NCERT Solutions for Maths Class 11 Chapter 2 Exercise 2.2 - FREE PDF Download
NCERT Solutions for Class 11 Maths Chapter 2 - Relations and Functions Exercise 2.2 focuses on understanding different types of relations and their properties. This exercise is important as it helps build a solid foundation for more advanced topics in mathematics. Students should focus on the concepts of reflexive, symmetric, and transitive relations, and practice problems related to these types.
- 3.1Exercise 2.2
By working through Exercise 2.2 Class 11 Maths, students will develop a deeper understanding of how relations and functions operate, which is essential for higher studies in mathematics. It is important to pay attention to the definitions and examples provided to master this topic effectively. Students can also visit our page to access as per the latest CBSE Syllabus for Class 11 Maths.
Glance on NCERT Solutions Maths Chapter 2 Exercise 2.2 Class 11 | Vedantu
This exercise explains the fundamental concepts of relations and functions in Class 11 Maths. It begins with the definition of a relation, which is a set of ordered pairs where the first element is related to the second element according to a specific rule and how the product of two sets is defined.
Students will learn about the set of related terms like image, domain, codomain, range, etc and how these terms are related with each other.
Sets are represented in two methods, Roster method and Set builder method.
The chapter also explains to find the number of relations exist from one set to another.
This Exercise 2.2 Class 11 Maths have 9 questions in total which will help students practise the above mentioned concepts.
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
1. How the cartesian product of two sets is defined in Class 11 Ex 2.2?
The cartesian product is a set formed from two or more given sets and contains all ordered pairs of elements such that the initial element of the pair is from the first set and the second is from the second set and the third is from the third set and so on.
The Cartesian product of two sets $A$ and $B$, denoted as $A \times B$, is defined as the set of all ordered pairs $(a,b)$ where $a \in A$ and $b \in B$. Formally, it is written as:
$A \times B = \{ (a, b) \ | \ a \in A \text{ and } b \in B \}$.
2. Why do we use cartesian products in Class 11th Maths Exercise 2.2 Answers?
The product of cartesian in computing is exactly the same as in mathematics. It will apply to matrix applications. An ordered pair means that two elements are taken from each set.
3. What are the four operations of sets in NCERT solution Class 11 Maths Chapter 2 Relation and Function Exercise 2.2?
Set operation is performed on two or more sets and gets a combination of elements according to the operation performed on them and there are four important basic operations of sets are:
Union of sets
Intersection of sets.
Complement of sets.
Cartesian product of sets.
4. How do we count the number of possible relations between two sets in Class 11 Chapter 2 Relations and Functions?
The number of possible relations between two sets $A$ and $B$ is $2^{m \times n}$, where $m$ is the number of elements in set $A$ and $n$ is the number of elements in set $B$.
5. What is the difference in Sets and Relations in reference to the Chapter 2 of Class 11 Maths?
A set is simply a collection of elements, whereas a relation is a set of ordered pairs that define a specific relationship between two sets. Also, the elements of a set are individual objects, while the elements of a relation are ordered pairs.
6. What is the range and codomain in a relation, and what is the relation between them?
The codomain of a relation is the set of all possible second elements (outputs) that can be paired with elements from the first set (domain).
The range of a relation is the set of all actual second elements (outputs) that are paired with elements from the first set (domain) in the relation.
The range is always a subset of the codomain. This means every element in the range is also an element of the codomain, but not every element in the codomain must be in the range.
7. What is the domain and range of a relation taught in Maths Ex 2.2 Class 11?
The domain of a relation is the set of all first elements (x-values) of the ordered pairs. The range is the set of all second elements (y-values) as per Ex 2.2 Class 11.
8. What are the types of relations as per the Class 11 Maths Chapter 2 Exercise 2.2 Syllabus ?
The types of relations include:
Reflexive Relation: Every element is related to itself.
Symmetric Relation: If $(a, b)$ is in R, then (b, a) is also in R.
Transitive Relation: If (a, b) and (b, c) are in R, then (a, c) is in R.
Equivalence Relation: A relation that is reflexive, symmetric, and transitive.
9. How many questions from Class 11 Maths Chapter 2 Relations appeared in previous year exams?
The number of questions on relations in previous year exams can vary. On average, 2-3 questions from the topic of relations are typically asked in Class 11 exams. It's important to review past papers for specific patterns. Student can visit Vedantu’s NCERT Solutions pages for detailed step-by-step solutions of NCERT questions.
10. Where can I find Exercise 2.2 Class 11 Maths Solutions from NCERT book?
You can easily find the NCERT solutions for Class 11th Maths Chapter 2 Relation and Function Exercise 2.2 on the Vedantu website. Our highly skilled and qualified teachers provide you with the best and easily understandable solutions that will help students in learning the concept in a better way.
