NCERT Solutions for Exercise 2.2 Class 11 Maths Chapter 2 - Relations and Functions

# NCERT Solutions for Exercise 2.2 Class 11 Maths Chapter 2 - Relations and Functions

Edited By Vishal kumar | Updated on Nov 03, 2023 07:59 AM IST

## NCERT Solutions for Class 11 Maths Chapter 2 - Relations And Functions Exercise 2.2- Download Free PDF

NCERT Solutions for Class 11 Maths Chapter 2: Relations and Functions Exercise 2.2- In the previous exercise, you have already learned about the cartesian product of sets and ordered pairs. In the NCERT solutions for Class 11 Maths chapter 2 exercise 2.2, you will learn about the relations. In our daily lives, we come across many relationships such as brother, sister, mother, father, son, daughter, teacher, etc. In Mathematics, we also have relationships such as 3 is less than 5, two lines are perpendicular, two triangles are symmetric, etc. In the Mathematics term, the relation R from a non-empty set A to a non-empty set B is a subset of the cartesian product A × B. In exercise 2.2 Class 11 Maths you need to define the relation from set A to set B in the given condition.

There are some questions related to depicting this relationship using set builder form and roster form in the Class 11 Maths chapter 2 exercise 2.2. The questions related to finding the domain, codomain, and range are also covered in the Class 11 Maths chapter 2 exercise 2.2 solutions. If you are looking for the NCERT Solutions, you can click on the given link where you will find the NCERT solutions from Classes 6 to 12 in a detailed manner.

The 11th class maths exercise 2.2 answers, also prepared by subject experts at Careers360, follow the same approach as in Exercise 2.1. They provide detailed, step-by-step explanations to help students grasp the concepts effectively. Similarly, a PDF version of the class 11 maths chapter 2 exercise 2.2 solution is available for easy access, and these valuable resources are provided free of charge. This makes them a convenient and cost-effective tool for students striving to enhance their mathematical understanding and excel in their studies.

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## Question:1 Let A = {1, 2, 3,...,14}. Define a relation R from A to A by $R = \left \{ ( x,y): 3x -y = 0 , where \: \: x , y \epsilon A \right \}$ .Write down its domain, codomain and range.

It is given that
$A = \left \{ 1, 2, 3, ..., 14 \right \} \ and \ R = \left \{ (x, y) : 3x - y = 0, \ where \ x, y \ \epsilon \ A \right \}$
Now, the relation R from A to A is given as
$R = \left \{ ( x,y): 3x -y = 0 , where \: \: x , y \epsilon A \right \}$
Therefore,
the relation in roaster form is , $,R = \left \{ (1, 3), (2, 6), (3, 9), (4, 12) \right \}$
Now,
We know that Domain of R = set of all first elements of the order pairs in the relation
Therefore,
Domain of $R = \left \{ 1, 2, 3, 4 \right \}$
And
Codomain of R = the whole set A
i.e. Codomain of $R = \left \{ 1, 2, 3, ..., 14 \right \}$
Now,
Range of R = set of all second elements of the order pairs in the relation.
Therefore,
range of $R = \left \{ 3, 6, 9, 12 \right \}$

As x is a natural number which is less than 4.
Therefore,
the relation in roaster form is, $R = \left \{ (1,6), (2,7), (3,8) \right \}$
As Domain of R = set of all first elements of the order pairs in the relation.
Therefore,
Domain of $R = \left \{ 1, 2, 3 \right \}$

Now,
Range of R = set of all second elements of the order pairs in the relation.
Therefore,
the range of $R = \left \{ 6, 7, 8 \right \}$

Therefore, domain and the range are $\left \{ 1,2,3 \right \} \ \ and \ \ \left \{ 6, 7, 8 \right \}$ respectively

It is given that
A = {1, 2, 3, 5} and B = {4, 6, 9}
And
$R = \left \{ ( x,y ) : the \: \: diffrence \: \: between \: \: x \: \: and \: \: y \: \: is \: \: odd ; x \epsilon A , y \epsilon B \right \}$
Now, it is given that the difference should be odd. Let us take all possible differences.
(1 - 4) = - 3, (1 - 6) = - 5, (1 - 9) = - 8(2 - 4) = - 2, (2 - 6) = - 4, (2 - 9) = - 7(3 - 4) = - 1, (3 - 6) = - 3, (3 - 9) = - 6(5 - 4) = 1, (5 - 6) = - 1, (5 - 9) = - 4
Taking the difference which are odd we get,

Therefore,
the relation in roaster form, $R = \left \{ (1,4), (1,6), (2,9), (3,4), (3,6), (5,4), (5,6) \right \}$

It is given in the figure that

P = {5,6,7}, Q = {3,4,5}

Therefore,
the relation in set builder form is ,
$R = \left \{ {(x, y): y = x-2; x \ \epsilon \ P} \right \}$
OR
$R = \left \{ {(x, y): y = x-2; \ for \ x = 5, 6, 7} \right \}$

From the given figure. we observe that

P = {5,6,7}, Q = {3,4,5}

And the relation in roaster form is , $R = \left \{ {(5,3), (6,4), (7,5)} \right \}$

As Domain of R = set of all first elements of the order pairs in the relation.
Therefore,
Domain of $R = \left \{ {5, 6, 7} \right \}$

Now,
Range of R = set of all second elements of the order pairs in the relation.
Therefore,
the range of $R = \left \{ {3, 4, 5} \right \}$

It is given that
A = {1, 2, 3, 4, 6}
And
$R = \left \{ ( a,b): a ,b \epsilon A , b\: \: is\: \: exactly \: \: divisible\: \: by \: \: a \right \}$

Therefore,
the relation in roaster form is , $R = \left \{ {(1,1), (1,2), (1,3), (1,4), (1,6), (2,2), (2,4), (2,6), (3,3), (3,6), (4,4), (6,6)} \right \}$

It is given that
A = {1, 2, 3, 4, 6}
And
$R = \left \{ ( a,b): a ,b \epsilon A , b\: \: is\: \: exactly \: \: divisible\: \: by \: \: a \right \}$
Now,
As Domain of R = set of all first elements of the order pairs in the relation.
Therefore,
Domain of $R = \left \{ {1, 2, 3, 4, 6} \right \}$

It is given that
A = {1, 2, 3, 4, 6}
And
$R = \left \{ ( a,b): a ,b \epsilon A , b\: \: is\: \: exactly \: \: divisible\: \: by \: \: a \right \}$
Now,
As the range of R = set of all second elements of the order pairs in the relation.
Therefore,
Range of $R = \left \{ {1, 2, 3, 4, 6} \right \}$

It is given that
$R = \left \{ ( x , x +5 ): x \epsilon \left \{ 0,1,2,3,,4,5 \right \} \right \}$

Therefore,
the relation in roaster form is , $R = \left \{ {(0,5), (1,6), (2,7), (3,8), (4,9), (5,10)} \right \}$

Now,
As Domain of R = set of all first elements of the order pairs in the relation.
Therefore,
Domain of $R =\left \{ {0, 1, 2, 3, 4, 5} \right \}$

Now,
As Range of R = set of all second elements of the order pairs in the relation.
Range of $R =\left \{ {5, 6, 7, 8, 9, 10} \right \}$

Therefore, the domain and range of the relation R is $\left \{ 0,1,2,3,4,5 \right \} \ \ and \ \ \left \{ {5, 6, 7, 8, 9, 10} \right \}$ respectively

It is given that
$R = \left \{ \right.(x, x^3) : x\: \: is\: \: a\: \: prime\: \: number \: \: less\: \: than\: \: 10\: \: \left. \right \}$
Now,
As we know the prime number less than 10 are 2, 3, 5 and 7.
Therefore,
the relation in roaster form is , $R =\left \{ {(2,8), (3,27), (5,125), (7,343)} \right \}$

It is given that
A = {x, y, z} and B = {1, 2}
Now,
$A \times B = \left \{ {(x,1), (x,2), (y,1), (y,2), (z,1), (z,2)} \right \}$
Therefore,
$n(A \times B) = 6$
Then, the number of subsets of the set $(A \times B) = 2^n = 2^6$

Therefore, the number of relations from A to B is $2^6$

It is given that
$R = \left \{ ( a,b) : a , b \epsilon Z , a-b\: \: is \: \: an \: \: integer \right \}$
Now, as we know that the difference between any two integers is always an integer.
And
As Domain of R = set of all first elements of the order pairs in the relation.
Therefore,
The domain of R = Z

Now,
Range of R = set of all second elements of the order pairs in the relation.
Therefore,
range of R = Z

Therefore, the domain and range of R is Z and Z respectively

## More About NCERT Solutions for Class 11 Maths Chapter 2 Exercise 2.2:-

Class 11 ex 2.2 consists of questions related to defining the relations between the given two sets, and finding the domain, co-domain, and range of the defined relation. There are three examples and a few definitions given in ex 2.2 class 11
before the NCERT book Class 11 Maths chapter 2 exercise 2.2 which you can solve to get more clarity. A total of nine questions given in this exercise 2.2 class 11 maths that you should try to solve by yourself. These questions are very basic based on the definition relation only. If you find any difficulty while solving them, you can take help from Class 11 Maths chapter 2 exercise 2.2 solutions.

Also Read| Relations And Functions Class 11th Notes

## Benefits of NCERT Solutions for Class 11 Maths Chapter 2 Exercise 2.2:-

• Relations is a very important concept useful to understand the functions, domain, co-domain, and range of the functions that you will learn in the next exercise.
• Exercise 2.2 Class 11 Maths solutions are very useful for the students to get in-depth knowledge of relations.
• NCERT syllabus Class 11 Maths chapter 2 exercise 2.2 solutions are helpful for the students in their homework.

## Key Features of 11th Class Maths Exercise 2.2 Answers

1. Detailed step-by-step solutions: The class 11 maths ex 2.2 solutions provide a comprehensive, step-by-step explanation for each problem, making it easier for students to understand the concepts and problem-solving techniques.

2. Clarity and accuracy: The ex 2.2 class 11 solutions are clear and accurate, ensuring that students can confidently prepare for their exams and grasp the concepts effectively.

3. Conceptual understanding: The class 11 ex 2.2 solutions aim to help students not only find the correct answers but also develop a deep understanding of the mathematical concepts covered in Chapter 2.

4. Alignment with NCERT curriculum: The solutions are specifically designed to align with the NCERT curriculum, ensuring that they cover the topics and concepts as per the official syllabus.

5. Free access: These solutions are often available for free, making them accessible to a wide range of students.

6. Format options: PDF versions of the ex 2.2 class 11 solutions is provided, allowing students to download and use them anytime and anywhere for their convenience.

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## Subject Wise NCERT Exampler Solutions

Happy learning!!!

1. What is relation ?

Relation R defined from a nonempty set A to nonempty set B is a subset of cartesian product A x B.

2. A relation R is defined from set A to set B than the what is the domain of the relation R ?

The relation R is a set of ordered pairs from cartesian product A x B and the set of all first elements of the ordered pairs in a relation R is called the domain of the relation R.

3. A relation R is defined from set A to set B than the what is the co-domain of the relation R ?

The relation R is a set of ordered pairs from cartesian product A x B and the whole set B is called the codomain of the relation R.

4. A relation R is defined from set A to set B than the what is the range of the relation R ?

The relation R is a set of ordered pairs from cartesian product A x B and the set of all second elements of the ordered pairs in a relation R is called the range of the relation R.

5. Does range is a subset of co-domain ?

Yes, the range is always a subset of co-domain.

6. Find the total number of relations can be defined from set A to set B where n(A) = 2 and n(B) = 3 ?

n(A) = 2 and n(B) = 3

n(A x B) = 2 x 3= 6

The total number of relations = 2^(6) = 64

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