Relations Test 17

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Relations Test 17

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Weekly Quiz Competition

Question 1
1 / -0

A relation $$R$$ is defined on the set $$Z$$ of integers as follows: R=$$(x,y)$$ $$\displaystyle \in {R}:x^{2}+y^{2}= 25$$. Express $$R$$ and $$\displaystyle R^{-1}$$ as the sets of ordered pairs and hence find their respective domains.

A
$$0$$

B
Domain of $$\displaystyle R= \left \{ 0, \pm 3 \right \}= $$ domain of $$\displaystyle R^{-1}.$$

C
Domain of $$\displaystyle R= \left \{ 0, \pm 3, \pm 4 \right \}= $$ domain of $$\displaystyle R^{-1}.$$

D
Domain of $$\displaystyle R= \left \{ 0, \pm 3, \pm 4, \pm 5 \right \}= $$ domain of $$\displaystyle R^{-1}.$$

Solution

$$R=\left\{ \left( 0,5 \right) ,\left( 0,-5 \right) ,\left( 3,4 \right) ,\left( -3,4 \right) ,\left( 3,-4 \right) ,\left( -3,-4 \right) ,\\ \left( 4,3 \right) ,\left( -4,3 \right) ,\left( 4,-3 \right) ,\left( -4,3 \right) ,\left( 5,0 \right) ,\left( -5,0 \right) \right\} \\ { R }^{- 1 }=\left\{ \left( 5,0 \right) ,\left( -5,0 \right) ,\left( 4,3 \right) ,\left( 4,-3 \right) ,\left( -4,3 \right) ,\left( -4,-3 \right) ,\\ \left( 3,4 \right) ,\left( 3,-4 \right) ,\left( -3,4 \right) ,\left( 3,-4 \right) ,\left( 0,5 \right) ,\left( 0,-5 \right) \right\} $$
Therefore domain of $$R\equiv \left\{ 0,3,-3,-4,4,-5,-5 \right\} =$$ domain of $${ R }^{ -1 }$$
Question 2
1 / -0

If $$A=\left\{ 2,3 \right\} $$ and $$B=\left\{ 1,2,3,4 \right\} $$, then which of the following is not a subset of $$A\times B$$

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

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

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

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

Solution

Given $$A=\left\{ 2,3 \right\} $$ and $$B=\left\{ 1,2,3,4 \right\} $$
$$\therefore A\times B =\left\{(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(3,4)\right\}$$
$$\therefore$$ options A,B,C are the subsets of the $$A \times B$$ except option D.
Question 3
1 / -0

In order that a relation $$R$$ defined in a non-empty set $$A$$ is an equivalence relation, it is sufficient that $$R$$

A
is reflexive

B
is symmetric

C
is transitive

D
possess all the above three properties

Solution

A relation R is said to be an equivalence relation if it is
(i)Reflexive
(ii)Symmetric
(iii)Transitive
Question 4
1 / -0

Which one of the following relations on $$R$$ is equivalence redlation-

A
$$x{R}_{1}y\Leftrightarrow \left| x \right| =\left| y \right| $$

B
$$x{R}_{2}y\Leftrightarrow \left| x \right| > \left| y \right| $$

C
$$x{R}_{3}y\Leftrightarrow x|y$$

D
$$x{R}_{4}y\Leftrightarrow x< y$$

Solution

Because $$\left| x \right| =\left| x \right| $$ (Reflexive)

$$\left| x \right| =\left| y \right| \Rightarrow \left| y \right| =\left| x \right| $$ (Symmetric)

$$\left| x \right| =\left| y \right| $$ and $$\left| y \right| =\left| z \right| \Rightarrow \left| x \right| =\left| z \right| $$ (Transitive)

Hence option (A) is equivalence.

None of the relations given in option (b),(c)and(d) are equivalence relations because none of them are symmetric.
Question 5
1 / -0

$$A$$ and $$B$$ are two sets having $$3$$ and $$4$$ elements respectively and having $$2$$ elements in common. The number of relations which can be defined from $$A$$ to $$B$$ is

A
$${2}^{5}$$

B
$${2}^{10}-1$$

C
$${2}^{12}-1$$

D
None of these

Solution

Given $$n(A)=3$$, $$n(B)=4$$

Now, number of elements (ordered pairs) in $$A\times B$$ is $$3 \times 4=12$$

Number of subsets of $$A\times B$$ is $$2^{12}$$ including empty set.

Since, every subset of $$A\times B$$ is a relation from $$A$$ to $$B.$$

Hence, number of relations from $$A$$ to $$B$$ is $$2^{12}-1$$
Question 6
1 / -0

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

A
$$6$$

B
$$9$$

C
$$3$$

D
$$0$$

Solution

Given $$n(A)=3,$$ and $$n(B) =3$$
Hence $$n(A\times B) = 3\times 3=9$$
Question 7
1 / -0

Let $$X=\left\{ 1,2,3,4 \right\} $$ and $$Y=\left\{ 1,2,3,4 \right\} $$. Which of the following is a relation from $$X$$ to $$Y$$.

A
$${R}_{1}=\left\{ (x,y)| y=2+x, x\in X, y\in Y \right\} $$

B
$${R}_{2}=\left\{ (1,1),(2,1),(3,3),(4,3),(5,5) \right\} $$

C
$${R}_{3}=\left\{ (1,1),(1,3),(3,5),(3,7),(5,7) \right\} $$

D
$${R}_{4}=\left\{ (1,3),(2,5),(2,4),(7,9) \right\} $$

Solution

Given $$X=\left\{ 1,2,3,4 \right\} $$ and $$Y=\left\{ 1,2,3,4 \right\} $$

$$X\times Y=\{ (1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)$$
$$(3,1),(3,2),(3,3),(3,4),(4,1),(4,2),(4,3),(4,4)\} $$

Now, $$R_1=\{(1,3),(2,2)\}$$

Since, $$R_1\subset X\times Y$$

Hence, $$R_1$$ is a relation from $$X$$ to $$Y$$

$${R}_{2}=\left\{ (1,1),(2,1),(3,3),(4,3),(5,5) \right\} $$

Since, $$(5,5) \in R_2$$ but $$\notin X\times Y$$

So, $$R_2$$ is not a relation from $$X$$ to$$Y$$

$${R}_{3}=\left\{ (1,1),(1,3),(3,5),(3,7),(5,7) \right\} $$

Since, $$(3,5),(3,7),(5,7) \in R_3$$ but $$\notin X\times Y$$

So, $$R_3$$ is not a relation from $$X$$ to $$Y$$

$${R}_{4}=\left\{ (1,3),(2,5),(2,4),(7,9) \right\} $$

Since, $$(2,5),(7,9) \in R_4$$ but $$\notin X\times Y$$

So, $$R_4$$ is not a relation from $$X$$ to $$Y$$
Question 8
1 / -0

If $$A=\left\{ 1,2,3 \right\} , B=\left\{ 1,4,6,9 \right\} $$ and $$R$$ is a relation from $$A$$ to $$B$$ defined by '$$x$$ is greater than $$y$$'. Then range of $$R$$ is

A
$$\left\{ 1,4,6,9 \right\} $$

B
$$\left\{ 4,6,9 \right\} $$

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

D
None of these

Solution

Here $$R$$ is a relation $$A$$ to $$B$$ defined by '$$x$$ is greater then $$y$$'
$$R=\left\{ (2,1);(3,1) \right\} $$.Hence, range of $$R=\left\{ 1 \right\} $$
Question 9
1 / -0

Let $$g\left( x \right)=1+x-\left[ x \right] $$ and $$f\left( x \right)=\begin{cases} \begin{matrix} -1 & x<0 \end{matrix} \\ \begin{matrix} 0 & x=0 \end{matrix} \\ \begin{matrix} 1 & x>0 \end{matrix} \end{cases}$$. Then for all $$x,f\left[ g\left( x \right) \right] $$ is equal to

A
$$x$$

B
$$1$$

C
$$f(x)$$

D
$$g(x)$$

Solution

We have,
$$g\left( x \right)=\begin{cases} 1+n-n=1,x=n\in Z \\ 1+n+k-n=1+k,x=n+k \end{cases}$$
where $$n\in Z,0<k<1$$
Now $$f\left[ g\left( x \right) \right] =\begin{cases} \begin{matrix} -1 & g\left( x \right)<0 \end{matrix} \\ \begin{matrix} 0 & g\left( x \right)=0 \end{matrix} \\ \begin{matrix} 1 & g\left( x \right)>1 \end{matrix} \end{cases}$$
Clearly, $$g(x)>0$$ for all $$x$$. So $$f\left[ g\left( x \right) \right] =1$$, for all $$x$$.
Question 10
1 / -0

If $$A = \left\{1,2,3\right\}$$, $$B = \left\{1,4,6,9\right\}$$ and $$R$$ is relation from $$A$$ to $$B$$ defined by $$'x'$$ is greater than $$'y'$$. Then range of $$R$$ is

A
$$\left\{1,4,6,9\right\}$$

B
$$\left\{4,6,9\right\}$$

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

D
None of these

Solution

Here $$R$$ is a relation $$A$$ to $$B$$ defined by $$x$$ is greater than $$y$$.
$$\therefore\quad R = \left\{(2,1); (3,1)\right\}$$. Hence, range of $$R = \left\{1\right\}$$

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Relations Test 17

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Relations Test 17

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Question 1

$$0$$

Domain of $$\displaystyle R= \left \{ 0, \pm 3 \right \}= $$ domain of $$\displaystyle R^{-1}.$$

Domain of $$\displaystyle R= \left \{ 0, \pm 3, \pm 4 \right \}= $$ domain of $$\displaystyle R^{-1}.$$

Domain of $$\displaystyle R= \left \{ 0, \pm 3, \pm 4, \pm 5 \right \}= $$ domain of $$\displaystyle R^{-1}.$$

Solution

Question 2

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

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

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

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

Solution

Question 3

is reflexive

is symmetric

is transitive

possess all the above three properties

Solution

Question 4

$$x{R}_{1}y\Leftrightarrow \left| x \right| =\left| y \right| $$

$$x{R}_{2}y\Leftrightarrow \left| x \right| > \left| y \right| $$

$$x{R}_{3}y\Leftrightarrow x|y$$

$$x{R}_{4}y\Leftrightarrow x< y$$

Solution

Question 5

$${2}^{5}$$

$${2}^{10}-1$$

$${2}^{12}-1$$

None of these

Solution

Question 6

$$6$$

$$9$$

$$3$$

$$0$$

Solution

Question 7

$${R}_{1}=\left\{ (x,y)| y=2+x, x\in X, y\in Y \right\} $$

$${R}_{2}=\left\{ (1,1),(2,1),(3,3),(4,3),(5,5) \right\} $$

$${R}_{3}=\left\{ (1,1),(1,3),(3,5),(3,7),(5,7) \right\} $$

$${R}_{4}=\left\{ (1,3),(2,5),(2,4),(7,9) \right\} $$

Solution

Question 8

$$\left\{ 1,4,6,9 \right\} $$

$$\left\{ 4,6,9 \right\} $$

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

None of these

Solution

Question 9

$$x$$

$$1$$

$$f(x)$$

$$g(x)$$

Solution

Question 10

$$\left\{1,4,6,9\right\}$$

$$\left\{4,6,9\right\}$$

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

None of these

Solution

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