# NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2 (Ex 1.2)

The Central Board of Secondary Education (CBSE) conducts the Class 12 examination every year. The students of Class 12 need to prepare well to score good marks in board exams. All the subjects in Class 12 are important and require thorough understanding and practice on a regular basis.

Mathematics can sometimes be challenging to understand, and students may come across difficulties while studying the Mathematics syllabus. It is essential to understand the concepts covered in detail and practice textbook exercises regularly. The NCERT syllabus needs to be referred for making a strategy for Class Mathematics.

Class 12 Mathematics is one of the critical subjects for getting admission into engineering colleges and preparing for various competitive exams like JEE Mains, Staff Selection Commission (SSC) exams, etc. Mathematics is crucial for students who want to become statisticians, scientists, mathematicians, etc.

The topic for Chapter 1 in Mathematics is Relations and Functions. It is one of the essential chapters from the exam point of view and Class 12 students need to prepare all the topics carefully. The NCERT textbook is the main source for learning Mathematics in Class 12. All the questions given in Chapter 1 of the NCERT textbook Mathematics are relevant for preparing students for the board exams.

Students must practice all the questions given in Exercise 1.2 and improve their understanding of the concerns of Chapter 1. It can be tough for students to solve Exercise 1.2 questions on their own, so they need to access NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2 from credible sources. Extramarks provides very authentic NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2 in PDF format. Students will find it easier to solve Exercise 1.2 questions with the help of expert-given solutions. The NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2 can also be accessed on Extramarks’ mobile application. These solutions are given by expert Mathematics teachers and are easy to understand. In a subject like Mathematics, it is critical to analyse one’s preparation level and improve methods of practising questions. Some questions can be solved in different ways, and it is necessary to get used to the accurate methods of solving them.

Students are advised to make use of NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2 to find the most efficient way to solve Exercise 1.2 questions.

Mathematics is a critical subject for getting higher marks on the Class 12 board exam. Students are advised to make use of NCERT solutions to get familiar with the technique of solving questions from the exam’s perspective. The NCERT textbook has so many questions related to each topic that is sufficient for preparing for the board exams, and students must practice them regularly. It is advisable to refer to NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2 while solving the questions of Class 12 Maths Chapter 1 Exercise 1.2. The questions asked in Exercise 1.2 are based on the topic of Types of Functions. It is necessary to understand the contents of the Types of Functions topic to solve Exercise 1.2 questions. This is available on the Extramarks’ website and mobile application. Practising with the help of NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2 is beneficial for preparing accurately for the board exam of Mathematics, and it also saves a student’s time.

The formulas and definitions learned in the chapters are to be applied correctly to get desired results in exercise problems. Students are advised to keep revising definitions given in the Types of Functions chapter to solve Exercise 1.2 questions.

Question 2 in Exercise 1.2 is related to checking the Injectivity and Subjectivity of given functions. To find the best solution to question 2 of Exercise 1.2 Class 12, students can refer to NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2. Some questions of Exercise 1.2 are a bit tricky to solve and require in-depth knowledge of concepts taught in  the chapter Types of Functions. The NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2 consist of solutions for each of the twelve questions asked in Exercise 1.2.

## NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2 (Ex 1.2)

Class 12 students of the Central Board of Secondary Education (CBSE) looking for the NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2 can access it on the Extramarks’ website and mobile application. All the study material provided by Extramarks is designed by expert teachers of each subject and are upgraded from time to time according to the changes in the syllabus.

The best way to practice questions given in Exercise 1.2, is by accessing NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2. All the questions of Class 12th Exercise 1.2 are to be practised regularly while preparing for the Class 12 Mathematics board exam. The Class 12 Maths NCERT Solutions Chapter 1 Exercise 1.2 helps give ideas about how to solve questions easily and effectively. The Extramarks website and mobile application have NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2 to help students in understanding the requirements of Exercise 1.2 questions and answering them accordingly. The Central Board of Secondary Education (CBSE) syllabus for Class 12 Mathematics is large and needs extra focus. It is necessary to access NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2 to guide students to prepare effectively for the Class 12 Mathematics board exam. The NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2 is available in PDF format on the Extramarks’ website and mobile application. The solutions given in NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2 are of high quality and are designed by expert Mathematics teachers with absolute precision. The students need to solve all the Exercise 1.2 questions one by one using NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2.

It is important to revise all the past topics given in Chapter 1 before trying to solve Exercise 1.2.

It is possible that many of the Class 12 students feel stuck while solving exercise questions in Mathematics subject. Class 12 Mathematics Chapters can be thoroughly understood when students focus on solving exercise questions. The Relations and Functions Chapter comprises four exercises and one miscellaneous exercise. Exercise 1.2 questions are to be solved for having a better understanding of the Types of Functions chapter. Accessing NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2 is helpful in self-assessment of the ongoing board exam preparation of the Mathematics subject. All the difficulties that students are facing in Exercise 1.2 can be solved with the help of Class 12 Maths NCERT Solutions Chapter 1 Exercise 1.2.

### Important Topics

Relations and Functions is one of the most important topics in the Class 12 Central Board of Secondary Education Mathematics syllabus. The topics covered under the Relations and Functions Chapter contain some significant questions that may be asked in the board examination. When solving Exercise 1.2 questions, it is necessary to use the NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2 for easier practice. Practising with NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2 helps improve the understanding of students in the Relations and Functions Chapter.

To prepare well for the Class 12 Mathematics CBSE syllabus, it is essential to thoroughly understand Chapter 1 topics and keep practising exercise questions. Relations and Functions topic is an essential chapter from the board exam perspective. The NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2 needs to be referred to for getting detailed and stepwise solutions to Exercise 1.2 problems.

The NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2 can be downloaded in PDF format from the Extramarks’ website and mobile application and can be viewed in offline mode.

### Important Points

Relations and Functions chapter exercise questions have to be solved regularly. Class 12 students can access solutions for Mathematics Chapter 1 from the Extramarks’ website and mobile application.

### NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2

The Chapters included in the Class 12 Mathematics syllabus are quite vast and are important for building a base to pursue a career in various Mathematics fields. Class 12 students need to learn all the topics and have proper revisions in each chapter in the Mathematics syllabus. Students can access NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2 to answer the questions of Exercise 1.2 efficiently. The solutions help improve students’ problem-solving skills and make them feel confident in the Chapter topics. The NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2 is an important resource for Class 12 Central Board of Secondary Education (CBSE) students. Extramarks provides the most reliable and accurate NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2. The solutions given by the Extramarks’ website and mobile application are carefully designed by expert Mathematics teachers.

Extramarks provides NCERT solutions for all the classes of the Central Board of Secondary Education. Students can access subject-wise solutions for all the following classes:

NCERT Solutions Class 12

NCERT Solutions Class 11

NCERT Solutions Class 10

NCERT Solutions Class 9

NCERT Solutions Class 8

NCERT Solutions Class 7

NCERT Solutions Class 6

NCERT Solutions Class 5

NCERT Solutions Class 4

NCERT Solutions Class 3

NCERT Solutions Class 2

NCERT Solutions Class 1

### NCERT Solutions for Class 12 Maths

Students can access solutions for all the chapters of the Class 12 NCERT Mathematics textbook from the Extramarks’ website and mobile application. Students can download the solutions for the exercises of each chapter in PDF format, and they can use them anytime during their exam preparation. Students are advised to download NCERT Solutions Class 12 Maths Chapter 1 Exercise 1.2 from the Extramarks’ website and mobile application to be able to solve Exercise 1.2 questions.

Relations and Functions chapter is the very first chapter in NCERT Class 12 Mathematics textbook. It is very important to practice its questions regularly to have a good start to Mathematics board exam preparations. Exercise 1.2 is one of the significant exercises of Chapter 1, and it should be practised along with the NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2. When going through the NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2, students will understand how to fulfil the requirements of each question efficiently.

#### Chapter 1 – Relations and Functions Exercises

Class 12 students belonging to the Central Board of Secondary Education (CBSE) can access exercise-wise NCERT Solutions for each chapter of Mathematics.

When solving Exercise questions given in the NCERT textbook of Class 12 Mathematics, it is critical to understand the concepts well. Revising concepts along with the questions helps in scoring well on the board exam of Mathematics. The syllabus is to be covered systemically and exercises are to be solved regularly so that concepts become strong.

The Relations and Functions chapter consists of topics like Types of Relations, Types of Functions, Composition of Functions and Invertible Functions and Binary Operations. All these topics are supposed to be thoroughly revised with the proper practice of exercise questions. It is recommended to use NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2 when solving Exercise 1.2 questions. Class 12 CBSE students can access the NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.2 from the Extramarks’ website and mobile application.

### NCERT Solutions for Class 12 Maths Chapter 1 Exercises

 Chapter 1 Relations And Functions Other Exercises Exercise 1.1 16 Questions And Solutions (3 Short Answer, 13 Long Answers) Exercise 1.3 14 Questions And Solutions (4 Short Answer, 10 Long Answers) Exercise 1.4 13 Questions And Solutions (6 Short Answer, 7 Long Answers)

Q.1

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{x},\mathrm{y}\in {\mathrm{R}}_{\mathrm{*}}\mathrm{ }\mathrm{such}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{y}\right)\\ \mathrm{one}-\mathrm{one}:\\ \mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{y}\right)\\ ⇒ \frac{1}{\mathrm{x}}=\frac{1}{\mathrm{y}}\\ ⇒ \mathrm{x}=\mathrm{y}\\ \mathrm{So}, \mathrm{f}:{\mathrm{R}}_{\mathrm{*}}\to {\mathrm{R}}_{\mathrm{*}}\mathrm{ }\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{.}\\ \mathrm{Onto}:\end{array}$

Letf( x )=ywhere y be any element of R * ( Codomain )

$\begin{array}{l}⇒ \frac{1}{\mathrm{x}}=\mathrm{y} \mathrm{or} \mathrm{x}=\frac{1}{\mathrm{y}}\\ \mathrm{Now},\mathrm{ }\mathrm{f}\left(\frac{1}{\mathrm{y}}\right)=\frac{1}{\left(\frac{1}{\mathrm{y}}\right)}=\mathrm{y}\\ \mathrm{Since}\mathrm{Range}=\mathrm{Co}–\mathrm{domain}\\ \therefore \mathrm{f}\mathrm{is}\mathrm{onto}\mathrm{.}\\ \mathrm{Thus},\mathrm{the}\mathrm{given}\mathrm{function}\left(\mathrm{f}\right)\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{and}\mathrm{onto}\mathrm{.}\\ \mathrm{Consider}\mathrm{function}\mathrm{f}:\mathrm{N}\to {\mathrm{R}}_{\mathrm{*}}\mathrm{ }\mathrm{defined}\mathrm{by}\\ \mathrm{f}\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}}\\ \mathrm{For}\mathrm{any}\mathrm{x},\mathrm{y}\in \mathrm{N}\mathrm{we}\mathrm{see}\mathrm{that}\\ \mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{y}\right)\\ ⇒ \mathrm{ }\frac{1}{\mathrm{x}}=\frac{1}{\mathrm{y}}\\ ⇒ \mathrm{x}=\mathrm{y}\\ \mathrm{So}\mathrm{f}:\mathrm{N}\to {\mathrm{R}}_{\mathrm{*}}\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{function}\\ \mathrm{Since}\mathrm{fractional}\mathrm{numbers}\mathrm{like}\frac{2}{3},\frac{2}{5}\mathrm{ }\mathrm{etc}.\mathrm{ }\mathrm{in}\mathrm{co}–\mathrm{domain}{\mathrm{R}}_{*}\mathrm{ }\mathrm{have}\\ {\mathrm{R}}_{\mathrm{*}}\mathrm{ }\mathrm{have}\mathrm{no}\mathrm{pre}\mathrm{image}\mathrm{in}\mathrm{ }\mathrm{domain}\mathrm{N}.\\ \mathrm{So}, \mathrm{f}:\mathrm{N}\to {\mathrm{R}}_{*}\mathrm{is}\mathrm{not}\mathrm{onto}\mathrm{.}\end{array}$

Q.2 (i) f: N N given by f(x)= x2

$\begin{array}{l}\text{\hspace{0.17em}}\left(ii\right)f:Z\to Z\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}givenbyf\left(x\right)={x}^{2}\\ \left(iii\right)f:R\to R\text{\hspace{0.17em}}\text{\hspace{0.17em}}givenbyf\left(x\right)={x}^{2}\\ \left(iv\right)f:N\to N\text{\hspace{0.17em}}\text{\hspace{0.17em}}givenbyf\left(x\right)={x}^{3}\\ \text{\hspace{0.17em}}\left(v\right)f:Z\to Zgivenbyf\left(x\right)={x}^{3}\end{array}$

Ans

$\begin{array}{l}\text{(i)}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{f:N}\to \text{N is given by,}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{f}\left(x\right)={x}^{2}\\ \text{To check for injectivity let f}\left(x\right)=\text{f}\left(y\right)\\ ⇒\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}^{2}={y}^{2}\\ ⇒\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=y\text{}\left(\begin{array}{l}There\text{are no negative}\\ \text{natural numbers}\text{.}\end{array}\right)\\ \therefore f\left(x\right)\text{is injective}\text{.}\\ \text{To check for onto}:\\ \text{Let}\text{\hspace{0.17em}}\text{f}\left(x\right)\text{=y}\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}^{2}\text{=y}\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{x=}\sqrt{y}\\ \therefore \text{f}\left(\sqrt{y}\right)\text{=y}\\ \text{Since range is not equal to co-domain}\text{.}\\ \text{f(x) is not onto or surjective}\\ \text{(ii) f: Z}\to \text{Z is given by,}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{f(x)}={\text{x}}^{\text{2}}\\ \text{For injectivity let}\text{\hspace{0.17em}}\text{f(x)}=\text{f(y)}\\ ⇒\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}^{2}={y}^{2}\\ ⇒\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=±y\\ \\ \therefore \text{f is not injective}\text{.}\\ \text{For onto:}\\ \text{Let}\text{}\text{}\text{}\text{}\text{(x)}=\text{y}\\ ⇒\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{x}}^{\text{2}}=\text{y}\\ ⇒\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{x}=\sqrt{\text{y}}\\ \therefore \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}f\left(\sqrt{y}\right)=y\\ \text{Hence, function f is neither injective nor surjective}\text{.}\\ \\ \left(iii\right)\text{\hspace{0.17em}}f:R\to R\text{\hspace{0.17em}}\text{is given by, f}\left(x\right)={x}^{2}\\ For\text{injectivity:}\\ \text{}\text{}\text{}\text{}\text{f}\left(x\right)=f\left(y\right)\\ ⇒\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}^{2}={y}^{2}\\ ⇒\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=±y\text{}\text{}\left(\begin{array}{l}x\text{can not take more than one}\\ \text{value for injectivity}\text{.}\end{array}\right)\\ \therefore f\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{is not injective}\text{.}\\ \text{For onto:}\\ \text{Let}\text{}\text{}\text{}\text{}\text{f}\left(x\right)=y⇒{x}^{2}=y\\ ⇒\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=\sqrt{y}\\ \therefore \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}f\left(\sqrt{y}\right)=y\\ \text{Since range is not equal to co-domain}.\\ \therefore f\text{\hspace{0.17em}}\text{is not surjective}\text{.}\\ \left(iv\right)\text{\hspace{0.17em}}f:N\to N\text{\hspace{0.17em}}\text{is given by, f}\left(x\right)={x}^{3}\\ For\text{injectivity:}\\ \text{}\text{}\text{}\text{}\text{f}\left(x\right)=f\left(y\right)\\ \\ ⇒\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}^{3}={y}^{3}\\ ⇒\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=y\text{}\text{}\\ \therefore f\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{is injective}\text{.}\\ \text{For onto:}\\ \text{Let}\text{}\text{}\text{}\text{}\text{f}\left(x\right)=y⇒{x}^{3}=y\\ ⇒\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=\sqrt{y}\\ \therefore \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}f\left(\sqrt{y}\right)=y\text{}\text{}\text{}\left(y\in N\text{\hspace{0.17em}}\text{\hspace{0.17em}}but\text{\hspace{0.17em}}\sqrt{y}\notin N\right)\\ \text{Since range is not equal to co-domain}.\\ \therefore f\text{\hspace{0.17em}}\text{is not surjective}\text{.}\\ \text{Thus, given function is injective but not surjective}\text{.}\\ \\ \left(v\right)\text{\hspace{0.17em}}f:Z\to Z\text{\hspace{0.17em}}\text{is given by, f}\left(x\right)={x}^{3}\\ For\text{injectivity:}\\ \text{}\text{}\text{}\text{}\text{f}\left(x\right)=f\left(y\right)\\ ⇒\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}^{3}={y}^{3}\\ ⇒\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=y\\ \therefore f\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{is injective}\text{.}\\ \text{For onto:}\\ \text{Let}\text{}\text{}\text{}\text{}\text{f}\left(x\right)=y⇒{x}^{3}=y\\ ⇒\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=\sqrt{y}\\ \therefore \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}f\left(\sqrt{y}\right)=y\text{}\text{}\text{}\left(y\in Z\text{\hspace{0.17em}}\text{\hspace{0.17em}}but\text{\hspace{0.17em}}\sqrt{y}\notin Z\right)\\ \text{Since range is not equal to co-domain}.\\ \therefore f\text{\hspace{0.17em}}\text{is not surjective}\text{.}\\ \text{Thus, given function is injective but not surjective}\text{.}\end{array}$

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

Ans

$\begin{array}{l}Since,\\ f:R\to R\text{is given by,}\\ \text{f}\left(x\right)=\left[x\right]\\ We\text{â€‹}\text{see that f}\left(0.1\right)=0,\text{f}\left(0.3\right)=0.\\ \therefore \text{f}\left(0.1\right)=\text{f}\left(0.3\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}but\text{0}\text{.1}\ne \text{0}\text{.3}\\ \text{So, f is not injective i}\text{.e}\text{., one-one}\text{.}\\ \text{Now, let 0}\text{.6}\in \text{R}\text{.}\\ \text{It is given that f}\left(\text{x}\right)\text{=}\left[x\right]\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{is always an integer}\text{. Thus, there does}\\ \text{not exist any element x}\in \text{R such that f}\left(\text{x}\right)\text{=0}\text{.6}\text{.}\\ \text{Then, f is not onto}\text{.}\\ \text{Therefore, the greatest integer function is neither one-one}\\ \text{nor onto}\text{.}\end{array}$

Q.4 Show that the Modul us Function f :RR  given by f(x)=|x|,is neither oneonen or onto, where|x|is  x,ifx  is positiveor 0 and |x| is x, if x  is negative.

Ans

$\begin{array}{l}\mathrm{f}:\mathrm{R}\to \mathrm{R}\mathrm{is}\mathrm{given}\mathrm{by},\\ \mathrm{Modulus}\mathrm{function}\mathrm{is}\mathrm{defined}\mathrm{by}\\ \mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}\mathrm{x},\mathrm{}\mathrm{if}\mathrm{x}\ge \mathrm{0}\\ -\mathrm{x},\mathrm{}\mathrm{if}\mathrm{x}<0\end{array}\\ \mathrm{Here},\mathrm{f}\left(-1\right)=|-1|=1,\mathrm{ }\mathrm{f}\left(1\right)=|1|=1,\mathrm{}\\ \therefore \mathrm{f}\left(-1\right)=\mathrm{f}\left(1\right), \mathrm{ }\mathrm{but}-\mathrm{1}\ne \mathrm{1}\\ \mathrm{So},\mathrm{f}\mathrm{is}\mathrm{not}\mathrm{one}–\mathrm{one}\mathrm{.}\\ \mathrm{Let}\mathrm{}-\mathrm{1}\in \mathrm{R}\mathrm{.}\\ \mathrm{Since},\mathrm{f}\left(\mathrm{x}\right)\mathrm{=}|\mathrm{x}|\mathrm{is}\mathrm{always}\mathrm{positive}.\mathrm{So},\mathrm{there}\mathrm{does}\mathrm{not}\mathrm{exist}\\ \mathrm{any}\mathrm{element}\mathrm{in}\mathrm{x}\mathrm{in}\mathrm{domain}\mathrm{R}\mathrm{such}\mathrm{that}\mathrm{f}\left(\mathrm{x}\right)=-1.\\ \therefore \mathrm{f}\left(\mathrm{x}\right)\mathrm{â€‹}\mathrm{is}\mathrm{not}\mathrm{onto}\mathrm{.}\\ \mathrm{Therefore},\mathrm{the}\mathrm{modulus}\mathrm{function}\mathrm{is}\mathrm{neither}\mathrm{one}–\mathrm{one}\mathrm{nor}\mathrm{onto}\mathrm{.}\end{array}$

Q.5

$\begin{array}{l}\mathrm{Show}\mathrm{that}\mathrm{the}\mathrm{Signum}\mathrm{Function}\mathrm{f}:\mathrm{R}\to \mathrm{R},\mathrm{given}\mathrm{by}\\ \mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l} \mathrm{ }1,\mathrm{ }\mathrm{if}\mathrm{x}>0\\ 0,\mathrm{ }\mathrm{if}\mathrm{x}=0\\ –1,\mathrm{ }\mathrm{if}\mathrm{x}<0\end{array}\\ \mathrm{is}\mathrm{neither}\mathrm{one}–\mathrm{one}\mathrm{noronto}.\end{array}$

Ans

$\begin{array}{l}\mathrm{f}:\mathrm{R}\to \mathrm{R}\mathrm{is}\mathrm{given}\mathrm{by},\\ \mathrm{Modulus}\mathrm{function}\mathrm{is}\mathrm{defined}\mathrm{by}\\ \mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}1,\mathrm{}\mathrm{if}\mathrm{x}>0\\ 0,\mathrm{}\mathrm{if}\mathrm{x}=0\\ -1,\mathrm{}\mathrm{if}\mathrm{x}<0\end{array}\\ \mathrm{Here},\mathrm{f}\left(1\right)=1,\mathrm{ }\mathrm{f}\left(3\right)=1,\mathrm{but}1\ne \mathrm{3}\\ \mathrm{So},\mathrm{f}\mathrm{is}\mathrm{not}\mathrm{one}–\mathrm{one}\mathrm{.}\\ \\ \mathrm{Since},\mathrm{range}\mathrm{of}\mathrm{f}\mathrm{is}\left(1,0,-1\right)\mathrm{for}\mathrm{element}-2\mathrm{in}\mathrm{co}–\mathrm{domain}\mathrm{R},\\ \mathrm{there}\mathrm{does}\mathrm{not}\mathrm{exist}\mathrm{any}\mathrm{value}\mathrm{in}\mathrm{domain}\mathrm{R}\mathrm{such}\mathrm{that}\\ \mathrm{f}\left(\mathrm{x}\right)=-2.\\ \therefore \mathrm{f}\mathrm{is}\mathrm{not}\mathrm{onto}\mathrm{.}\\ \mathrm{Hence},\mathrm{the}\mathrm{signum}\mathrm{function}\mathrm{is}\mathrm{neither}\mathrm{one}–\mathrm{one}\mathrm{nor}\mathrm{onto}\mathrm{.}\end{array}$

Q.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.

Ans

$\begin{array}{l}\text{It is given that}\\ \text{A}=\text{}\left\{\text{1},\text{2},\text{3}\right\},\text{B}=\left\{\text{4},\text{5},\text{6},\text{7}\right\}\text{}\text{\hspace{0.17em}}\\ \text{f}:\text{A}\to \text{B is defined as}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{f = {(1, 4), (2, 5), (3, 6)}}\text{.}\\ \therefore \text{f (1) = 4, f (2) = 5, f (3) = 6}\\ \text{Since every element has a unique value}\text{.}\\ \text{The given function f is one-one}\text{.}\end{array}$

Q.7

$\begin{array}{l}Ineachofthefollowingcases,statewhetherthefunctionis\\ one–one,ontoorbijective.Justifyyouranswer.\\ \left(i\right)f:R\to Rdefinedbyf\left(x\right)=3–4x\\ \left(ii\right)f:R\to Rdefinedbyf\left(x\right)=1+{x}^{2}\end{array}$

Ans

$\begin{array}{l}\text{(i) f: R}\to \text{R is defined as f(x) = 3}-\text{4x}\text{.}\\ \text{For one-one}\\ \text{Let}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{f(x)}=\text{f(y)}\\ ⇒\text{3}-\text{4x}=\text{3}-\text{4y}\\ \\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{x}=\text{y}\\ \therefore \text{f is one-one}\text{.}\\ \text{for onto}\\ \text{Let f(x)=y}\\ ⇒\text{3}-\text{4x=y}\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{x=}\frac{\text{3}-y}{4}\\ ⇒\text{f}\left(\frac{\text{3}-y}{4}\right)\text{=3}-\text{4}\left(\frac{\text{3}-y}{4}\right)\text{=y}\\ \text{Since Range = Co-domain}\\ \therefore \text{f is onto}\text{.}\\ \text{Hence, f is bijective}\text{.}\\ \\ \text{(ii) f: R}\to {\text{R is defined as f(x)=1+x}}^{\text{2}}\\ \text{For one-one}\\ \text{Let f(x)=f(y )}\\ ⇒{\text{1+x}}^{\text{2}}={\text{1+y}}^{\text{2}}\\ ⇒\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{x}}^{\text{2}}={\text{y}}^{\text{2}}\\ ⇒\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{x}=±\text{y}\\ \therefore \text{f is not one-one}\text{.}\\ \text{For onto:}\\ \text{Let}\text{\hspace{0.17em}}\text{f}\left(x\right)=y\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{1+x}}^{\text{2}}=\text{y}\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{x}=±\sqrt{y-1}\\ \text{Since Range is not equal to Co-domain}\text{.}\\ \text{Then given function f(x) is not onto}\text{.}\end{array}$

Q.8

$\begin{array}{l}\mathrm{Let}\mathrm{A}\mathrm{and}\mathrm{B}\mathrm{be}\mathrm{sets}.\mathrm{Show}\mathrm{that}\mathrm{f}:\mathrm{A}×\mathrm{B}\to \mathrm{B}×\mathrm{A} \mathrm{such}\mathrm{that}\\ \mathrm{f}\left(\mathrm{a},\mathrm{b}\right)=\mathrm{f}\left(\mathrm{b},\mathrm{a}\right) \mathrm{is}\mathrm{bijective}\mathrm{function}.\end{array}$

Ans

$\begin{array}{l}\mathrm{f}:\mathrm{A}×\mathrm{B}\to \mathrm{B}×\mathrm{A}\mathrm{is}\mathrm{defined}\mathrm{as}\mathrm{f}\left(\mathrm{a},\mathrm{b}\right) = \left(\mathrm{b},\mathrm{a}\right). \\ \mathrm{Let}\mathrm{}\left({\mathrm{a}}_{1}, {\mathrm{b}}_{1}\right),\left({\mathrm{a}}_{2}, {\mathrm{b}}_{2}\right)\in \mathrm{A}×\mathrm{B}\mathrm{such}\mathrm{that}\mathrm{f}\left({\mathrm{a}}_{1}, {\mathrm{b}}_{1}\right)=\mathrm{f}\left({\mathrm{a}}_{2}, {\mathrm{b}}_{2}\right)\\ ⇒\left({\mathrm{b}}_{1}, {\mathrm{a}}_{1}\right)=\left({\mathrm{b}}_{2}, {\mathrm{a}}_{2}\right)\\ ⇒{\mathrm{b}}_{1}={\mathrm{b}}_{2}\mathrm{ }\mathrm{and}\mathrm{ }{\mathrm{a}}_{1}={\mathrm{a}}_{2}\\ ⇒\left({\mathrm{a}}_{1},{\mathrm{b}}_{1}\right)=\left({\mathrm{a}}_{2},{\mathrm{b}}_{2}\right)\\ \therefore \mathrm{f}\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{.}\\ \mathrm{Now},\mathrm{let}\left(\mathrm{b},\mathrm{a}\right)\in \mathrm{B}×\mathrm{A}\mathrm{be}\mathrm{any}\mathrm{element}\mathrm{.}\\ \mathrm{Then},\mathrm{there}\mathrm{exists}\left(\mathrm{a},\mathrm{b}\right)\in \mathrm{A}×\mathrm{B}\mathrm{such}\mathrm{that}\mathrm{f}\left(\mathrm{a},\mathrm{b}\right) = \left(\mathrm{b},\mathrm{a}\right).\\ \left[\mathrm{By}\mathrm{definition}\mathrm{of}\mathrm{f}\right]\\ \therefore \mathrm{ }\mathrm{f}\mathrm{is}\mathrm{onto}\mathrm{.}\\ \mathrm{Hence},\mathrm{f}\mathrm{is}\mathrm{bijective}\mathrm{.}\end{array}$

Q.9

$\begin{array}{l}Letf:N\to Nbedefinedby\\ f\left(n\right)=\left\{\begin{array}{l}\frac{n+1}{2},\text{}ifnisodd\\ \frac{n}{2},\text{}\text{}ifniseven\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}foralln\in N.\\ Statewhetherthefunctionfisbijective.\\ Justifyyouranswer.\end{array}$

Ans

$\begin{array}{l}Letf:N\to Nisdefinedas\\ \\ f\left(n\right)=\left\{\begin{array}{l}\frac{n+1}{2},\text{}ifnisodd\\ \frac{n}{2},\text{}\text{}ifniseven\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}foralln\in N.\\ \text{It can be observed that:}\\ f\left(1\right)=\frac{1+1}{2}=1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}f\left(2\right)=\frac{2}{2}=1\text{}\text{}\left[By\text{definition of f}\right]\\ \therefore f\left(1\right)=f\left(2\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}where\text{\hspace{0.17em}}\text{\hspace{0.17em}}1\ne 2.\\ \therefore \text{f is not one-one}\text{.}\\ \text{Consider a natural number (n) in co-domain N}\text{.}\\ \text{Case I: n is odd}\\ \therefore \text{n = 2m + 1 for some m}\in \text{N}\text{. Then,there exists 4m + 1}\in \text{N}\\ \text{such that}\\ \text{f}\left(4m+1\right)=\frac{4m+1+1}{2}=2m+1\\ Case\text{\hspace{0.17em}}II:\text{n is even}\\ \therefore \text{\hspace{0.17em}}\text{n=2}m\text{\hspace{0.17em}}\text{for some}m\in \text{N}\text{. Then,there exists 4}m\in \text{N such that}\\ \text{f}\left(4m\right)=\frac{4m}{2}=2m\\ \therefore \text{f is onto}\text{.}\\ \text{Hence, f is not a bijective function}\text{.}\end{array}$

Q.10

$\begin{array}{l}LetA=R-\left\{3\right\}andB=R-\left\{1\right\}.Considerthefunction\\ f:A\to Bdefinedby\text{\hspace{0.17em}}f\left(x\right)=\frac{x–2}{x–3}.Isfone-oneandonto?\\ Justifyyouranswer.\end{array}$

Ans

$\begin{array}{l}A=R-\left\{3\right\}andB=R-\left\{1\right\}\\ \text{f: A}\to \text{B is defined as f}\left(x\right)=\frac{x-2}{x-3}\\ Let\text{x,y}\in \text{A such that}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{f}\left(x\right)=f\left(y\right).\\ ⇒\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{x-2}{x-3}=\frac{y-2}{y-3}\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(x-2\right)\left(y-3\right)=\left(x-3\right)\left(y-2\right)\\ ⇒\overline{)xy}-3x-2y+\overline{)6}=\overline{)xy}-2x-3y+\overline{)6}\\ ⇒\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-3x-2y=-2x-3y\\ ⇒\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=y\\ \therefore \text{\hspace{0.17em}}f\text{if one-one}\text{.}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{y}\in \text{B}=\text{R-}\left\{1\right\}.\text{\hspace{0.17em}}Then,\text{\hspace{0.17em}}y\ne 1\\ \text{The function f is onto if there exists x}\in \text{A such that}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{f(x) = y}\\ ⇒\text{}\text{}\frac{x-2}{x-3}=y\\ ⇒\text{}\text{}x-2=y\left(x-3\right)\\ ⇒\text{}\text{}x-2=yx-3y\\ ⇒\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x-yx=2-3y\\ ⇒\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\left(1-y\right)=2-3y\\ ⇒\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=\frac{2-3y}{\left(1-y\right)}\in A\text{}\text{}\left[y\ne 1\right]\\ \text{Thus, for any y}\in \text{B, there exists}\text{\hspace{0.17em}}\frac{2-3y}{\left(1-y\right)}\in A\text{\hspace{0.17em}}such\text{\hspace{0.17em}}\text{\hspace{0.17em}}that\\ \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}f\left(\frac{2-3y}{1-y}\right)=\frac{\frac{2-3y}{\left(1-y\right)}-2}{\frac{2-3y}{\left(1-y\right)}-3}\\ \text{}\text{}\text{}=\frac{\frac{2-3y-2\left(1-y\right)}{1-y}}{\frac{2-3y-3\left(1-y\right)}{1-y}}\\ \text{}\text{}\text{}=\frac{2-3y-2+2y}{2-3y-3+3y}\\ \text{}\text{}\text{}=\frac{-y}{-1}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}f\left(\frac{2-3y}{1-y}\right)=y\\ \therefore \text{\hspace{0.17em}}f\text{is onto}\text{.}\\ \text{Hence, function f is one-one and onto}\text{.}\end{array}$

Q.11

$\begin{array}{l}\mathrm{Letf}:\mathrm{R}\to \mathrm{R}\mathrm{be}\mathrm{defined}\mathrm{as}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{4}.\mathrm{Choose}\mathrm{the}\mathrm{correct}\\ \mathrm{answer}.\\ \left(\mathrm{A}\right)\mathrm{f}\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{onto} \mathrm{ }\left(\mathrm{B}\right)\mathrm{fismany}–\mathrm{one}\mathrm{onto}.\\ \left(\mathrm{C}\right)\mathrm{f}\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{but}\mathrm{not}\mathrm{onto} \left(\mathrm{D}\right)\mathrm{f}\mathrm{is}\mathrm{n}\mathrm{either}\mathrm{one}–\mathrm{one}\\ \mathrm{noronto}.\end{array}$

Ans

$\begin{array}{l}\mathrm{f}:\mathrm{R}\to \mathrm{R}\mathrm{is}\mathrm{defined}\mathrm{as}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{4}.\\ \mathrm{Let}\mathrm{x},\mathrm{y}\in \mathrm{R}\mathrm{such}\mathrm{that}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{y}\right).\\ \mathrm{Then}, \mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{y}\right)\\ ⇒ \mathrm{ }{\mathrm{x}}^{4}={\mathrm{y}}^{4}\\ ⇒ \mathrm{ }\mathrm{x}=±\mathrm{y}\\ \therefore \mathrm{ }\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{y}\right)\mathrm{ }\mathrm{but}\mathrm{ }\mathrm{x}\ne \mathrm{y}\\ \mathrm{So},\mathrm{ }\mathrm{f}\mathrm{is}\mathrm{not}\mathrm{one}–\mathrm{one}\mathrm{.}\\ \mathrm{Consider}\mathrm{an}\mathrm{element}3\mathrm{in}\mathrm{co}–\mathrm{domain}\mathrm{R}.\mathrm{It}\mathrm{is}\mathrm{clear}\mathrm{that}\mathrm{there}\\ \mathrm{does}\mathrm{not}\mathrm{exist}\mathrm{any}\mathrm{x}\mathrm{in}\mathrm{domain}\mathrm{R}\mathrm{such}\mathrm{that}\mathrm{f}\left(\mathrm{x}\right) = 3\mathrm{.}\\ \therefore \mathrm{f}\mathrm{is}\mathrm{not}\mathrm{onto}\mathrm{.}\\ \mathrm{Hence},\mathrm{function}\mathrm{f}\mathrm{is}\mathrm{neither}\mathrm{one}–\mathrm{one}\mathrm{nor}\mathrm{onto}\mathrm{.}\\ \mathrm{The}\mathrm{correct}\mathrm{answer}\mathrm{is}\mathrm{D}\mathrm{.}\end{array}$

Q.12

$\begin{array}{l}\mathrm{Let}\mathrm{f}:\mathrm{R}\to \mathrm{R}\mathrm{be}\mathrm{defined}\mathrm{asf}\left(\mathrm{x}\right)=3\mathrm{x}.\mathrm{Choose}\mathrm{the}\mathrm{correct}\\ \mathrm{answer}.\\ \left(\mathrm{A}\right)\mathrm{f} \mathrm{isone}–\mathrm{one}\mathrm{onto} \mathrm{ }\left(\mathrm{B}\right)\mathrm{f} \mathrm{is}\mathrm{many}–\mathrm{one}\mathrm{on}\mathrm{to}\\ \left(\mathrm{C}\right)\mathrm{f} \mathrm{isone}–\mathrm{one}\mathrm{but}\mathrm{not}\mathrm{onto}\left(\mathrm{D}\right)\mathrm{f}\mathrm{ }\mathrm{is}\mathrm{neither}\mathrm{one}–\mathrm{one}\mathrm{nor}\mathrm{onto}.\end{array}$

Ans

$\begin{array}{l}\mathrm{f}:\mathrm{R}\to \mathrm{Rbedefinedasf}\left(\mathrm{x}\right)=3\mathrm{x}.\\ \mathrm{Let}\mathrm{x},\mathrm{y}\in \mathrm{R}\mathrm{such}\mathrm{that}\mathrm{f}\left(\mathrm{x}\right) =\mathrm{f}\left(\mathrm{y}\right)\mathrm{.}\\ ⇒3\mathrm{x}=3\mathrm{y}\\ ⇒ \mathrm{x}=\mathrm{y}\\ \therefore \mathrm{f}\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{.}\\ \mathrm{Also},\mathrm{for}\mathrm{any}\mathrm{real}\mathrm{number}\left(\mathrm{y}\right)\mathrm{in}\mathrm{co}–\mathrm{domain}\mathrm{R},\mathrm{there}\mathrm{exists}\\ \frac{\mathrm{y}}{3} \mathrm{in}\mathrm{R}\mathrm{such}\mathrm{that} \mathrm{f}\left(\frac{\mathrm{y}}{3}\right)=3\left(\frac{\mathrm{y}}{3}\right)=\mathrm{y}\\ \therefore \mathrm{f}\mathrm{is}\mathrm{onto}\mathrm{.}\\ \\ \mathrm{Therefore},\mathrm{function}\mathrm{f}\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{and}\mathrm{onto}\mathrm{.}\\ \mathrm{The}\mathrm{correct}\mathrm{answer}\mathrm{is}\mathrm{A}\mathrm{.}\end{array}$