NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions

 

NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions: The performance of students in the 12th board exams is very important to determine their future so that they can take admission into a good college. NCERT solutions for class 12 maths chapter 1 Relations and Functions will help you to understand the concepts and score well in CBSE 12th board exam. This chapter is not only important for mathematics it is also important in real life. We have relations like father, mother, brother, sister, husband, wife. Relation becomes function when there is only one output for every input. In CBSE class 11 maths you have already learnt in brief about relations and functions, range, domain and co-domain with different types of specific real-valued functions and their graphs. In solutions of NCERT for class 12 maths chapter 1 Relations and Functions, you are going to learn about different types of relations and functions, invertible functions, the composition of functions, and binary operations. Concepts of this chapter Relations and Functions are very useful in various other topics of calculus and are also very important from the exam point of view. Unit "Relation and Function" of NCERT class 12th includes two chapters i.e. relation and function, and inverse trigonometry which together has 10 % weightage in the CBSE class 12th final examination. This is the reason you should study this chapter carefully, and solve every problem on your own including solved examples. In this article, you will find CBSE NCERT solutions for class 12 maths chapter 1 Relations and Functions including miscellaneous exercise which will help you to score more marks in the exam. Here you will find all NCERT solutions at a single place which will be helpful when you are not able to solve the NCERT questions.

In this chapter, there are 4 exercises with 55 questions and one miscellaneous exercise with 19 questions. In this article, you will find the detailed solutions of NCERT for class maths 12 chapter 1 Relations and functions. These CBSE NCERT solutions for class 12 maths chapter 1 Relations and Functions are explained in a step-by-step method, so it will be very easy to understand the concepts. Still if you are in a doubt anywhere, you can contact our subject matter experts who are available to help you out and make learning easier for you.

What is the Relation?

The meaning of the term ‘relation’ in mathematics is the same as the meaning of 'relation' in the English language. Relation means two quantities or objects are related if there is a link between them. In other words, we can say that it is a connection between or among things.

 

Let's understand with an example - let A is the set of students of class XII of a school and B is the set of students of class XI of the same school. Then some of the examples of relations from A to B are-

(i) {(a, b) ∈A × B: a is a brother of b},

(ii) {(a, b) ∈A × B: a is a sister of b},

(iii) {(a, b) ∈A × B: age of a is less than the age of b}.

If (a, b) ∈ R, we can say that ‘a’ is related to ‘b’ under the relation ‘R’ and we write as  ‘a R b’. To understand the topic in-depth, after every concept, some topic wise questions are given in the textbook of CBSE class 12. In this article, you will find solutions of NCERT for class 12 maths chapter 1 Relations and Functions for such type of questions also.

Topics of NCERT Grade 12 Maths Chapter-1  Relations and Functions

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

CBSE NCERT solutions for class 12 maths chapter 1 Relations and Functions: Exercise 1.1

Question1(i). Determine whether each of the following relations are reflexive, symmetric and transitive:

(i) Relation R in the set A = \{1,2,3 ...,13 ,14\} defined asR = \{(x,y): 3x - y = 0\}

Answer:

A = \{1,2,3 ...,13 ,14\}

R = \{(x,y): 3x - y = 0\} = \left \{ \left ( 1,3 \right ),\left ( 2,6 \right ),\left ( 3,9 \right ),\left ( 4,12 \right ) \right \}

Since,  \left ( 1,1 \right ),\left ( 2,2 \right ),\left ( 3,3 \right ),\left ( 4,4 \right ),\left ( 5,5 \right )\cdot \cdot \cdot \cdot \cdot \cdot \left ( 14,14 \right ) \notin R so R is not reflexive.

Since, \left ( 1,3 \right ) \in R but  \left ( 3,1 \right ) \notin R so R is not symmetric.

Since, \left ( 1,3 \right ),\left ( 3,9 \right ) \in R but \left ( 1,9 \right ) \notin R so R is not transitive.

Hence, R is neither reflexive nor symmetric and nor transitive.

Question 1(ii). Determine whether each of the following relations are reflexive, symmetric and
transitive:

(ii) Relation R in the set N of natural numbers defined as
R = \{(x,y): y = x + 5 \;\textup{and}\;x<4\}

Answer:

R = \{(x,y): y = x + 5 \;\textup{and}\;x<4\} = \left \{ \left ( 1,6 \right ),\left ( 2,7 \right ),\left ( 3,8 \right ) \right \}

Since, \left ( 1,1 \right ) \notin R

 so R is not  reflexive.

Since, \left ( 1,6 \right )\in R but \left ( 6,1 \right )\notin R

 so R is not symmetric.

Since there is no pair in  R such that \left ( x,y \right ),\left ( y,x \right )\in R so this is not transitive.

Hence, R is neither reflexive nor symmetric and
nor transitive.

Question1(iii) Determine whether each of the following relations are reflexive, symmetric and
transitive:

(iii) Relation R in the set A = \{1,2,3,4,5,6\} as R = \{(x,y) : y \; \textup{is} \; divisible \; by\; x\}

Answer:

A = \{1,2,3,4,5,6\}

R = \left \{ \left ( 2,4 \right ),\left ( 3,6 \right ),\left ( 2,6 \right ),\left ( 1,1 \right ),\left ( 2,2 \right ),\left ( 3,3 \right ),\left ( 4,4 \right ),\left ( 5,5 \right ),\left ( 6,6 \right )\right \}

Any number is divisible by itself  and  \left ( x,x \right ) \in R.So it is  reflexive.

\left ( 2,4 \right ) \in R but \left ( 4,2 \right ) \notin R .Hence,it is not symmetric.

\left ( 2,4 \right ),\left ( 4,4 \right ) \in R and  4 is divisible by 2 and 4 is divisible by 4.

Hence, it is transitive.

Hence, it is reflexive and transitive but not symmetric.

Question.1(iv) Determine whether each of the following relations are reflexive, symmetric and
transitive:

(iv). Relation R in the set Z of all integers defined as R = \{(x,y): x - y \;is\;an\;integer\}

Answer:

R = \{(x,y): x - y \;is\;an\;integer\} 

For x \in Z\left ( x,x \right ) \in R as x-x = 0 which is an integer.

So,it is reflexive.

 For x,y \in Z ,\left ( x,y \right ) \in R and \left ( y,x \right ) \in R because x-y \, \, and \, \, y-x  are both integers.

So, it is symmetric.

 For x,y,z \in Z ,\left ( x,y \right ),\left ( y,z \right ) \in R as x-y \, \, and \, \, y-z are both integers.

Now, x-z = \left ( x-y \right )+\left ( y-z \right ) is also an integer.

So,\left ( x,z \right ) \in R and hence it is transitive.

Hence, it is reflexive, symmetric and transitive.

Question:1(v) Determine whether each of the following relations are reflexive, symmetric and
transitive:

(v) Relation R in the set A of human beings in a town at a particular time given by

(a) R = \{(x,y) : x \;and\; y\;work\;at\;the\;same\;place\}

Answer:

R = \{(x,y) : x \;and\; y\;work\;at\;the\;same\;place\}

\left ( x,x \right )\in R,so it is reflexive

\left ( x,y \right )\in R means x \;and\; y\;work\;at\;the\;same\;place .

y \;and\; x\;work\;at\;the\;same\;place i.e. \left ( y,x \right )\in R so it is symmetric.

\left ( x,y \right ),\left ( y,z \right )\in R means x \;and\; y\;work\;at\;the\;same\;place also y \;and\; z\;work\;at\;the\;same\;place.It states that x \;and\; z\;work\;at\;the\;same\;place i.e. \left ( x,z \right )\in R.So, it is transitive.

Hence, it is reflexive, symmetric and transitive.

Question:1(v) Determine whether each of the following relations are reflexive, symmetric and
transitive:

(v) Relation R in the set A of human beings in a town at a particular time given by

(b) R = \{(x,y): x\;and\;y\;live\;in\;the\;same\;locality\}

Answer:

R = \{(x,y): x\;and\;y\;live\;in\;the\;same\;locality\}

\left ( x,x \right )\in R as x and x is same human being.So, it is reflexive.

\left ( x,y \right )\in R means  x\;and\;y\;live\;in\;the\;same\;locality.

It is same as y\;and\;x\;live\;in\;the\;same\;locality  i.e. \left ( y,x \right )\in R.

So,it is symmetric.

\left ( x,y \right ),\left ( y,z \right )\in R means  x\;and\;y\;live\;in\;the\;same\;locality and y\;and\;z\;live\;in\;the\;same\;locality.

It implies that x\;and\;z\;live\;in\;the\;same\;locality i.e. \left ( x,z \right )\in R.

Hence, it is reflexive, symmetric and
transitive.

Question:1(v) Determine whether each of the following relations are reflexive, symmetric and
transitive:

(v) Relation R in the set A of human beings in a town at a particular time given by 

(c) R = \{(x, y) : x\;is\;exactly\;7\;cm\;taller\;than\;y\}

Answer:

R = \{(x, y) : x\;is\;exactly\;7\;cm\;taller\;than\;y\}

\left ( x,y\right )\in R means x\;is\;exactly\;7\;cm\;taller\;than\;y but x\;is\;not\;\;taller\;than\;x i.e. \left ( x,x \right )\notin R.So, it is not reflexive.

\left ( x,y\right )\in R means x\;is\;exactly\;7\;cm\;taller\;than\;y  but y\;is\;not\;\;taller\;than\;x i.e  \left ( y,x \right )\notin R.So, it is not symmetric.

\left ( x,y\right ),\left ( y,z \right )\in R means x\;is\;exactly\;7\;cm\;taller\;than\;y and y\;is\;exactly\;7\;cm\;taller\;than\;z.

x\;is\;exactly\;14\;cm\;taller\;than\;z  i.e.  \left ( x,z \right )\notin R.

Hence, it is not reflexive,not symmetric and
not transitive.

Question:1(v) Determine whether each of the following relations are reflexive, symmetric and
transitive:

(v). Relation R in the set A of human beings in a town at a particular time given by

(d) R = \{(x, y) : x\;is\;wife\;of\;y\}

Answer:

R = \{(x, y) : x\;is\;wife\;of\;y\}

\left ( x,y \right ) \in R means x\;is\;wife\;of\;y but  x\;is\;not\, wife\;of\;x i.e.\left ( x,x \right ) \notin R.

So, it is not reflexive.

\left ( x,y \right ) \in R means x\;is\;wife\;of\;y but  y\;is\;not\, wife\;of\;x i.e.\left ( y,x \right ) \notin R.

So, it is not symmetric.

Let, \left ( x,y \right ),\left ( y,z \right ) \in R means x\;is\;wife\;of\;y and y\;is\;wife\;of\;z.

This case is not possible so it is not transitive.

Hence, it is not  reflexive, symmetric and
transitive.

Question:1(v) Determine whether each of the following relations are reflexive, symmetric and
transitive:

(v) Relation R in the set A of human beings in a town at a particular time given by

(e) R = \{(x, y) : x \;is \;father \;of \;y \}

Answer:

R = \{(x, y) : x \;is \;father \;of \;y \}

(x, y) \in R means x \;is \;father \;of \;y than x \;cannot \, be \;father \;of \;x  i.e. (x, x) \notin R.So, it is not reflexive..

(x, y) \in R means x \;is \;father \;of \;y than  y \;cannot \, be \;father \;of \;x i.e. (y, x) \notin R.So, it is not symmetric.

Let, (x, y),\left ( y,z \right )\in R means x \;is \;father \;of \;y and y \;is \;father \;of \;z than x \;cannot \, be \;father \;of \;z i.e. (x, z) \notin R.

So, it is not transitive.

Hence, it is neither reflexive nor symmetric and nor transitive.

Question:2 Show that the relation R in the set R of real numbers defined as
R = \{(a, b) : a \leq b^2 \} is neither reflexive nor symmetric nor transitive.

Answer:

R = \{(a, b) : a \leq b^2 \}

Taking  

\left ( \frac{1}{2},\frac{1}{2} \right )\notin R

and 

\left ( \frac{1}{2} \right )> \left ( \frac{1}{2} \right )^{2}

So, R is not reflexive.

Now,

\left ( 1,2 \right )\in R because    1< 4.

But, 4\nless 1  i.e. 4 is not less than 1 

So,\left ( 2,1 \right )\notin R

Hence, it is not symmetric.

\left ( 3,2 \right )\in R\, \, and \, \, \left ( 2,1.5 \right )\in R   as 3< 4\, \, and \, \, 2< 2.25

 Since \left ( 3,1.5 \right )\notin R  because 3\nless 2.25

Hence, it is not transitive.

Thus, we can conclude that it is neither reflexive, nor symmetric, nor transitive. 

Question:3 Check whether the relation R defined in the set \{1, 2, 3, 4, 5, 6\} as
R = \{(a, b) : b = a + 1\} is reflexive, symmetric or transitive.

Answer:

R defined in the set \{1, 2, 3, 4, 5, 6\}

R = \{(a, b) : b = a + 1\}

R=\left \{ \left ( 1,2 \right ),\left ( 2,3 \right ),\left ( 3,4 \right ),\left ( 4,5 \right ),\left ( 5,6 \right ) \right \}

Since, \left \{ \left ( 1,1 \right ),\left ( 2,2 \right ),\left ( 3,3 \right ),\left ( 4,4 \right ),\left ( 5,5 \right ),\left ( 6,6 \right ) \right \}\notin R so it is not reflexive.

\left \{ \left ( 1,2 \right ),\left ( 2,3 \right ),\left ( 3,4 \right ),\left ( 4,5 \right ),\left ( 5,6 \right ) \right \}\in R   but \left \{ \left ( 2,1 \right ),\left ( 3,2 \right ),\left ( 4,3 \right ),\left ( 5,4 \right ),\left ( 6,5 \right ) \right \}\notin R

So, it is not symmetric

\left \{ \left ( 1,2 \right ),\left ( 2,3 \right ),\left ( 3,4 \right ),\left ( 4,5 \right ),\left ( 5,6 \right ) \right \}\in R   but \left \{ \left ( 1,3 \right ),\left ( 2,4 \right ),\left ( 3,5 \right ),\left ( 4,6 \right )\right \}\notin R

So, it is not transitive.

Hence, it is neither reflexive, nor symmetric, nor transitive.

Question:4 Show that the relation R in R defined as R = \{(a, b) : a \leq b\}, is reflexive and

transitive but not symmetric.

Answer:

R = \{(a, b) : a \leq b\}

As \left ( a,a \right )\in R so it is reflexive.

Now we take an  example 

                                       \left ( 2,3 \right )\in R    as 2< 3

But \left ( 3,2 \right )\notin R  because 2 \nless 3.

So,it is not symmetric.

Now if we take,\left ( 2,3 \right )\in R\, \, and\, \, \left ( 3,4 \right )\in R

Than, \left ( 2,4 \right )\in R because 2< 4

So, it is transitive.

Hence, we can say that it is reflexive and transitive but not symmetric.

Question:5 Check whether the relation R in R defined by R = \{(a, b) : a \leq b^3 \} is reflexive,
symmetric or transitive.

Answer:

R = \{(a, b) : a \leq b^3 \}

\left ( \frac{1}{2},\frac{1}{2} \right )\notin R     because     \frac{1}{2}\nleqslant (\frac{1}{2}) ^{3}

So, it is not symmetric

Now, \left ( 1,2 \right ) \in R    because 1< 2^{3}

but \left ( 2,1 \right )\notin R  because 2\nleqslant 1^{3}

It is not symmetric

\left ( 3,1.5 \right ) \in R\, \, and \, \, \left ( 1.5,1.2 \right ) \in R   as   3< 1.5^{3} \, \, and \, \, 1.5< 1.2^{3}.

But, \left ( 3,1.2 \right )\notin R   because 3 \nleqslant 1.2^{3}

So it is not transitive

Thus, it is neither reflexive, nor symmetric, nor transitive.

Question:6 Show that the relation R in the set \{1, 2, 3\}given by R = \{(1, 2), (2, 1)\} is
symmetric but neither reflexive nor transitive.

Answer:

Let A= \{1, 2, 3\}

R = \{(1, 2), (2, 1)\}

We can see \left ( 1,1 \right ),\left ( 2,2 \right ),\left ( 3,3 \right )\notin R  so it is not reflexive.

As \left ( 1,2 \right )\in R \, and \, \left ( 2,1 \right )\in R so it is symmetric.

(1, 2) \in R \, and\, (2, 1)\in R

But  (1, 1)\notin R so it is not transitive.

Hence, R is symmetric but neither reflexive nor transitive.

Question:7 Show that the relation R in the set A of all the books in a library of a college,
given by R = \{(x, y) : x \;and\;y\;have\;same\;number\;of\;pages\} is an equivalence
relation.?

Answer:

 A = all the books in a library of a college

R = \{(x, y) : x \;and\;y\;have\;same\;number\;of\;pages\}

(x,x) \in R  because x and x have the same number of pages so it is reflexive.

Let  (x,y) \in R  means x and y have same number of pages.

Since y and x have the same number of pages so  (y,x) \in R  .

Hence, it is symmetric.

Let  (x,y) \in R  means x and y have the same number of pages.

 and  (y,z) \in R  means y and z have the same number of pages.

This states,x and z also have the same number of pages i.e.(x,z) \in R

Hence, it is transitive.

Thus, it is reflexive, symmetric and transitive i.e. it is an equivalence
relation.?

Question:8 Show that the relation R in the set A = \{1, 2, 3, 4, 5\} given by R = \{(a, b) : |a - b| \;is\;even\}, is an equivalence relation. Show that all the elements of \{1, 3, 5\} are related to each other and all the elements of \{2, 4\}are related to each other. But no element of \{1, 3, 5\} is related to any element of \{2, 4\}.

Answer:

A = \{1, 2, 3, 4, 5\}

R = \{(a, b) : |a - b| \;is\;even\}

R=\left \{ \left ( 1,1 \right ),\left ( 2,2 \right ),\left ( 3,3 \right ),\left ( 4,4 \right ),\left ( 5,5 \right ),\left ( 1,3 \right ),\left ( 2,4 \right ),\left ( 3,5 \right ),\left ( 3,1 \right ),\left ( 5,1 \right ),\left ( 4,2 \right ),\left ( 5,3 \right )\right \}

Let there be a\in A then (a,a)\in R as \left | a-a \right |=0  which is even number. Hence, it is reflexive

Let  (a,b)\in R where a,b\in A then (b,a)\in R as \left | a-b \right |=\left | b-a \right |

Hence, it is symmetric

Now, let (a,b)\in R \, and\, (b,c)\in R

\left | a-b \right | \, and \, \left | b-c \right |   are even number i.e. (a-b)\, and\,(b-c) are even

then, (a-c)=(a-b)+(b-c) is even                  (sum of even integer is even)

So, (a,c)\in R. Hence, it is transitive.

Thus, it is reflexive, symmetric and transitive i.e. it is an equivalence relation.

The elements of \{1, 3, 5\} are related to each other because the difference of odd numbers gives even number and in this set all numbers are odd.

The elements of \{2, 4\}are related to each other because the difference of even number is even number and in this set, all numbers are even.

The element of \{1, 3, 5\} is not related to  \{2, 4\} because a difference of odd and even number is not even.

Question:9(i) Show that each of the relation R in the set A = \{x \in Z : 0 \leq x \leq 12\}, given by

(i) R = \{(a, b) : |a - b|\; is\;a\;multiple \;of\; 4\} is an equivalence relation. Find the set of all elements related to 1 in each case.

Answer:

A = \{x \in Z : 0 \leq x \leq 12\}

A=\left \{ 0,1,2,3,4,5,6,7,8,9,10,11,12 \right \}

R = \{(a, b) : |a - b|\; is\;a\;multiple \;of\; 4\}

For a\in A , (a,a)\in R  as \left | a-a \right |=0 which is multiple of 4.

Henec, it is reflexive.

Let, (a,b)\in R i.e. \left | a-b \right | is multiple of 4.

then  \left | b-a \right | is also multiple of 4 because \left | a-b \right |  =  \left | b-a \right |  i.e.(b,a)\in R

Hence, it is symmetric.

Let, (a,b)\in R i.e. \left | a-b \right | is multiple of 4   and    (b,c)\in R  i.e. \left | b-c \right | is multiple of 4 .

 ( a-b ) is multiple of 4  and   (b-c) is multiple of 4

(a-c)=(a-b)+(b-c)  is multiple of 4

\left | a-c \right | is multiple of 4 i.e. (a,c)\in R

Hence, it is transitive.

Thus, it is reflexive, symmetric and transitive i.e. it is an equivalence relation.

The set of all elements related to 1 is  \left \{1,5,9 \right \}

\left | 1-1 \right |=0 is multiple of 4.

\left | 5-1 \right |=4 is multiple of 4.

\left | 9-1 \right |=8 is multiple of 4.

Question:9(ii) Show that each of the relation R in the set A = \{x \in Z : 0 \leq x \leq 12\}, given by

(ii)  R = \{(a, b) : a = b\} is an equivalence relation. Find the set of all elements related to 1 in each case.

Answer:

A = \{x \in Z : 0 \leq x \leq 12\}

A=\left \{ 0,1,2,3,4,5,6,7,8,9,10,11,12 \right \}

R = \{(a, b) : a = b\}

For a\in A , (a,a)\in R  as  a=a

Henec, it is reflexive.

Let, (a,b)\in R i.e. a=b 

 a=b \Rightarrow  b=a  i.e.(b,a)\in R

Hence, it is symmetric.

Let, (a,b)\in R i.e. a=b   and    (b,c)\in R  i.e. b=c

       \therefore  a=b=c

a=c i.e. (a,c)\in R

Hence, it is transitive.

Thus, it is reflexive, symmetric and transitive i.e. it is an equivalence relation.

The set of all elements related to 1 is {1}

Question:10(i) Give an example of a relation.

(i) Which is Symmetric but neither reflexive nor transitive.

Answer:

Let 

A = \left \{ 1,2,3 \right \}

R = \left \{ \left ( 1,2 \right ),\left ( 2,1 \right )\right \}

\left ( 1,1 \right ),\left ( 2,2 \right ),(3,3) \notin R so it is not reflexive.

(1,2)\in R   and  (2,1)\in R  so it is symmetric.

(1,2)\in R \, and\, (2,1)\in R  but  (1,1)\notin R so it is not transitive.

Hence, symmetric but neither reflexive nor transitive. 

Question:10(ii) Give an example of a relation.

(ii) Which is transitive but neither reflexive nor symmetric.

Answer:

Let  

R = \left \{ \left ( x,y \right ): x> y \right \}

Now for x\in R ,(x,x)\notin R so it is not reflexive.

Let (x,y) \in R  i.e. x> y

Then y> x is not possible i.e. (y,x) \notin R . So it is not symmetric.

Let (x,y) \in R  i.e. x> y    and  (y,z) \in R i.e.y> z

we can write this as x> y> z

Hence,x> z  i.e. (x,z)\in R. So it is transitive.

Hence, it is transitive but neither reflexive nor symmetric.

Question:10(iii) Give an example of a relation.

(iii) Which is Reflexive and symmetric but not transitive.

Answer:

Let 

A = \left \{ 1,2,3 \right \}

Define a relation R on A as

R= \left \{ (1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2) \right \}

If   x\in A ,(x,x)\in R i.e.\left \{ (1,1),(2,2),(3,3)\right \} \in R. So it is reflexive.

If  x,y\in A  ,  (x,y)\in R   and  (y,x)\in R i.e.\left \{(1,2),(2,1),(2,3),(3,2) \right \}\in R. So it is symmetric.

(x,y)\in R  and (y,z)\in R  i.e. (1,2)\in R.  and (2,3)\in R 

But (1,3)\notin R So it is not transitive.

Hence, it  is Reflexive and symmetric but not transitive.

Question:10(iv) Give an example of a relation.

(iv) Which is Reflexive and transitive but not symmetric.

Answer:

Let there be a relation R in R

R=\left \{ (a,b):a\leq b \right \}

(a,a)\in R  because a=a

Let (a,b)\in R  i.e.a\leq b

But (b,a)\notin R i.e.b\nleqslant a

So it is not symmetric.

Let (a,b)\in R  i.e.a\leq b   and (b,c)\in R  i.e. b\leq c

This can be written as a\leq b\leq c  i.e. a\leq c  implies (a,c)\in R

Hence, it is transitive.

Thus, it  is Reflexive and transitive but not symmetric.

Question:10(v) Give an example of a relation.

(v) Which is Symmetric and transitive but not reflexive.

Answer:

Let there be a relation A in R

A= \left \{ 1,2 \right \}

R=\left \{ (1,2),(2,1),(2,2)\right \}

(1,1)\notin R  So R is not reflexive.

We can see  (1,2)\in R  and  (2,1)\in R 

So it is symmetric.

Let (1,2)\in R     and  (2,1)\in R  

Also  (2,2)\in R

Hence, it is transitive.

Thus, it  Symmetric and transitive but not reflexive.

Question:11 Show that the relation R in the set A of points in a plane given by
R = \{(P, Q) : \;distance \;of \;the\; point\; P\; from \;the \;origin \;is \;same \;as \;the\; distance \;of \;the \;point \;Q \;from \;the \;origin\}, is an equivalence relation. Further, show that the set of
all points related to a point P \neq (0, 0) is the circle passing through P with origin as
centre.

Answer:

R = \{(P, Q) : \;distance \;of \;the\; point\; P\; from \;the \;origin \;is \;same \;as \;the\; distance \;of \;the \;point \;Q \;from \;the \;origin\}

The distance of point P from the origin is always the same as the distance of same point P from origin i.e.(P,P)\in R

 \therefore R is reflexive.

Let (P,Q)\in R i.e. the distance of the point P from the origin is the same as the distance of the point Q from the origin.

this is the same as distance of the point Q from the origin is the same as the distance of the point P from the origin i.e. (Q,P)\in R

\thereforeR is symmetric.

Let    (P,Q)\in R     and    (Q,S)\in R 

       i.e. the distance of point P from the origin is the same as the distance of point Q from the origin, and also the distance of point Q from the origin is the same as the distance of the point S from the origin.

We can say that the distance of point P, Q, S  from the origin is the same. Means distance of point P from the origin is the same as the distance of point S from origin i.e.  (P,S)\in R

\therefore  R is transitive.

Hence, R  is an equivalence relation.

The set of all points related to a point P \neq (0, 0) are points whose distance from the origin is the same as the distance of point P from the origin.

In other words, we can say there be a point O(0,0) as origin and distance between point O and point P be k=OP then set of all points related to P is at distance k from the origin.

Hence, these sets of points form a circle with the centre as the origin and this circle passes through the point.

Question:12 Show that the relation R defined in the set A of all triangles as R = \{(T_1 , T_2 ) : T_1 \;is\; similar \;to\; T_2 \}, is equivalence relation. Consider three right angle triangles Twith sides 3, 4, 5, T2 with sides 5, 12, 13 and T3 with sides 6, 8, 10. Which triangles among T1 , T2 and T3 are related?

Answer:

R = \{(T_1 , T_2 ) : T_1 \;is\; similar \;to\; T_2 \}

All triangles are similar to itself, so it is reflexive.

Let,

(T_1,T_2) \in R  i.e.T1 is similar to T2

T1 is similar to T2 is the same asT2 is similar to T1 i.e. (T_2,T_1) \in R

Hence, it is symmetric.

Let,

(T_1,T_2) \in R  and  (T_2,T_3) \in R  i.e. T1 is similar to T2  and T2 is similar toT3 .

\RightarrowT1 is similar toT3   i.e. (T_1,T_3) \in R

Hence, it is transitive,

Thus,  R = \{(T_1 , T_2 ) : T_1 \;is\; similar \;to\; T_2 \}, is equivalence relation.

Now, we see the ratio of sides of triangle T1 andT3 are as shown

 \frac{3}{6}=\frac{4}{8}=\frac{5}{10}=\frac{1}{2}

i.e. ratios of sides of T1 and T3 are equal.Hence, T1 and T3 are related.

Question:13 Show that the relation R defined in the set A of all polygons as R = \{(P _1 , P _2 ) : P_1 \;and\; P_2 \;have \;same\; number \;of\; sides\}, is an equivalence relation. What is the set of all elements in A related to the right angle triangle T with sides 3, 4 and 5?

Answer:

R = \{(P _1 , P _2 ) : P_1 \;and\; P_2 \;have \;same\; number \;of\; sides\}

The same polygon has the same number of sides with itself,i.e. (P_1,P_2) \in R, so it is reflexive.

Let,

(P_1,P_2) \in R  i.e.P1 have same number of sides as  P2

P1 have the same number of sides as Pis the same as P2 have same number of sides as P1 i.e. (P_2,P_1) \in R

Hence,it is symmetric.

Let,

(P_1,P_2) \in R  and  (P_2,P_3) \in R  i.e. P1 have the same number of sides as P2  and Phave same number of sides as P3

\Rightarrow Phave same number of sides as P3   i.e. (P_1,P_3) \in R

Hence, it is transitive,

Thus,  R = \{(P _1 , P _2 ) : P_1 \;and\; P_2 \;have \;same\; number \;of\; sides\}, is an equivalence relation.

The elements in A related to the right angle triangle T with sides 3, 4 and 5 are those polygons which have 3 sides.

Hence, the set of all elements in A related to the right angle triangle T is set of all triangles.

Question:14 Let L be the set of all lines in XY plane and R be the relation in L defined as R = \{(L_1 , L_2 ) : L_1\;is\;parallel\;to\;L_2 \}. Show that R is an equivalence relation. Find the set of all lines related to the line y = 2x + 4.

Answer:

R = \{(L_1 , L_2 ) : L_1\;is\;parallel\;to\;L_2 \}

All lines are parallel to itself, so it is reflexive.

Let,

(L_1,L_2) \in R  i.e.L1 is parallel to L2.

L1 is parallel to L2 is same as L2 is parallel to L1 i.e. (L_2,L_1) \in R

Hence, it is symmetric.

Let,

(L_1,L_2) \in R  and  (L_2,L_3) \in R  i.e. L1 is parallel to L2  and L2 is parallel  to L3 .

\RightarrowL1 is parallel to L3   i.e. (L_1,L_3) \in R

Hence, it is transitive,

Thus,  R = \{(L_1 , L_2 ) : L_1\;is\;parallel\;to\;L_2 \} , is equivalence relation.

The set of all lines related to the line y = 2x + 4. are lines parallel to y = 2x + 4.

Here,  Slope = m = 2  and constant = c = 4

 It is known that the slope of parallel lines are equal.

Lines parallel to this ( y = 2x + 4. )  line  are y = 2x + c  , c \in R

Hence, set of all parallel lines to y = 2x + 4. are y = 2x + c.

Question:15 Let R be the relation in the set \{}1, 2, 3, 4\} given by R = \{(1, 2), (2, 2), (1, 1), (4,4), (1, 3), (3, 3), (3, 2)\}. Choose the correct answer.

(A)  R is reflexive and symmetric but not transitive.

(B) R is reflexive and transitive but not symmetric.

(C) R is symmetric and transitive but not reflexive.

(D) R is an equivalence relation.

Answer:

A = \{}1, 2, 3, 4\}

R = \{(1, 2), (2, 2), (1, 1), (4,4), (1, 3), (3, 3), (3, 2)\}

For every  a \in A  there is  (a,a) \in R

\therefore R is reflexive.

Given, (1,2) \in R  but  (2,1) \notin R 

\therefore R is not symmetric.

For  a,b,c \in A there are (a,b) \in R \, and \, (b,c) \in R  \Rightarrow (a,c) \in R

\therefore R is transitive.

Hence, R  is reflexive and transitive but not symmetric.

The correct answer is option B.

Question:16 Let R be the relation in the set N given by R = \{(a, b) : a = b - 2, b > 6\}. Choose the correct answer. 

(A) (2, 4) \in R
(B) (3,8) \in R
(C) (6,8) \in R
(D) (8,7) \in R

Answer:

R = \{(a, b) : a = b - 2, b > 6\}

(A) Since, b< 6 so (2, 4) \notin R

(B) Since, 3\neq 8-2  so  (3,8) \notin R

(C) Since, 8> 6  and  6=8-2  so  (6,8) \in R

(d) Since,8\neq 7-2 so  (8,7) \notin R

The correct answer is option C.

NCERT solutions for class 12 maths chapter 1 Relations and Functions: Exercise 1.2

Question:1 Show that the function f: R_* \longrightarrow R_{*} defined by f(x) = \frac{1}{x} is one-one and onto,where R is the set of all non-zero real numbers. Is the result true, if the domain R is replaced by N with co-domain being same as R?

Answer:

Given, f: R_* \longrightarrow R_{*} is defined by  f(x) = \frac{1}{x}.

One - One :

             f(x)=f(y)

                  \frac{1}{x}=\frac{1}{y}

                  x=y

       \therefore  f is one-one.

Onto:

We have  y \in R_*, then there exists x=\frac{1}{y} \in R_*     ( Here y\neq 0) such that 

f(x)= \frac{1}{(\frac{1}{y})} = y             

\therefore f is \, \, onto.

Hence, the function is one-one and onto.

If the domain R is replaced by N with co-domain being same as R∗   i.e.  g: N \longrightarrow R_{*}  defined by

                                  g(x)=\frac{1}{x}

                                g(x_1)=g(x_2) 

                                      \frac{1}{x_1}=\frac{1}{x_2}

                                       x_1=x_2

                             \therefore  g is one-one.

 For   1.5 \in R_* ,    

   g(x) = \frac{1}{1.5}   but there does not exists any x in N.

Hence, function g is one-one but not onto.

Question:2(i) Check the injectivity and surjectivity of the following functions:

(i) f : N\rightarrow N given by f(x) = x^2

Answer:

f : N\rightarrow N

f(x) = x^2

One- one:

x,y \in N  then f(x)=f(y)

                                  x^{2}=y^{2}

                                   x=y

\therefore  f is one- one i.e. injective.

For 3 \in N there is no x in N such that f(x)=x^{2}=3

\therefore  f is not onto i.e. not surjective.

Hence, f is injective but not surjective.

Question:2(ii) Check the injectivity and surjectivity of the following functions:

(ii) f : Z \rightarrow Z given by f(x) = x^2

Answer:

f : Z \rightarrow Z

f(x) = x^2

One- one:

For   -1,1 \in Z  then f(x) = x^2

                                  f(-1)= (-1)^{2}

                                  f(-1)= 1   but  -1 \neq 1 

\therefore  f is not one- one i.e. not injective.

For  -3 \in Z  there is no x in Z such that f(x)=x^{2}= -3

\therefore  f is not onto i.e. not surjective.

Hence, f is neither injective nor surjective.

Question:2(iii) Check the injectivity and surjectivity of the following functions:

(iii) f: R \rightarrow R given by f(x) = x^2

Answer:

f: R \rightarrow R

f(x) = x^2

One- one:

For   -1,1 \in R  then f(x) = x^2

                                  f(-1)= (-1)^{2}

                                  f(-1)= 1   but  -1 \neq 1 

\therefore  f is not one- one i.e. not injective.

For  -3 \in R  there is no x in R such that f(x)=x^{2}= -3

\therefore  f is not onto i.e. not surjective.

Hence, f is not injective and not surjective.

Question:2(iv) Check the injectivity and surjectivity of the following functions:

(iv) f: N \rightarrow N given by f(x) = x^3

Answer:

f : N\rightarrow N

f(x) = x^3

One- one:

x,y \in N  then f(x)=f(y)

                                  x^{3}=y^{3}

                                   x=y

\therefore  f is one- one i.e. injective.

For 3 \in N there is no x in N such that f(x)=x^{3}=3

\therefore  f is not onto i.e. not surjective.

Hence, f is injective but not surjective.

Question:2(v) Check the injectivity and surjectivity of the following functions:

(v)  f : Z \rightarrow Z given by f(x) = x^3

Answer:

f : Z \rightarrow Z

f(x) = x^3

One- one:

For   (x,y) \in Z  then f(x) = f(y)

                                            x^{3}=y^{3}

                                            x=y 

\therefore  f is one- one i.e. injective.

For  3 \in Z  there is no x in Z such that f(x)=x^{3}= 3

\therefore  f is not onto i.e. not surjective.

Hence, f is injective but not surjective.

Question:3 Prove that the Greatest Integer Function f : R\longrightarrow R, given by f (x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.

Answer:

 f : R\longrightarrow R

 f (x) = [x]

One- one:

For   1.5,1.7 \in R   then f(1.5)=\left [ 1.5 \right ] = 1     and    f(1.7)=\left [ 1.7 \right ] = 1

                                 but   1.5\neq 1.7 

\therefore  f is not one- one i.e. not injective.

For  0.6 \in R  there is no x in R such that f(x)=\left [ 0.6 \right ]

\therefore  f is not onto i.e. not surjective.

Hence, f is not injective but not surjective.

Question:4 Show that the Modulus Function f : R → R, given by f (x) = | x |, is neither one-one nor onto, where | x | is x, if x is positive or 0 and | x |  is - x,  if x is negative.

Answer:

f : R \rightarrow R

f (x) = | x |

f (x) = | x | = x \, if\, x\geq 0 \,\, and \, \, -x\, if\, x< 0

One- one:

For   -1,1 \in R   then f (-1) = | -1 |= 1 

                                         f (1) = | 1 |= 1

                                           -1\neq 1

\therefore  f is not one- one i.e. not injective.

For  -2 \in R,

We know f (x) = | x |   is always positive there is no x in R such that f (x) = | x |=-2

\therefore  f is not onto i.e. not surjective.

Hence,  f (x) = | x |, is neither one-one nor onto.

Question:5 Show that the Signum Function f : R \rightarrow R, given by

                            f (x) = \left\{\begin{matrix} 1 & if\;x>0 \\ 0& if\;x=0 \\ -1& if\;x<0 \end{matrix}\right. is neither one-one nor onto.

Answer:

f : R \rightarrow R  is given by 

f (x) = \left\{\begin{matrix} 1 & if\;x>0 \\ 0& if\;x=0 \\ -1& if\;x<0 \end{matrix}\right.

As we can see    f(1)=f(2)=1   ,  but 1\neq 2

So it is not one-one.

Now, f(x) takes only 3 values (1,0,-1) for the element -3 in codomain R ,there does not exists x in domain R such that f(x)= -3.

So it is not onto.

Hence, signum function is neither one-one nor onto.

Question:6 Let A = \{1, 2, 3\},B = \{4, 5, 6, 7\} and let f = \{(1, 4), (2, 5), (3, 6)\} be a function from A to B. Show that f is one-one.

Answer:

A = \{1, 2, 3\}

B = \{4, 5, 6, 7\}

f = \{(1, 4), (2, 5), (3, 6)\}

f : A \rightarrow B

\therefore  f(1)=4,f(2)=5,f(3)=6

Every element of A has a distant value in f.

Hence, it is one-one.

Question:7(i) In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.

(i) f: R\rightarrow R defined by f(x) = 3 -4x

Answer:

f: R\rightarrow R

f(x) = 3 -4x

Let  there  be  (a,b) \in R  such that f(a)=f(b)

                                                          3-4a = 3 -4b

                                                               -4a = -4b

                                                                    a = b

\therefore f is one-one.

Let there be y \in R,    y = 3 -4x

                                    x = \frac{(3-y)}{4}

                                 f(x) = 3 -4x

Puting value of x,    f(\frac{3-y}{4}) = 3 - 4(\frac{3-y}{4})

                               f(\frac{3-y}{4}) = y

                           \therefore  f is onto.

f is both one-one and onto hence, f is bijective.

Question:7(ii) In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.

(ii) f : R\rightarrow R defined by f(x) = 1 + x^2

Answer:

f : R\rightarrow R

f(x) = 1 + x^2 

Let  there  be  (a,b) \in R  such that f(a)=f(b)

                                                          1+a^{2} = 1 +b^{2}

                                                               a^{2}=b^{2}

                                                                 a = \pm b

For   f(1)=f(-1)=2   and  1\neq -1

\therefore f is not one-one.

Let there be -2 \in R   (-2 in codomain of R)

                                   f(x) = 1 + x^2 = -2 

        There does not exists any x in domain R  such that  f(x) = -2                        

                           \therefore  f is not onto.

Hence, f is neither one-one nor onto.

Question:8 Let A and B be sets. Show that f : A \times B \rightarrow B \times A such that f (a, b) = (b, a) is
bijective function.

Answer:

f : A \times B \rightarrow B \times A

f (a, b) = (b, a)

Let (a_1,b_1),(a_2,b_2) \in A\times B

such that  f (a_1, b_1) = f(a_2, b_2)

                     (b_1,a_1)=(b_2,a_2)

      \Rightarrow     b_1= b_2   and   a_1= a_2

     \Rightarrow          (a_1,b_1) = (a_2,b_2)

\therefore    f is one- one

Let,  (b,a) \in B\times A

then there exists (a,b) \in A\times B  such that  f (a, b) = (b, a)

\therefore f is onto.

Hence, it is bijective.

Question:9 Let f : N \rightarrow N be defined by f(n) = \left\{\begin{matrix} \frac{n+1}{2} & if\;n\;is\;odd \\ \frac{n}{2} & if\;n\;is\;even \end{matrix}\right.   for all n\in N. State whether the function f is bijective. Justify your answer.

Answer:

f : N \rightarrow N   ,  n\in N

    f(n) = \left\{\begin{matrix} \frac{n+1}{2} & if\;n\;is\;odd \\ \frac{n}{2} & if\;n\;is\;evem \end{matrix}\right.

Here we can observe,

 f(2)=\frac{2}{2}=1           and       f(1)=\frac{1+1}{2}=1

As we can see f(1)=f(2)=1  but 1\neq 2

\therefore     f is not one-one.

Let,n\in N    (N=co-domain)

case1   n be even

   For r \in N,      n=2r   

then there is 4r \in N such that f(4r)=\frac{4r}{2}=2r

case2   n be odd

For  r \in N,   n=2r+1

then there is 4r+1 \in N such that f(4r+1)=\frac{4r+1+1}{2}=2r +1

\therefore  f is onto.

f is not one-one but onto

hence, the function f is not bijective.

Question:10 Let A = R - \{3\} and B = R - \{1\}. Consider the functionf : A\rightarrow B defined by f(x) = \left (\frac{x-2}{x-3} \right ) . Is f one-one and onto? Justify your answer.

Answer:

A = R - \{3\}

B = R - \{1\}

f : A\rightarrow B

f(x) = \left (\frac{x-2}{x-3} \right )

Let a,b \in A such that  f(a)=f(b)

                               \left (\frac{a-2}{a-3} \right ) = \left ( \frac{b-2}{b-3} \right )

                       (a-2)(b-3)=(b-2)(a-3)

                ab-3a-2b+6=ab-2a-3b+6

                          -3a-2b=-2a-3b

                              3a+2b= 2a+3b

                             3a-2a= 3b-2b

                                       a=b

                    \therefore  f is one-one.

Let,  b \in B = R - \{1\}      then   b\neq 1

      a \in A such that   f(a)=b

                        \left (\frac{a-2}{a-3} \right ) =b

                        (a-2)=(a-3)b

                          a-2 = ab -3b                              

                          a-ab = 2 -3b

                         a(1-b) = 2 -3b

                                a= \frac{2-3b}{1-b}\, \, \, \, \in A

For any b \in B there exists  a= \frac{2-3b}{1-b}\, \, \, \, \in A   such that 

                        f(\frac{2-3b}{1-b}) = \frac{\frac{2-3b}{1-b}-2}{\frac{2-3b}{1-b}-3}

                                              =\frac{2-3b-2+2b}{2-3b-3+3b}

                                             =\frac{-3b+2b}{2-3}

                                             = b

                              \therefore  f is onto

Hence, the function is one-one and onto. 

Question:11 Let f : R \rightarrow R be defined as f(x) = x^4 . Choose the correct answer.

(A) f is one-one onto

(B) f is many-one onto

(C) f is one-one but not onto

(D) f is neither one-one nor onto.

Answer:

f : R \rightarrow R

f(x) = x^4

One- one:

For   a,b \in R  then f(a) = f(b)

                                  a^{4}=b^{4}

                                  a=\pm b

  \therefore f(a)=f(b) does not imply that a=b

example: and 2\neq -2 

\therefore  f is not  one- one 

For  2\in R  there is no x in R such that f(x)=x^{4}= 2

\therefore  f is not onto.

Hence, f is neither one-one nor onto. 

Option D is correct.

Question:12 Letf : R\rightarrow R be defined asf(x) = 3x. Choose the correct answer.

(A) f is one-one onto

(B) f is many-one onto

(C) f is one-one but not onto

(D) f is neither one-one nor onto.

Answer:

f : R\rightarrow R

f(x) = 3x

One - One :

Let \left ( x,y \right ) \in R

             f(x)=f(y)

                  3x=3y

                  x=y

       \therefore  f is one-one.

Onto:

We have  y \in R, then there exists x=\frac{y}{3} \in R    such that 

f(\frac{y}{3})= 3\times \frac{y}{3} = y             

\therefore f is \, \, onto.

Hence, the function is one-one and onto.

The correct answer is A .

CBSE NCERT solutions for class 12 maths chapter 1 Relations and Functions: Exercise: 1.3

Question:1 Let f : \{1, 3, 4\}\rightarrow \{1, 2, 5\}and g : \{1, 2, 5\} \rightarrow \{1, 3\} be given byf = \{(1, 2), (3, 5), (4, 1)\} and g = \{(1, 3), (2, 3), (5, 1)\}. Write down gof.

Answer:

Given :       f : \{1, 3, 4\}\rightarrow \{1, 2, 5\}             and       g : \{1, 2, 5\} \rightarrow \{1, 3\}

                  f = \{(1, 2), (3, 5), (4, 1)\}         and      g = \{(1, 3), (2, 3), (5, 1)\}

gof(1) = g(f(1))=g(2) = 3                   \left [ f(1)=2 \, and\, g(2)=3 \right ] 

gof(3) = g(f(3))=g(5) = 1                   \left [ f(3)=5 \, and\, g(5)=1 \right ]

gof(4) = g(f(4))=g(1) = 3                 \left [ f(4)=1 \, and\, g(1)=3 \right ]

Hence,   gof = \left \{ (1,3),(3,1),(4,3) \right \}

Question:2 Let f, g and h be functions from R to R. Show that  \\(f + g) o h = foh + goh\\ (f \cdot g) o h = (foh) \cdot (goh)

Answer:

To prove : \\(f + g) o h = foh + goh

                   ((f + g) o h)(x)

                  =(f + g) ( h(x) )

                 =f ( h(x) ) +g(h(x))

                 =(f o h)(x) +(goh)(x) 

               =\left \{ (f o h) +(goh) \right \}(x)                      x\forall R

Hence, \\(f + g) o h = foh + goh

To prove:(f \cdot g) o h = (foh) \cdot (goh)

                         ((f . g) o h)(x)

                  =(f . g) ( h(x) )

                 =f ( h(x) ) . g(h(x))

                 =(f o h)(x) . (goh)(x) 

              =\left \{ (f o h) .(goh) \right \}(x)                      x\forall R

           \therefore   (f \cdot g) o h = (foh) \cdot (goh)

Hence, (f \cdot g) o h = (foh) \cdot (goh)

Question:3(i) Find gof and fog, if 

(i) f (x) = | x |  and  g(x) = \left | 5x-2 \right |

Answer:

f (x) = | x | and g(x) = \left | 5x-2 \right | 

gof = g(f(x))

         = g( | x |)

          = |5 | x |-2|

fog = f(g(x))

         =f( \left | 5x-2 \right |)

          =\left \| 5x-2 \right \|

            =\left | 5x-2 \right |

Question:3(ii) Find gof and fog, if 

(ii) f (x) = 8x^{3}and g(x) = x^{\frac{1}{3}}

Answer:

The solution is as follows

(ii)  f (x) = 8x^{3}and g(x) = x^{\frac{1}{3}}

     gof  = g(f(x))

           = g( 8x^{3})

            = ( 8x^{3})^{\frac{1}{3}}

             =2x

fog = f(g(x))

         =f(x^{\frac{1}{3}} )

          =8((x^{\frac{1}{3}} )^{3})

           =8x

Question:4 If f(x) = \frac{4x + 3}{6x - 4}, x \neq \frac{2}{3} show that fof (x) = x,  for all x \neq\frac{2}{3}. What is the inverse of f?

Answer:

f(x) = \frac{4x + 3}{6x - 4}, x \neq \frac{2}{3}

fof (x) = x

(fof) (x) = f(f(x))

                    =f( \frac{4x + 3}{6x - 4})

                  =\frac{4( \frac{4x + 3}{6x - 4}) +3}{6( \frac{4x + 3}{6x - 4}) -4}

                = \frac{16x+12+18x-12}{24x+1824x+16}

                 = \frac{34x}{34}

                 \therefore fof(x) = x                ,  for all  x \neq \frac{2}{3}

\Rightarrow fof=Ix

Hence,the given function fis invertible and the inverse of f is f itself.

Question:5(i)  State with reason whether following functions have inverse

(i) f : \{1, 2, 3, 4\} \rightarrow\{10\}

with f = \{(1, 10), (2, 10), (3, 10), (4, 10)\}

Answer:

(i) f : \{1, 2, 3, 4\} \rightarrow\{10\} with
   f = \{(1, 10), (2, 10), (3, 10), (4, 10)\}

From the given definition,we have:

f\left ( 1 \right )=f\left ( 2 \right )=f\left ( 3 \right )=f(4)=10

\therefore f is not one-one.

Hence, f do not have an inverse function.

Question:5(ii) State with reason whether following functions have inverse

(ii)  g : \{5, 6, 7, 8\} \rightarrow \{1, 2, 3, 4\} with
     g = \{(5, 4), (6, 3), (7, 4), (8, 2)\}

Answer:

(ii)  g : \{5, 6, 7, 8\} \rightarrow \{1, 2, 3, 4\} with
     g = \{(5, 4), (6, 3), (7, 4), (8, 2)\}

From the definition, we can conclude :

g(5)=g(7)=4

\therefore g is not one-one.

Hence, function g does not have inverse function.

Question:5(iii) State with reason whether following functions have inverse 

(iii)   h : \{2, 3, 4, 5\}\rightarrow \{7, 9, 11, 13\} with
       h = \{(2, 7), (3, 9), (4, 11), (5, 13)\}

Answer:

(iii)   h : \{2, 3, 4, 5\}\rightarrow \{7, 9, 11, 13\} with
       h = \{(2, 7), (3, 9), (4, 11), (5, 13)\}

From the definition, we can see the set \left \{ 2,3,4,5 \right \} have distant values under h.

\therefore h is one-one.

For every element y of set \left \{ 7,9,11,13 \right \},there exists an element x  in \left \{ 2,3,4,5 \right \} such that  h(x)=y

\therefore h is onto

Thus, h is one-one and onto so h has an inverse function.

Question:6  Show that f : [-1, 1] \rightarrow R, given by f(x) = \frac{x}{(x + 2)} is one-one. Find the inverse of the functionf : [-1, 1] \rightarrow Range f

Answer:

f : [-1, 1] \rightarrow R

f(x) = \frac{x}{(x + 2)}

One -one:

            f(x)=f(y)

        \frac{x}{x+2}=\frac{y}{y+2}

     x(y+2)=y(x+2)

   xy+2x=xy+2y 

            2x=2y

             x=y

\therefore   f is one-one.

It is clear that  f : [-1, 1] \rightarrow Range f is onto.

Thus,f is one-one and onto so inverse of f exists.

Let g be inverse function of f in  Range f\rightarrow [-1, 1]

g: Range f\rightarrow [-1, 1]

let y be an arbitrary element of range f

Since, f : [-1, 1] \rightarrow R is onto, so

        y=f(x)  for x \in \left [ -1,1 \right ]

          y=\frac{x}{x+2}

       xy+2y=x

       2y=x-xy

     2y=x(1-y)

    x = \frac{2y}{1-y}     ,y\neq 1

 

g(y) = \frac{2y}{1-y}

 

f^{-1}=\frac{2y}{1-y},y\neq 1

Question:7 Consider f : R \rightarrow R given by f (x) = 4x + 3. Show that f is invertible. Find the inverse of f.

Answer:

f : R \rightarrow R  is given by  f (x) = 4x + 3

One-one :

Let  f(x)=f(y)

    4x + 3 = 4y+3

            4x=4y

              x=y

\therefore f is one-one function.

Onto:

y=4x+3\, \, \, , y \in R

\Rightarrow x=\frac{y-3}{4} \in R

So, for y \in R there is  x=\frac{y-3}{4} \in R   ,such that 

f(x)=f(\frac{y-3}{4})=4(\frac{y-3}{4})+3

                                    = y-3+3 

                                    = y

 \therefore f  is onto.

Thus, f is one-one and onto so f^{-1} exists.

Let, g:R\rightarrow R by g(x)=\frac{y-3}{4}

Now, 

(gof)(x)= g(f(x))= g(4x+3)

                                           =\frac{(4x+3)-3}{4}

                                          =\frac{4x}{4}

                                          =x

(fog)(x)= f(g(x))= f(\frac{y-3}{4})

                                         = 4\times \frac{y-3}{4}+3

                                         = y-3+3

                                         = y

(gof)(x)= x             and     (fog)(x)= y

Hence, function f is invertible and inverse of f is g(y)=\frac{y-3}{4}.

Question:8 Consider f : R+ → [4, ∞) given by f(x) = x^2+4. Show that f is invertible with the inverse f^{-1} of f given by  f^{-1}(y)= \sqrt{y-4} , where R+ is the set of all non-negative real numbers.

Answer:

It is given that
f : R^+ \rightarrow [4,\infty)  ,  f(x) = x^2+4  and 

Now, Let f(x) = f(y)

⇒ x2 + 4 = y2 + 4

⇒ x2 = y2

⇒ x = y

⇒ f is one-one function.

Now, for y \epsilon[4, ∞), let y = x2 + 4.

⇒ x2 = y -4 ≥ 0

⇒ for any y \epsilon R, there exists x =  \epsilon R such that

 = y -4 + 4 = y.

⇒ f is onto function.

Therefore, f is one–one and onto function, so f-1 exists.

Now, let us define g: [4, ∞) → R+ by,

g(y) = 

Now, gof(x) = g(f(x)) = g(x2 + 4) = 

And, fog(y) = f(g(y)) =  = 

Therefore, gof = gof = IR.

Therefore, f is invertible and the inverse of f is given by

f-1(y) = g(y) = 

Question:9 Consider f : R_+ \rightarrow [- 5, \infty) given by f (x) = 9x^2 + 6x - 5. Show that f is invertible with f^{-1} (y) = \left (\frac{(\sqrt{y + 6}) - 1}{3} \right )

Answer:

f : R_+ \rightarrow [- 5, \infty)

f (x) = 9x^2 + 6x - 5

One- one:

Let       f(x)=f(y)    for \, \, x,y\in R

            9x^{2}+6x-5=9y^{2}+6y-5

                   9x^{2}+6x=9y^{2}+6y

\Rightarrow              9(x^{2}-y^{2})=6(y-x)

             9(x+y)(x-y)+6(x-y)= 0

        (x-y)(9(x+y)+6)=0

Since, x and y are positive. 

         (9(x+y)+6)> 0

\therefore x=y

 \therefore   f is one-one.

Onto:

Let for  y \in [-5,\infty)  , y=9x^{2}+6x-5

                  \Rightarrow         y=(3x+1)^{2}-1-5

                  \Rightarrow             y=(3x+1)^{2}-6

                \Rightarrow                  y+6=(3x+1)^{2}

                                         (3x+1)=\sqrt{y+6}

                                           x = \frac{\sqrt{y+6}-1}{3}                                      

\therefore f is onto  and range is y \in [-5,\infty).

Since f is one-one and onto so it is invertible.

Let  g : [-5,\infty)\rightarrow R_+    by  g(y) = \frac{\sqrt{y+6}-1}{3}  

(gof)(x)=g(f(x))=g(9x^{2}+6x)-5=g((3x+1)^{2}-6)\\=\sqrt{ { (3x+1)^{2} }-6+6} -1

       (gof)(x)=\frac{3x+1-1}{3}=\frac{3x}{3}= x        

(fog)(x)=f(g(x))=f(\frac{\sqrt{y+6}-1}{3})

                                          =[3(\frac{\sqrt{y+6}-1}{3})+1]^{2}-6

                                           =(\sqrt{y+6})^{2}-6

                                         =y+6-6

                                        =y

               \therefore gof=fog=I_R

Hence,  f is invertible with the inverse f^{-1} of f given by  f^{-1} (y) = \left (\frac{(\sqrt{y + 6}) - 1}{3} \right )

Question:10 Let f : X \rightarrow Ybe an invertible function. Show that f has a unique inverse. (Hint: suppose g_1 and g_2 are two inverses of f. Then for all y \in Y,
fog_1 (y) = I_Y (y) = fog_2 (y). Use one-one ness of f).

Answer:

Let f : X \rightarrow Ybe an invertible function

Also, suppose f has two inverse g_1 and g_2

For y \in Y, we have

     fog_1(y) = I_y(y)=fog_2(y)

 \Rightarrow     f(g_1(y))=f(g_2(y))  

\Rightarrow             g_1(y)=g_2(y)                     [f is invertible implies f is one - one]

\Rightarrow                     g_1=g_2                        [g is one-one]

Thus,f has a unique inverse.

Question:11 Consider f : \{1, 2, 3\} \rightarrow \{a, b, c\} given by f (1) = a,f (2) = b  and f (3) = c. Find f^{-1} and show that  (f^{-1})^{-1} = f.

Answer:

It is given that
f : \left \{ 1,2,3 \right \}\rightarrow \left \{ a,b,c \right \}

f(1) = a, f(2) = b \ and \ f(3) = c

Now,, lets define a function g :
\left \{ a,b,c \right \}\rightarrow \left \{ 1,2,3 \right \}  such that

g(a) = 1, g(b) = 2 \ and \ g(c) = 3
Now,

(fog)(a) = f(g(a)) = f(1) = a
Similarly,

(fog)(b) = f(g(b)) = f(2) = b

(fog)(c) = f(g(c)) = f(3) = c

And

(gof)(1) = g(f(1)) = g(a) = 1

(gof)(2) = g(f(2)) = g(b) = 2

(gof)(3) = g(f(3)) = g(c) = 3

Hence, gof = I_X andfog = I_Y, where X = \left \{ 1,2,3 \right \} andY = \left \{ a,b,c \right \}

Therefore, the inverse of f exists and f^{-1} = g

Now, 
f^{-1} : \left \{ a,b,c \right \}\rightarrow \left \{ 1,2,3 \right \}  is given by

f^{-1}(a) = 1, f^{-1}(b) = 2 \ and \ f^{-1}(c) = 3

Now, we need to  find the inverse of  f^{-1},

Therefore, lets defineh: \left \{ 1,2,3 \right \}\rightarrow \left \{ a,b,c \right \}  such that

h(1) = a, h(2) = b \ and \ h(3) = c  

Now,

(goh)(1) = g(h(1)) = g(a) = 1

(goh)(2) = g(h(2)) = g(b) = 2

(goh)(3) = g(h(3)) = g(c) = 3

Similarly,

(hog)(a) = h(g(a)) = h(1) = a

(hog)(b) = h(g(b)) = h(2) = b

(hog)(c) = h(g(c)) = h(3) = c

Hence,  goh = I_X and hog = I_Y, whereX = \left \{ 1,2,3 \right \} andY = \left \{ a,b,c \right \}

Therefore,  inverse of  g^{-1} = (f^{-1})^{-1}  exists and  g^{-1} = (f^{-1})^{-1} = h

\Rightarrow h = f

Therefore,  (f^{-1})^{-1} = f

Hence proved

Question:12 Let f : X \rightarrow Y be an invertible function. Show that the inverse of f^{-1} is f, i.e., (f^{-1})^{-1} = f

Answer:

f : X \rightarrow Y

To prove: (f^{-1})^{-1} = f

Let  f:X\rightarrow Y  be a invertible function.

  Then there is    g:Y\rightarrow X  such that   gof =I_x  and fog=I_y

    Also,       f^{-1}= g  

     gof =I_x  and fog=I_y     

   \Rightarrow        f^{-1}of = I_x            and         fof^{-1} = I_y

Hence, f^{-1}:Y\rightarrow X    is invertible function and f is inverse of f^{-1}.

i.e. (f^{-1})^{-1} = f

Question:13 If f : R \rightarrow R be given by f(x) = (3 - x^3)^{\frac{1}{3}}, then fof(x) is 

(A) x^{\frac{1}{3}}

(B) x^3

(C) x

(D) (3 - x^3)

Answer:

f(x) = (3 - x^3)^{\frac{1}{3}}

fof(x)=f(f(x))=f((3-x^{3})^{\frac{1}{3}})

                                     =[3- ((3-x^{3})^{\frac{1}{3}})^{3}]^{\frac{1}{3}}

                                =([3- ((3-x^{3})]^{\frac{1}{3}})

                                   = (x^{3})^{\frac{1}{3}}

                                   =x

Thus,fof(x) is x.

Hence, option c is correct answer.

Question:14 Let f: R - \left\{-\frac{4}{3}\right\} \rightarrow R be a function defined as f(x) = \frac{4x}{3x + 4}. The inverse of f is the map g : Range\;f \rightarrow R - \left \{-\frac{4}{3} \right \}given by 

(A) g(y) = \frac{3y}{3 -4y}

(B) g(y) = \frac{4y}{4 -3y}

(C) g(y) = \frac{4y}{3 -4y}

(D) g(y) = \frac{3y}{3 -4y}

Answer:

f: R - \left\{-\frac{4}{3}\right\} \rightarrow R

f(x) = \frac{4x}{3x + 4}

Let f inverse  g : Range\;f \rightarrow R - \left \{-\frac{4}{3} \right \}

Let y be the element of range f.

Then there is x \in R - \left\{-\frac{4}{3}\right\}    such that    

                 y=f(x)

               y=\frac{4x}{3x+4}

          y(3x+4)=4x

          3xy+4y=4x

        3xy-4x+4y=0

        x(3y-4)+4y=0

                       x= \frac{-4y}{3y-4}

                   x= \frac{4y}{4-3y}

 

Now , defineg : Range\;f \rightarrow R - \left \{-\frac{4}{3} \right \}     as g(y)= \frac{4y}{4-3y}

 

gof(x)= g(f(x))= g(\frac{4x}{3x+4})

                             = \frac{4(\frac{4x}{3x+4})}{4-3(\frac{4x}{3x+4})}

                             =\frac{16x}{12x+16-12x}

                            =\frac{16x}{16}

                            =x

fog(y)=f(g(y))=f(\frac{4y}{4-3y})

                                       = \frac{4(\frac{4y}{4-3y})}{3(\frac{4y}{4-3y}) + 4}

                                       =\frac{16y}{12y+16-12y}=\frac{16y}{16}

                                         =y

Hence, g is inverse of f and f^{-1}=g

The inverse of f is given by   g(y)= \frac{4y}{4-3y}.

The correct option is B.

Solutions of NCERT for class 12 maths chapter 1 Relations and Functions: Exercise 1.4

Question:1(i) Determine whether or not each of the definition of ∗ given below gives a binary operation. In the event that ∗ is not a binary operation, give justification for this.

(i) On Z^+ , define ∗ by a * b = a - b

Answer:

(i) On Z^+ , define ∗ by a * b = a - b

 It is not a binary operation as the image of (1,2) under * is 1\ast 2=1-2 =-1 \notin Z^{+}.                                                                                       

Question:1(ii) Determine whether or not each of the definition of ∗ given below gives a binary operation. In the event that ∗ is not a binary operation, give justification for this.

(ii) On Z^+ , define ∗ by a * b = ab

Answer:

(ii) On Z^+ , define ∗ by a * b = ab

 We can observe that for a,b \in Z^+,there is a unique element ab in Z^+.

This means * carries each pair (a,b)  to a unique element a * b = ab in Z^+.

Therefore,* is a binary operation.

Question:1(iii) Determine whether or not each of the definition of ∗ given below gives a binary operation. In the event that ∗ is not a binary operation, give justification for this.

(iii) On R, define ∗ by a * b = ab^2

Answer:

(iii) On R, define ∗ by a * b = ab^2

 We can observe that for a,b \in R,there is a unique element ab^{2}  in R.

This means * carries each pair (a,b)  to a unique element a * b = ab^{2} in R.

Therefore,* is a binary operation.

Question:1(iv) Determine whether or not each of the definition of ∗ given below gives a binary operation. In the event that ∗ is not a binary operation, give justification for this.

(iv) On Z^+, define ∗ by a * b = | a - b |

Answer:

(iv) On Z^+, define ∗ by a * b = | a - b |

 We can observe that for a,b \in Z^+,there is a unique element | a - b |  in Z^+.

This means * carries each pair (a,b)  to a unique element a * b = | a - b | in Z^+.

Therefore,* is a binary operation.

Question:1(v) Determine whether or not each of the definition of ∗ given below gives a binary operation. In the event that ∗ is not a binary operation, give justification for this

(v) On  Z^+ , define ∗ by a * b = a

Answer:

(v) On  Z^+ , define ∗ by a * b = a

* carries each pair (a,b)  to a unique element a * b = a in  Z^+.

Therefore,* is a binary operation.

Question:2(i) For each operation ∗ defined below, determine whether ∗ is binary, commutative or associative.

(i)On Z, define a * b = a-b

Answer:

a*b=a-b

b*a=b-a

a*b\neq b*a

so * is not commutative

(a*b)*c=(a-b)-c

a*(b*c)=a-(b-c)=a-b+c

(a*b)*c not equal to a*(b*c), so * is not associative