { 2 }^{ 2x }+{ 2 }^{ 2y })+({ 2 }^{ -2y }+{ 2 }^{ -2x } \right) \right]$

$=\dfrac { 1 }{ 4 } \left[ 2f(2x)+2f(2y) \right]$

$=\dfrac { 1 }{ 2 } \left[ f(2x)+f(2y) \right]$
Question 4
1 / -0

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

A
$20$

B
$10$

C
$30$

D
None of the above

Solution

If $n(A)=m$ ,and $n(B)=n$ ,then $n(A\times B)=mn$
so $n(A\times B)=5.4=20$
Question 5
1 / -0

n (A $\times$ B) =

A
n(A) x n(B)

B
n(A $\bigcap$ B)

C
n(A $\bigcup$ B)

D
all of these

Solution
Question 6
1 / -0

Let A and B be finite sets containing m and n elements respectively. The number of relations that can be defined from A and B is:

A
mn

B
$2^{mn}$

C
$2^{m+n}$

D
$n^{m}$

Solution

Here, $O(A)=m$ and $O(B)=n$ .
Hence $O(A×B)=mn$
Since every subset of $A×B$ is a relation from $A$ to $B$ , therefore, number of relations from A to B is equal to the number of the subsets of $A×B$ , i.e., $2^{m{n}}$
Question 7
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Which of the following statements is true?

A
The x-axis is a vertical line

B
The y-axis is a horizontal line

C
The scale on both axes must be the same in a Cartesian plane

D
The point of intersection between the x-axis and the y-axis is called the origin

Solution

Since, Origin is the point of intersection of $x$ and $y,$ we can say that option $D$ is correct.
Question 8
1 / -0

Let A be a finite set containing n distinct elements. The number of relations that can be defined on A is:

A
$2^{n}$

B
$n^{2}$

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

D
2n

Solution

Since $A$ contains $n$ distinct elements, therefore, $A×A$ contains $n×n=n^2$ distinct elements. since every subset of $A×A$ is a relation on $A$ ., therefore, number of relations on $A$ is equal to the order of the power set of $A×A$ , is equal to the order of the power set of $A×A$ , i.e., $2^{n^{2}}$
Question 9
1 / -0

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

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

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

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

D
All the above

Solution

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

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

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

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

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

${\textbf{Hence, option A. }}\left\{ \mathbf{\left( {1,3} \right),\left( {1,4} \right),\left( {2,3} \right),\left( {2,4} \right)} \right\}$ ${\textbf{is correct answer.}}$
Question 10
1 / -0

Which one of the statement is false ?

A
$\phi \times A = \phi$

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

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

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

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Relations and Functions Test - 23

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Relations and Functions Test - 23

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Question 1

(A×B)∩(B×C)\left (A \times B \right ) \cap \left (B \times C \right )(A×B)∩(B×C)

(A×C)∩(B×C)\left (A \times C \right ) \cap \left (B \times C \right )(A×C)∩(B×C)

(A×B)∪(B×C)\left (A \times B \right ) \cup \left (B \times C \right )(A×B)∪(B×C)

(A×B)∪(A×C)\left (A \times B \right ) \cup \left (A \times C \right )(A×B)∪(A×C)

Solution

Question 2

{5,6}\{5,6\}{5,6}

{5,8}\{5,8\}{5,8}

{5,10}\{5,10\}{5,10}

All of the above

Solution

Question 3

14[f(2x)+f(2y)]\displaystyle \frac{1}{4} [f(2x)+f(2y)]41​[f(2x)+f(2y)]

12[f(2x)+f(2y)]\displaystyle \frac{1}{2} [f(2x)+f(2y)]21​[f(2x)+f(2y)]

12[f(2x)−f(2y)]\displaystyle \frac{1}{2} [f(2x)-f(2y)]21​[f(2x)−f(2y)]

14[f(2x)−f(2y)]\displaystyle \frac{1}{4} [f(2x)-f(2y)]41​[f(2x)−f(2y)]

Solution

Question 4

202020

101010

303030

None of the above

Solution

Question 5

n(A) x n(B)

n(A ⋂\bigcap⋂ B)

n(A ⋃\bigcup⋃ B)

all of these

Solution

Question 6

mn

2mn2^{mn}2mn

2m+n2^{m+n}2m+n

nmn^{m}nm

Solution

Question 7

The x-axis is a vertical line

The y-axis is a horizontal line

The scale on both axes must be the same in a Cartesian plane

The point of intersection between the x-axis and the y-axis is called the origin

Solution

Question 8

2n2^{n}2n

n2n^{2}n2

2n22^{n^{2}}2n2

2n

Solution

Question 9

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

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

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

All the above

Solution

Question 10

ϕ×A=ϕ\phi \times A = \phiϕ×A=ϕ

A ×\times× B = B ×\times× A

A ×\times× B = {(x ×\times× y) : x A and y B}

R−1R^{-1}R−1 = {(y, x) : (x, y) R}

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

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

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

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

$\{5,6\}$

$\{5,8\}$

$\{5,10\}$

$\displaystyle \frac{1}{4} [f(2x)+f(2y)]$

$\displaystyle \frac{1}{2} [f(2x)+f(2y)]$

$\displaystyle \frac{1}{2} [f(2x)-f(2y)]$

$\displaystyle \frac{1}{4} [f(2x)-f(2y)]$

$20$

$10$

$30$

n(A $\bigcap$ B)

n(A $\bigcup$ B)

$2^{mn}$

$2^{m+n}$

$n^{m}$

$2^{n}$

$n^{2}$

$2^{n^{2}}$

$\phi \times A = \phi$

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

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

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