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NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions

NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions

Edited By Ramraj Saini | Updated on Sep 12, 2023 09:06 PM IST | #CBSE Class 12th

Relations And Functions Class 12 Questions And Answers

Relations and Functions class 12 solutions are provided here. These NCERT solutions are created by expert team at careers360 keeping in mind of latest syllabus of CBSE 2023-24. 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. 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.

This Story also Contains
  1. Relations And Functions Class 12 Questions And Answers
  2. Relations and Functions Class 12 Questions And Answers PDF Free Download
  3. Relations and Functions Class 12 Solutions - Important Formulae
  4. Relations and Functions Class 12 NCERT Solutions (Intext Questions and Exercise)
  5. NCERT Exemplar Class 12th Maths Solutions
  6. Class 12 Maths chapter 1 ncert solutions - Topics
  7. Class 12 maths chapter 1 NCERT Solutions - Chapter wise
  8. Class 12 maths chapter 1 NCERT solutions - Subject Wise
  9. NCERT Solutions Class wise
  10. NCERT Books and NCERT Syllabus
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions

In Relations and Functions class 12 maths chapter 1 question answer, 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.

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Relations and Functions Class 12 Questions And Answers PDF Free Download

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Relations and Functions Class 12 Solutions - Important Formulae

>> Relations: A relation R is a subset of the cartesian product of A × B, where A and B are non-empty sets.

R-1, the inverse of relation R, is defined as R-1 = {(b, a) : (a, b) ∈ R}

Domain of R = Range of R-1

Range of R = Domain of R-1

>> Functions: A relation f from set A to set B is a function if every element in A has one and only one image in B.

A × B = {(a, b): a ϵ A, b ϵ B}

If (a, b) = (x, y), then a = x and b = y

n(A × B) = n(A) * n(B), where n(A) is the cardinality of set A.

A × ϕ = ϕ (where ϕ is the empty set)

A function f: A → B is denoted as f(x) = y.

Algebra of functions:

  • (f + g)(x) = f(x) + g(x)

  • (f - g)(x) = f(x) - g(x)

  • (f * g)(x) = f(x) * g(x)

  • (kf)(x) = k * f(x), where k is a real number

  • {f/g}(x) = f(x)/g(x), where g(x) ≠ 0

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Free download NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions for CBSE Exam.

Relations and Functions Class 12 NCERT Solutions (Intext Questions and Exercise)

NCERT Solutions for Class 12 relations and functions NCERT solutions: 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.


Relations and Functions Class 12 NCERT Solutions: 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 .


Relation and Function Class 12 maths chapter 1 question answer: 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.


Class 12 maths chapter 1 NCERT solutions: 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 .

It can be observed that

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

g(-1) = 2\left |-1 - \frac{1}{2} \right | - 1= 2\left | \frac{-3}{2} \right |-1=3-1=2

f(-1)=g(-1)

f(0) = (0)^2 -(0)=0+0=0

g(0) = 2\left |0 - \frac{1}{2} \right | - 1= 2\left | \frac{-1}{2} \right |-1=1-1=0

f(0)=g(0)

f(1) = (1)^2 -(1)=1-1=0

g(1) = 2\left |1 - \frac{1}{2} \right | - 1= 2\left | \frac{1}{2} \right |-1=1-1=0

f(1)=g(1)

f(2) = (2)^2 -(2)=4-2=2

g(2) = 2\left |2 - \frac{1}{2} \right | - 1= 2\left | \frac{3}{2} \right |-1=3-1=2

f(2)=g(2)

\therefore \, \, \, f(a)=g(a) \forall a\in A

Hence, f and g are equal functions.

Question:16 Let A = \{1, 2, 3\} . Then number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is

(A) 1
(B) 2
(C) 3
(D) 4

Answer:

A = \{1, 2, 3\}

The smallest relations containing (1, 2) and (1, 3) which are
reflexive and symmetric but not transitive is given by

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

(1,1),(2,2),(3,3) \in R , so relation R is reflexive.

(1,2),(2,1) \in R and (1,3),(3,1) \in R , so relation R is symmetric.

(2,1),(1,3) \in R but (2,3) \notin R , so realation R is not transitive.

Now, if we add any two pairs (2,3) and (3,2) to relation R, then relation R will become transitive.

Hence, the total number of the desired relation is one.

Thus, option A is correct.

Question:17 Let A = \{1, 2, 3\} . Then number of equivalence relations containing (1, 2) is
(A) 1
(B) 2
(C) 3
(D) 4

Answer:

A = \{1, 2, 3\}

The number of equivalence relations containing (1, 2) is given by

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

We are left with four pairs (2,3) , (3,2) , (1,3),(3,1) .

(1,1),(2,2),(3,3) \in R , so relation R is reflexive.

(1,2),(2,1) \in R and (2,3),(3,2) \notin R , so relation R is not symmetric.

(1,3),(3,1) \notin R , so realation R is not transitive.

Hence , equivalence relation is bigger than R is universal relation.

Thus the total number of equivalence relations cotaining (1,2) is two.

Thus, option B is correct.

Question:18 Let f : R \rightarrow R be the Signum Function defined as f(x) = \left\{\begin{matrix} 1 & x> 0 \\ 0 &x = 0 \\ -1& x < 0 \end{matrix}\right. and g : R \rightarrow R be the Greatest Integer Function given by g (x) = [x] , where [x] is greatest integer less than or equal to x . Then, does fog and gof coincide in (0, 1] ?

Answer:

f : R \rightarrow R is defined as f(x) = \left\{\begin{matrix} 1 & x> 0 \\ 0 &x = 0 \\ -1& x < 0 \end{matrix}\right.

g : R \rightarrow R is defined as g (x) = [x]

Let x \in (0, 1]

Then we have , [1]=1 if x=1 and [x]=0\, \, \, \, \, if\, 0< x< 1

1627294208851

\therefore \, \, gof(x)=g(f(x))=g(1)=\left [ 1 \right ]=1\, \, \, \,(since \, \, x> 0)

Hence,for x \in (0, 1] , fog(x)=0 and gof(x)=1 .

Hence , gof and fog do not coincide with (0, 1] .

Question:19 Number of binary operations on the set are(A) 10(B) 16(C) 20(D ) 8

Answer:

Binary operations on the set \{a, b\} are is a function from \{a, b\}\times \{a, b\}\rightarrow \{a, b\}

i.e. * is a function from \left \{ (a,b),(b,a),(a,a),(b,b) \right \} \rightarrow \{a, b\}

Hence, the total number of binary operations on set \{a, b\} is 2^{4}=16.

Hence, option B is correct

If you are looking for relation and function class 12 ncert solutions of exercises then these are listed below.

Relation And Function Class 12 Ncert Solutions Exercise 1.1

Relation And Function Class 12 Ncert Solutions Exercise 1.2

Relation And Function Class 12 Ncert Solutions Exercise 1.3

Relation And Function Class 12 Ncert Solutions Exercise 1.4

Relation And Function Class 12 Ncert Solutions Miscellaneous Exercise

We have relations like father, mother, brother, sister, husband, wife. Relation becomes a function when there is only one output for every input. In NCERT class 11 maths solutions 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 Class 12 maths chapter 1 question answer, you will learn about different types of relations and functions, invertible functions, the composition of functions, and binary operations. Also you can find ncert solutions for class 12 chapter 1 by careers360 expert team.

Concepts of this chapter 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 maths syllabus includes two chapters i.e. relation and function, and inverse trigonometry which together has 10 % weightage in the CBSE class 12th final examination. So, you should study class 12 maths ch 1 question answer carefully, and solve every question on your own including solved examples.

NCERT Exemplar Class 12th Maths Solutions

NCERT Exemplar Solutions Chapter 1 - Relations and Functions

NCERT Exemplar Solutions Chapter 2 - Inverse Trigonometric Functions

NCERT Exemplar Solutions Chapter 3 - Matrices

NCERT Exemplar Solutions Chapter 4 - Determinants

NCERT Exemplar Solutions Chapter 5 - Continuity and Differentiability

NCERT Exemplar Solutions Chapter 6 - Application of Derivatives

NCERT Exemplar Solutions Chapter 7 - Integrals

NCERT Exemplar Solutions Chapter 8 - Application of Integrals

NCERT Exemplar Solutions Chapter 9 - Differential Equations

NCERT Exemplar Solutions Chapter 10 - Vector Algebra

NCERT Exemplar Solutions Chapter 11 - Three Dimensional Geometry

NCERT Exemplar Solutions Chapter 12 - Linear Programming

NCERT Exemplar Solutions Chapter 13 - Probability

These Relations and Functions class 12 NCERT solutions 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.

Class 12 Maths chapter 1 ncert solutions - Topics

The main topics covered in chapter 1 maths class 12 are:

  • Type of Relation
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In this ch 1 maths class 12 topics discuss different types of relations namely reflexive, symmetric, and transitive. we also study the concept of empty relation, universal relation, trivial relation, and equivalence relation in the chapter relations and functions. there are good quality questions in functions and relations class 12 solutions.

  • type of functions

This ch 1 maths class 12 concerns different types of functions like constant function, polynomial function, identity function, rational function, modulus function, signum function, etc. this chapter also contains the concept of one-one (or injective), onto (or subjective), one-one and onto (or bijective) functions. The concept of addition, subtraction, multiplication, and division of two functions have also been discussed. to get command on these concepts you can refer to NCERT solutions for class 12 maths chapter 1.

  • composition of function and Invertible function

we understand the concept of composition of a function in this chapter of class 12 NCERT. also we get a good hold of invertible functions concepts in this chapter. for questions, you can browse class 12 NCERT solutions.

  • Binary operations

this ch 1 maths class 12 also includes concepts of binary operations. terms like commutative, associative invertible, inverse, identity are also discussed in class 12 NCERT. you can refer to class 12 NCERT solutions for questions about these concepts.

Topics enumerated in class 12 NCERT are very important and students are advised to go through all the concepts discussed in the topics. Questions related to all the above topics are covered in the class 12 maths ch 1 question answer.

Also read,

Class 12 maths chapter 1 NCERT Solutions - Chapter wise

Chapter 1
Relations And Functions

Chapter 2

Inverse Trigonometric Functions

Chapter 3

Matrices

Chapter 4

Determinants

Chapter 5

Differentiability

Chapter 6

Application of Derivatives

Chapter 7

Integrals

Chapter 8

Application of Integrals

Chapter 9

Differential Equations

Chapter 10

Vector Algebra

Chapter 11

Three Dimensional Geometry

Chapter 12

Linear Programming

Chapter 13

Probability

Class 12 maths chapter 1 NCERT solutions - Subject Wise

NCERT Solutions Class wise

NCERT Books and NCERT Syllabus

Guidelines to use NCERT Class 12 solutions Maths Chapter 1

Class 12 relations and functions NCERT solutions is helpful for the students who wish to perform well in the CBSE 12 board examination. Some guidelines to follow to make the best use of NCERT solutions:

  • Before solving an exercise, first, go through the examples that are given in the NCERT class 12 maths textbook.
  • Also, try to solve every exercise including miscellaneous exercise, NCERT chapter examples, miscellaneous examples on your own, if you are not able to do it, then you can take help of NCERT solutions for class 12 maths chapter 1 Relations and functions.
  • Reading the solutions is not enough, you have to solve it on your own, even after reading the solutions
  • Stick to the syllabus that is provided by NCERT and solve it completely including all examples and all exercises
  • If you have solved all NCERT questions, then you can solve previous years papers of CBSE board to get familiar with the pattern of the exam.
  • NCERT solutions for class 12 maths chapter 1 pdf download will also be made available soon.

Happy learning !!!

Frequently Asked Question (FAQs)

1. How the NCERT solutions are helpful in the board exam ?

As CBSE board exam paper is designed entirelly based on NCERT textbooks and most of the questions in CBSE board exam are directly asked from NCERT textbook, students must know the relations and functions class 12 questions and answers very well to perform well in the exam. NCERT solutions are not only important when you stuck while solving the problems but students will get how to answer in the board exam in order to get good marks in the board exam.

2. What are the important topics in chapter relations and functions?

Definitions of relations and functions, types of relations, types of functions, composition of functions, invertible function and binary operations are the important topics in this maths chapter 1 class 12. these topics are important because concepts are used in calculus and other topics as well as exams therefore students are recommended ncert solutions and ncert exercise to get command on the concepts.

3. What concepts are covered in the class 12 maths chapter 1 NCERT solutions?

The NCERT Solutions for maths chapter 1 class 12 provide in-depth explanations of several important concepts, including types of relations, different types of functions, composition of functions, invertible functions, and binary operations. These solutions  for class 12 maths ch 1 are created by a team of highly qualified and experienced teachers and their primary goal is to assist students in achieving a high score on their Class 12 Maths board exams.

4. What are the key attributes of chapter 1 maths class 12 ncert solutions?

 Following are some key attributes of relation and function class 12 solutions.

  • The maths chapter 1 class 12 are created by experienced subject matter experts who possess a deep understanding of the key concepts. 

  • The solutions for relation function class 12 are presented in a clear and straightforward language to make it easy for students to grasp the methods for solving complex problems.

  • The step-by-step solutions  for class 12 relation and function are designed based on the marks weightage assigned by the CBSE exam, ensuring that students can maximize their scores on the exam.

5. Is relying solely on NCERT Solutions for Class 12 Maths Chapter 1 sufficient for preparing for the CBSE exams?

The Maths class 12 relations and functions ncert solutions are updated with the latest CBSE guidelines, ensuring that all the important topics are covered. The chapter contains four exercises, offering students a variety of problems to solve on their own. The class 12 maths ch 1 ncert solutions are structured to build confidence in students ahead of the CBSE exams. For ease, students can study relations and functions class 12 ncert pdf online and offline in both modes.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?
I am interested in graphic designing and coding. how do I make a career .which stream should I go.what engineering subject to take

Hello aspirant,

The purpose of graphic design extends beyond the brand's look. Nevertheless, by conveying what the brand stands for, it significantly aids in the development of a sense of understanding between a company and its audience. The future in the field of graphic designing is very promising.

There are various courses available for graphic designing. To know more information about these courses and much more details, you can visit our website by clicking on the link given below.

https://www.careers360.com/courses/graphic-designing-course

Thank you

Hope this information helps you.

Read Complete Answer
Sajal Trivedi 29 January,2024
i couldn't clear 1st compartment (which held in August 2022) cbse class 12th so can I give improvement in all subjects which to be held on March 2023. tell me a solution i need more than 75% in class 12th

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

Read Complete Answer
Upasana Barman 24 August,2022
i was not able to clear 1st compartment and now I have to give 2nd compartment so can I give improvement exam next year? plz tell me very much tensed(cbse class 12th)

Hello dear,

If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.


As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.


Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.


Believe in Yourself! You can make anything happen


All the very best.

Read Complete Answer
Mayankjha.student2007 23 August,2022
if i am giving improvement in one subject this year, will i be able to give improvement in more subjects next year?

Hello Student,

I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects  and  we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

All the best.

Read Complete Answer
Nuthalapati Vyshnavi 21 August,2022
I got an RT for 2 subjects, so do I have to give exams for all the subjects in the coming year or just the 2. pls help me

Hi,

You just need to give the exams for the concerned two subjects in which you have got RT. There is no need to give exam for all of your subjects, you can just fill the form for the two subjects only.

Read Complete Answer
Muskan Dhalwal 17 August,2022
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Popular CBSE Class 12th Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

 Option 2)

0.16\; J
Option 3)

1.00\; J

 Option 4)

0.67\; J
Read More

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

 Option 2)

 6.45×10−3 kg
Option 3)

 9.89×10−3 kg

 Option 4)

12.89×10−3 kg

 
Read More

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

 Option 2)

200 \, \, J - 500 \, \, J
Option 3)

2\times 10^{5}J-3\times 10^{5}J

 Option 4)

20,000 \, \, J - 50,000 \, \, J
Read More

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

 Option 2)

\; K\;
Option 3)

zero\;

 Option 4)

K/4
Read More

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

 Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced
Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

 Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .
Read More

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

 Option 2)

3.125 × 10-2
Option 3)

1.25 × 10-2

 Option 4)

2.5 × 10-2
Read More

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

 Option 2)

increase two fold
Option 3)

remain unchanged

 Option 4)

be a function of the molecular mass of the substance.
Read More

With increase of temperature, which of these changes?

Option 1)

Molality

 Option 2)

Weight fraction of solute
Option 3)

Fraction of solute present in water

 Option 4)

Mole fraction.
Read More

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

 Option 2)

6.023 × 1022
Option 3)

half that in 8 g He

 Option 4)

558.5 × 6.023 × 1023
Read More

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

 Option 2)

more than 3 but less than 6
Option 3)

more than 6 but less than 9

 Option 4)

more than 9
Read More

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