Browse by Stream
My Companion New
Search Colleges, Exams, Schools & more

NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions

Edited By Ramraj Saini | Updated on Feb 10, 2023 - 2:22 p.m. IST | #CBSE Class 12th
subscribe Subscribe to Premium
Download Guide Download PDF

NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions :- This is the first chapter of Class 12 math. NCERT solutions class 12 maths chapter 1 Relations and Functions contains the answer and step-by-step solution to each question asked in the exercise of NCERT Class 12 maths book. NCERT Class 12 maths solutions Chapter 1 will help you to understand the concepts and score well in CBSE 12th board exam.

Latest Updates for CBSE Class 12th

Stay up-to date with CBSE Class 12th News

Get Updates
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions

This chapter 1 12th class is not only important for mathematics, but also it is important in real life too. In this article, you will find NCERT class 12 maths chapter 1 solutions including miscellaneous exercise which will help you to score more marks in the exam. Here you will find all NCERT solutions of chapter 1 maths class 12 at a single place which will be helpful when you are not able to solve the NCERT questions.

In NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions, there are four exercises with 55 questions and one miscellaneous exercise with 19 questions. relations and functions class 12 solutions are very important for students because they comprise quality practice questions. In this article, you will find the detailed NCERT solutions for class 12 maths chapter 1. Here you will get NCERT solutions for class 12 also.

Also Read:

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 as R = \{(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

\therefore R 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 T 1 with sides 3, 4, 5, T 2 with sides 5, 12, 13 and T 3 with sides 6, 8, 10. Which triangles among T 1 , T 2 and T 3 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.T 1 is similar to T2

T 1 is similar to T2 is the same asT2 is similar to T 1 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. T 1 is similar to T2 and T2 is similar toT 3 .

\Rightarrow T 1 is similar toT 3 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 T 1 andT 3 are as shown

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

i.e. ratios of sides of T 1 and T 3 are equal.Hence, T 1 and T 3 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.P 1 have same number of sides as P 2

P 1 have the same number of sides as P 2 is the same as P 2 have same number of sides as P 1 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. P 1 have the same number of sides as P 2 and P 2 have same number of sides as P 3

\Rightarrow P 1 have same number of sides as P 3 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.L 1 is parallel to L 2 .

L1 is parallel to L 2 is same as L 2 is parallel to L 1 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 L 2 and L 2 is parallel to L 3 .

\Rightarrow L 1 is parallel to L 3 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 A= {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 function f : 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 Let f : R\rightarrow R be defined as f(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 .


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 by f = \{(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 f is 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 function f : [-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)

⇒ x 2 + 4 = y 2 + 4

⇒ x 2 = y 2

⇒ x = y

⇒ f is one-one function.

Now, for y \epsilon [4, ∞), let y = x 2 + 4.

⇒ x 2 = y -4 ≥ 0

1516353559244276

⇒ for any y \epsilon R, there exists x = 1516353559980485 \epsilon R such that

1516353560716145 = 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) = 151635356145214

Now, gof(x) = g(f(x)) = g(x 2 + 4) = 15163535621863

And, fog(y) = f(g(y)) = 1516353562911365 = 1516353563643843

Therefore, gof = gof = I R .

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

f-1(y) = g(y) = 1516353564378360

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 Y be 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 Y be 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 and fog = I_Y , where X = \left \{ 1,2,3 \right \} and Y = \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 define h: \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 , where X = \left \{ 1,2,3 \right \} and Y = \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 , define g : 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.


NCERT Solutions 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

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

(ii) On Q , define a * b = ab + 1

Answer:

(ii) On Q , define a * b = ab + 1

ab = ba for all a,b \in Q

ab+1 = ba + 1 for all a,b \in Q

\Rightarrow a\ast b=b\ast a for a,b \in Q

(1*2)*3 = (1\times 2+1) * 3 = 3 * 3 = 3\times 3+1 = 10

1*(2*3) = 1 * (2\times 3+1) = 1 * 7 = 1\times 7+1 = 8

\therefore (1\ast 2)\ast 3\neq 1\ast (2\ast 3) ; where 1,2,3 \in Q

\therefore operation * is not associative.

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

(iii) On Q , define a * b = \frac{ab}{2}

Answer:

(iii) On Q , define a * b = \frac{ab}{2}

ab = ba for all a,b \in Q

\frac{ab}{2}=\frac{ba}{2} for all a,b \in Q

\Rightarrow a\ast b=b\ast a for a,b \in Q

\therefore operation * is commutative.

(a*b)*c = \frac{ab}{2}*c = \frac{(\frac{ab}{2})c}{2} = \frac{abc}{4}

a*(b*c) = a*\frac{bc}{2} = \frac{a(\frac{bc}{2})}{2} = \frac{abc}{4}

\therefore (a*b)*c=a*(b*c) ;

\therefore operation * is associative.

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

(iv) On Z^+ , define a * b = 2^{ab}

Answer:

(iv) On Z^+ , define a * b = 2^{ab}

ab = ba for all a,b \in Z^{+}

2ab = 2ba for all a,b \in Z^{+}

\Rightarrow a\ast b=b\ast a for a,b \in Z^{+}

\therefore the operation is commutative.

(1*2)*3 = 2^{1\times 2} * 3 = 4 * 3 = 2^{4\times 3} = 2^{12}

1*(2*3) = 1 * 2^{2\times 3} = 1 * 64 = 2^{1\times 64}=2^{64}

\therefore (1\ast 2)\ast 3\neq 1\ast (2\ast 3) ; where 1,2,3 \in Z^{+}

\therefore operation * is not associative.

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

(v) On Z^+ , define a * b = a^b

Answer:

(v) On Z^+ , define a * b = a^b

1\ast 2 = 1^{2}= 1 and 2\ast 1 = 2^{1}= 2

\Rightarrow 1\ast 2\neq 2\ast 1 for 1,2 \in Z^{+}

\therefore the operation is not commutative.

(2\ast 3)\ast 4 = 2^{3} \ast 4 = 8 \ast 4 = 8^{ 4}=2^{12}

2\ast (3\ast 4) = 2 \ast 3^{ 4} = 2 \ast 81 = 2^{81}

\therefore (2\ast 3)\ast 4\neq 2\ast (3\ast 4) ; where 2,3,4 \in Z^{+}

\therefore operation * is not associative.

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

(vi) On R - \{-1 \} , define a * b = \frac{a}{b +1}

Answer:

(iv) On R - \{-1 \} , define a * b = \frac{a}{b +1}


1\ast 2 = \frac{1}{2+1}=\frac{1}{3} and 2\ast 1 = \frac{2}{2+1}= \frac{2}{3}

\Rightarrow 1\ast 2\neq 2\ast 1 for 1,2 \in R - \{-1 \}

\therefore the operation is not commutative.

(1\ast 2)\ast 3 = (\frac{1}{2+1}) \ast 3 = \frac{1}{3} \ast 3 = \frac{\frac{1}{3}}{3+1}= \frac{1}{12}

1\ast (2\ast 3) = 1 \ast (\frac{2}{3+1}) = 1 \ast \frac{2}{4} = 1 \ast \frac{1}{2} = \frac{1}{\frac{1}{2}+1} = \frac{2}{3}

\therefore (1\ast 2)\ast 3\neq 1\ast (2\ast 3) ; where 1,2,3 \in R - \{-1 \}

\therefore operation * is not associative.

Question:3

Consider the binary operation Λ on the set {1, 2, 3, 4, 5} defined by

a Λ b = min {a, b}. Write the operation table of the operation Λ .

Answer:

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

a \wedge b = min \{a, b\} for a,b \in \{1, 2, 3, 4, 5\}

The operation table of the operation \wedge is given by :

\wedge

1

2

3

4

5

1

1

1

1

1

1

2

1

2

2

2

2

3

1

2

3

3

3

4

1

2

3

4

4

5

1

2

3

4

5

Question:4(i) Consider a binary operation ∗ on the set \{1, 2, 3, 4, 5\} given by the following multiplication table (Table 1.2).

(i) Compute (2 * 3) * 4 and 2 * (3 * 4)

(Hint: use the following table)


1643786590966

Answer:

(i)

(2 * 3) * 4 = 1*4 =1

2 * (3 * 4) = 2*1=1

Question:4(ii) Consider a binary operation ∗ on the set \{1, 2, 3, 4, 5\} given by the following multiplication table (Table 1.2).

(ii) Is ∗ commutative?

(Hint: use the following table) 1643786526935

Answer:

(ii)

For every a,b \in\{1, 2, 3, 4, 5\} , we have a*b = b*a . Hence it is commutative.

Question:4(iii) Consider a binary operation ∗ on the set { \{1, 2, 3, 4, 5\} given by the following multiplication table (Table 1.2).

(iii) Compute (2 ∗ 3) ∗ (4 ∗ 5).

(Hint: use the following table)


1643786571713

Answer:

(iii) (2 ∗ 3) ∗ (4 ∗ 5).

from the above table

(2*3)*(4*5)=1*1=1

Question:5 Let ∗′ be the binary operation on the set \{1, 2, 3, 4, 5\} defined by a *' b = H.C.F. \;of\;a\;and\;b . Is the operation ∗′ same as the operation ∗ defined
in Exercise 4 above? Justify your answer.

Answer:

a *' b = H.C.F. \;of\;a\;and\;b for a,b \in \{1, 2, 3, 4, 5\}

The operation table is as shown below:

\ast

1

2

3

4

5

1

1

1

1

1

1

2

1

2

1

2

1

3

1

1

3

1

1

4

1

2

1

4

1

5

1

1

1

1

5

The operation ∗′ same as the operation ∗ defined in Exercise 4 above.

Question:6(i) let ∗ be the binary operation on N given by . Find

(i) 5 ∗ 7, 20 ∗ 16

Answer:

a*b=LCM of a and b

(i) 5 ∗ 7, 20 ∗ 16

5*7 = L.C.M \, of\, 5\, \, and \, 7=35

20*16 = L.C.M \, of\, 20\, \, and \, 16 =80

Question:6(ii) Let ∗ be the binary operation on N given by a * b = L.C.M. \;of \;a\; and \;b . Find

(ii) Is ∗ commutative?

Answer:

a * b = L.C.M. \;of \;a\; and \;b

(ii) L.C.M. \;of \;a\; and \;b = L.C.M. \;of \;b\; and \;a for all a,b \in N

\therefore \, \, \, \, \, \, \, a*b = b*a

Hence, it is commutative.

Question:6(iii) Let ∗ be the binary operation on N given by a ∗ b = L.C.M. of a and b. Find

(iii) Is ∗ associative?

Answer:

a * b = L.C.M. of a and b

(iii) a,b,c \in N

(a*b)*c = (L.C.M \, of\, a\, and \, b)*c= L.C.M\, of \, a,b\, and\, c

a*(b*c) = a*(L.C.M \, of\, b\, and \, c)= L.C.M\, of \, a,b\, and\, c

\therefore \, \, \, \, \, \, \, \, \, (a*b)*c=a*(b*c)

Hence, the operation is associative.

Question:6(iv) Let ∗ be the binary operation on N given by a* b = L.C.M.\; of \;a\; and \;b . Find

(iv) the identity of ∗ in N

Answer:

a* b = L.C.M.\; of \;a\; and \;b

(iv) the identity of ∗ in N

We know that L.C.M.\; of \;a\; and \;1 = a = L.C.M.\; of 1\, \, and \;a\;

\therefore a*1=a=1*a for a \in N

Hence, 1 is the identity of ∗ in N.

Question 6(v) Let ∗ be the binary operation on N given by a ∗ b = L.C.M. of a and b. Find

(v) Which elements of N are invertible for the operation ∗?

Answer:

a* b = L.C.M.\; of \;a\; and \;b

An element a is invertible in N

if a*b=e=b*a

Here a is inverse of b.

a*b=1=b*a

a*b=L.C.M. od a and b

a=b=1

So 1 is the only invertible element of N

Question:7 Is ∗ defined on the set \{1, 2, 3, 4, 5\} by a * b = L.C.M. \;of \;a\; and \;b a binary operation? Justify your answer.

Answer:

a * b = L.C.M. \;of \;a\; and \;b

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

Operation table is as shown below:

*

1

2

3

4

5

1

1

2

3

4

5

2

2

2

6

4

10

3

3

6

3

12

15

4

4

4

12

4

20

5

5

10

15

20

5

From the table, we can observe that

2*3=3*2=6 \notin A

2*5=5*2=10 \notin A

3*4=4*3=12 \notin A

3*5=5*3=15 \notin A

4*5=5*4=20 \notin A

Hence, the operation is not a binary operation.

Question:8 Let ∗ be the binary operation on N defined by a ∗ b = H.C.F. of a and b. Is ∗ commutative? Is ∗ associative? Does there exist identity for this binary
operation on N?

Answer:

a ∗ b = H.C.F. of a and b for all a,b \in A

H.C.F. of a and b = H.C.F of b and a for all a,b \in A

\therefore \, \, \, \, a*b=b*a

Hence, operation ∗ is commutative.

For a,b,c \in N ,

(a*b)*c = (H.C.F \, of\, a\, and\, b)*c= H.C.F\, of \, a,b,c.

a*(b*c )= a*(H.C.F \, of\, b\, and\, c)= H.C.F\, of \, a,b,c.

\therefore (a*b)*c=a*(b*c)

Hence, ∗ is associative.

An element c \in N will be identity for operation * if a*c=a= c*a for a \in N .

Hence, the operation * does not have any identity in N.

Question:9(i) Let ∗ be a binary operation on the set Q of rational numbers as follows:

(i) a * b = a - b Find which of the binary operations are commutative and which are associative.

Answer:

On the set Q ,the operation * is defined as a * b = a - b .It is observed that:

\frac{1}{2}*\frac{1}{3}= \frac{1}{2}-\frac{1}{3}=\frac{1}{6}

\frac{1}{3}*\frac{1}{2}= \frac{1}{3}-\frac{1}{2}=\frac{-1}{6}

\therefore \frac{1}{2}*\frac{1}{3}\neq \frac{1}{3}*\frac{1}{2} here \frac{1}{2},\frac{1}{3} \in Q

Hence, the * operation is not commutative.

It can be observed that

(\frac{1}{2}*\frac{1}{3})*\frac{1}{4} = \left ( \frac{1}{2}-\frac{1}{3}\right )*\frac{1}{4}=\frac{1}{6}*\frac{1}{4}=\frac{1}{6}-\frac{1}{4}=\frac{-1}{12}

\frac{1}{2}*(\frac{1}{3}*\frac{1}{4})= \frac{1}{2}*\left ( \frac{1}{3} - \frac{1}{4}\right ) = \frac{1}{2}*\frac{1}{12} = \left ( \frac{1}{2} - \frac{1}{12} \right ) = \frac{5}{12}

\left ( \frac{1}{2}*\frac{1}{3} \right )*\frac{1}{4}\neq \frac{1}{2}*(\frac{1}{3}*\frac{1}{4}) for all \frac{1}{2},\frac{1}{3}, \frac{1}{4} \in Q

The operation * is not associative.

Question:9(ii) Let ∗ be a binary operation on the set Q of rational numbers as follows:

(ii) a*b = a^2 + b^2 Find which of the binary operations are commutative and which are associative.

Answer:

On the set Q ,the operation * is defines as a*b = a^2 + b^2 .It is observed that:

For a,b \in Q

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

\therefore a*b=b*a

Hence, the * operation is commutative.

It can be observed that

(1*2)*3 =(1^{2}+2^{2})*3 = 5*3 = 5^{2}+3^{2} = 25+9 =34

1*(2*3) =1*(2^{2}+3^{2}) = 1*13 = 1^{2}+13^{2} = 1+169 =170

(1*2)*3 \neq 1*(2*3) for all 1,2,3 \in Q

The operation * is not associative.

Question:9(iii) Let ∗ be a binary operation on the set Q of rational numbers as follows:

(iii) a * b = a + ab Find which of the binary operations are commutative and which are associative.

Answer:

On the set Q ,the operation * is defines as a * b = a + ab .It is observed that:

For a,b \in Q

1 * 2 = 1+1\times 2 =1 + 2 = 3

2 * 1= 2+2\times 1 =2 + 2 = 4

\therefore 1*2\neq 2*1 for 1,2 \in Q

Hence, the * operation is not commutative.

It can be observed that

1*(2*3) =1*(2+3\times 2) = 1*8 = 1+1\times 8 = 1+8 =9

(1*2)*3 \neq 1*(2*3) for all 1,2,3 \in Q

The operation * is not associative.

Question:9(iv) Let ∗ be a binary operation on the set Q of rational numbers as follows:

(iv) a * b = (a-b)^2 Find which of the binary operations are commutative and which are associative.

Answer:

On the set Q ,the operation * is defined as a * b = (a-b)^2 .It is observed that:

For a,b \in Q

a * b = (a-b)^2

b* a = (b-a)^2 = \left [ -\left ( a-b \right ) \right ]^{2} = (a-b)^{2}

\therefore a*b = b* a for a,b \in Q

Hence, the * operation is commutative.

It can be observed that

(1*2)*3 =(1-2)^{2}*3 = 1*3 =(1-3)^{2}= (-2)^{2} =4

1*(2*3) =1*(2-3)^{2} = 1*1 =(1-1)^{2} = 0^{2} =0

(1*2)*3 \neq 1*(2*3) for all 1,2,3 \in Q

The operation * is not associative.

Question:9(v) Let ∗ be a binary operation on the set Q of rational numbers as follows:

(v) a * b = \frac{ab}{4} Find which of the binary operations are commutative and which are associative.

Answer:

On the set Q ,the operation * is defines as a * b = \frac{ab}{4} .It is observed that:

For a,b \in Q

a * b = \frac{ab}{4}

b* a = \frac{ba}{4}

\therefore a*b = b* a for a,b \in Q

Hence, the * operation is commutative.

It can be observed that

(a*b)*c =(\frac{ab}{4})*c = \frac{\frac{ab}{4}c}{4}=\frac{abc}{16}

a*(b*c) =a*(\frac{bc}{4}) = \frac{\frac{bc}{4}a}{4}=\frac{abc}{16}

(a*b)*c = a*(b*c) for all a,b,c \in Q

The operation * is associative.

Question:9(vi) Let ∗ be a binary operation on the set Q of rational numbers as follows:

(vi) a* b = ab^2 Find which of the binary operations are commutative and which are associative.

Answer:

On the set Q ,the operation * is defines as a* b = ab^2 .It is observed that:

For a,b \in Q

1* 2 = 1\times 2^2=1\times 4=4

2* 1 = 2\times 1^2=2\times 1=2

\therefore 1*2\neq 2*1 for 1,2 \in Q

Hence, the * operation is not commutative.

It can be observed that

(1*2)*3 = (1\times 2^{2})*3 = 4*3 = 4\times 3^{2}=4\times 9=36

1*(2*3) = 1*(2\times 3^{2}) = 1*18 = 1\times 18^{2}=1\times 324=324

(1*2)*3\neq 1*(2*3) for all 1,2,3 \in Q

The operation * is not associative.

Question:10 Find which of the operations given above has identity.

Answer:

An element p \in Q will be identity element for operation *

if a*p = a = p*a for all a \in Q

(v) a * b = \frac{ab}{4}

a * p = \frac{ap}{4}

p * a = \frac{pa}{4}

a*p = a = p*a when p=4 .

Hence, (v) a * b = \frac{ab}{4} has identity as 4.

However, there is no such element p \in Q which satisfies above condition for all rest five operations.

Hence, only (v) operations have identity.

Question:11 Let A = N \times N and ∗ be the binary operation on A defined by (a, b) * (c, d) = (a + c, b + d) Show that ∗ is commutative and associative. Find the identity element for ∗ on A, if any.

Answer:

A = N \times N and ∗ be the binary operation on A defined by

(a, b) * (c, d) = (a + c, b + d)

Let (a,b),(c,d) \in A

Then, a,b,c,d \in N

We have

(a, b) * (c, d) = (a + c, b + d)

(c,d)*(a,b) = (c+a,d+b)= (a+c,b+d)

\therefore \, \, \, \, \, (a,b)*(c,d)=(c,d)*(a,b)

Thus it is commutative.

Let (a,b),(c,d),(e,f) \in A

Then, a,b,c,d,e,f \in N

[(a, b) * (c, d)]*(e,f)= [(a + c, b + d)]*(e,f) =[(a+c+e),(b+d+f)]

(a, b) * [(c, d)*(e,f)]= (a,b)*[(c + e, d + f)] =[(a+c+e),(b+d+f)]

\therefore \, \, \, \, \, [(a,b)*(c,d)]*(e,f)=(a, b) * [(c, d)*(e,f)]

Thus, it is associative.

Let e= (e1,e2) \in A will be a element for operation * if (a*e)=a=(e*a) for all a= (a1,a2) \in A .

i.e. (a1+e1,a2+e2)= (a1,a2)= (e1+a1,e2+a2)

This is not possible for any element in A .

Hence, it does not have any identity.

Question:12(i) State whether the following statements are true or false. Justify.

(i) For an arbitrary binary operation ∗ on a set N, a*a = a,\; \forall a\in N

Answer:

(i) For an arbitrary binary operation ∗ on a set N, a*a = a,\; \forall a\in N

An operation * on a set N as a*b=a+b\, \, \, \, \forall\, \, a,b \in N

Then , for b=a=2

2*2= 2+2 = 4\neq 2

Hence, statement (i) is false.

Question:12(ii) State whether the following statements are true or false. Justify.

(ii) If ∗ is a commutative binary operation on N, then a *(b*c) =( c*b)*a

Answer:

(ii) If ∗ is a commutative binary operation on N, then a *(b*c) =( c*b)*a

R.H.S =(c*b)*a

=(b*c)*a (* is commutative)

= a*(b*c) ( as * is commutative)

= L.H.S

\therefore a *(b*c) =( c*b)*a

Hence, statement (ii) is true.

Question:13 Consider a binary operation ∗ on N defined as a * b = a^3 + b^3 . Choose the correct answer.

(A) Is ∗ both associative and commutative?
(B) Is ∗ commutative but not associative?
(C) Is ∗ associative but not commutative?
(D) Is ∗ neither commutative nor associative?

Answer:

A binary operation ∗ on N defined as a * b = a^3 + b^3 .

For a,b \in N

a * b = a^3 + b^3 = b^{3}+a^{3}=b*a

Thus, it is commutative.

(1*2)*3 = (1^{3}+2^{3})*3=9*3 =9^{3}+3^{3}=729+27=756

1*(2*3) = 1*(2^{3}+3^{3})=1*35 =1^{3}+35^{3}=1+42875=42876

\therefore \, \, \, \, \, (1*2)*3 \neq 1*(2*3) where 1,2,3 \in N

Hence, it is not associative.

Hence, B is the correct option.


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

Question:1 Let f : R \rightarrow R be defined as f (x) = 10x + 7 . Find the function g : R \rightarrow R such that g o f = f o g = I_R .

Answer:

f : R \rightarrow R

f (x) = 10x + 7

g : R \rightarrow R and g o f = f o g = I_R

For one-one:

f(x)=f(y)

10x+7=10y+7

10x=10y

x=y

Thus, f is one-one.

For onto:

For y \in R , y=10x+7

x= \frac{y-7}{10} \in R

Thus, for y \in R , there exists x= \frac{y-7}{10} \in R such that

f(x) = f(\frac{y-7}{10})=10(\frac{y-7}{10})+7=y-7+7=y

Thus, f is onto.

Hence, f is one-one and onto i.e. it is invertible.

Let g : R \rightarrow R as f(y)=\frac{y-7}{10}

gof(x)=g(f(x))= g(10x+7) =\frac{(10x+7)-7}{10}=\frac{10x}{10}=x

fog(x)=f(g(x))= f(\frac{y-7}{10}) =10\frac{y-7}{10}+7=y-7+7=y


\therefore gof(x)=I_R and fog(x)=I_R

Hence, g : R \rightarrow R defined as g(y)=\frac{y-7}{10}

Question:2 Let f : W \rightarrow W be defined as f (n) = n -1 , if n is odd and f (n) = n +1 , if n is even. Show that f is invertible. Find the inverse of f. Here, W is the set of all whole numbers.

Answer:

f : W \rightarrow W

f (n) = n -1 if n is odd

f (n) = n +1 if n is even.

For one-one:

Taking x as odd number and y as even number.

f(x)=f(y)

x-1=y+1

x-y=2

Now, Taking y as odd number and x as even number.

f(x)=f(y)

x+1=y-1

x-y = - 2

This is also impossible.

If both x and y are odd :

f(x)=f(y)

x-1=y-1

x=y

If both x and y are even :

f(x)=f(y)

x+1=y+1

x=y

\therefore f is one-one.

Onto:

Any odd number 2r+1 in codomain of N is an image of 2r in domain N and any even number 2r in codomain N is the image of 2r+1 in domain N.

Thus, f is onto.

Hence, f is one-one and onto i.e. it is invertible.

Sice, f is invertible.

Let g : W \rightarrow W as m+1 if m is even and m-1 if m is odd.

When x is odd.

gof(x)=g(f(x))= g(n-1) =n-1+1=n

When x is even

gof(x)=g(f(x))= g(n+1) =n+1-1=n

Similarly, m is odd

fog(x)=f(g(x))= f(m-1) =m-1+1=m

m is even ,

fog(x)=f(g(x))= f(m+1) =m-1+1=m

\therefore gof(x)=I_W and fog(x)=I_W

Hence, f is invertible and the inverse of f is g i.e. f^{-1}=g , which is the same as f.

Hence, inverse of f is f itself.

Question:3 If f : R → R is defined by f(x) = x 2 – 3x + 2, find f (f (x)).

Answer:

This can be solved as following

f : R → R

f(x) = x^{2}-3x+2

f(f(x)) = f(x^{2}-3x+2) = (x^{2}-3x+2)^{2} - 3(x^{2}-3x+2)+2

= (x^{4}+9x^{2}+4-6x^{3}-12x+4x^{2})-3x^{2}+9x-6+2

= x^{4} - 6x^{3}+10x^{2} - 3x

Question:4 Show that the function f : R \rightarrow \{x \in R : - 1 < x < 1\} defined by f(x) = \frac{x}{1 + |x|} x \in R is one one and onto function.

Answer:

The function f : R \rightarrow \{x \in R : - 1 < x < 1\} defined by

f(x) = \frac{x}{1 + |x|} , x \in R

One- one:

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

\frac{x}{1+\left | x \right |}=\frac{y}{1+\left | y \right |}

It is observed that if x is positive and y is negative.

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

Since x is positive and y is negative.

x> y\Rightarrow x-y> 0 but 2xy is negative.

x-y\neq 2xy

Thus, the case of x is positive and y is negative is removed.

Same happens in the case of y is positive and x is negative so this case is also removed.

When x and y both are positive:

f(x)=f(y)

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

x(1+y)=y(1+x)

x+xy=y+xy

x=y

When x and y both are negative : f(x)=f(y)

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

x(1-y)=y(1-x)

x-xy=y-xy

x=y

\therefore f is one-one.

Onto:

Let y \in R such that -1< y< 1

If y is negative, then x= \frac{y}{y+1} \in R

f(x)=f(\frac{y}{y+1} )= \frac{\frac{y}{1+y}}{1+ | \frac{y}{1+y}|} = \frac{\frac{y}{1+y}}{1+ \frac{-y}{1+y}}=\frac{y}{1+y-y}=y

If y is positive, then x= \frac{y}{1-y} \in R

f(x)=f(\frac{y}{1-y} )= \frac{\frac{y}{1-y}}{1+ | \frac{y}{1-y}|} = \frac{\frac{y}{1-y}}{1+ \frac{-y}{1-y}}=\frac{y}{1-y+y}=y

Thus, f is onto.

Hence, f is one-one and onto.

Question:5 Show that the function f : R \rightarrow R given by f (x) = x ^3 is injective.

Answer:

f : R \rightarrow R

f (x) = x ^3

One-one:

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

x^{3}=y^{3}

We need to prove x=y .So,

  • Let x\neq y then there cubes will not be equal i.e. x^{3}\neq y^{3} .

  • It will contradict given condition of cubes being equal.

  • Hence, x=y and it is one -one which means it is injective

Question:6 Give examples of two functions f : N \rightarrow Z and g : Z \rightarrow Z such that gof is injective but g is not injective. (Hint : Consider f (x) = x and g (x) = | x | ).

Answer:

f : N \rightarrow Z

g : Z \rightarrow Z

f (x) = x

g (x) = | x |

One - one:

Since g (x) = | x |

f(1)=\left | 1 \right | = 1

f(-1)=\left |- 1 \right | = 1

As we can see f(1)=f(-1) but 1\neq -1 so g(x) is not one-one.

Thus , g(x) is not injective.

gof : N \rightarrow Z

gof(x) = g(f(x)) = g(x) = \left | x \right |

Let gof(x)=gof(y)\, \, \, \, \, x,y \in N

\left | x \right |=\left | y \right |

Since, x,y \in N so x and y are both positive.

\therefore x=y

Hence, gof is injective.

Question:7 Give examples of two functions f : N\rightarrow N and g : N\rightarrow N such that gof is onto but f is not onto.

(Hint : Consider f(x) = x + 1 and g(x) = \left\{\begin{matrix} x -1 & if x > 1\\ 1 & if x = 1 \end{matrix}\right.

Answer:

f : N\rightarrow N and g : N\rightarrow N

f(x) = x + 1 and g(x) = \left\{\begin{matrix} x -1 & if x > 1\\ 1 & if x = 1 \end{matrix}\right.

Onto :

f(x) = x + 1

Consider element in codomain N . It is clear that this element is not an image of any of element in domain N .

\therefore f is not onto.

gof : N\rightarrow N

gof(x)= g(f(x))= g(x+1)= x+1-1 =x \, \, \, \, \, \, \, \, \, since\, x \in N\Rightarrow x+1> 1

Now, it is clear that y \in N , there exists x=y \in N such that gof(x)=y .

Hence, gof is onto.

Question:8 Given a non empty set X, consider P(X) which is the set of all subsets of X. Define the relation R in P(X) as follows: For subsets A, B in P(X), ARB if and only if A \subset B . Is R an equivalence relation on P(X)? Justify your answer.

Answer:

Given a non empty set X, consider P(X) which is the set of all subsets of X.

Since, every set is subset of itself , ARA for all A \in P(x)

\therefore R is reflexive.

Let ARB \Rightarrow A\subset B

This is not same as B\subset A

If A =\left \{ 0,1 \right \} and B =\left \{ 0,1,2 \right \}

then we cannot say that B is related to A.

\therefore R is not symmetric.

If ARB \, \, \, and \, \, \, BRC, \, \, then \, \, A\subset B \, \, \, and \, \, B\subset C

this implies A\subset C = ARC

\therefore R is transitive.

Thus, R is not an equivalence relation because it is not symmetric.

Question:9 Given a non-empty set X, consider the binary operation * : P(X) \times P(X) \rightarrow P(X) given by A * B = A \cap B\;\; \forall A ,B\in P(X) , where P(X) is the power set of X. Show that X is the identity element for this operation and X is the only invertible element in P(X) with respect to the operation ∗.

Answer:

Given * : P(X) \times P(X) \rightarrow P(X) is defined as A * B = A \cap B\;\; \forall A ,B\in P(X) .

As we know that A \cap X=A=X\cap A\forall A \in P(X)

\Rightarrow A*X =A=X*A \forall A in P(X)

Hence, X is the identity element of binary operation *.

Now, an element A \in P(X) is invertible if there exists a B \in P(X) ,

such that A*B=X=B*A (X is identity element)

i.e. A\cap B=X=B\cap A

This is possible only if A=X=B .

Hence, X is only invertible element in P(X) with respect to operation *

Question:10 Find the number of all onto functions from the set \{1, 2, 3, ... , n\} to itself.

Answer:

The number of all onto functions from the set \{1, 2, 3, ... , n\} to itself is permutations on n symbols 1,2,3,4,5...............n.

Hence, permutations on n symbols 1,2,3,4,5...............n = n

Thus, total number of all onto maps from the set \{1, 2, 3, ... , n\} to itself is same as permutations on n symbols 1,2,3,4,5...............n which is n.

Question:11(i) Let S = \{a, b, c\} and T = \{1, 2, 3\} . Find F^{-1} of the following functions F from S to T, if it exists.

(i) F = \{(a, 3), (b, 2), (c, 1)\}

Answer:

F:S\rightarrow T

F : \{a, b, c\}\rightarrow \left \{ 1,2,3 \right \} is defined as F = \{(a, 3), (b, 2), (c, 1)\}

F(a)=3,F(b)=2,F(c)=1

\therefore \, \, F^{-1}:T\rightarrow S is given by

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

Question:11(ii) Let S = \{a, b, c\} and T = \{1, 2, 3\} . Find F^{-1} of the following functions F from S to T, if it exists.

(ii) F = \{(a, 2), (b, 1), (c, 1)\}

Answer:

F:S\rightarrow T

F : \{a, b, c\}\rightarrow \left \{ 1,2,3 \right \} is defined as F = \{(a, 2), (b, 1), (c, 1)\}

F(a)=2,F(b)=1,F(c)=1 , F is not one-one.

So inverse of F does not exists.

Hence, F is not invertible i.e. F^{-1} does not exists.

Question:12 Consider the binary operations * : R \times R \rightarrow R and \circ : R \times R \rightarrow R defined as a *b = |a - b| and a \circ b = a \;\forall a \in R . Show that ∗ is commutative but not associative, \circ is associative but not commutative. Further, show that \forall a,b,c \in R , a*(b\circ c) = (a*b)\circ (a*c) . [If it is so, we say that the operation ∗ distributes over the operation \circ ]. Does \circ distribute over ∗? Justify your answer.

Answer:

Given * : R \times R \rightarrow R and \circ : R \times R \rightarrow R is defined as
a *b = |a - b| and a \circ b = a \;\forall a,b \in R

For a,b \in R , we have

a *b = |a - b|

b *a = |b - a| = \left | -(a-b) \right |=\left | a-b \right |

\therefore a*b = b *a

\therefore the operation is commutative.

(1*2)*3 = (\left | 1-2 \right |)*3=1*3=\left | 1-3 \right |=2

1*(2*3) = 1*(\left | 2-3 \right |)=1*1=\left | 1-1 \right |=0

\therefore (1*2)*3\neq 1*(2*3) where 1,2,3 \in R

\therefore the operation is not associative

Let a,b,c \in R . Then we have :

a*(b \circ c) = a *b =\left | a-b \right |

(a*b )\circ(a* c) = \left | a-b \right | \circ \left | a-c \right | = \left | a-b \right |

Hence, a*(b \circ c)=(a*b )\circ(a* c)

Now,

1\circ (2*3) = 1\circ(\left | 2-3 \right |)=1\circ1=1

(1\circ 2)*(1 \circ 3) =1*1=\left | 1-1 \right |=0

\therefore 1 \circ(2*3)\neq (1\circ 2)*(1 \circ 3) for 1,2,3 \in R

Hence, operation o does not distribute over operation *.

Question:13 Given a non-empty set X, let ∗ : P(X) × P(X) → P(X) be defined as A * B = (A – B) ∪ (B – A), ∀ A, B ∈ P(X). Show that the empty set φ is the identity for the operation ∗ and all the elements A of P(X) are invertible with A–1 = A. (Hint : (A – φ) ∪ (φ – A) = A and (A – A) ∪ (A – A) = A ∗ A = φ).

Answer:

It is given that

*: P(X) \times P(X) \rightarrow P(X) be defined as

A * B = (A - B) \cup (B - A), A, B \ \epsilon \ P(X).

Now, let A \ \epsilon \ P(X) .
Then,

A * \phi = (A - \phi) \cup (\phi - A) = A \cup \phi = A
And

\phi *A = (\phi - A) \cup (A - \phi) = \phi \cup A = A

Therefore,

A * \phi = \phi *A = A , A \ \epsilon \ P(X)

Therefore, we can say that \phi is the identity element for the given operation *.

Now, an element A \epsilon P(X) will be invertible if there exists B \epsilon P(X) such that

A*B = \phi = B*A \ \ \ \ \ \ \ \ \ \ \ \ (\because \phi \ is \ an \ identity \ element)

Now, We can see that

???????A * A = (A -A) \cup (A - A) = \phi \cup \phi = \phi , A \ \epsilon \ P(X). A * A = \left ( A - A \right ) \cup \left ( A-A \right ) = \phi \cup \phi = \phi , such that A \ \epsilon \ P(X)

Therefore, by this we can say that all the element A of P(X) are invertible with A^{-1}= A

Question:14 Define a binary operation ∗ on the set \{0, 1, 2, 3, 4, 5\} as a * b = \left\{\begin{matrix} a + b &if\;a+b < 6 \\ a+ b -6 & if\;a+b\geq6 \end{matrix}\right. Show that zero is the identity for this operation and each element a\neq 0 of the set is invertible with 6 - a being the inverse of a .

Answer:

X = \{0, 1, 2, 3, 4, 5\} as

a * b = \left\{\begin{matrix} a + b &if\;a+b < 6 \\ a+ b -6 & if\;a+b\geq6 \end{matrix}\right.

An element c \in X is identity element for operation *, if a*c=a=c*a \, \, \forall \, \, a \in X

For a \in X ,

a *0 = a+0=a\, \, \, \, \, \, \, \, \, \, \, \, \left [ a \in X \Rightarrow a+0< 6 \right ]

0 *a = 0+a=a\, \, \, \, \, \, \, \, \, \, \, \, \left [ a \in X \Rightarrow a+0< 6 \right ]

\therefore \, \, \, \, \, \, \, \, \, a*0=a=0 *a \forall a \in X

Hence, 0 is identity element of operation *.

An element a \in X is invertible if there exists b \in X ,

such that a*b=0=b*a i.e. \left\{\begin{matrix} a + b =0=b+a &if\;a+b < 6 \\ a+ b -6=0=b+a-6 & if\;a+b\geq6\end{matrix}\right.

means a=-b or b=6-a

But since we have X = \{0, 1, 2, 3, 4, 5\} and a,b \in X . Then a\neq -b .

\therefore b=a-x is inverse of a for a \in X .

Hence, inverse of element a \in X , a\neq 0 is 6-a i.e. , a^{-1} = 6-a

Question:15 Let A = \{- 1, 0, 1, 2\} , B = \{- 4, - 2, 0, 2\} and f, g : A \rightarrow B be functions defined by f(x) = x^2 -x, x\in A and g(x) = 2\left |x - \frac{1}{2} \right | - 1, x\in A . Are f and g equal? Justify your answer. (Hint: One may note that two functions f : A \rightarrow B and g : A \rightarrow B such that f (a) = g (a) \;\forall a \in A , are called equal functions).

Answer:

Given :

A = \{- 1, 0, 1, 2\} , B = \{- 4, - 2, 0, 2\}

f, g : A \rightarrow B are defined by f(x) = x^2 -x, x\in A and g(x) = 2\left |x - \frac{1}{2} \right | - 1, x\in A .