Relations and Functions Test

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Relations and Functions Test - 16

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Weekly Quiz Competition

Question 1
1 / -0

Let the number of elements of the sets $$A$$ and $$B$$ be $$p$$ and $$q$$ respectively. Then, the number of relations from the set $$A$$ to the set $$B$$ is

A
$${ 2 }^{ p+q }$$

B
$${ 2 }^{ pq }$$

C
$$p+q$$

D
$$pq$$

Solution

Given, number of elements of $$A$$ and $$B$$ are $$p$$ and $$q$$ respectively.
$$\therefore$$ The number of relations from the set $$A$$ to the set $$B$$ is $${2}^{pq}$$.
Question 2
1 / -0

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

A
$$2$$

B
$$3$$

C
$$0$$

D
$$-3$$

Solution

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 $$-3$$.
Question 3
1 / -0

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.

A
cartesian

B
coordinate

C
simple

D
discrete

Solution

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, $$A \times B = {(1,2), (2, 2)}$$
Question 4
1 / -0

Let R be the relation in the set N given by = {(a, b): a = b - 2, b > 6}. Choose the correct answer

A
$$(2, 4) \epsilon R$$

B
$$(3, 8) \epsilon R$$

C
$$(6, 8) \epsilon R$$

D
$$(8, 7) \epsilon R$$

Solution

Firstly b should be greater than 6 & also difference between b& a should be 2 , option C is satisfing both the conditions hence the correct answer is C
Question 5
1 / -0

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

A
$$A \times B$$

B
$$B \times A$$

C
$$A \times A$$

D
$$B \times B$$

Solution

Cartesian product of Set $$A$$ and $$B$$ is denoted by $$A\times B$$.
Question 6
1 / -0

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

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

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

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

D
None of these

Solution

$$ (A\times B) = \left \{ (1,3),(1,4),(2,3),(2,4),(3,3),(3,4) \right \} $$
$$ (B\times A) = \left \{ (3,1),(3,2),(3,3),(4,1),(4,2),(4,3) \right \} $$
$$ \Rightarrow (A\times B)\cup (B\times A)= \left \{ (1,3),(1,4),(2,3),(2,4),(3,3),(3,4),(3,1),(3,2),(4,1),(4,2),(4,3) \right \} $$
Question 7
1 / -0

Which of the following functions is/are constant ?

A
$$f(x)=x^{2}+2$$

B
$$f(x)=x+\dfrac{1}{x}$$

C
$$f(x)=7$$

D
$$f(x)=6+x$$

Solution

$$f(x)=7$$ is constant function as its values do not depend on the variable $$x$$.
Its value is $$7$$ for any value of $$x$$.
Option $$C$$ is correct.
Question 8
1 / -0

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

A
{(0, 1), (1, 0)}

B
{(0, 0), (1, 1)}

C
{(0, 1), (1, 0), (1, I)}

D
A X A

Solution

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

So, $$A\times B=A\times A$$
Hence, D is correct.
Question 9
1 / -0

Let $$f:\mathbb{R}\rightarrow \mathbb{R}$$ be a function such that for any irrational number $$r,$$ and any real number $$x$$ we have $$f(x)=f(x+r)$$. Then, $$f$$ is

A
an identity function

B
a constant function

C
a zero function

D
onto function

Solution

A constant function is a function that has the same output value no matter what your input value is. Because of this, a constant function has the form y=k where k is a constant(a single value that doesn't change).
So in here our input is $$x$$ i f we change our input to say $$x+r$$
and still output does'nt change i.e.
$$f(x)=f(x+r)$$
it means it is a constant function,
example:
$$ f(x)=5=5(x)^{0} $$
$$ f(x+r)=5(x+r)^{0}$$
$$ f(x+r)=5\times 1 $$
$$ f(x+r)=5=f(x) $$
Question 10
1 / -0

If $$(x)={ x }^{ 2 }-2x+4$$ then the set of values of $$x$$ satisfying $$f\left( x-1 \right) =f\left( x+1 \right) $$ is

A
$$\left\{ -1 \right\} $$

B
$$\left\{ -1,1 \right\} $$

C
$$\left\{ 1 \right\} $$

D
$$\left\{ 1,2 \right\} $$

Solution

$$f(x-1) = f(x+1) $$
$$\implies (x-1)^{2} -2(x-1) +4 = (x+1)^2-2(x+1) +4$$
$$\implies x^2-2x+1 -2x+2 +4 = x^2+2x+1 -2x-2 +4 $$
$$\implies 4x = 4 $$
$$\implies x = 1 $$
Therefore, the answer is option (c)

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Relations and Functions Test - 16

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Relations and Functions Test - 16

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Question 1

$${ 2 }^{ p+q }$$

$${ 2 }^{ pq }$$

$$p+q$$

$$pq$$

Solution

Question 2

$$2$$

$$3$$

$$0$$

$$-3$$

Solution

Question 3

cartesian

coordinate

simple

discrete

Solution

Question 4

$$(2, 4) \epsilon R$$

$$(3, 8) \epsilon R$$

$$(6, 8) \epsilon R$$

$$(8, 7) \epsilon R$$

Solution

Question 5

$$A \times B$$

$$B \times A$$

$$A \times A$$

$$B \times B$$

Solution

Question 6

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

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

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

None of these

Solution

Question 7

$$f(x)=x^{2}+2$$

$$f(x)=x+\dfrac{1}{x}$$

$$f(x)=7$$

$$f(x)=6+x$$

Solution

Question 8

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

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

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

A X A

Solution

Question 9

an identity function

a constant function

a zero function

onto function

Solution

Question 10

$$\left\{ -1 \right\} $$

$$\left\{ -1,1 \right\} $$

$$\left\{ 1 \right\} $$

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

Solution

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