Sets and Relations — Page 4 | JEE Mathematics

Q31

Let the relations

R_1

and

R_2

on the set

X=\{1,2,3, \ldots, 20\}

be given by

R_1=\{(x, y): 2 x-3 y=2\}

and

R_2=\{(x, y):-5 x+4 y=0\}

. If

M

and

N

be the minimum number of elements required to be added in

R_1

and

R_2

, respectively, in order to make the relations symmetric, then

M+N

equals

A 16

B 12

C 8

D 10

Correct Answer

Option D

Solution

\begin{aligned} & R_1=\{(x, y): 2 x-3 y=2\} \\ & R_2=\{(x, y):-5 x+4 y=0\} \\ & 2 x-3 y=2 \end{aligned}

So

2 x

and

3 y

both has to be even or odd simultaneously and

2 x

can't be odd so

2 x

and

3 y

both will be even

R_1=\{(4,2),(7,4),(10,6),(13,8),(16,10),(19,12)\}

For symmetric we need to add 6 elements as

\begin{aligned} & (2,4),(4,7),(6,10),(8,13),(10,16),(12,19) \\ & M=6 \end{aligned}

For

R_2-5 x+4 y=0

5 x

and

4 y

has to be equal

4 y

is always even so

5 x

will also be even

R_2=\{(4,5),(8,10),(12,15),(16,20)\}

For symmetric we need to add 4 element as

\begin{aligned} & (5,4)(10,8)(15,12)(20,16) \\ & N=4 \\ & M+N=6+4=10 \end{aligned}

Q32

The relation

R=\{(x, y): x, y \in \mathbb{Z}

and

x+y

is even

\}

is:

A reflexive and transitive but not symmetric

B reflexive and symmetric but not transitive

C an equivalence relation

D symmetric and transitive but not reflexive

Correct Answer

Option C

Solution

$R=\{(x, y): x, y \in z$ and $x+y$ is even $\}$ reflexive $x+x=2 x$ even symmetric of $x+y$ is even, then $(y+x)$ is also even transitive of $\mathrm{x}+\mathrm{y}$ is even $\& \mathrm{y}+\mathrm{z}$ is even then $x+z$ is also even So, relation is an equivalence relation.

Q33

Let A = { (

\alpha, \beta

)

\in \mathbb{R} \times \mathbb{R}

: |

\alpha

- 1|

\leq 4

and |

\beta

- 5|

\leq 6

} and B = { (

\alpha, \beta

)

\in \mathbb{R} \times \mathbb{R}

: 16(

\alpha

-

2)^2

+ 9(

\beta

-

6)^2

\leq 144

}. Then

A A

\subset

B

B B

\subset

A

C neither A

\subset

B nor B

\subset

A

D

A \cup B=\{(x, y):-4 \leqslant x \leqslant 4,-1 \leqslant y \leqslant 11\}

Correct Answer

Option B

Solution

\begin{aligned} & \text{ A: }|x-1| \leq 4 \text{ and }|y-5| \leq 6 \\ & \Rightarrow-4 \leq x-1 \leq 4 \Rightarrow-6 \leq y-5 \leq 6 \\ & \Rightarrow-3 \leq x \leq 5 \quad \Rightarrow-1 \leq y \leq 11 \\ & \text{ B : } 16(x-2)^2+9(y-6)^2 \leq 144 \\ & \text{ B : } \frac{(x-2)^2}{9}+\frac{(y-6)^2}{16} \leq 1 \end{aligned}

From Diagram $\mathrm{B} \subset \mathrm{A}$

Q34

Let

R

be a relation on

\mathbb{R}

, given by

R=\{(a, b): 3 a-3 b+\sqrt{7}

is an irrational number

\}

. Then

R

is

A an equivalence relation

B reflexive and symmetric but not transitive

C reflexive and transitive but not symmetric

D reflexive but neither symmetric nor transitive

Correct Answer

Option D

Solution

For reflexive : $3 a-3 a+\sqrt{7}$ is an irrational number $\forall a \in R R$ is reflexive For symmetric : Let $3 a-3 b+\sqrt{7}$ is an irrational number $\Rightarrow 3 b-3 a+\sqrt{7}$ is an irrational number For example, Let $3 a-3 b=\sqrt{7}$ $\sqrt{7}+\sqrt{7}$ is irrational but $-\sqrt{7}+\sqrt{7}$ is not. $\therefore R$ is not symmetric For transitive : Let $3 a-3 b+\sqrt{7}$ is irrational and $3 b-3 c+\sqrt{7}$ is irrational. $\Rightarrow 3 a-3 c+\sqrt{7}$ is irrational.

For example, take $a=0, b=-\sqrt{7}, c=\dfrac{\sqrt{7}}{3}$ $R$ is not transitive.

Q35

Let A be the set of all functions

f: \mathbf{Z} \rightarrow \mathbf{Z}

and R be a relation on A such that

\mathrm{R}=\{(\mathrm{f}, \mathrm{g}): f(0)=\mathrm{g}(1)

and

f(1)=\mathrm{g}(0)\}

. Then R is :

A Symmetric and transitive but not reflective

B Symmetric but neither reflective nor transitive

C Transitive but neither reflexive nor symmetric

D Reflexive but neither symmetric nor transitive

Correct Answer

Option B

Solution

For $R$ to be reflexive, $(f, f)$ must be in $R$ .

The means $f(0)=f(1)$ and $f(1)=f(0)$ must be true for all $f$ .

But $f(0) \neq f(1)$ always Therefore, $R$ is not reflexive If $(f, g) \in R$ , then $f(0)=g(1)$ and $f(1)=g(0)$

\begin{aligned} & \because \quad f(0)=g(1) \Rightarrow g(1)=f(0) \\ & \text{ and } f(1)=g(0) \Rightarrow g(0)=f(1) \end{aligned}

$R$ is symmetric If $(f, g) \in R$ and $(g, h) \in \mathrm{R}$ , then $f(0)=g(1)$ , $f(1)=g(0), g(0)=n(1) \& g(1)=h(0)$ Since, $f(0)=g(1)$ and $g(1)=h(0)$ , then $f(0)$ is not necessarily equal to $h(0)$ .

Therefore, $R$ is not transitive.

$\therefore \quad$ The relation $R$ is symmetric but not reflexive or transitive.

Q36

The number of elements in the set {x

\in

R : (|x|

-

3) |x + 4| = 6} is equal to :

A 4

B 2

C 3

D 1

Correct Answer

Option B

Solution

Case 1 : x $\le$ $-$ 4 ( $-$ x $-$ 3)( $-$ x $-$ 4) = 6 $\Rightarrow$ (x + 3)(x + 4) = 6 $\Rightarrow$ x2 + 7x + 6 = 0 $\Rightarrow$ x = $-$ 1 or $-$ 6 but x $\le$ $-$ 4 x = $-$ 6 Case 2 : x

\in

( $-$ 4, 0) ( $-$ x $-$ 3)(x + 4) = 6 $\Rightarrow$ $-$ x2 $-$ 7x $-$ 12 $-$ 6 = 0 $\Rightarrow$ x2 + 7x + 18 = 0 D < 0 No solution Case 3 : x $\ge$ 0 (x $-$ 3)(x + 4) = 6 $\Rightarrow$ x2 + x $-$ 12 $-$ 6 = 0 $\Rightarrow$ x2 + x $-$ 18 = 0 x =

{{ - 1 \pm \sqrt {1 + 72} } \over 2}

$\therefore$ x =

{{\sqrt {73} - 1} \over 2}

only

Q37

Let

\mathrm{A}=\{-3,-2,-1,0,1,2,3\}

and R be a relation on A defined by

x \mathrm{R} y

if and only if

2 x-y \in\{0,1\}

. Let

l

be the number of elements in

R

. Let

m

and

n

be the minimum number of elements required to be added in R to make it reflexive and symmetric relations, respectively. Then

l+\mathrm{m}+\mathrm{n}

is equal to:

A 17

B 18

C 15

D 16

Correct Answer

Option A

Solution

\begin{aligned} &\begin{aligned} & x R y \Leftrightarrow 2 x-y \in\{0,1\} \\ & \Rightarrow \quad y=2 x \text{ or } y=2 x-1 \\ & A=\{-3,-2,-1,0,1,2,3\} \\ & \mathrm{R}=\{(-1,-2),(0,0),(1,2),(-1,-3),(0,-1),(1,1), \\ & (2,3)\} \\ & \Rightarrow \quad I=7 \end{aligned}\\ &\text{ For } R \text{ to be reflexive }(0,0),(1,1) \in R \end{aligned}

But other $(a, a)$ such that $2 a-a \in\{0,1\}$

\Rightarrow \quad a \in\{0,1\}

5 other pairs needs to be added $\Rightarrow m=5$ $x R y \Rightarrow y R x$ to be symmetric $(-1,-2) \Rightarrow(-2,-1)$ $(1,2) \Rightarrow(2,1)$ $(-1,-3) \Rightarrow(-3,-1)$ $(0,-1) \Rightarrow(-1,0)$ $(2,3) \Rightarrow(3,2) \Rightarrow 5$ needs to be added, $n=5$ $\Rightarrow \quad l+m+n=17$

Q38

Let R be the real line. Consider the following subsets of the plane

R \times R

:

S = \left\{ {(x,y):y = x + 1\,\,and\,\,0 < x < 2} \right\}

T = \left\{ {(x,y): x - y\,\,\,is\,\,an\,\,{\mathop{\rm \int}} eger\,} \right\}

, Which one of the following is true ?

A Neither S nor T is an equivalence relation on R

B Both S and T are equivalence relation on R

C S is an equivalence relation on R but T is not

D T is an equivalence relation on R but S is not

Correct Answer

Option D

Solution

Given

S = \left\{ {\left( {x,y} \right):y = x + 1\,\,} \right.\,

and

\,\,\,\left. {0 < x < 2} \right\}

As

\,\,\,\,x \ne x + 1\,\,\,

for any

\,\,\,x \in \left( {0,2} \right) \Rightarrow \left( {x,x} \right) \notin S

$\therefore$

S

is not reflexive. Hence

S

in not an equivalence relation. Also

\,\,\,T = \left\{ {x,\left. y \right)} \right.:x - y

is an integer

\left. {} \right\}

as

x - x = 0

is an integer

\forall x \in R

$\therefore$

T

is reflexive. If

x-y

is an integer then

y-x

is also an integer $\therefore$

T

is symmetric If

x-y

is an integer and

y - z

is an integer then

(x-y)+(y-z)=x-z

is also an integer. $\therefore$

T

is transitive Hence

T

is an equivalence relation

Q39

Let

\mathrm{A}=\{2,3,4\}

and

\mathrm{B}=\{8,9,12\}

. Then the number of elements in the relation

\mathrm{R}=\left\{\left(\left(a_{1}, \mathrm{~b}_{1}\right),\left(a_{2}, \mathrm{~b}_{2}\right)\right) \in(A \times B, A \times B): a_{1}\right.

divides

\mathrm{b}_{2}

and

\mathrm{a}_{2}

divides

\left.\mathrm{b}_{1}\right\}

is :

A 18

B 24

C 36

D 12

Correct Answer

Option C

Solution

Given sets : $A = {2,3,4}$ $B = {8,9,12}$ We want to find the number of elements of the form $( (a_1, b_1), (a_2, b_2) )$ such that : $a_1$ divides $b_2$ $a_2$ divides $b_1$ For the first condition : $a_1$ divides $b_2$ Given $a_1 \in A$ and $b_2 \in B$ , we can list the pairs: $(a_1, b_2) \in {(2,8),(2,12),(3,9),(3,12),(4,8),(4,12)}$ This gives 6 pairs.

For the second condition, the pairs are the same, because it's just the reversed relation.

So : $a_2$ divides $b_1$ Again has 6 valid pairs.

Now, for every pair from the first condition, we can have any pair from the second condition.

This leads to : $6 \times 6 = 36$ relations.

Q40

Among the relations

\mathrm{S}=\left\{(\mathrm{a}, \mathrm{b}): \mathrm{a}, \mathrm{b} \in \mathbb{R}-\{0\}, 2+\dfrac{\mathrm{a}}{\mathrm{b}}>0\right\}

and

\mathrm{T}=\left\{(\mathrm{a}, \mathrm{b}): \mathrm{a}, \mathrm{b} \in \mathbb{R}, \mathrm{a}^{2}-\mathrm{b}^{2} \in \mathbb{Z}\right\}

,

A

\mathrm{S}

is transitive but

\mathrm{T}

is not

B both

\mathrm{S}

and

\mathrm{T}

are symmetric

C neither

S

nor

T

is transitive

D

T

is symmetric but

S

is not

Correct Answer

Option D

Solution

For relation $\mathrm{T}=\mathrm{a}^{2}-\mathrm{b}^{2}=-\mathrm{I}$ Then, $(\mathrm{b}, \mathrm{a})$ on relation $\mathrm{R}$ $\Rightarrow \mathrm{b}^{2}-\mathrm{a}^{2}=-\mathrm{I}$ $\therefore \mathrm{T}$ is symmetric $\mathrm{S}=\left\{(\mathrm{a}, \mathrm{b}): \mathrm{a}, \mathrm{b} \in \mathrm{R}-\{0\}, 2+\dfrac{\mathrm{a}}{\mathrm{b}}>0\right\}$ $2+\dfrac{\mathrm{a}}{\mathrm{b}}>0 \Rightarrow \dfrac{\mathrm{a}}{\mathrm{b}}>-2, \Rightarrow \dfrac{\mathrm{b}}{\mathrm{a}}<\dfrac{-1}{2}$ If $(b, a) \in \mathbf{S}$ then $2+\dfrac{\mathrm{b}}{\mathrm{a}}$ not necessarily positive $\therefore \mathrm{S}$ is not symmetric