GSEB Class 11 Maths Solutions Chapter 2 Relations and Functions Exercise 2.2

Question 1. Let A = {1, 2, 3, ..., 14}. Define a relation R from A to A by R = {(x, y) : 3x − y = 0, where x, y ∈ A}. Write down its domain, co-domain and range?
Answer: The given set A is \( \{1, 2, 3, ..., 14\} \). The relation R from A to A is defined as \( R = \{(x, y) : 3x - y = 0, \text{ where } x, y \in A\} \). We can express this rule as \( 3x - y = 0 \)
\( \implies y = 3x \).
Since both x and y need to be part of set A:
If \( x = 1 \), then \( y = 3(1) = 3 \). This means \( (1, 3) \) is in R.
If \( x = 2 \), then \( y = 3(2) = 6 \). This means \( (2, 6) \) is in R.
If \( x = 3 \), then \( y = 3(3) = 9 \). This means \( (3, 9) \) is in R.
If \( x = 4 \), then \( y = 3(4) = 12 \). This means \( (4, 12) \) is in R.
If \( x = 5 \), then \( y = 3(5) = 15 \). Because \( 15 \) is not in A, \( (5, 15) \) is not in R. No more pairs will meet the requirement.
So, the relation is \( R = \{(1, 3), (2, 6), (3, 9), (4, 12)\} \).
(i) The relation in roster form is \( R = \{(1, 3), (2, 6), (3, 9), (4, 12)\} \).
(ii) The domain of R involves all the first elements of the ordered pairs, which is \( \{1, 2, 3, 4\} \).
(iii) The co-domain of R is the whole set A, which is \( \{1, 2, 3, ..., 14\} \).
(iv) The range of R is the set of all second elements of the ordered pairs, which is \( \{3, 6, 9, 12\} \).
In simple words: First, we find all the pairs (x, y) where y is three times x, and both x and y are numbers from 1 to 14. Then, the domain is all the 'x' values in these pairs, the co-domain is the entire set A, and the range is all the 'y' values in these pairs.

Exam Tip: Always list out the possible pairs for a given relation carefully, making sure both elements satisfy the specified conditions of belonging to the set.

Question 2. Define a relation R on the set N of natural numbers by R = {x, y) : y = x + 5, x is a natural number less than 4; x, y ∈ N}. Depict this relationship using (i) roster form (ii) an arrow diagram. Write down the domain and range?
Answer: The provided relation R on natural numbers (N) is \( R = \{(x, y) : y = x + 5, x \text{ is a natural number smaller than } 4; x, y \in N\} \). The natural numbers that are less than 4 are 1, 2, and 3. We determine the pairs:
If \( x = 1 \), then \( y = 1 + 5 = 6 \). This means \( (1, 6) \) is in R.
If \( x = 2 \), then \( y = 2 + 5 = 7 \). This means \( (2, 7) \) is in R.
If \( x = 3 \), then \( y = 3 + 5 = 8 \). This means \( (3, 8) \) is in R.
(i) The relation in roster form is \( R = \{(1, 6), (2, 7), (3, 8)\} \).
(ii) The arrow diagram illustrating this relationship is shown below.
(iii) The domain of the relation is \( \{1, 2, 3\} \).
The range of the relation is \( \{6, 7, 8\} \).
In simple words: To get the pairs, add 5 to each natural number smaller than 4. The domain is the set of original numbers, and the range is the set of numbers you get after adding 5.

Exam Tip: When dealing with relations defined by conditions, test each element from the domain set to ensure it satisfies the criteria and generate all valid pairs.

Question 3. A = {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by R = {(x, y) : the difference between x and y is odd, x ∈ A, y ∈ B}. Write R in roster form?
Answer: The given sets are \( A = \{1, 2, 3, 5\} \) and \( B = \{4, 6, 9\} \). The relation R from A to B is defined as \( R = \{(x, y) : \text{the difference between x and y is an odd number}, x \in A, y \in B\} \). We will examine the absolute difference \( |x - y| \) for each potential pair:
When \( x = 1 \): \( |1 - 4| = 3 \) (odd), \( |1 - 6| = 5 \) (odd), \( |1 - 9| = 8 \) (even). The valid pairs are \( (1, 4), (1, 6) \).
When \( x = 2 \): \( |2 - 4| = 2 \) (even), \( |2 - 6| = 4 \) (even), \( |2 - 9| = 7 \) (odd). The valid pair is \( (2, 9) \).
When \( x = 3 \): \( |3 - 4| = 1 \) (odd), \( |3 - 6| = 3 \) (odd), \( |3 - 9| = 6 \) (even). The valid pairs are \( (3, 4), (3, 6) \).
When \( x = 5 \): \( |5 - 4| = 1 \) (odd), \( |5 - 6| = 1 \) (odd), \( |5 - 9| = 4 \) (even). The valid pairs are \( (5, 4), (5, 6) \).
Thus, the relation in roster form becomes \( R = \{(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)\} \).
In simple words: For each number in set A, find numbers in set B where subtracting them gives an odd number. List all these pairs as the relation.

Exam Tip: To verify if a difference is odd, remember that the difference of an odd and an even number is always odd. The difference of two odd or two even numbers is always even.

Question 4. Figure given below shows a relationship between the sets P and Q. Write this relation (i) in set builder form (ii) roster form. What is its domain and range?
Answer: From the presented figure, we can clearly see the sets P and Q, and the connection between their elements.
Set P is \( \{5, 6, 7\} \). Set Q is \( \{3, 4, 5\} \).
The arrows demonstrate the subsequent pairs:
\( (5, 3), (6, 4), (7, 5) \).
(i) For every pair \( (x, y) \), we discover that \( x - y = 2 \). Additionally, the \( x \) values are \( 5, 6, 7 \), which meet the condition \( 4 < x < 8 \). Therefore, the relation in set-builder form is \( R = \{(x, y) : x - y = 2, 4 < x < 8, x, y \in N\} \).
(ii) The relation in roster form is \( R = \{(5, 3), (6, 4), (7, 5)\} \).
The domain of this relation is \( \{5, 6, 7\} \).
The range of this relation is \( \{3, 4, 5\} \).
In simple words: Look at the diagram to find pairs where numbers from P connect to numbers in Q. Then, describe these pairs as a list, as a rule, and identify all the starting numbers (domain) and all the ending numbers (range).

Exam Tip: When converting a visual diagram into set-builder form, first list out the pairs in roster form, then find a mathematical rule that connects the x and y values in each pair.

Question 5. Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by {(a, b) : a ∈ A, b ∈ A, a divides b}.
1. Write R in roster form.
2. Find the domain of R.
3. Find the range of R.
Answer: The specified set A is \( \{1, 2, 3, 4, 6\} \). The relation R on A is described by \( \{(a, b) : a \in A, b \in A, a \text{ divides } b\} \). We need to locate all pairs \( (a, b) \) where 'a' evenly divides 'b':
When \( a = 1 \): The pairs are \( (1, 1), (1, 2), (1, 3), (1, 4), (1, 6) \).
When \( a = 2 \): The pairs are \( (2, 2), (2, 4), (2, 6) \).
When \( a = 3 \): The pairs are \( (3, 3), (3, 6) \).
When \( a = 4 \): The pair is \( (4, 4) \).
When \( a = 6 \): The pair is \( (6, 6) \).
1. The relation presented in roster form is \( R = \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)\} \).
2. The domain of R contains all the first elements: \( \{1, 2, 3, 4, 6\} \).
3. The range of R contains all the second elements: \( \{1, 2, 3, 4, 6\} \).
In simple words: To find this relation, check every possible pair of numbers from set A. If the first number can divide the second number exactly, then that pair is part of the relation. Then list the unique first numbers as the domain and the unique second numbers as the range.

Exam Tip: For 'a divides b' relations, systematically check each element 'a' from the domain and find all 'b' elements it divides, ensuring 'b' is also in the set.

Question 6. Determine the domain and the range of the relation R defined by R = {x, x + 5) : x ∈ (0, 1, 2, 3, 4, 5}}
Answer: In this case, the relation R is given by \( R = \{(x, x + 5) : x \in \{0, 1, 2, 3, 4, 5\}\} \). We will list the values of x and their matching \( x + 5 \) values:
When \( x = 0 \), \( x + 5 = 5 \). The pair is \( (0, 5) \).
When \( x = 1 \), \( x + 5 = 6 \). The pair is \( (1, 6) \).
When \( x = 2 \), \( x + 5 = 7 \). The pair is \( (2, 7) \).
When \( x = 3 \), \( x + 5 = 8 \). The pair is \( (3, 8) \).
When \( x = 4 \), \( x + 5 = 9 \). The pair is \( (4, 9) \).
When \( x = 5 \), \( x + 5 = 10 \). The pair is \( (5, 10) \).
The domain of R represents all the first numbers: \( \{0, 1, 2, 3, 4, 5\} \).
The range of R signifies all the second numbers: \( \{5, 6, 7, 8, 9, 10\} \).
In simple words: For each given x value, calculate y by adding 5 to x. The x values form the domain, and the y values form the range.

Exam Tip: When a relation is defined by a formula like \( y = x + 5 \), substitute each allowed x-value to find its corresponding y-value to correctly identify the ordered pairs, domain, and range.

Question 7. Write the relation R = {(x, x³) : x is a prime number less than 10} in roster form?
Answer: The relation R is described by \( R = \{(x, x^3) : x \text{ is a prime number smaller than } 10\} \). The prime numbers that are less than 10 include 2, 3, 5, and 7. We will compute \( x^3 \) for each of these:
When \( x = 2 \), then \( x^3 = 2^3 = 8 \). The resulting pair is \( (2, 8) \).
When \( x = 3 \), then \( x^3 = 3^3 = 27 \). The resulting pair is \( (3, 27) \).
When \( x = 5 \), then \( x^3 = 5^3 = 125 \). The resulting pair is \( (5, 125) \).
When \( x = 7 \), then \( x^3 = 7^3 = 343 \). The resulting pair is \( (7, 343) \).
Consequently, the relation in roster form is \( R = \{(2, 8), (3, 27), (5, 125), (7, 343)\} \).
In simple words: First, list all prime numbers less than 10. Then, for each prime number, find its cube. The pairs of (prime number, its cube) make up the relation in roster form.

Exam Tip: Remember the definition of a prime number (a natural number greater than 1 that has no positive divisors other than 1 and itself) to correctly list the prime numbers.

Question 8. Let A as {x, y, z} and B = {1, 2}. Find the number of relations from A into B?
Answer: The given sets are \( A = \{x, y, z\} \) and \( B = \{1, 2\} \). The count of elements in set A is \( n(A) = 3 \). The count of elements in set B is \( n(B) = 2 \). Initially, we need to determine the Cartesian product \( A \times B \). \( A \times B = \{(x, 1), (x, 2), (y, 1), (y, 2), (z, 1), (z, 2)\} \). The quantity of elements in \( A \times B \) is \( n(A \times B) = n(A) \times n(B) = 3 \times 2 = 6 \). The total number of relations from A to B is the same as the number of subsets of \( A \times B \). This value is \( 2^{n(A \times B)} = 2^6 = 64 \). Consequently, there are 64 relations from A into B.
In simple words: To find the number of relations from set A to set B, first multiply the number of elements in A by the number of elements in B. Then, raise 2 to the power of that product. This tells you how many different ways elements from A can relate to elements in B.

Exam Tip: Remember that the number of relations from set A to set B is \( 2^{n(A) \times n(B)} \), where \( n(A) \) and \( n(B) \) are the number of elements in sets A and B, respectively.

Question 9. Let R be the relation on Z defined by R = {(a, b) : a ∈ Z, b ∈ Z, a − b is an integer}. Find the domain and range of R?
Answer: The relation R on integers (Z) is described by \( R = \{(a, b) : a \in Z, b \in Z, a - b \text{ is an integer}\} \). Given that 'a' and 'b' are integers, their subtraction \( a - b \) will consistently result in an integer. This implies that all pairs of integers \( (a, b) \) will fulfill the condition. As a result, the domain of R comprises the set of all integers, Z. The range of R also comprises the set of all integers, Z.
In simple words: Because subtracting any two whole numbers always gives another whole number, every pair of whole numbers (a, b) fits this rule. So, both the starting numbers (domain) and the ending numbers (range) can be any whole number.

Exam Tip: For relations involving integers, remember that the sum, difference, and product of any two integers are always integers. This property is crucial for determining domain and range.