Relations and Functions Test

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Relations and Functions Test - 23

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Weekly Quiz Competition

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
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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
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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
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$$n(A)=4 $$ 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
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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
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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
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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
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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
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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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SHARING IS CARING

Question 1

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

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

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

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

Solution

Question 2

$$\{5,6\}$$

$$\{5,8\}$$

$$\{5,10\}$$

All of the above

Solution

Question 3

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

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

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

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

Solution

Question 4

$$20$$

$$10$$

$$30$$

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

$$2^{mn}$$

$$2^{m+n}$$

$$n^{m}$$

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

$$2^{n}$$

$$n^{2}$$

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

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

$$\phi \times A = \phi$$

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

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

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

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