Relations And Functions Chapter 1 Class 12 Maths NCERT Textbook PDF

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NCERT Class 12 Maths Textbook Chapter 1 With Answer PDF Free Download

Relations and Functions class 12th

Chapter 1: Relations and Functions

1.1 Introduction

Recall that the notion of relations and functions, domain, co-domain, and range have been introduced in Class XI along with different types of specific real-valued functions and their graphs.

The concept of the term ‘relation’ in mathematics has been drawn from the meaning of relationships in the English language, according to which two objects or quantities are related if there is a recognizable connection or link between the two objects or quantities.

1.2 Types of Relations

In this section, we would like to study different types of relations. We know that a relation in a set A is a subset of A × A. Thus, the empty set φ and A × A are two extreme relations.

For illustration, consider a relation R in the set A = {1, 2, 3, 4} given by R = {(a, b): a – b = 10}. This is the empty set, as no pair (a, b) satisfies the condition a – b = 10.

Similarly, R′ = {(a, b) : | a – b | ≥ 0} is the whole set A × A, as all pairs (a, b) in A × A satisfy | a – b | ≥ 0. These two extreme examples lead us to the following definitions.

1.3 Types of Functions

The notion of a function along with some special functions like identity function, constant function, polynomial function, rational function, modulus function, signum function, etc. along with their graphs has been given in Class XI.

Addition, subtraction, multiplication, and division of two functions have also been studied.

As the concept of function is of paramount importance in mathematics and among other disciplines as well, we would like to extend our study about function from where we finished earlier.

In this section, we would like to study different types of functions.

AuthorNCERT
Language English
No. of Pages32
PDF Size663 KB
CategoryMathematics
Source/Creditsncert.nic.in

NCERT Solutions Class 12 Maths Chapter 1 Relations and Functions

1. Determine whether each of the following relations is reflexive, symmetric, and transitive.

  1. Relation R in the set A = {1, 2, 3…13, 14} defined as R = {(x, y): 3x – y = 0}

Ans: The given relation is: R = {(1, 3), (2, 6), (3, 9), (4, 12)}

Since (1, 1), (2, 2) … and (14, 14)∉R.

We conclude that R is not reflexive.

Since (1, 3)∈R, but (3, 1)∉R. (since 3(3)-1≠0)

We conclude that R is not symmetric.

Since (1, 3) and (3, 9)∈R, but(1, 9)∉R. [3(1)-9≠0].

We conclude that R is not transitive.

Therefore, the relation R is not reflexive, symmetric, or transitive.

  1. Relation R in the set N of natural numbers defined as R = { (x, y): y = x + 5 and x4 } 

Ans: The given relation is: R = {(1, 6), (2, 7), (3, 8)}.

Since (1, 1)∉R.

We conclude that R is not reflexive.

Since (1, 6)∈R but (6, 1)∉R.

We conclude that R is not symmetric.

In the given relation R there is not any ordered pair such that (x, y) and (y, z) are both ∈R, therefore we can say that (x, z) cannot belong to R.

Therefore R is not transitive.

Hence, the given relation R is not reflexive, symmetric, or transitive.

  1. Relation Rin the set A = { 1, 2, 3, 4, 5, 6 }  as R= { (x, y): yis divisible by x } 

Ans: The given relation is R = { (x, y) : y is divisible by x } 

As we know that any number except 0 is divisible by itself, therefore (x, x)∈R.

We conclude that R is reflexive.

Since (2, 4)∈R (because 4 is divisible by 2), but (4, 2)∉R (since 2 is not divisible by 4).

We conclude that R is not symmetric.

Assuming that (x, y) and (y, z)∈R, y is divisible by x and z is divisible by y. Hence z is divisible by x⇒(x, z)∈R.

We conclude that R is transitive.

Hence, the given relation R is reflexive and transitive but it is not symmetric.

  1. Relation R in the set Z of all integers defined as R = { (x, y): x – y }  is as integer

Ans: The given relation is R = { (x, y): x – y is an integer } 

If x∈Z, (x, x)∈R because x-x = 0 is an integer.

Hence, we conclude that R is reflexive.

For x, y∈Z, if (x, y)∈R, then x – y is an integer, and therefore (y-x) is also an integer.

Therefore, we conclude that (y, x)∈Rand hence R is symmetric.

Assuming that (x, y) and (y, z)∈R, where x, y, z∈Z.

We can say that (x-y) and (y-z) are integers.

so, (x, z)∈R 

Hence, we conclude that R is transitive.

Therefore the given relation R is reflexive, symmetric, and transitive.

  1. Relation R in the set A of human beings in a town at a particular time given by
  2. The relation is: R = { (x, y): x and y work at the same place}

Ans: The given relation is: R = { (x, y): x and y work at the same place } 

This implies that (x, x)∈R.

Hence, we conclude that R is reflexive.

Now, (x, y)∈R, then x and y work at the same place, which means y and x also work at the same place. Therefore, (y, x)∈R.

Hence, we conclude that R is symmetric.

Let us assume that (x, y), (y, z)∈R. 

Then, we can say that x and y work at the same place and y and z work at the same place. This means that x and z also work at the same place.

Therefore, (x, z)∈R.

Hence, we conclude that R is transitive.

Therefore, the given relation R is reflexive, symmetric, and transitive.

NCERT Class 12 Maths Textbook Chapter 1 With Answer Book PDF Free Download

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