Let A = { 1 , 3 } and B = { 5 , 9 } . Let R = { ( a , b ) : a ∈ A , b ∈ B , a - b is an odd integer } . Show that R is an empty relation from A to B .

A relation is a set of ordered pairs. If A and B be any two non-empty sets, then a relation R from the set A to the set B is a subset of A × B . In this case, an element a of A is said to be related to an element b of B by R and we express this by the symbol a R b , where ( a , b ) ∈ R . If R is a relation from A to B , then the domain of R is the set of all first components of the ordered pairs in R and the range of R is the set of all second components of all ordered pairs in R . Thus, the domain of R = { a : ( a , b ) ∈ R } and range of R = { b : ( a , b ) ∈ R } . An empty relation (or void relation) is one in which there is no relation between any elements of a set. Given, A = { 1 , 3 } and B = { 5 , 9 } . And, R = { ( a , b ) : a ∈ A , b ∈ B , a - b is an odd integer } . For a = 1 , b = 5 , a - b = 1 - 5 = - 4 = not an odd integer. For a = 1 , b = 9 , a - b = 1 - 9 = - 8 = not an odd integer. For a = 3 , b = 5 , a - b = 3 - 5 = - 2 = not an odd integer. For a = 3 , b = 9 , a - b = 3 - 9 = - 6 = not an odd integer. Therefore, R is an empty relation from A to B . Hence, proved.

Let R be the relation on Z defined by R = { ( x , y ) : x , y ∈ Z , x - y is an integer } . Find the domain and range of R .

A relation is a set of ordered pairs. If A and B be any two non-empty sets, then a relation R from the set A to the set B is a subset of A × B . In this case, an element a of A is said to be related to an element b of B by R and we express this by the symbol a R b , where ( a , b ) ∈ R . If R is a relation from A to B , then the domain of R is the set of all first components of the ordered pairs in R and the range of R is the set of all second components of all ordered pairs in R . Thus, the domain of R = { a : ( a , b ) ∈ R } and range of R = { b : ( a , b ) ∈ R } . Given, R be the relation on Z defined by R = { ( x , y ) : x , y ∈ Z , x - y is an integer } . Since x , y ∈ Z , x and y are both integers and therefore, x - y is an integer for all x , y ∈ Z . Hence, the domain of R = Z and the range of R = Z .

A relation R is defined as R = { ( x , y ) : x ∈ N , y ∈ N and 2 x + y = 14 } . Find R as the set of ordered pairs and hence, find its domain and range.

A relation is a set of ordered pairs. If A and B be any two non-empty sets, then a relation R from the set A to the set B is a subset of A × B . In this case, an element a of A is said to be related to an element b of B by R and we express this by the symbol a R b , where ( a , b ) ∈ R . If R is a relation from A to B , then the domain of R is the set of all first components of the ordered pairs in R and the range of R is the set of all second components of all ordered pairs in R . Thus, the domain of R = { a : ( a , b ) ∈ R } and range of R = { b : ( a , b ) ∈ R } . Given, a relation R is defined as R = { ( x , y ) : x ∈ N , y ∈ N and 2 x + y = 14 } . Here, the values of x are 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , . . . . . . . We have, 2 x + y = 14 ⇒ y = 14 - 2 x . For x = 1 , y = 14 - 2 x = 14 - 2 × 1 = 14 - 2 = 12 = a natural number. For x = 2 , y = 14 - 2 x = 14 - 2 × 2 = 14 - 4 = 10 = a natural number. For x = 3 , y = 14 - 2 x = 14 - 2 × 3 = 14 - 6 = 8 = a natural number. For x = 4 , y = 14 - 2 x = 14 - 2 × 4 = 14 - 8 = 6 = a natural number. For x = 5 , y = 14 - 2 x = 14 - 2 × 5 = 14 - 10 = 4 = a natural number. For x = 6 , y = 14 - 2 x = 14 - 2 × 6 = 14 - 12 = 2 = a natural number. For x = 7 , y = 14 - 2 x = 14 - 2 × 7 = 14 - 14 = 0 = not a natural number. For x = 8 , y = 14 - 2 x = 14 - 2 × 8 = 14 - 16 = - 2 = not a natural number. For x = 9 , 10 , 11 , . . . . . , the values of y will not be natural numbers. Hence, R = 1 , 12 , 2 , 10 , 3 , 8 , 4 , 6 , 5 , 4 , 6 , 2 . The domain of R is 1 , 2 , 3 , 4 , 5 , 6 and the range of R is 12 , 10 , 8 , 6 , 4 , 2 .

A relation R is defined on the set of natural numbers N as R = { ( x , y ) : x ∈ N , y ∈ N and y = x + 3 , x ≤ 22 } . Find R as the set of order pairs and hence, find its domain and range.

A relation is a set of ordered pairs. If A and B be any two non-empty sets, then a relation R from the set A to the set B is a subset of A × B . In this case, an element a of A is said to be related to an element b of B by R and we express this by the symbol a R b , where ( a , b ) ∈ R . If R is a relation from A to B , then the domain of R is the set of all first components of the ordered pairs in R and the range of R is the set of all second components of all ordered pairs in R . Thus, the domain of R = { a : ( a , b ) ∈ R } and range of R = { b : ( a , b ) ∈ R } . Given, a relation R is defined on the set of natural numbers N as R = { ( x , y ) : x ∈ N , y ∈ N and y = x + 3 , x ≤ 22 } . Here, the values of x are 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , . . . . . . , 22 . For x = 1 , y = x + 3 = 1 + 3 = 4 = ± 2 . 2 is a natural number, but - 2 is not a natural number. For x = 2 , 3 , 4 , 5 , the value of y will be irrational numbers and hence, not natural numbers. For x = 6 , y = x + 3 = 6 + 3 = 9 = ± 3 . 3 is a natural number, but - 3 is not a natutal number. For x = 7 , 8 , 9 , 10 , 11 , 12 , the value of y will be irrational numbers and hence, not natural numbers. For x = 13 , y = x + 3 = 13 + 3 = 16 = ± 4 . 4 is a natural number, but - 4 is not a natural number. For x = 14 , 15 , 16 , 17 , 18 , 19 , 20 , 21 , the value of y will be irrational numbers and hence, not natural numbers. For x = 22 , y = x + 3 = 22 + 3 = 25 = ± 5 . 5 is a natural number, but - 5 is not a natural number. Hence, R = 1 , 2 , 6 , 3 , 13 , 4 , 22 , 5 . The domain of R is 1 , 6 , 13 , 22 and the range of R is 2 , 3 , 4 , 5 .

If R 1 defined on R by the relation R ( ( a , b ) : 1 + a b > 0 , for a , b ∈ R 1 and b , c ∈ R 1 ⇒ a , c ∈ R 1 is not true for all a , b , c ∈ R .

Given, for the relation R 1 defined on R by the rule a , b ∈ R 1 ⇔ 1 + a b > 0 and a , b ∈ R Let 2 , - 1 4 ∈ R 1 ⇒ 1 + 2 × - 1 4 = 1 2 > 0 - 1 4 , - 3 ∈ R 1 ⇒ 1 + - 1 4 - 3 > 0 But 2 , - 3 ∉ R 1 ∵ 1 + 2 - 3 < 0 Hence proved.

E M B I B E

I.S.C MATHEMATICS FOR CLASS XI>Chapter 2 - Relations and Functions>EXERCISE-2(II)

SURANJAN SAHA and SABITA SAHA Solutions for Chapter: Relations and Functions, Exercise 2: EXERCISE-2(II)

Author:SURANJAN SAHA & SABITA SAHA

SURANJAN SAHA Mathematics Solutions for Exercise - SURANJAN SAHA and SABITA SAHA Solutions for Chapter: Relations and Functions, Exercise 2: EXERCISE-2(II)

Attempt the free practice questions on Chapter 2: Relations and Functions, Exercise 2: EXERCISE-2(II) with hints and solutions to strengthen your understanding. I.S.C MATHEMATICS FOR CLASS XI solutions are prepared by Experienced Embibe Experts.

Questions from SURANJAN SAHA and SABITA SAHA Solutions for Chapter: Relations and Functions, Exercise 2: EXERCISE-2(II) with Hints & Solutions

HARD

11th ICSE

IMPORTANT

I.S.C MATHEMATICS FOR CLASS XI>Chapter 2 - Relations and Functions>EXERCISE-2(II)>Q 7

Let $A = {1, 3}$ and $B = {5, 9}$ . Let $R = {(a, b) : a \in A, b \in B, a - b$ is an odd integer $a = 1, b = 9, a - b = 1 - 9 = - 8 =$ . Show that $R$ is an empty relation from $A$ to $B$ .

MEDIUM

11th ICSE

IMPORTANT

I.S.C MATHEMATICS FOR CLASS XI>Chapter 2 - Relations and Functions>EXERCISE-2(II)>Q 7

Let $R$ be the relation on $Z$ defined by $R = {(x, y) : x, y \in Z, x - y$ is an integer $}$ . Find the domain and range of $R$ .

HARD

11th ICSE

IMPORTANT

I.S.C MATHEMATICS FOR CLASS XI>Chapter 2 - Relations and Functions>EXERCISE-2(II)>Q 8

A relation $R$ is defined as $A$ and $2 x + y = 14}$ . Find $R$ as the set of ordered pairs and hence, find its domain and range.

HARD

11th ICSE

IMPORTANT

I.S.C MATHEMATICS FOR CLASS XI>Chapter 2 - Relations and Functions>EXERCISE-2(II)>Q 9

A relation $R$ is defined on the set of natural numbers $N$ as $R = {(x, y) : x \in N, y \in N$ and $y = \sqrt{x + 3}, x \leq 22}$ . Find $R$ as the set of order pairs and hence, find its domain and range.

HARD

11th ICSE

IMPORTANT

I.S.C MATHEMATICS FOR CLASS XI>Chapter 2 - Relations and Functions>EXERCISE-2(II)>Q 10

A relation $R$ is defined on the set $Z$ of integers as $R = \{(x, y) : x \in Z, y \in Z$ and $x^{2} + y^{2} = 100\}$ , where $Z$ is the set of all integers. Find $R$ as the set of ordered pairs and hence, find its domain and range. What is the set of ordered pairs for $R^{- 1}$ of $R$ ?

EASY

11th ICSE

IMPORTANT

I.S.C MATHEMATICS FOR CLASS XI>Chapter 2 - Relations and Functions>EXERCISE-2(II)>Q 11

Let $R_{1}$ be a relation on the set $R$ of all real numbers defined by $R_{1} = [(a, b) : 1 + a b > 0$ for all $a, b \in R]$ .

Show that $(a, a) \in R$ , for all $a \in R$ .

EASY

11th ICSE

IMPORTANT

I.S.C MATHEMATICS FOR CLASS XI>Chapter 2 - Relations and Functions>EXERCISE-2(II)>Q 11

Let $R_{1}$ be a relation on the set $R$ of all real numbers defined by $R_{1} = [(a, b) : 1 + a b > 0$ for all $a, b \in R]$ .

Show that $(a, b) \in R_{1}, (b, a) \in R_{1}$ for all $a, b \in R$ .

EASY

11th ICSE

IMPORTANT

I.S.C MATHEMATICS FOR CLASS XI>Chapter 2 - Relations and Functions>EXERCISE-2(II)>Q 12

If $R_{1}$ defined on $R$ by the relation $R ((a, b) : 1 + a b > 0,$ for $(a, b) \in R_{1}$ and $(b, c) \in R_{1} \Rightarrow (a, c) \in R_{1}$ is not true for all $a, b, c \in R$ .

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