**Sub-topics covered under NCERT Solutions for Class 12 Maths Chapter 1**

- 1.1 Introduction
- 1.2 Types of Relations
- 1.3 Types of Functions
- 1.4 Composition of Functions and Invertible Function
- 1.5 Binary Operations

**NCERT Solutions for Class 12 Maths Chapter 1**

NCERT solutions for class 12 Maths for Chapter 1 Relations and Functions elaborate each concept sequentially from relations to functions. Various theorems and operations based on them will give the real concept and inner view of the topic. This chapter explains all major aspects with the help of real-life examples and Vein diagram. Plotting of the functions will give the exact presentations about functions between the independent variable and dependent variable.

Let us discuss the sub-topics in detail.

**1.1 Introduction**

In the previous class, students have got concepts about the notion of relations and functions, domain, co-domain, and range along with different types of specific real-valued functions and their graphs. Now here they will get expanded concepts in that continuation. Also, types of relations and functions will give much strong base.

**1.2 Types of Relations**

This topic gives definitions and examples of various relations such as Empty relation, Universal relation, reflexive relation, symmetric relation, and transitive relations. Also, Equivalence relation will give a combined view of reflexive, symmetric and transitive relations. Different notations will elaborate on the questions and their solutions. The student will learn to prove various theorems and lemmas.

**1.3 Types of Functions**

In class XI students has studied the notion of a function and some functions like identity function, constant function, polynomial function, rational function, modulus function, etc. They also have studied their graphs.

Now, this chapter will discuss Addition, subtraction, multiplication, and division of two functions. Also, it explores different types of functions like into, onto, one-one, etc. The student will prove the injectivity and surjectivity also.

**1.4 Composition of Functions and Invertible Function**

In this section, the student will study the composition of functions and the inverse of a bijective function.Â This topic will explain another concept i.e. invertible function.

**1.5 Binary Operations**

It explains the binary operations which need two operands for it. Any binary operationÂ âˆ—Â on a set A is given as a functionÂ âˆ—Â : A Ã— A â†’ A. Here we denoteÂ âˆ—Â (a, b)Â by aÂ âˆ—Â b.Â A binary operation follows the rule of symmetry, commutativity, and associativity for addition and multiplication both.

**Solved Questions for You**

**Question 1: Show that the relativeÂ RÂ in the setÂ {1,2,3}Â given byÂ R={(1,2),(2,1)}Â is symmetric but neither reflexive nor transitive.**

**Answer:Â **LetÂ *A*={1,2,3}

A relationÂ *R*Â onÂ *A*Â is defined asÂ *R*={(1,2),(2,1)}.

It is seen thatÂ (1,1),(2,2),(3,3)âˆˆ/â€‹*R*.

âˆ´*R*Â is not reflexive.

Now, asÂ (1,2)âˆˆ*R*Â andÂ (2,1)âˆˆ*R*, thenÂ *R*Â is symmetric.

Now,Â (1,2)Â andÂ (2,1)âˆˆ*R*

However,

(1,1)âˆˆ/â€‹*R*

âˆ´*R*Â is not transitive.

Hence,Â *R*Â is symmetric but neither reflexive nor transitive.

**Question 2: Given an example of a relation. Which isÂ Reflexive and symmetric but not transitive.**

**Answer:Â **LetÂ *A*={4,6,8}

Define a relationÂ *R*Â onÂ *A*Â as:

*A*={(4,4),(6,6),(8,8),(4,6),(6,4),(6,8),(8,6)}

RelationÂ *R*Â is reflexive since for everyÂ {*a*âˆˆ *A*,(*a*,*a*)âˆˆ*R i*.*e*.,(4,4),(6,6),(8,8)}âˆˆ*R*

RelationÂ *R*Â is symmetric sinceÂ (*a*,*b*)âˆˆ*R*â‡’(*b*,*a*)âˆˆ*R* for allÂ *a*,*b*âˆˆ*R*.

RelationÂ *R* is not transitive sinceÂ (4,6),(6,8)âˆˆÂ *R*, butÂ (4,8)âˆˆ/â€‹*R*.

Hence, relationÂ *R*Â is reflexive and symmetric but not transitive.

**Question 3: RelationÂ RÂ in the setÂ AÂ of human beings in a town at aÂ particular time given byÂ R={(x,y):x is father of y}**

**Â 1-reflexive and transitive but not symmetric**

**Â Â Â Â Â 2-reflexive only**

**Â Â Â Â Â 3-Transitive only**

**Â Â Â Â Â 4-Equivalence**

**Â Â Â Â Â 5-Neither reflexive, nor symmetric, nor transitive**

**Answer:Â ***R*={(*x*,*y*):xÂ isÂ theÂ fatherÂ ofÂ y}

(*x*,*x*)âˆˆ/â€‹*R*

AsÂ *x*Â cannot be the father of himself.

âˆ´*R*Â is not reflexive.

Now, letÂ (*x*,*y*)âˆˆ*R*

â‡’*x*Â is the father ofÂ *y*.

â‡’*y*Â cannot be the father ofÂ *y*.

Indeed,Â *y*Â is the son or the daughter ofÂ *y*.

âˆ´(*y*,*x*)âˆˆ/â€‹*R*

âˆ´*R*Â is not symmetric.

Now, letÂ (*x*,*y*)âˆˆ*R*Â andÂ (*y*,*z*)âˆˆ*R*.

â‡’*x*Â is the father ofÂ *y*Â andÂ *y*Â is the father ofÂ *z*.

â‡’*x*Â is not the father ofÂ *z*.

Indeed,Â *x*Â is the grandfather ofÂ *z*.

âˆ´(*x*,*z*)âˆˆ/â€‹*R*

âˆ´*R*Â is not transitive.

Hence,Â *R*Â is neither reflexive, nor symmetric, nor transitive.

**Question 4: LetÂ A={1,2,3},Â B={4,5,6,7} and letÂ f=(1,4),(2,5),(3,6)Â be a function fromÂ AÂ toÂ B. Show thatÂ fÂ is one-one.**

**Answer**: It is given thatÂ *A*={1,2,3},Â *B*={4,5,6,7}.

*f*:*A*â†’*B*Â is defined asÂ *f*=(1,4),(2,5),(3,6).

âˆ´*f*(1)=4,*f*(2)=5,*f*(3)=6

It is seen that the images of distinct elements ofÂ *A*

underÂ *f*Â are distinct

Hence, functionÂ *f*Â is one-one.