Relations and Functions Test

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Relations and Functions Test - 25

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Weekly Quiz Competition

Question 1
1 / -0

If $$A=\{2, 4\}$$ and $$B=\{3, 4, 5\}$$ then $$(A\cap B)\times (A\cup B)$$ is

A
$$\{(2, 2), (3, 4), (4, 2), (5, 4)\}$$

B
$$\{(2, 3), (4, 3), (4, 5)\}$$

C
$$\{(2, 4), (3, 4), (4, 4), (4, 5)\}$$

D
$$\{(4, 2), (4, 3), (4, 4), (4, 5)\}$$

Solution

$$A\cap B=\{4\}$$
$$A\cup B=\{2,3,4,5\}$$
Hence
$$(A\cap B)\times(A\cup B)$$
$$=\{(4,2),(4,3),(4,4),(4,5)\}$$
Question 2
1 / -0

$$R$$ is a relation from $$\displaystyle \left \{ 11,12,13 \right \}$$ to $$\displaystyle \left \{ 8,10,12 \right \}$$ defined by $$y=x-3$$ then the $$R^{-1}$$ .

A
$$\displaystyle \left \{ (11,8),(13,10) \right \}$$

B
$$\displaystyle \left \{ (8,11),(10,13) \right \}$$

C
$$\displaystyle \left \{ (8,11),(9,12),(10,13) \right \}$$

D
$$None\ of\ these$$

Solution

$$x=\{11,12,13\}, y=\{8,10,12\}$$
Now, $$y=x-3$$
So, the relation $$R$$ becomes $$\{(11,8),(13,10)\}$$
So, $$R^{-1}$$ becomes $$\{(8,11),(10,13)\}$$
Question 3
1 / -0

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

A
$$\{(3, 1), (3, 2), (3, 3), (3, 8)\}$$

B
$$\{(1, 3), (2, 3), (3, 3), (8, 3)\}$$

C
$$\{(1, 2), (2, 2), (3, 3), (8, 8)\}$$

D
$$\{(8, 3), (8, 2), (8, 1), (8, 8)\}$$

Solution

$$A\cup B=\{1,2,3,8\}$$
$$A\cap B=\{3\}$$
$$\therefore (A\cup B)\times (A\cap B)$$
$$=\{(1,3),(2,3),(3,3),(8,3)\}$$
Question 4
1 / -0

If $$\displaystyle :n(A)= m, $$ then number of relations in $$A$$ are

A
$$\displaystyle 2^{m}$$

B
$$\displaystyle 2^{m}-2 $$

C
$$\displaystyle 2^{m^{2}} $$

D
None of these

Solution

Given $$\displaystyle :n(A)= m $$ Relation in a set $$A$$ is subset of

$$\displaystyle A\times A $$ For Number of subset of $$A$$ We have

$$\displaystyle n(A\times A) = n(A)\times n(A)= m^{2}$$
$$\displaystyle

\therefore$$ Number of subset of $$\displaystyle A\times A = 2^{m^{2}}$$

every subset of $$\displaystyle A\times A $$ is a relation

$$\displaystyle \therefore$$ Number of relations in A is $$\displaystyle

2^{m^{2}}= 2^{m\times m}$$
Question 5
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Let $$x,y$$ be real numbers. Let $$f(x,y)= |x+y| ;\ F [f(x,y)]= -f(x,y) $$ and $$\displaystyle G [f(x,y)]= -F [f(x,y)]$$ then which of the following is true ?

A
$$\displaystyle F [f(x,y)]\cdot G [f(x,y)]= -F [f(x,y)]\cdot G [f(x,y)]$$

B
$$\displaystyle F [f(x,y)]\cdot G [f(x,y)]> -F [f(x,y)]\cdot G [f(x,y)]$$

C
$$\displaystyle F [f(x,y)]\cdot G [f(x,y)]\neq G [f(x,y)]\cdot F [f(x,y)]$$

D
$$\displaystyle F [f(x,y)]+ G [f(x,y)]+ f(x,y)= f(-x,-y)$$

Solution

Given that,
$$\displaystyle f\left ( x, y \right ) = |x+y| $$
$$\displaystyle F\left [ f\left ( x, y \right ) \right ]= -f\left ( x, y \right ) \dots (1)$$
$$\displaystyle G\left [ f\left ( x, y \right ) \right ]= -F\left [ f\left ( x, y \right ) \right ] \dots (2)$$

Considering $$(1)$$

$$\displaystyle F\left [ f\left ( x, y \right ) \right ] -f\left ( x, y \right )=0 \dots (3)$$

Considering $$(2)$$ and $$(1)$$

$$\displaystyle G\left [ f\left ( x, y \right ) \right ]= f\left ( x, y \right ) = f\left ( -x, -y \right )\left ( \because f\left ( x, y \right )= \left | x+y \right | \right )$$

$$\therefore \displaystyle G\left [ f\left ( x, y \right ) \right ]= f\left ( -x, -y \right ) \dots (4)$$

Adding $$(3)$$ and $$(4)$$. We get,

$$\displaystyle F [f(x,y)]+ G [f(x,y)]+ f(x,y)= f(-x,-y)$$

$$\therefore Option (D) $$ is correct
Question 6
1 / -0

If $$\displaystyle f(x) = \cos(\log x) $$ then the value of $$\displaystyle f(x)\cdot f(y)-\frac{1}{2}\left [ f\left ( \frac{x}{y} \right )+f(xy) \right ]$$ equals

A
$$\displaystyle x^{2}$$

B
$$\displaystyle x^{2}+2x+1$$

C
$$0$$

D
None of these

Solution

Given $$\displaystyle f\left ( x \right )= \cos \left ( \log x \right )$$
$$\displaystyle

\therefore f\left ( y \right )= \cos \left ( \log y \right )$$
Now
$$\displaystyle f\left ( x \right )f\left ( y \right )-\frac{1}{2}\left [

f\left ( \frac{x}{y} \right )+f\left ( xy \right ) \right ]$$
$$\displaystyle

= \cos \left ( \log x \right )\cos \left ( \log y \right

)-\frac{1}{2}\left [ \cos \log \left ( \frac{x}{y} \right )+\cos \log

\left ( xy \right ) \right ]$$
$$\displaystyle = \cos \left ( \log x

\right )\cos \left ( \log y \right )-\frac{1}{2}\left [ \cos \left (

\log x-\log y \right )+\cos \left ( \log x+\log y \right ) \right ]$$
$$\displaystyle

= \cos A \cos B-\frac{1}{2}\left [ \cos \left ( A-B \right )+\cos \left

( A+B \right ) \right ]$$, where A$$\displaystyle = \log x, B= \log y$$
$$\displaystyle = \cos A \cos B-\frac{1}{2}\left ( 2\cos A \cos B \right )= 0$$
Question 7
1 / -0

If $$\displaystyle A=\left \{ 0,1,2,3,4,5 \right \}$$ and relation $$R$$ defined by $$a R b$$ such that $$2a+b=10$$ then $$ R^{-1}$$ equals

A
$$\displaystyle \left \{ (4,3),(2,4) ,(5,0) \right \}$$

B
$$\displaystyle \left \{ (3,4),(4,2) ,(5,0) \right \}$$

C
$$\displaystyle \left \{ (4,3),(2,4) ,(0,5) \right \}$$

D
$$\displaystyle \left \{ (4,3),(4,2) ,(5,0) \right \}$$

Solution

$$2a+b=10$$, $$2a$$ is always even and $$10$$ is even.So, $$b$$ should also be even.
Hence the solution set is $$\{(0,10),(1,8),(2,6),(3,4),(4,2),(5,0)\}$$
But $$A=\{0,1,2,3,4,5\}$$
So $$R$$ becomes $$\{(3,4),(4,2),(5,0)\}$$
Hence $$R^{-1}$$ is $$\{(4,3),(2,4),(0,5)\}$$
Question 8
1 / -0

If $$\displaystyle R=\{ (x,y):x, y \in Z ,x^{2}+y^{2}\leq 4 \}$$ is a relation in $$Z$$ then domain $$D$$ is

A
$$\displaystyle \left \{ -2,-1,0,1,2\right \}$$

B
$$\displaystyle \left \{ -2,-1,0 \right \}$$

C
$$\displaystyle \left \{ 0,1,2\right \}$$

D
None of these

Solution

Relation $$Z$$ consists of all the integral points inside and on the circle of radius $$2$$ and center $$(0,0)$$.

For all the points $$(x,y)$$ which lie inside and on the circle $$x\epsilon [-2,2]$$ and $$y\epsilon [-2,2]$$. Hence the domain is the integer values of $$x$$ i.e $$-2,-1,0,1,2.$$
Question 9
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Let $$\displaystyle A= \left \{ a,b,c \right \} $$ and $$\displaystyle B= \left \{ 4,5 \right \} $$ Consider a relation $$R$$ defined from set $$A$$ to set $$B$$ then $$R$$ is subset of

A
$$A$$

B
$$B$$

C
$$\displaystyle A\times B $$

D
$$\displaystyle B\times A $$

Solution

If A and B are two non empty sets, the relation from set $$A$$ to set $$B$$ is denoted as $$\displaystyle A\times B $$
Question 10
1 / -0

Let $$A$$ and $$B$$ be two finite sets having $$m$$ and $$n$$ elements respectively. Then the total number of mapping from $$A$$ to $$B$$ is

A
$$mn$$

B
$$\displaystyle 2^{mn}$$

C
$$\displaystyle m^{n}$$

D
$$\displaystyle n^{m}$$

Solution

Consider an element $$\displaystyle a\in A,$$ it can be assigned to any of the $$n $$ elements of $$B$$ i.e. it has $$n$$ images. Similarly each of the $$m$$ elements of $$A$$ can have $$n$$ images in $$B.$$ Hence the number of mapping is
$$\displaystyle n\times n\times n\times ...m \mbox{times}=n^{m}.$$

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Relations and Functions Test - 25

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Question 1

$$\{(2, 2), (3, 4), (4, 2), (5, 4)\}$$

$$\{(2, 3), (4, 3), (4, 5)\}$$

$$\{(2, 4), (3, 4), (4, 4), (4, 5)\}$$

$$\{(4, 2), (4, 3), (4, 4), (4, 5)\}$$

Solution

Question 2

$$\displaystyle \left \{ (11,8),(13,10) \right \}$$

$$\displaystyle \left \{ (8,11),(10,13) \right \}$$

$$\displaystyle \left \{ (8,11),(9,12),(10,13) \right \}$$

$$None\ of\ these$$

Solution

Question 3

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

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

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

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

Solution

Question 4

$$\displaystyle 2^{m}$$

$$\displaystyle 2^{m}-2 $$

$$\displaystyle 2^{m^{2}} $$

None of these

Solution

Question 5

$$\displaystyle F [f(x,y)]\cdot G [f(x,y)]= -F [f(x,y)]\cdot G [f(x,y)]$$

$$\displaystyle F [f(x,y)]\cdot G [f(x,y)]> -F [f(x,y)]\cdot G [f(x,y)]$$

$$\displaystyle F [f(x,y)]\cdot G [f(x,y)]\neq G [f(x,y)]\cdot F [f(x,y)]$$

$$\displaystyle F [f(x,y)]+ G [f(x,y)]+ f(x,y)= f(-x,-y)$$

Solution

Question 6

$$\displaystyle x^{2}$$

$$\displaystyle x^{2}+2x+1$$

$$0$$

None of these

Solution

Question 7

$$\displaystyle \left \{ (4,3),(2,4) ,(5,0) \right \}$$

$$\displaystyle \left \{ (3,4),(4,2) ,(5,0) \right \}$$

$$\displaystyle \left \{ (4,3),(2,4) ,(0,5) \right \}$$

$$\displaystyle \left \{ (4,3),(4,2) ,(5,0) \right \}$$

Solution

Question 8

$$\displaystyle \left \{ -2,-1,0,1,2\right \}$$

$$\displaystyle \left \{ -2,-1,0 \right \}$$

$$\displaystyle \left \{ 0,1,2\right \}$$

None of these

Solution

Question 9

$$A$$

$$B$$

$$\displaystyle A\times B $$

$$\displaystyle B\times A $$

Solution

Question 10

$$mn$$

$$\displaystyle 2^{mn}$$

$$\displaystyle m^{n}$$

$$\displaystyle n^{m}$$

Solution

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