Download Free PDF of Relations and Functions Exercise 2.2 Solutions for Class 11 Maths
You’re about to work through NCERT Solutions for Class 11 Maths Chapter 2 Exercise 2.2, focusing on Relations and Functions—a key foundation for advanced mathematics in the CBSE syllabus. This exercise introduces essential concepts like domain, range, ordered pairs, and function mapping, which will support your understanding of crucial topics such as function types and mathematical mapping as you progress in your studies.
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Mastering topics like one-one and onto functions is vital because Chapter 2 carries significant weightage within the Sets & Functions unit, often appearing in board questions year after year. If you ever wonder “how can I easily identify domain and range in NCERT Exercise 2.2 problems?”, you’re not alone—these are some of the most searched student concerns for this chapter.
Every solution here is designed to simplify relation types and stepwise analysis, making it easier to differentiate between relations and functions—something many students find confusing right before exams. Use these stepwise explanations, curated by Vedantu’s experienced team, to develop strong problem-solving skills for your board assessments and to stay fully aligned with the latest CBSE guidelines.
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 CBSE Class 11 Maths Chapter 2 Relations and Functions – NCERT Solutions 2025-26
1. What are the key differences between a relation and a function in Class 11 Maths?
The main difference between a relation and a function in Class 11 Maths is that every function is a relation, but not every relation is a function.
Key differences:
- Relation is any set of ordered pairs from two sets, whereas a function assigns exactly one output to each input.
- In a function, no two ordered pairs can have the same first element with different second elements.
- Functions are a special type of relation that satisfy the uniqueness condition.
2. How can I easily identify domain and range in NCERT Exercise 2.2 problems?
To find the domain and range in NCERT Exercise 2.2 problems:
- Domain: List all first elements from the given set of ordered pairs or relation. This is the set of all inputs.
- Range: List all second elements from the ordered pairs. This is the set of all possible outputs.
- Check for duplicates and list each value only once in both sets.
Mastering this helps in quick revision and scoring full marks in board exams.
3. Where can I download the PDF solutions for Class 11 Maths Chapter 2 Exercise 2.2?
You can download the NCERT Solutions for Class 11 Maths Chapter 2 Exercise 2.2 PDF from Vedantu’s official solution pages.
- Look for a bold PDF download button or link labeled "Class 11 Maths Ex 2.2 Solutions PDF" on the NCERT Solutions section.
- These PDFs are aligned with the latest CBSE syllabus 2024-25 and cover all stepwise solutions.
- The PDFs are exam-oriented and suitable for revision and last-minute practice.
4. Does solving all NCERT Ex 2.2 problems help in board exams?
Yes, completing all Class 11 Maths Chapter 2 Exercise 2.2 questions is highly beneficial for board exams.
- Builds strong conceptual understanding of relations, functions, domain & range.
- Many board exam and entrance questions are directly based on NCERT Exercise 2.2.
- Improves speed and accuracy for MCQs and reasoning problems.
- Boosts confidence and helps you revise all types of questions (one-one, onto, etc.).
5. How do Vedantu NCERT Solutions explain one-one and onto functions?
Vedantu's NCERT Solutions for Class 11 Maths Exercise 2.2 provide student-friendly explanations with examples for one-one (injective) and onto (surjective) functions.
- Stepwise answers show how to test if a function is one-one (no two inputs map to the same output) or onto (every element in the codomain is covered).
- Solutions use mapping diagrams and solved ordered pairs to visually clarify these concepts.
- Exam tips highlight how to quickly identify these properties for board questions.
6. What is the best strategy to master Relations and Functions for Class 11?
To master Relations and Functions in Class 11 Maths, follow these steps:
- Start with NCERT theory for basic definitions and types (relation, function, domain, range).
- Solve all examples and Exercise 2.2 step by step, focusing on mapping, one-one, onto, etc.
- Create revision notes with key points and mapping diagrams (mind maps help).
- Practice MCQ sheets and board-style sample questions.
- Use downloadable PDFs for last-minute revision and doubt-solving.
7. Is Class 11 Maths Chapter 2 Exercise 2.2 tough to solve?
Exercise 2.2 of Class 11 Maths Chapter 2 can be challenging due to abstract concepts like relations and functions, but it becomes straightforward with practice.
- Key areas like ordered pairs, domain, range, and types of functions require careful attention.
- Using stepwise solutions, mapping diagrams, and revision notes simplifies complex problems.
- Practising sample papers and solved NCERT questions improves understanding and exam performance.
8. What topics are included in the CBSE 2024-25 syllabus for Relations and Functions?
The CBSE Class 11 Syllabus 2024-25 for Relations and Functions includes:
- Definition and types of relations (reflexive, symmetric, transitive, equivalence).
- Definition and types of functions (one-one, onto, invertible).
- Domain, codomain, and range of functions.
- Ordered pairs, Cartesian product, mapping diagrams.
- Composition of functions, practical examples, and solved exercises such as Exercise 2.2.
9. How does solving NCERT Class 11 Maths Exercise 2.2 help in JEE or competitive exams?
Practising NCERT Exercise 2.2 develops your foundational understanding of sets, relations, and functions, which is essential for entrance exams like JEE.
- Improves problem-solving skills for functional equations, domain and range, and mapping questions.
- Covers important definitions and properties tested in competitive MCQs.
- Reinforces logical thinking that's important for advanced algebra topics in JEE.
10. How can I avoid common mistakes in Class 11 Maths Relations and Functions questions?
You can avoid common errors in Relations and Functions by following these tips:
- Carefully differentiate between a relation and a function by checking the uniqueness condition.
- List all unique elements when finding domain and range—don’t repeat values.
- Read definitions for one-one and onto functions closely before answering.
- Double-check mapping diagrams for accuracy.
- Practise with stepwise solutions to see each stage of the answer.
