NCERT Solutions for Class 11 – Mathematics – Chapter 2 Relations and Functions – Exercise 2.2
This section provides detailed solutions to the problems in Exercise 2.2 of Chapter 2, “Relations and Functions,” from the Class 11 NCERT Mathematics textbook. The exercise focuses on different types of functions, including one-to-one, onto, and bijective functions, as well as the methods to determine and prove these properties.
Table of Contents
Exercise 2.2
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, codomain and range.
Solution:
The relation R from A to A is given as:
R = {(x, y): 3x – y = 0, where x, y ∈ A}
= {(x, y): 3x = 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 is its 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, x3): x is a prime number less than 10} in roster form.
Solution:
Given,
Relation R = {(x, x3): 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 26.
Thus, the number of relations from A to B is 26.
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
FAQs on Relation And Functions
What is the importance of functions in mathematics?
Functions are fundamental in mathematics because they describe the relationship between variables. They are used in various branches of mathematics, including calculus, algebra, and geometry, to model real-world phenomena and solve problems.
Can a function be both one-to-one and onto?
Yes, a function can be both one-to-one and onto. Such a function is called bijective, meaning there is a perfect pairing between the elements of the domain and the codomain.
What is the difference between injective, surjective, and bijective functions?
- Injective (One-to-One) Function: A function is injective if different elements of the domain map to different elements of the codomain.
- Surjective (Onto) Function: A function is surjective if every element of the codomain has a preimage in the domain.
- Bijective Function: A function is bijective if it is both injective and surjective, meaning it has a one-to-one correspondence between elements of the domain and codomain.
Relations and Functions NCERT Solutions for Class 11 Maths Chapter 2 Free PDF Download
NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions are solved in detail in the PDF given below. All the solutions to the problems in the exercises are created in such a way that it enables the students to prepare for the exam and ace it. The NCERT Solutions are prepared by the most experienced teachers in the education space, making the explanation of each solution simple, understandable, and according to the latest CBSE Syllabus. The solution helps Class 11 students to master the concept of Relations and Functions.
The solutions provide a good understanding of the fundamental concepts before they solve the equations. Through regular practice, students will know the difference between relations and functions, which are included under the syllabus, and become well-versed in its concepts. Numerous examples are present in the textbook before the exercise questions to help them understand the methodologies to be followed while solving the problems. Referring to the NCERT Class 11 Solutions PDF, students can get a glimpse of the important concepts before facing their final exams.
