Class 9 Advanced Maths Chapter 3 Relations and Functions Ex 3.2 Solutions

Students often refer to NCERT Class 9 Advanced Maths Solutions Chapter 3 Relations and Functions Ex 3.2 to verify their answers.

Advanced Maths Class 9 Exercise 3.2 Solutions

Class 9 Advanced Maths Ex 3.2 Solutions

Question 1.
Let A = {1, 2, 3}, B = {4, 5, 6, 7}. Define a relation R from A to B by R = {(a, b): a + b = 7; a ∈ A, b ∈ B}. Write the relation R in roster form and hence find its domain and range.
Solution:
For a + b = 7:

• If a = 1, then b = 6
• If a = 2, then b = 5
• If a = 3, then b = 4

So, R = {(1, 6), (2, 5), (3, 4)}
Domain = set of first elements
∴ Domain = {1, 2, 3}
Range = set of second elements
∴ Range = {4, 5, 6}

Question 2.
Given A = {2, 3, 4, 5}, B = (3, 6, 7, 10}. Define R = {(a, b): a divides b; a ∈ A, b ∈ B}. Write R in roster form hence find its domain and range.
Solution:
Check divisibility:

• 2 | 6, 2 | 10
• 3|3, 3|6
• 4 divides none
• 5 | 10

Hence, R = {(2, 6), (2, 10), (3, 3), (3, 6), (5, 10)}
Domain = set of all first elements
∴ Domain = {2, 3,5}
Range = set of all second elements
∴ Range = {3, 6, 10}

Question 3.
Let R = {(a, b): a + 2b = 12, a, b ∈ N }. Write R in roster form and hence find its domain and range.
Solution:
Consider R = {(a, b): a + 2b = 12, a, b ∈ N}, where N denotes the natural numbers.
Using: a = 12 – 2b
Taking natural number values of b:

1. b = 1 ∈ a = 10
2. b = 2 ∈ a = 8
3. b = 3 ∈ a = 6
4. b = 4 ∈ a = 4
5. b = 5 ∈ a = 2
6. If b = 6, a = 12 – 2(6) = 0 (not a natural number)

So, R = {(10, 1), (8, 2), (6, 3), (4, 4), (2, 5)}
Domain = set of all first elements
∴ Domain = {10, 8, 6, 4, 2}
Range = set of all second elements
∴ Range = {1, 2, 3, 4, 5}

Question 4.
Write R’ = {(x, x²): x is a prime number less than 10} in roster form. Also find the range of R.
Solution:
Prime numbers less than 10 are: 2, 3, 5, 7
Their squares:
2² = 4, 3² = 9, 5² = 25, 7² = 49
Hence,
R = {(2, 4), (3, 9), (5, 25), (7, 49)}
Range = set of all second elements
∴ Range = {4, 9, 25, 49}

Question 5.
Let A = {p, q, r, s} and B = {1, 2}. How many relations can be defined from set A to set B? List any four of them.
Solution:
Number of elements in A × B : n(A × B) = 4 × 2 = 8
Number of possible relations: 2ⁿ = 2⁸ = 256
Any four relations are:
R₁ = {(p, 1)}
R₂ = {(q, 2), (r, 1)}
R₃ = {(p, 1),(q, 1), (r, 2)}
R₄ = {(S, 2)}

Question 6.
Let A = {1, 2, 3, 4, 5}. Define a relation R on A by R = {(a, b): | a – b | = 2}. Write R in roster form and hence find its domain and range.
Solution:
Pairs with difference 2:
(1, 3), (3, 1)
(2, 4), (4, 2)
(3, 5), (5, 3)
Hence, R = {(1, 3), (3, 1), (2, 4), (4, 2), (3, 5), (5, 3)}
Domain = set of all first elements
∴ Domain ={1, 2, 3, 4, 5}
Range = set of all second elements
∴ Range = {1, 2, 3, 4, 5}