NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.2 are part of NCERT Solutions for Class 11 Maths. Here we have given NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.2.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 11 |

Subject |
Maths |

Chapter |
Chapter 2 |

Chapter Name |
Relations and Functions |

Exercise |
Ex 2.2 |

Number of Questions Solved |
9 |

Category |
NCERT Solutions |

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

**Question 1.**

Let 4 = {1,2,3, ,14}. Define a relation R from A to A by R = {(x,y): 3x – y = 0, where x, y ∈ 4}.

Write down its domain, codomain and range.

**Solution.**

We have A = (1, 2, 3,……..,14)

Given relation R = {(x, y) : 3x – y = 0, where x, y ∈ A}

= {(x, y): y = 3x, where x, y ∈ A)

= {(x, 3x), where x, 3x ∈ A}

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

[∵ 1 ≤ 3x ≤ 14, ∴ ⇒ x = 1, 2, 3, 4 ]

Domain of R = {1, 2, 3, 4}

Codomain of R = {1, 2,……, 14}

Range of R = {3, 6, 9, 12}.

**Question 2.**

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.

**Solution.**

Given relation R = {(x, y): y = x + 5, x < 4 and x, y ∈ N)

= {(x, y): y = x + 5, x ∈ (1, 2, 3) & y ∈ N}

= {(x, x + 5): x = 1, 2, 3}

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

Domain of R = {1, 2, 3}, Range of R = {6, 7, 8}.

**Question 3.**

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 ∈ 4, y ∈B}. Write R in roster form.

**Solution.**

We have, A = {1, 2, 3, 5} and B = {4, 6, 9} R = {(x, y): difference between x and y is odd;

x ∈ A, y ∈B}

= {(x, y): y – x = odd; x ∈ A, y ∈ B}

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

**Question 4.**

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

(i) in set-builder form

(ii) roster form.

What is its domain and range?

**Solution.**

**(i)** Its set builder form is

R = {(x, y): x – y = 2; x ∈ P, y ∈ Q}

i.e., R = {(x, y): y = x – 2 for x = 5, 6, 7)}

**(ii)** Roster form is R = {(5, 3), (6, 4), (7, 5)}

Domain of R = {5, 6, 7} = P,

Range of R = {3, 4, 5} = Q.

**Question 5.**

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?.

**Solution.**

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

Given relation is R = {(a, b):a,b ∈ A, b is exactly divisible by a}

**(i)** Roster form of 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)** Domain of R = {1, 2, 3, 4, 6} = A.

**(iii)** Range of R = {1, 2, 3, 4, 6} = A.

**Question 6.**

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

**Solution.**

Given relation is R = {(x, x + 5): x ∈ {0, 1, 2, 3, 4, 5)}

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

∴Domain of R = {0,1, 2, 3, 4, 5} and

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

**Question 7.**

Write the relation R = {(x, x^{3}): x is a prime number less than 10} in roster form.

**Solution.**

Given relation is R = {(x, x^{3}): x is a prime number less than 10)

= {(x, x^{3}): x ∈ {2, 3, 5, 7}}

= {(2, 23), (3, 33), (5, 53), (7, 73)}

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

**Question 8.**

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

**Solution.**

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

∴ n(A) = 3 & n(B) = 2

Since n(A x B) = n(A) x n(B)

∴ n(A x B) = 3 x 2 = 6

Number of relations from A to B is equal to the number of subsets of A x B.

Since A x B contains 6 elements.

⇒ Number of subsets of A x B = 2^{6} = 64

So, there are 64 relations from A to B.

**Question 9.**

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.

**Solution.**

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

If a, b ∈ Z, then a- b ∈ Z ⇒ Every ordered pair of integers is contained in R.

R = {(a, b) :a,b ∈ Z}

So, Range of R = Domain of R = Z.

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