NCERT solutions for class 11 exercise 2.2 of chapter 2-Relations and Functions

NCERT solutions for class 11 exercise 2.2 of chapter 2-Relations and Functions are presented here for clearing the maths concepts concerning exercise 2.2 of chapter 2-Relations and Functions. The purpose of these NCERT solutions of class 11 maths exercises 2.2 is to help the students in their preparation for the exams and class tests. All these NCERT solutions are explained beautifully by an expert of maths by a step-by-step method so that every student could understand it properly. Here you can also study NCERT solutions of other chapters of maths.

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PDF-NCERT Solutions for class 11 exercise 2.2 of chapter 2-Relations and Functions

Exercise 2.1 -Relations and Functions

Exercise 2.3-Relations and Functions

NCERT Solutions for class 11 maths

Chapter 1-Sets	Chapter 9-Sequences and Series
Chapter 2- Relations and functions	Chapter 10- Straight Lines
Chapter 3- Trigonometry	Chapter 11-Conic Sections
Chapter 4-Principle of mathematical induction	Chapter 12-Introduction to three Dimensional Geometry
Chapter 5-Complex numbers	Chapter 13- Limits and Derivatives
Chapter 6- Linear Inequalities	Chapter 14-Mathematical Reasoning
Chapter 7- Permutations and Combinations	Chapter 15- Statistics
Chapter 8- Binomial Theorem	Chapter 16- Probability

CBSE Class 11-Question paper of maths 2015

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NCERT solutions for class 11 exercise 2.2 of chapter 2-Relations and Functions

Q1.Let A = {1,2,3………14}.Define a relation R from A to A by R= {(x,y):3x – y = 0,where x,y ∈ A},write down its domain ,co-domain and range.

Ans. We are given a relation R from A to A as below.

R= {(x,y):3x – y = 0,where x,y ∈ A}

⇒3x – y = 0⇒y = 3x, where x,y ∈ A

Putting the value of x(1,2,3….14} such that x ∈ A and y∈ A

The ordered pair of the relation R

R={(1,3),(2,6),(3,9),(4,12)}

The domain is the set of all first elements of the set R

∴ The domain of relation R = {1,2,3,4}

The co-domain of relation R is the set of all elements existing in A

∴The co-domain of relation R ={1,2,3………14}

The range is the set of all elements of the set R{(1,3),(2,6),(3,9),(4,12)}

∴The range of relation R is = {3,6,9,12}

Q2. Define a relation R on the set N of natural numbers by R = {(x,y) : y = x + 5} ,x is a natural number less than 4 ; x,y ∈ N}.Depict this relationship using roster form .Write down the domain and the range.

Ans. We are given a relation R on the set N of natural numbers as below.

R = {(x,y) : y = x + 5, x is a natural number less than 4 ; x,y ∈ N}

y = x + 5

Putting the value of x in above equation such that x < 4(i.e 1,2 and 3) for every value of x,y ∈ N

R = {(1,6),(2,7),(3,8)}

The domain is the set of all first elements of the set R

∴ The domain of relation R = {1,2,3,}

The range is the set of all elements of the set R

∴The range of relation R is = {6,7,8}

Q3. A={1,2,3,5} and B = {4,6,9}. Define a relation R from A to B by R = {(x,y): the difference between x and y is odd ; x∈A and y∈ B}. Write R in roster form.

Ans. A={1,2,3,5} and B = {4,6,9}

We are given a relation R from A to B as below.

R = {(x,y): the difference between x and y is odd; x∈A and y∈ B}

Computing the relationship R such that x – y = odd number

∴ R = {(1,4),(1,6),(2,9),(3,4),(3,6),(5,4),(5,6)}

Q4.The given figure shows the relationship between the sets P and Q .Write this relation .

(i) In set builder form (ii) In roster form

What is its domain and range.

Ans. According to the figure the set P and Q are as follows

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

Observing the relationship between P and Q, we get that if x ∈ P and y ∈ Q then x -y =2

Therefore, the relationship R between P and Q in set-builder form would be written as

R = { (x,y): x – y = 2; x ∈ P and y ∈ Q}

(ii) In roster form, the relationship between P and Q is written as following

R = {(5,3),(6,4),(7,5)}

Domain of the relationship R between P and Q = {5,6,7}

Range of the relationship R between P and Q = {3,4,5}

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Q5. Let A ={1,2,3,4,6}.Let R be the relation on A defined by

{(a,b): a,b ∈ A ,b is exactly divisible by A}

(i) Write R in roster form

(ii) Find the domain of R

(iii) Find the range of R

Ans.(i) We are given that

A ={1,2,3,4,6}.

R is the relation on A defined by

{(a,b): a,b ∈ A ,b is exactly divisible by A}

Therefore,R is written in roster form as follows

∴ R = {(1,1),(1,2),(1,3),(1,4),(1,6),(2,2),(2,4),(2,6),(3,3),(3,6),(4,4),(6,6)}

(ii) The domain of R

= {1,2,3,4,6}

(iii) The range of R = {1,2,3,4,6}

Q6. Determine the domain and range of the relation R defined by R = {(x,x +5): x ∈ {0,1,2,3,4,5}}.

Ans.(i) We are given the relation R defined by

R = {(x,x +5): x ∈ {0,1,2,3,4,5}}.

It is written as in the roster form

R = {(0,5),(1,6),(2,7),(3,8),(4,9),(5,10)}

∴ Domain of the relation R = {0,1,2,3,4,5}

Range of the relation R = {5,6,7,8,9,10}

Q7. Write the relation R = {(x,x³): x is a prime number less than 10} in roster form.

Ans. We are given the relation R defined by

R = {(x,x³): x is a prime number less than 10}

Prime number less than 10 are 2,3,5 and 7

∴ R = {(2,2³),(3,3³),(5,5³),(7,7³)}

= {(2,8),(3,27),(5,125),(7,343)}

Q8. Let A = {x,y,z} and B = {1,2}. Find the number of relations from A to B.

Ans. We are given that

A = {x,y,z} and B = {1,2}

A × B = {(x,1),(x,2),(y,1),(y,2),(z,1),(z,2)}

n(A × B) shows the number of elements in A × B

n(A × B) = 6

The number of relations from A to B is = Total subsets of A × B

If the number of elements in a set is n then number of subsets is =2ⁿ= 2⁶= 64

Hence the number of relations from A to B is = 64

Q9. Let R be the relation on Z defined by R = {(a,b): a,b ∈ Z, (a – b) is an integer}.Find the domain and range of R.

Ans. It is given to us that R is the relation on Z defined by

R = {(a,b): a,b ∈ Z, (a – b) is an integer}

Since the difference between the two integers is always an integer

Set of first elements of the ordered pairs shows domain

a ∈ Z

Therefore domain of the given relation is = Z (i.e all integers)

Set of second elements of the ordered pairs shows range

b ∈ Z

Therefore the range of the given relation is = Z (i.e all integers)

Summary of exercise 2.2 of chapter 2

Class 11 maths exercise 2.2 of chapter 2 : Relations and functions is based on the cartesian product of sets and relations from one set to other.

Cartesion Product: Let there is a set A={a,b,c) and set B = (x,y) then mathematically the cartesion product A × B is the set of ordered pair {(a,x),(a,y),(b,x),(b,y),(c,x),(c,y)}.

Relations:Let two sets A ={1,3,5) and B =(2,6,10), then here the relation R is defined from A to B is R ={(x,y): y =2x,where x ∈A and y∈ B}.The first element of the ordered pair is the domain of the relation R, the whole set of B is codomain and range is the set of second elements in the ordered pair of R.

NCERT Solutions of Science and Maths for Class 9,10,11 and 12

NCERT Solutions for class 9 maths

Chapter 1- Number System	Chapter 9-Areas of parallelogram and triangles
Chapter 2-Polynomial	Chapter 10-Circles
Chapter 3- Coordinate Geometry	Chapter 11-Construction
Chapter 4- Linear equations in two variables	Chapter 12-Heron’s Formula
Chapter 5- Introduction to Euclid’s Geometry	Chapter 13-Surface Areas and Volumes
Chapter 6-Lines and Angles	Chapter 14-Statistics
Chapter 7-Triangles	Chapter 15-Probability
Chapter 8- Quadrilateral

NCERT Solutions for class 9 science

Chapter 1-Matter in our surroundings	Chapter 9- Force and laws of motion
Chapter 2-Is matter around us pure?	Chapter 10- Gravitation
Chapter3- Atoms and Molecules	Chapter 11- Work and Energy
Chapter 4-Structure of the Atom	Chapter 12- Sound
Chapter 5-Fundamental unit of life	Chapter 13-Why do we fall ill ?
Chapter 6- Tissues	Chapter 14- Natural Resources
Chapter 7- Diversity in living organism	Chapter 15-Improvement in food resources
Chapter 8- Motion	Last years question papers & sample papers

NCERT Solutions for class 10 maths

Chapter 1-Real number	Chapter 9-Some application of Trigonometry
Chapter 2-Polynomial	Chapter 10-Circles
Chapter 3-Linear equations	Chapter 11- Construction
Chapter 4- Quadratic equations	Chapter 12-Area related to circle
Chapter 5-Arithmetic Progression	Chapter 13-Surface areas and Volume
Chapter 6-Triangle	Chapter 14-Statistics
Chapter 7- Co-ordinate geometry	Chapter 15-Probability
Chapter 8-Trigonometry

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NCERT Solutions for Class 10 Science

Chapter 1- Chemical reactions and equations	Chapter 9- Heredity and Evolution
Chapter 2- Acid, Base and Salt	Chapter 10- Light reflection and refraction
Chapter 3- Metals and Non-Metals	Chapter 11- Human eye and colorful world
Chapter 4- Carbon and its Compounds	Chapter 12- Electricity
Chapter 5-Periodic classification of elements	Chapter 13-Magnetic effect of electric current
Chapter 6- Life Process	Chapter 14-Sources of Energy
Chapter 7-Control and Coordination	Chapter 15-Environment
Chapter 8- How do organisms reproduce?	Chapter 16-Management of Natural Resources

NCERT Solutions for class 11 maths

Chapter 1-Sets	Chapter 9-Sequences and Series
Chapter 2- Relations and functions	Chapter 10- Straight Lines
Chapter 3- Trigonometry	Chapter 11-Conic Sections
Chapter 4-Principle of mathematical induction	Chapter 12-Introduction to three Dimensional Geometry
Chapter 5-Complex numbers	Chapter 13- Limits and Derivatives
Chapter 6- Linear Inequalities	Chapter 14-Mathematical Reasoning
Chapter 7- Permutations and Combinations	Chapter 15- Statistics
Chapter 8- Binomial Theorem	Chapter 16- Probability

CBSE Class 11-Question paper of maths 2015

CBSE Class 11 – Second unit test of maths 2021 with solutions

NCERT solutions for class 12 maths

Chapter 1-Relations and Functions	Chapter 9-Differential Equations
Chapter 2-Inverse Trigonometric Functions	Chapter 10-Vector Algebra
Chapter 3-Matrices	Chapter 11 – Three Dimensional Geometry
Chapter 4-Determinants	Chapter 12-Linear Programming
Chapter 5- Continuity and Differentiability	Chapter 13-Probability
Chapter 6- Application of Derivation	CBSE Class 12- Question paper of maths 2021 with solutions
Chapter 7- Integrals
Chapter 8-Application of Integrals