NCERT Solutions for Class 11 Maths Chapter 2 Exercise 2.2: Here are the NCERT Solutions for Class 11 Maths Chapter 2 Exercise 2.2, which focus on understanding and solving problems related to relations. A relation R from a non-empty set A to another set B is a subset of the Cartesian product A×B, where each ordered pair(a,b) describes a connection between elements of A and B.
The solutions cover important concepts such as domain, range, and codomain, along with methods for representing relations, including the roster method, set-builder method, and arrow diagrams. These step-by-step solutions help students grasp the fundamentals of relations and their practical applications.Key Concepts Covered:
Relations :
Domain, Range, and Codomain :
Representation of Relations :
Class 11 Maths Chapter 2 Exercise 2.2 Questions and Answers PDF
1. Let A = {1, 2, 3, … , 14}. Define a relation R from A to A by R = {( x , y ): 3 x – y = 0, where x , y ∈ A}. Write down its domain, codomain and range.
Solution:
The relation R from A to A is given as R = {( x , y ): 3 x – y = 0, where x , y ∈ A} = {( x , y ): 3 x = y , where x , y ∈ A} So, R = {(1, 3), (2, 6), (3, 9), (4, 12)} Now, The domain of R is the set of all first elements of the ordered pairs in the relation. Hence, Domain of R = {1, 2, 3, 4} The whole set A is the codomain of the relation R. Hence, Codomain of R = A = {1, 2, 3, …, 14} The range of R is the set of all second elements of the ordered pairs in the relation. Hence, Range of R = {3, 6, 9, 12}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 roster form. Write down the domain and the range.
Solution:
The relation R is given by
R = {( x , y ): y = x + 5, x is a natural number less than 4, x , y ∈ N } The natural numbers less than 4 are 1, 2, and 3. So, R = {(1, 6), (2, 7), (3, 8)} Now, The domain of R is the set of all first elements of the ordered pairs in the relation. Hence, Domain of R = {1, 2, 3} The range of R is the set of all second elements of the ordered pairs in the relation. Hence, Range of R = {6, 7, 8}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.
Solution:
Given, A = {1, 2, 3, 5} and B = {4, 6, 9} The relation from A to B is given as R = {( x , y ): the difference between x and y is odd; x ∈ A, y ∈ B} Thus, R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}4. The figure shows a relationship between the sets P and Q. Write this relation
(i) in set-builder form (ii) in roster form.
What are their domain and range?
Solution:
From the given figure, it’s seen that P = {5, 6, 7}, Q = {3, 4, 5} The relation between P and Q Set-builder form (i) R = {( x, y ): y = x – 2; x ∈ P} or R = {( x, y ): y = x – 2 for x = 5, 6, 7} Roster form (ii) R = {(5, 3), (6, 4), (7, 5)} Domain of R = {5, 6, 7} Range of R = {3, 4, 5}5. Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by
{( a , b ): a , b ∈ A, b is exactly divisible by a }.
(i) Write R in roster form.
(ii) Find the domain of R.
(iii) Find the range of R.
Solution:
Given, A = {1, 2, 3, 4, 6} and relation R = {( a , b ): a , b ∈ A, b is exactly divisible by a } Hence, (i) 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)} (ii) Domain of R = {1, 2, 3, 4, 6} (iii) Range of R = {1, 2, 3, 4, 6}6. Determine the domain and range of the relation R defined by R = {( x , x + 5): x ∈ {0, 1, 2, 3, 4, 5}}.
Solution:
Given, Relation R = {( x , x + 5): x ∈ {0, 1, 2, 3, 4, 5}} Thus, R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)} So, Domain of R = {0, 1, 2, 3, 4, 5} and Range of R = {5, 6, 7, 8, 9, 10}7. Write the relation R = {( x , x 3 ): x is a prime number less than 10} in roster form.
Solution:
Given, Relation R = {( x , x 3 ): x is a prime number less than 10} The prime numbers less than 10 are 2, 3, 5, and 7. Therefore, R = {(2, 8), (3, 27), (5, 125), (7, 343)}8. Let A = { x , y , z} and B = {1, 2}. Find the number of relations from A to B.
Solution:
Given, A = { x , y , z} and B = {1, 2} Now, A × B = {( x , 1), ( x , 2), ( y , 1), ( y , 2), ( z , 1), ( z , 2)} As n (A × B) = 6, the number of subsets of A × B will be 2 6 . Thus, the number of relations from A to B is 2 6 .9. Let R be the relation on Z defined by R = {( a , b ): a , b ∈ Z, a – b is an integer}. Find the domain and range of R.
Solution:
Given, Relation R = {( a , b ): a , b ∈ Z, a – b is an integer} We know that the difference between any two integers is always an integer. Therefore, Domain of R = Z and Range of R = Z