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 is the best step-by-step approach to solve NCERT Class 11 Maths Chapter 2 Relations and Functions Exercise 2.2 using the official solutions?
Begin by understanding the definitions of relation, function, domain, and range from the NCERT textbook. For each question, systematically:
- List all sets and elements given.
- Identify the rule or condition for the relation/function.
- Write ordered pairs satisfying the condition.
- List all first elements (domain) and all second elements (range), ensuring no duplicates.
- Check if the relation is a function by ensuring each input maps to one output.
2. How can students distinguish between a relation and a function when solving NCERT Solutions for Class 11 Maths Chapter 2?
Every function is a relation, but not every relation is a function. In a function, each element of the domain is paired with exactly one element in the range, meaning no input has more than one output. To check, look for repeated first elements with different second elements in ordered pairs—if they exist, it's a relation but not a function.
3. What is the most effective way to find the domain and range in Relations and Functions problems in Class 11?
To determine the domain, extract all unique first elements (inputs) from the set of ordered pairs. To find the range, list all unique second elements (outputs). Always remove duplicates and cross-check with the original set definitions to ensure completeness.
4. Why is stepwise NCERT Solution methodology important for scoring high marks in CBSE exams for Chapter 2?
CBSE evaluation emphasizes clarity of steps. Writing each step—such as defining sets, constructing ordered pairs, and explicitly stating domain and range—shows your understanding and ensures full marks. This methodical approach minimizes common mistakes and enhances conceptual clarity, as required in board assessments.
5. How can one recognise one-one (injective) and onto (surjective) functions in NCERT Class 11 Maths solutions?
A function is one-one if every element of the range comes from a unique element in the domain (no two inputs share the same output). It's onto if every element in the codomain has a pre-image in the domain. In solutions, use mapping diagrams or check the sets to verify these properties for each function-related question.
6. What are the typical mistakes students make while writing relations or functions in roster and set-builder forms?
Common errors include:
- Omitting some ordered pairs or duplicating them.
- Mixing up domain and range elements.
- Using incorrect symbols or incomplete set notation.
- Failing to specify conditions clearly in set-builder form.
7. In what scenarios can a relation fail to be a function as per CBSE guidelines?
If any element of the domain is associated with more than one element in the codomain (i.e., two ordered pairs share the same first element but have different second elements), then the relation is not a function. Always check the uniqueness condition for each input when using NCERT Solutions.
8. What is the significance of mastering domain, range, and ordered pairs for higher-level Maths and entrance exams?
Understanding domain, range, and ordered pairs forms the basis of advanced topics like calculus, functions, and mappings. These concepts are tested in MCQs and functional equations in entrance exams (like JEE), and mastery ensures readiness for complex problem-solving in advanced Mathematics.
9. How does the official NCERT Class 11 Maths Chapter 2 Solutions help in preventing misconceptions about relations and functions?
The official NCERT Solutions provide clear definitions, visual mapping, and systematically worked examples. They highlight the uniqueness condition for functions, clarify the construction of sets, and show domain and range identification, reducing misconceptions and ensuring conceptual accuracy as per CBSE syllabus.
10. What strategies can students use to self-check their answers when practicing Relations and Functions through NCERT Solutions?
After solving, students should:
- Cross-verify all ordered pairs as per the given relation or function rule.
- Ensure the domain and range sets have no missing elements or duplicates.
- Check the uniqueness condition for functions.
- Compare answers with stepwise NCERT Solutions for consistency.
