Sets and Relations — Page 3 | JEE Mathematics

Q21

If R = {(x, y) : x, y

\in

Z, x2 + 3y2

\le

8} is a relation on the set of integers Z, then the domain of R–1 is :

A {0, 1}

B {–2, –1, 1, 2}

C {–1, 0, 1}

D {–2, –1, 0, 1, 2}

Correct Answer

Option C

Solution

Given R = {(x, y) : x, y

\in

Z, x2 + 3y2 $\le$ 8} So R = {(0,1), (0,–1), (1,0), (–1,0), (1,1), (1,-1) (-1,1), (-1,-1), (2,0), (-2,0), (-2,0), (2,1), (2,-1), (-2,1), (-2,-1)} $\Rightarrow$ R : { -2, -1, 0, 1, 2} $\to$ {-1, 0, 1} $\therefore$ R-1 : {-1, 0, 1} $\to$ { -2, -1, 0, 1, 2} $\therefore$ Domain of R–1 = {-1, 0, 1}

Q22

Let R1 and R2 be two relation defined as follows : R1 = {(a, b)

\in

R2 : a2 + b2

\in

Q} and R2 = {(a, b)

\in

R2 : a2 + b2

\notin

Q}, where Q is the set of all rational numbers. Then :

A Neither R1 nor R2 is transitive.

B R2 is transitive but R1 is not transitive.

C R1 and R2 are both transitive.

D R1 is transitive but R2 is not transitive.

Correct Answer

Option A

Solution

For R1 : Let a = 1 +

\sqrt 2

, b = 1 $-$

\sqrt 2

, c =

{8^{{1 \over 4}}}

aR1b : a2 + b2 = 6

\in

Q bR1c : b2 + c2 = 3 $-$ 2

\sqrt 2

+ 2

\sqrt 2

= 3

\in

Q aR1c : a2 + c2 = 3 + 2

\sqrt 2

+ 2

\sqrt 2

$\notin$ Q $\therefore$ R1 is not transitive. For R2 : Let a = 1 +

\sqrt 2

, b =

\sqrt 2

, c = 1 $-$

\sqrt 2

aR2b : a2 + b2 = 5 + 2

\sqrt 2

$\notin$ Q bR2c : b2 + c2 = 5 $-$ 2

\sqrt 2

$\notin$ Q aR2c : a2 + c2 = 3 + 2

\sqrt 2

+ 3 $-$ 2

\sqrt 2

= 6

\in

Q $\therefore$ R2 is not transitive.

Q23

Let R = {(P, Q) | P and Q are at the same distance from the origin} be a relation, then the equivalence class of (1,

-

1) is the set :

A

S = \{ (x,y)|{x^2} + {y^2} = \sqrt 2 \}

B

S = \{ (x,y)|{x^2} + {y^2} = 2\}

C

S = \{ (x,y)|{x^2} + {y^2} = 1\}

D

S = \{ (x,y)|{x^2} + {y^2} = 4\}

Correct Answer

Option B

Solution

Given R = {(P, Q) | P and Q are at the same distance from the origin}.

Then equivalence class of (1, $-$ 1) will contain al such points which lies on circumference of the circle of centre at origin and passing through point (1, $-$ 1). i.e., radius of circle =

\sqrt {{1^2} + {1^2}} = \sqrt 2

$\therefore$ Required equivalence class of (S)

= \{ (x,y)|{x^2} + {y^2} = 2\}

.

Q24

Let

R_{1}

and

R_{2}

be two relations defined on

\mathbb{R}

by

a \,R_{1} \,b \Leftrightarrow a b \geq 0

and

a \,R_{2} \,b \Leftrightarrow a \geq b

Then,

A

R_{1}

is an equivalence relation but not

R_{2}

B

R_{2}

is an equivalence relation but not

R_{1}

C both

R_{1}

and

R_{2}

are equivalence relations

D neither

R_{1}

nor

R_{2}

is an equivalence relation

Correct Answer

Option D

Solution

a\,{R_1}\,b \Leftrightarrow ab \ge 0

So, definitely

(a,a) \in {R_1}

as

{a^2} \ge 0

If

(a,b) \in {R_1} \Rightarrow (b,a) \in {R_1}

But if

(a,b) \in {R_1},(b,c) \in {R_1}

$\Rightarrow$ Then

(a,c)

may or may not belong to R1 {Consider

a = - 5,b = 0,c = 5

so

(a,b)

and

(b,c) \in {R_1}

but

ac So, R1 is not equivalence relation

a\,{R_2}\,b \Leftrightarrow a \ge b

(a,a) \in {R_2} \Rightarrow

so reflexive relation If

(a,b) \in {R_2}

then

(b,a)

may or may not belong to R2

\Rightarrow$$ So not symmetric Hence it is not equivalence relation

Q25

Which of the following is not correct for relation R on the set of real numbers ?

A (x, y)

\in

R

\Leftrightarrow

0 < |x|

-

|y|

\le

1 is neither transitive nor symmetric.

B (x, y)

\in

R

\Leftrightarrow

0 < |x

-

y|

\le

1 is symmetric and transitive.

C (x, y)

\in

R

\Leftrightarrow

|x|

-

|y|

\le

1 is reflexive but not symmetric.

D (x, y)

\in

R

\Leftrightarrow

|x

-

y|

\le

1 is reflexive nd symmetric.

Correct Answer

Option B

Solution

Note that (a, b) and (b, c) satisfy 0 < |x $-$ y| $\le$ 1 but (a, c) does not satisfy it so 0 $\le$ |x $-$ y| $\le$ 1 is symmetric but not transitive.

For example, x = 0.2, y = 0.9, z = 1.5 0 ≤ |x – y| = 0.7 ≤ 1 0 ≤ |y – z| = 0.6 ≤ 1 But |x – z| = 1.3 > 1 So, (b) is correct.

Q26

The minimum number of elements that must be added to the relation

\mathrm{R}=\{(\mathrm{a}, \mathrm{b}),(\mathrm{b}, \mathrm{c})\}

on the set

\{a, b, c\}

so that it becomes symmetric and transitive is :

A 7

B 3

C 4

D 5

Correct Answer

Option A

Solution

For symmetric

(b,a),(c,b)\in R

For transitive

(a,c)\in R

\Rightarrow (c,a)\in R

\therefore (a,b),(b,a)\in R

\Rightarrow (a,a)\in R

(b,c),(c,b)\in R

\Rightarrow (b,b)\in R,(c,c)\in R

7 elements must be added

Q27

Let

P(S)

denote the power set of

S=\{1,2,3, \ldots ., 10\}

. Define the relations

R_{1}

and

R_{2}

on

P(S)

as

\mathrm{AR}_{1} \mathrm{~B}

if

\left(\mathrm{A} \cap \mathrm{B}^{\mathrm{c}}\right) \cup\left(\mathrm{B} \cap \mathrm{A}^{\mathrm{c}}\right)=\emptyset

and

\mathrm{AR}_{2} \mathrm{~B}

if

\mathrm{A} \cup \mathrm{B}^{\mathrm{c}}=\mathrm{B} \cup \mathrm{A}^{\mathrm{c}}, \forall \mathrm{A}, \mathrm{B} \in \mathrm{P}(\mathrm{S})

. Then :

A only

R_{2}

is an equivalence relation

B both

R_{1}

and

R_{2}

are not equivalence relations

C both

R_{1}

and

R_{2}

are equivalence relations

D only

R_{1}

is an equivalence relation

Correct Answer

Option C

Solution

$\begin{aligned} & \mathrm{S}=\{1,2,3, \ldots \ldots 10\} \\\\ & \mathrm{P}(\mathrm{S})=\text{ power set of } \mathrm{S} \\\\ & \mathrm{AR}_1 \mathrm{B} \Rightarrow(\mathrm{A} \cap \overline{\mathrm{B}}) \cup(\overline{\mathrm{A}} \cap \mathrm{B})=\phi \\\\ & \mathrm{R}_1 \text{ is reflexive, symmetric } \\\\ & \text{ For transitive } \\\\ & (\mathrm{A} \cap \overline{\mathrm{B}}) \cup(\overline{\mathrm{A}} \cap \mathrm{B})=\phi ;\{\mathrm{a}\}=\phi=\{\mathrm{b}\} \therefore \mathrm{A}=\mathrm{B} \\\\ & (\mathrm{B} \cap \overline{\mathrm{C}}) \cup(\overline{\mathrm{B}} \cap \mathrm{C})=\phi \therefore \mathrm{B}=\mathrm{C} \\\\ & \therefore \mathrm{A}=\mathrm{C} \text{ equivalence. }\end{aligned}$

\mathrm{R}_2 \equiv \mathrm{A} \cup \overline{\mathrm{B}}=\overline{\mathrm{A}} \cup \mathrm{B}

$\mathrm{R}_2 \rightarrow$ Reflexive, symmetric For transitive : $\begin{aligned} & \mathrm{A} \cup \overline{\mathrm{B}}=\overline{\mathrm{A}} \cup \mathrm{B} \Rightarrow\{\mathrm{a}, \mathrm{c}, \mathrm{d}\}=\{\mathrm{b}, \mathrm{c}, \mathrm{d}\} \\\\ & \{\mathrm{a}\}=\{\mathrm{b}\} \therefore \mathrm{A}=\mathrm{B} \\\\ & \mathrm{B} \cup \overline{\mathrm{C}}=\overline{\mathrm{B}} \cup \mathrm{C} \Rightarrow \mathrm{B}=\mathrm{C} \\\\ & \therefore \mathrm{A}=\mathrm{C} \quad \therefore \mathrm{A} \cup \overline{\mathrm{C}}=\overline{\mathrm{A}} \cup \mathrm{C} \therefore \text{ Equivalence }\end{aligned}$

Q28

Let R be a relation defined on

\mathbb{N}

as

a\mathrm{R}b

if

2a+3b

is a multiple of

5,a,b\in \mathbb{N}

. Then R is

A an equivalence relation

B non reflexive

C symmetric but not transitive

D transitive but not symmetric

Correct Answer

Option A

Solution

a R b if 2a + 3b = 5m, m

\in

l

(1)

(a,a) \in R

as

2a + 3a = 5a,a \in N

Hence, R is reflexive (2) If

(a,b) \in R

then

2a + 3 = 5m

Now,

5(a + b) = 5n

3a + 2b + 2a + 3b = 5n

$\therefore$

3a + 2b = 5(n - m)

$\therefore$

(b,a) \in R

$\therefore$ R is symmetric (3) If

(a,b) \in R

and

(b,c) \in R

then

2a + 3b = 5m,2b + 3c = 5n

\Rightarrow 2a + 5b + 3c = 5(m + n)

\Rightarrow 2a + 3c = 5(m = n - b)

$\therefore$

(a,c) \in R

$\therefore$ R is transitive Hence, R is equivalence relation. Option (1) is correct.

Q29

The relation

\mathrm{R = \{ (a,b):\gcd (a,b) = 1,2a \ne b,a,b \in \mathbb{Z}\}}

is :

A reflexive but not symmetric

B transitive but not reflexive

C symmetric but not transitive

D neither symmetric nor transitive

Correct Answer

Option D

Solution

Given, (a, b) belongs to relation R if

\gcd (a,b) = 1, 2a \ne b

. Here

\gcd

means greatest common divisor.

\gcd

of two numbers is the largest number that divides both of them. (1) For Reflexive, In

aRa,\,\gcd (a,a) = a

$\therefore$ This relation is not reflexive.

(2) For Symmetric: Take $a=2, b=1 \Rightarrow \operatorname{gcd}(2,1)=1$ Also $2 a=4 \neq b$ Now

\gcd (b,a) = 1

$\Rightarrow \operatorname{gcd}(1,2)=1$ and 2b should not be equal to a But here, $2 b=2=a$ $\Rightarrow \mathrm{R}$ is not Symmetric (3) For Transitive: Let $\mathrm{a}=14, \mathrm{~b}=19, \mathrm{c}=21$ $\operatorname{gcd}(\mathrm{a}, \mathrm{b})=1, 2a \ne b$ $\operatorname{gcd}(\mathrm{b}, \mathrm{c})=1, 2b \ne c$ $\operatorname{gcd}(\mathrm{a}, \mathrm{c})=7, 2a \ne c$ Hence not transitive $\Rightarrow R$ is neither symmetric nor transitive.

Q30

Let

\mathrm{A}=\{1,2,3,4,5,6,7\}

. Then the relation

\mathrm{R}=\{(x, y) \in \mathrm{A} \times \mathrm{A}: x+y=7\}

is :

A reflexive but neither symmetric nor transitive

B transitive but neither symmetric nor reflexive

C symmetric but neither reflexive nor transitive

D an equivalence relation

Correct Answer

Option C

Solution

Here, $A=\{1,2,3,4,5,6,7\}$ Since, $x+y=7 \Rightarrow y=7-x$ So, $\mathrm{R}=\{(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\}$ $\because(a, b) \in \mathrm{R} \Rightarrow(b, a) \in \mathrm{R}$ $\therefore \mathrm{R}$ is symmetric only.