Relations and Functions Exercise 2.2 NCERT Maths class 11

NCERT Mathematics class 11 chapter 2 Relations and Functions Exercise 2.2 solutions are given.

You should study the textbook lesson Relations and Functions very well.

You should also practice all example problems and solutions are given in the textbook.

Observe the solutions and try then in your own method.

NCERT Maths solutions for class 11 for some chapters

NCERT Solutions for class 11 maths exercise 2.2

Class 11 maths NCERT solutions

Relations and Functions

Exercise 2.1

Exercise 2.2

Exercise 2.3

Class 11 NCERT Maths chapter 2 Relations and Functions Exercise 2.2

Class 11 maths NCERT solutions

Chapter 2

Relations and Functions

Exercise 2.2

CBSE class 11 maths solutions exercise 2.2 problem 1

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

CBSE class 11 maths solutions exercise 2.2 problem 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 belongs to N. defict this relationship using roster form. Write down the domain and range.

CBSE class 11 maths solutions exercise 2.2 problem 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 belongs to A, y belongs to B}. Write R in roster form.

11th maths NCERT solutions exercise 2.2 problem 4

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

i. in set builder form

ii in roster form

What are their domains and range?

11th maths NCERT solutions exercise 2.2 problem 5

Let A = {1, 2, 3, 4, 6}, let R be the relation on A defined by { (a, b), a. b belongs to A, b is exactly divisible by a }

i. Write R in roster form

ii Find domain of R

iii. Find range of R.

Relations and Functions Exercise 2.2 chapter 2 NCERT Maths Class 11 Solutions

11th maths NCERT solutions exercise 2.2 problem 6
Determin the domain and range of the relation R defined by R = { (x, x + 5) : x belongs to 1, 2, 3, 4, 5}