Relations And Functions Questions and Answers - Free PDF Download
In NCERT Solutions Class 11 Maths Chapter 2 Exercise 2 2, you’ll explore the basics of Relations and Functions—two concepts that help us make connections between different sets in Maths. This exercise will teach you how to find domains, ranges, ordered pairs, and understand how sets relate to each other in step-by-step, easy-to-understand ways.
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If you ever get confused about finding relations, looking for the difference between a relation and a function, or need help with domains and ranges, Vedantu’s clear solutions have your back. The free PDF download of NCERT Solutions is designed to solve your common doubts and make exam preparations much smoother. If you want to check out the syllabus, you can find it in the Class 11 Maths CBSE Syllabus.
Studying these solutions will not only help with your board exams but also build a strong foundation for higher-level maths. Need all NCERT Solutions for your class? Head over to the complete NCERT Solutions for Class 11 Maths to get started right away!
Exercise 2.2
1. Let \[{\rm{A = }}\left\{ {{\rm{1,2,3}}...{\rm{14}}} \right\}\]. Define a relation \[{\rm{R}}\] from \[{\rm{A}}\]to \[{\rm{A}}\]by \[{\rm{R = }}\left\{ {\left( {{\rm{x,y}}} \right){\rm{:3x - y = 0}}} \right\}\],where \[{\rm{x,y}} \in {\rm{A}}\]. Write down its domain, codomain and range.
Ans: The relation \[{\rm{R}}\] from \[{\rm{A}}\]to \[{\rm{A}}\]is given as \[{\rm{R = }}\left\{ {\left( {{\rm{x,y}}} \right){\rm{:3x - y = 0}}} \right\}\] where \[{\rm{x,y}} \in {\rm{A}}\]
i.e., \[{\rm{R = }}\left\{ {\left( {{\rm{x,y}}} \right){\rm{:3x = y}}} \right\}\] where \[{\rm{x,y}} \in {\rm{A}}\]
\[\therefore {\rm{R = }}\left\{ {\left( {{\rm{1,3}}} \right){\rm{,}}\left( {{\rm{2,6}}} \right){\rm{,}}\left( {{\rm{3,9}}} \right){\rm{,}}\left( {{\rm{4,12}}} \right)} \right\}\]
The domain of \[{\rm{R}}\] is the set of all the first elements of the ordered pairs in the relation.
Therefore, domain of \[{\rm{R = }}\left\{ {{\rm{1,2,3,4}}} \right\}\].
The whole set \[{\rm{A}}\]is the codomain of the relation \[{\rm{R}}\].
Therefore, codomain of \[{\rm{R = }}\left\{ {{\rm{1,2,3}}...{\rm{14}}} \right\}\]
The range of \[{\rm{R}}\]is the set of all second elements of the ordered pairs in the relation.
Therefore, range of \[{\rm{R = }}\left\{ {{\rm{3,6,9,12}}} \right\}\]
2. Define a relation \[{\rm{R}}\]on the set \[{\rm{N}}\] of natural numbers by\[{\rm{R = }}\left\{ {\left( {{\rm{x,y}}} \right){\rm{:y = x + 5, \text{x is a natural number less than 4}; x,y}} \in {\rm{N}}} \right\}\]. Depict this relationship using roster form. Write down the domain and the range.
Ans: We are given that,
\[{\rm{R = }}\left\{ {\left( {{\rm{x,y}}} \right){\rm{:y = x + 5,\text{x is a natural number less than 4}; x,y}} \in {\rm{N}}} \right\}\]
The natural numbers less than 4 are 1,2 and 3.
Therefore, \[{\rm{R = }}\left\{ {\left( {{\rm{1,6}}} \right){\rm{,}}\left( {{\rm{2,7}}} \right){\rm{,}}\left( {{\rm{3,8}}} \right)} \right\}\]
The domain of \[{\rm{R}}\]is the set of all first elements of the ordered pairs in the relation.
Therefore, domain of \[{\rm{R = }}\left\{ {{\rm{1,2,3}}} \right\}\]
The range of \[{\rm{R}}\] is the set of all second elements of the ordered pairs in the relation.
Therefore, range of \[{\rm{R = }}\left\{ {{\rm{6,7,8}}} \right\}\]
3.\[{\rm{A = }}\left\{ {{\rm{1,2,3,5}}} \right\}\]and \[{\rm{B = }}\left\{ {{\rm{4,6,9}}} \right\}\]. Define a relation \[{\rm{R}}\]from \[{\rm{A}}\] to \[{\rm{B}}\] by \[{\rm{R = }}\left\{ {\left( {{\rm{x,y}}} \right){\rm{\text{: the difference between x and y is odd; x}}} \in {\rm{A, y}} \in {\rm{B}}} \right\}\]. Write \[{\rm{R}}\] in roster from.
Ans: We are given that,
\[{\rm{A = }}\left\{ {{\rm{1,2,3,5}}} \right\}\] and \[{\rm{B = }}\left\{ {{\rm{4,6,9}}} \right\}\]
The relation is given by,
\[{\rm{R = }}\left\{ {\left( {{\rm{x,y}}} \right){\rm{\text{: the difference between x and y is odd; x}}} \in {\rm{A, y}} \in {\rm{B}}} \right\}\]
Therefore, \[{\rm{R = }}\left\{ {\left( {{\rm{1,4}}} \right){\rm{,}}\left( {{\rm{1,6}}} \right){\rm{,}}\left( {{\rm{2,9}}} \right){\rm{,}}\left( {{\rm{3,4}}} \right){\rm{,}}\left( {{\rm{3,6}}} \right){\rm{,}}\left( {{\rm{5,4}}} \right){\rm{,}}\left( {{\rm{5,6}}} \right)} \right\}\]
4. The given figure shows a relationship between the sets \[{\rm{P}}\] and \[{\rm{Q}}\]. Write this relation
(i) In set-builder form
(ii) In roster form
What is its domain and range?
Ans:
(i) According to the given diagram, \[{\rm{P = }}\left\{ {{\rm{5,6,7}}} \right\}\]
And \[{\rm{Q = }}\left\{ {{\rm{3,4,5}}} \right\}\]
Therefore, the set builder form of relation is
\[{\rm{R = }}\left\{ {\left( {{\rm{x,y}}} \right){\rm{: y = x - 2; x}} \in {\rm{P}}} \right\}\] or
\[{\rm{R = }}\left\{ {\left( {{\rm{x,y}}} \right){\rm{: y = x - 2 \text{for} x = 5,6,7}}} \right\}\]
(ii) According to the given diagram, \[{\rm{P = }}\left\{ {{\rm{5,6,7}}} \right\}\]
And \[{\rm{Q = }}\left\{ {{\rm{3,4,5}}} \right\}\]
Therefore, the roster form of relation is
\[{\rm{R = }}\left\{ {\left( {{\rm{5,3}}} \right){\rm{,}}\left( {{\rm{6,4}}} \right){\rm{,}}\left( {{\rm{7,5}}} \right)} \right\}\]
5. Let \[{\rm{A = }}\left\{ {{\rm{1,2,3,4,6}}} \right\}\]. Let \[{\rm{R}}\] be the relation on \[{\rm{A}}\] defined by \[\left\{ {\left( {{\rm{a,b}}} \right){\rm{: a,b}} \in {\rm{A, \text{ b is exactly divisible by a}}}} \right\}\].
(i) Write \[{\rm{R}}\] in roster form
(ii) Find the domain of \[{\rm{R}}\]
(iii) Find the range of \[{\rm{R}}\]
Ans:
(i) We are given that \[{\rm{A = }}\left\{ {{\rm{1,2,3,4,6}}} \right\}\]
and \[{\rm{R}} = \left\{ {\left( {{\rm{a,b}}} \right){\rm{: a,b}} \in {\rm{A, \text{b is exactly divisible by a}}}} \right\}\]
Therefore, the roster form of relation \[{\rm{R}}\] is
\[{\rm{R = }}\left\{ {\left( {{\rm{1,1}}} \right){\rm{,}}\left( {{\rm{1,2}}} \right){\rm{,}}\left( {{\rm{1,3}}} \right){\rm{,}}\left( {{\rm{1,4}}} \right){\rm{,}}\left( {{\rm{1,6}}} \right){\rm{,}}\left( {{\rm{2,2}}} \right){\rm{,}}\left( {{\rm{2,4}}} \right){\rm{,}}\left( {{\rm{2,6}}} \right){\rm{,}}\left( {{\rm{3,3}}} \right){\rm{,}}\left( {{\rm{3,6}}} \right){\rm{,}}\left( {{\rm{4,4}}} \right){\rm{,}}\left( {{\rm{6,6}}} \right)} \right\}\]
(ii) The domain of \[{\rm{R}}\] is \[\left\{ {{\rm{1,2,3,4,6}}} \right\}\]
(iii) The range of \[{\rm{R}}\] is \[\left\{ {{\rm{1,2,3,4,6}}} \right\}\]
6. Determine the domain and range of the relation \[{\rm{R}}\] defined by \[{\rm{R = }}\left\{ {\left( {{\rm{x,x + 5}}} \right){\rm{:x}} \in \left\{ {{\rm{0,1,2,3,4,5}}} \right\}} \right\}\]
Ans: We are given that the relation \[{\rm{R}}\] is given by
\[{\rm{R = }}\left\{ {\left( {{\rm{x,x + 5}}} \right){\rm{:x}} \in \left\{ {{\rm{0,1,2,3,4,5}}} \right\}} \right\}\]
Therefore, \[{\rm{R = }}\left\{ {\left( {{\rm{0,5}}} \right){\rm{,}}\left( {{\rm{1,6}}} \right){\rm{,}}\left( {{\rm{2,7}}} \right){\rm{,}}\left( {{\rm{3,8}}} \right){\rm{,}}\left( {{\rm{4,9}}} \right){\rm{,}}\left( {{\rm{5,10}}} \right)} \right\}\]
Domain of \[{\rm{R = }}\left\{ {{\rm{0,1,2,3,4,5}}} \right\}\]
Range of \[{\rm{R = }}\left\{ {{\rm{5,6,7,8,9,10}}} \right\}\]
7.Write the relation \[{\rm{R = }}\left\{ {\left( {{\rm{x,}}{{\rm{x}}^{\rm{3}}}} \right){\rm{\text{:x is a prime number less than 10}}}} \right\}\] in roster form.
Ans: We are given that \[{\rm{R = }}\left\{ {\left( {{\rm{x,}}{{\rm{x}}^{\rm{3}}}} \right){\rm{\text{:x is a prime number less than 10}}}} \right\}\]
The prime numbers less than 10 are 2,3,5 and 7.
Therefore, \[R = \left\{ {\left( {{\rm{2,8}}} \right){\rm{,}}\left( {{\rm{3,27}}} \right){\rm{,}}\left( {{\rm{5,125}}} \right){\rm{,}}\left( {{\rm{7,343}}} \right)} \right\}\] is the roster form.
8. Let \[{\rm{A = }}\left\{ {{\rm{x,y,z}}} \right\}\] and \[{\rm{B = }}\left\{ {{\rm{1,2}}} \right\}\]. Find the number of relations from \[{\rm{A}}\]to \[{\rm{B}}\].
Ans: It is given that \[{\rm{A = }}\left\{ {{\rm{x,y,z}}} \right\}\] and \[{\rm{B = }}\left\{ {{\rm{1,2}}} \right\}\].
Therefore, \[{\rm{A \times B = }}\left\{ {\left( {{\rm{x,1}}} \right){\rm{,}}\left( {{\rm{x,2}}} \right){\rm{,}}\left( {{\rm{y,1}}} \right){\rm{,}}\left( {{\rm{y,2}}} \right){\rm{,}}\left( {{\rm{z,1}}} \right){\rm{,}}\left( {{\rm{z,2}}} \right)} \right\}\]
Since, \[{\rm{n(A \times B) = 6}}\], the number of subsets of \[{\rm{A \times B}}\] is \[{{\rm{2}}^{\rm{6}}}\].
9. Let \[{\rm{R}}\] be the relation on \[{\rm{Z}}\] defined by \[{\rm{R = }}\left\{ {\left( {{\rm{a,b}}} \right){\rm{: a,b}} \in {\rm{Z, a - b \text{is an integer}}}} \right\}\]. Find the domain and range of \[{\rm{R}}\].
Ans: We are given that \[{\rm{R = }}\left\{ {\left( {{\rm{a,b}}} \right){\rm{: a,b}} \in {\rm{Z, a - b \text{ is an integer}}}} \right\}\]
It is known that the difference between any two integers is always an integer.
Therefore, domain of \[{\rm{R = Z}}\]
And the range of \[{\rm{R = Z}}\]
Conclusion
In Exercise 2.2 Class 11 Maths, understanding relations is crucial as it forms the foundation for various advanced topics. Relations help in defining the connections between different sets of elements, which is essential for studying functions. Focus on understanding the types of relations and their properties. In previous years, about 2-3 questions from this chapter have been asked in exams, highlighting its importance. Mastering this topic will aid in solving complex problems in higher classes. Vedantu provides detailed explanations and practice problems to enhance your understanding.
Class 11 Maths Chapter 2: Exercises Breakdown
CBSE Class 11 Maths Chapter 2 Other Study Materials
Chapter-Specific NCERT Solutions for Class 11 Maths
Given below are the chapter-wise NCERT Solutions for Class 11 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.
Important Related Links for CBSE Class 11 Maths
FAQs on NCERT Solutions For Class 11 Maths Chapter 2 Relations And Functions Exercise 2.2 - 2025-26
1. Is A × B the same as B × A in relations?
No, the Cartesian product A × B is not the same as B × A unless A = B. The confusion arises because multiplication of numbers is commutative. For sets, the order matters. For example, if A = {1} and B = {2}, then A × B = {(1, 2)} while B × A = {(2, 1)}.
2. Are all relations also functions?
No, not every relation is a function. A relation is a function only if every element in the domain has exactly one unique image in the co-domain. A common mistake is assuming any set of ordered pairs is a function. For example, R = {(1, 2), (1, 3)} is a relation but not a function.
3. Do NCERT Solutions for Class 11 Maths just give the final answers?
No, quality NCERT Solutions provide detailed, step-by-step explanations for every question, not just the final answer. They are designed to show the correct methodology for arriving at the solution, which is crucial for learning and exams, covering all questions and answers from Class 11 exercises.
4. Are the domain and range of a relation always the same?
No, the domain and range of a relation are not always the same. The domain is the set of all first elements in the ordered pairs, while the range is the set of all second elements. For the relation R = {(1, a), (2, b)}, the domain is {1, 2} and the range is {a, b}.
5. Is the Relations and Functions Class 11 NCERT PDF free to download?
Yes, the Free PDF for Relations and Functions Class 11 NCERT solutions is typically available at no cost from educational platforms. This ensures all students can access high-quality study materials to understand concepts and solve problems effectively from the Relations and Functions NCERT PDF.
6. Is every element of the co-domain part of the range in a function?
No, the range of a function is a subset of its co-domain, but they are not always identical. The confusion often stems from not distinguishing between possible outputs (co-domain) and actual outputs (range).
7. How can you tell if a graph represents a function?
A graph represents a function if any vertical line drawn on it intersects the graph at most at one point. This is known as the Vertical Line Test. Students sometimes get confused and use a horizontal line, which tests if a function is one-to-one, not if it is a function in the first place.
8. Are Relations and Functions for Class 11 too difficult to learn?
Relations and Functions Class 11 is a foundational chapter that becomes manageable with a systematic approach. The perception of it being difficult often arises from weak basics in the chapter on Sets or trying to memorise rules without understanding the core logic of mapping elements. The key is to visualise the concepts.
9. What is included in the NCERT Solutions for Class 11 Maths Chapter 2?
NCERT Solutions for Class 11 Maths Chapter 2 provide complete, step-by-step answers for all questions in every exercise of the textbook. These solutions cover everything, from introductory questions defining ordered pairs and Cartesian products in Exercise 2.1 to more complex problems on domain and range in Exercise 2.2 and algebraic operations on functions in Exercise 2.3.
10. Is using a Relations and Functions Class 11 ncert pdf enough for exam preparation?
The Relations and Functions Class 11 NCERT PDF is the most crucial resource, but combining it with NCERT Solutions is the most effective strategy. A common myth is that simply reading the textbook chapter is sufficient for scoring well. The NCERT textbook is excellent for building foundational concepts and providing solved examples.
