**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**

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**NCERT solutions of 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 |

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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 =2^{n}= 2^{6}= 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.

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