# NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.3 (Ex 1.3)

Relations and Functions chapter in Class 12 Central Board of Secondary Education (CBSE) Mathematics is a continuation of the Class 11 chapter Relations and Functions. The Class 11 Relations and Functions chapter should be reviewed before moving on to the Class 12 chapters. Students are expected to regularly practice exercise questions in order to score higher marks in the Class 12 board exam of Mathematics. Firstly, students need to learn the concepts given in chapter 1 before starting to solve exercise questions. All the formulas and definitions are to be revised from time to time.

Class 12 students facing issues while solving Exercise 1.3 can make use of NCERT Solutions Class 12 Maths Chapter 1 Exercise 1.3. In order to understand the right method of solving the questions of Exercise 1.3, it is recommended to download the NCERT Solutions Class 12 Maths Chapter 1 Exercise 1.3 from the Extramarks’ website and mobile application. The questions of Exercise 1.3 need to be practised a few times so that all the concepts of Composition of Functions and Invertible Functions are understood.

It is recommended that students make a dedicated study plan for Class 12 Mathematics. Each chapter has its own requirements in Class 12 Mathematics CBSE syllabus. The students are required to go through the syllabus and get an overview of the topics covered in Mathematics.

Class 12 students are advised to make use of the NCERT Solutions Class 12 Maths Chapter 1 Exercise 1.3 if they feel stuck in any of the Exercise 1.3 questions. The NCERT Solutions Class 12 Maths Chapter 1 Exercise 1.3 is available on the Extramarks’ website and mobile application in PDF format.

Question 1 in Class 12 Maths Chapter 1 Exercise 1.3 can sometimes be difficult to solve for students. Each solution given in the NCERT Solutions Class 12 Maths Chapter 1 Exercise 1.3 lies just below each question so that it can be easily understood. The NCERT Solutions Class 12 Maths Chapter 1 Exercise 1.3 given on the Extramarks’ website and mobile application is very reliable, and its quality is maintained. The solutions given in NCERT Solutions Class 12 Maths Chapter 1 Exercise 1.3 are designed by expert Mathematics teachers.

The chapter Composition of Functions and Invertible Functions is really important from the exam perspective. It is necessary to solve all the questions given in Exercise 1.3 with the help of NCERT Solutions Class 12 Maths Chapter 1 Exercise 1.3. Class 12 students of the Central Board of Secondary Education (CBSE) can access NCERT Solutions Class 12 Maths Chapter 1 Exercise 1.3 from the Extramarks’ website and mobile application.

Question 4 in Exercise 1.3 is related to finding the inverse of a function. If Class 12 students feel any difficulty in solving Question 4, they can make use of NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.3. The solutions are given in a stepwise manner, and it helps improve the answer-writing skills of students.

The students need to get enough practice before their board exams.

All the 14 problems given in Exercise 1.3 are sufficient for understanding the Composition of Functions and Invertible Functions chapter. Students are advised to learn the concepts thoroughly and revise the definitions and theorems under the chapter. To get an accurate answer for Question 1 Exercise 1.3, students can refer to the NCERT Solutions Class 12 Maths Chapter 1 Exercise 1.3 given on the Extramarks’ website and mobile application. The questions of Exercise 1.3 are important for preparing students for the Central Board of Secondary Education (CBSE) Class 12 Mathematics exam. The NCERT Solutions Class 12 Maths Chapter 1 Exercise 1.3 is a very essential tool for helping students in solving all the 14 questions of Exercise 1.3. Students are required to go through each of the questions given in the NCERT Solutions Class 12 Maths Chapter 1 Exercise 1.3 carefully and enhance their exam preparations.

Class 12 students can access the NCERT Solutions Class 12 Maths Chapter 1 Exercise 1.3 from the Extramarks’ website and mobile application. Each concept in Mathematics must be practised step-by-step and students must revise definitions, formulas, theorems, etc. consistently during preparation.

The chapter Composition of Functions and Invertible Functions consists of many solved examples in the NCERT textbook itself to help students in understanding the concepts. The solutions of Chapter 1 Exercise 1.3 Class 12 are not available in the NCERT textbook, therefore it is advisable to refer to NCERT Solutions Class 12 Maths Chapter 1 Exercise 1.3. Class 12 CBSE students can obtain the NCERT Solutions Class 12 Maths Chapter 1 Exercise 1.3 from the Extramarks website and mobile application.

There is no doubt that Mathematics is a high-scoring subject, and with proper practice, a student can score well in it. A separate study plan should be developed by students for Class 12 Mathematics. The study plan needs to be regularly followed, and students are expected to solve past years’ question papers and sample question papers at regular intervals.

When preparing for Chapter 1, it is required to solve exercise questions along with a revision of concepts.

It is necessary to get a deeper understanding of each chapter and develop a separate approach for each of them. To solve questions, it is advisable to have an assessment of solved examples given in the NCERT textbook. Students having difficulty in solving Exercise 1.3 can take the help of NCERT Solutions Class 12 Maths Chapter 1 Exercise 1.3. All the solutions present in the NCERT Solutions Class 12 Maths Chapter 1 Exercise 1.3 are prepared by expert Mathematics teachers and are given in stepwise and easy format.

## NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.3 (Ex 1.3)

To enhance their learning experience and get better results in their upcoming exams, students are advised to access study materials from the Extramarks’ website and mobile application.

### List of Topics Covered Under NCERT Solutions for Class 12 Maths Chapter 1

The topics covered in Chapter 1 Relations and Functions are Introduction, Types of Relations, Types of Functions, Composition of Functions and Invertible Functions and Binary Operations. All topics need to be covered well and should be revised regularly for getting good marks in the Mathematics board exam.

### Introduction About Types of Functions

The topic of Types of Functions is one of the important topics in the Relations and Functions Chapter. The questions related to Types of Functions are asked in Exercise 1.2. Students are advised to solve all the questions given in Exercise 1.2 and prepare well for the exams. To solve Exercise 1.3 questions, Class 12 students can download the NCERT Solutions Class 12 Maths Chapter 1 Exercise 1.3 from the Extramarks’ website and mobile application.

### Types of functions

A function can be of three types, one-one, onto, and one-one and onto. The function one-one is called Injective Function, onto is called Surjective and one-one and onto function is called Bijective. Questions given in Exercise 1.2 are entirely based on these three types of functions. Types of functions are critical for learning Composition of Functions and Invertible Functions. Exercise 1.3 is based on Composition of Functions and Invertible Functions.

Students having difficulty in Exercise 1.3 can access the NCERT Solutions Class 12 Maths Chapter 1 Exercise 1.3 from the Extramarks’ website and mobile application. The NCERT Solutions Class 12 Maths Chapter 1 Exercise 1.3 is accessible in PDF format and can be used anytime after being downloaded.

### Ncert Solutions for Class 12 Maths Chapter 1 Includes the Following Topics

Chapter 1 topic in Class 12 Mathematics syllabus is Relations and Functions. The students are expected to learn all the subtopics given in the chapter for increasing their preparation level in the Class 12 Mathematics subject. The exercises are given just after each topic in the chapter. To solve questions given in Exercise 1.3 students must download NCERT Solutions Class 12 Maths Chapter 1 Exercise 1.3 and enhance their learning process.

### Introduction to the Topic:

The topics given in Chapter 1 Relations and Functions of Class 12 can be understood well if the Class 11 topics are revised well. Students are required to understand the topics and practice the exercises after that. To solve the questions of Exercise 1.3, it is advisable to access the NCERT Solutions Class 12 Maths Chapter 1 Exercise 1.3 from the Extramarks website and mobile application.

### Types of Relations:

The Types of Relations topic consists of definitions that are important for solving questions given in Exercise 1.1. It is necessary to understand all the Types of Relations before trying to solve Exercise 1.1 questions.

### Types of Functions:

The questions given in Exercise 1.2 are asked based on Types of Functions and all the definitions are to be thoroughly understood before practising Exercise 1.2 questions.

### Overview of the Chapter:

All the questions in the Relations and Functions Chapter are important from the exam perspective and students are advised to solve all the exercise questions that are included in Chapter 1. It is necessary to make use of the NCERT Solutions Class 12 Maths Chapter 1 Exercise 1.3 to solve Exercise 1.3 questions.

Extramarks is a learning platform designed to help students in their learning process. It is advisable to use study material provided by Extramarks to enhance their ongoing preparations. Class 12 students can access the NCERT Solutions Class 12 Maths Chapter 1 Exercise 1.3 to continue their Mathematics board exam preparation.

The solved NCERT Solutions Class 12 Maths Chapter 1 Exercise 1.3 are given by expert Mathematics teachers and are of the best quality. To practice questions effectively, it is recommended to make use of study materials given by the Extramarks learning platform. Extramarks has live lectures, which students can utilise to understand the concepts in a very efficient manner. Students can also ask their doubts during the live lectures and get instant solutions from the expert teachers of each subject. Extramarks provides study resources for all the classes from 1 to 12. All the teachers connected with Extramarks are experts in different subjects and are available to help students with any issues during their exam preparation.

Students need to have some important study resources that help get good marks in exams. They can download past years’ papers and sample papers from the Extramarks website and mobile application. Students of all the classes can obtain the syllabus for each subject. The syllabus helps students in getting an outline of the topics covered in each subject. Students will be able to formulate a study timetable with the help of the syllabus. The Extramarks’ website and mobile application also have mock tests related to the topics in subjects, which students can use to assess their preparation level.

### Class 12 Maths Chapter 1

The students of Class 12 Central Board of Secondary Education (CBSE) are advised to solve exercise questions of Chapter 1 Relations and Functions. To get detailed solutions for Exercise 1.3, students can refer to the NCERT Solutions Class 12 Maths Chapter 1 Exercise 1.3. It is important to solve all the questions related to Composition of Functions and Invertible Functions to make the concept strong and feel confident in the topic.

Practice is necessary for a subject like Mathematics. Students who want to practice Exercise 1.3 are advised to make use of the NCERT Solutions Class 12 Maths Chapter 1 Exercise 1.3.

### 1.3 Maths Class 12

The questions asked in Exercise 1.3 are framed using the topic Composition of Functions and Invertible Functions. The questions that require deep understanding of the topic and definition are to be revised again and again. It is significant to make use of solutions given by expert Mathematics teachers. The students are advised to use the NCERT Solutions Class 12 Maths Chapter 1 Exercise 1.3 provided by the Extramarks’ website and mobile application.

There are fourteen questions to be answered in Exercise 1.3. All the questions need to be answered properly with the use of correct definitions and theorems. Practising with the NCERT Solutions Class 12 Maths Chapter 1 Exercise 1.3 is helpful for Class 12 students in understanding the methods used to solve questions. Students can improve their answers with the help of NCERT Solutions Class 12 Maths Chapter 1 Exercise 1.3.

To get all the solutions for the exercise of Relations and Functions Chapter, Class 12 students can take the help of the Extramarks’ learning platform. The solutions for each exercise of Chapter 1 are available in PDF format on the Extramarks’ website and mobile application. When solving Exercise 1.3, students may come across some problems. Class 12 Students are required to access the NCERT Solutions Class 12 Maths Chapter 1 Exercise 1.3 from the Extramarks’ website and mobile application to get detailed solutions to each question of Exercise 1.3.

Central Board of Secondary Education (CBSE) provides students with the NCERT Solutions for all the following classes, for each of their subjects from the Extramarks’ website and mobile application:

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 Chapter 1 Exercises

 Chapter 1 Relations And Functions Other Exercises Exercise 1.1 16 Questions And Solutions (3 Short Answer, 13 Long Answers) Exercise 1.2 12 Questions And Solutions (5 Short Answer, 7 Long Answers) Exercise 1.4 13 Questions And Solutions (6 Short Answer, 7 Long Answers)

Q.1

$\begin{array}{l}\mathrm{Let}\mathrm{f}:\left\{1,3,4\right\}\to \left\{1,2,5\right\}\mathrm{and}\mathrm{g}:\left\{1,2,5\right\}\to \left\{1,3\right\}\mathrm{be}\mathrm{given}\\ \mathrm{by}\mathrm{f}=\left\{\left(1,2\right),\left(3,5\right),\left(4,1\right)\right\}\mathrm{and}\mathrm{g}=\left\{\left(1,3\right),\left(2,3\right),\left(5,1\right)\right\}.\\ \mathrm{Write}\mathrm{down}\mathrm{go}\mathrm{f}.\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\mathrm{functions}\mathrm{f}: \left\{1, 3, 4\right\}\to \left\{1, 2, 5\right\}\mathrm{and}\mathrm{g}: \left\{1, 2, 5\right\}\to \left\{1, 3\right\}\\ \mathrm{are}\mathrm{defined}\mathrm{as}\mathrm{f}= \left\{\left(1, 2\right), \left(3, 5\right), \left(4, 1\right)\right\}\mathrm{and}\\ \mathrm{g}= \left\{\left(1, 3\right), \left(2, 3\right), \left(5, 1\right)\right\}\mathrm{.}\\ \mathrm{gof}\left(1\right)=\mathrm{g}\left\{\mathrm{f}\left(1\right)\right\}=\mathrm{g}\left(2\right)=3\left[\mathrm{f}\left(1\right)=2,\mathrm{g}\left(2\right)=3\right]\\ \mathrm{gof}\left(3\right)=\mathrm{g}\left\{\mathrm{f}\left(3\right)\right\}=\mathrm{g}\left(5\right)=1\left[\mathrm{f}\left(3\right)=5,\mathrm{g}\left(5\right)=1\right]\\ \mathrm{gof}\left(4\right)=\mathrm{g}\left\{\mathrm{f}\left(4\right)\right\}=\mathrm{g}\left(1\right)=3\left[\mathrm{f}\left(4\right)=1,\mathrm{g}\left(1\right)=3\right]\\ \therefore \mathrm{gof}=\left\{\left(1,3\right),\left(3,1\right),\left(4,3\right)\right\}\end{array}$

Q.2 Let f, g and h be functions from R to R. Show that

(f + g)oh = foh + goh

(f. g)oh = (foh) . (goh)

Ans

$\begin{array}{l}\text{Let}\left(\left(\text{f}+\text{g}\right)\text{oh}\right)\left(\text{x}\right)=\left(\text{foh}\right)\left(\text{x}\right)+\left(\text{goh}\right)\left(\text{x}\right)\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}=\text{f}\left\{\text{h}\left(\text{x}\right)\right\}+\text{g}\left\{\text{h}\left(\text{x}\right)\right\}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}=\text{}\left(\text{foh}\right)\left(\text{x}\right)+\left(\text{goh}\right)\left(\text{x}\right)\\ \\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}=\text{}\left\{\left(\text{f}+\text{g}\right)\text{oh}\right\}\left(\text{x}\right)\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}=\text{}\left\{\left(\text{foh}\right)+\left(\text{goh}\right)\right\}\left(\text{x}\right)\text{}\forall x\in R\\ Hence,\text{}\left(f+g\right)oh=\left(\text{foh}\right)+\left(\text{goh}\right).\\ Now,\\ \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}}\left\{\left(f.g\right)oh\right\}\left(x\right)=\left(f.g\right)\left(h\left(x\right)\right)\\ \text{}\text{}\text{}\text{}=f\left(h\left(x\right)\right).g\left(h\left(x\right)\right)\\ \text{}\text{}\text{}\text{}=\left(foh\right)\left(x\right).\left(goh\right)\left(x\right)\\ \text{}\text{}\text{}\text{}=\left\{\left(foh\right).\left(goh\right)\right\}\left(x\right)\\ \therefore \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}}\left\{\left(f.g\right)oh\right\}\left(x\right)=\left\{\left(foh\right).\left(goh\right)\right\}\left(x\right)\text{}\text{}\forall x\in R\\ Hence,\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}}\left(f.\text{\hspace{0.17em}}g\right)oh=\left(foh\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\left(goh\right)\end{array}$

Q.3

$\begin{array}{l}Findgofandfog,if\\ \left(i\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}f\left(x\right)=|x|\text{\hspace{0.17em}}and\text{\hspace{0.17em}}g\left(x\right)=|5x–2|\\ \left(ii\right)\text{\hspace{0.17em}}f\left(x\right)=8{x}^{3}\text{\hspace{0.17em}}\text{\hspace{0.17em}}and\text{\hspace{0.17em}}\text{\hspace{0.17em}}g\left(x\right)={x}^{\frac{1}{3}}\end{array}$

Ans

$\begin{array}{l}\left(i\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}f\left(x\right)=|x|\text{\hspace{0.17em}}and\text{\hspace{0.17em}}\text{\hspace{0.17em}}g\left(x\right)=|5x-2|\\ \therefore \left(gof\right)\left(x\right)=g\left(f\left(x\right)\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=g\left(|x|\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=|5|x|-2|\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(fog\right)\left(x\right)=f\left(g\left(x\right)\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=f\left(|5x-2|\right)\\ \\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=||5x-2||\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=|5x-2|\\ \left(ii\right)\text{\hspace{0.17em}}f\left(x\right)=8{x}^{3}\text{\hspace{0.17em}}\text{\hspace{0.17em}}and\text{\hspace{0.17em}}\text{\hspace{0.17em}}g\left(x\right)={x}^{\frac{1}{3}}\\ \therefore \left(gof\right)\left(x\right)=g\left(f\left(x\right)\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=g\left(8{x}^{3}\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={\left(8{x}^{3}\right)}^{\frac{1}{3}}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2x\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(fog\right)\left(x\right)=f\left(g\left(x\right)\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=f\left({x}^{\frac{1}{3}}\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=8{\left({x}^{\frac{1}{3}}\right)}^{3}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=8x\end{array}$

Q.4

$\begin{array}{l}Iff\left(x\right)=\frac{4x+3}{6x–4},\text{\hspace{0.17em}}x\ne \frac{2}{3},showthatfof\left(x\right)=x,\text{\hspace{0.17em}}forallx\ne \frac{2}{3}.\\ Whatistheinverseoff?\end{array}$

Ans

$\begin{array}{l}It\text{is given that}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}f\left(x\right)=\frac{4x+3}{6x–4},\text{\hspace{0.17em}}then\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}fof\left(x\right)=f\left(f\left(x\right)\right)\\ \text{}\text{}\text{}=f\left(\frac{4x+3}{6x–4}\right)\\ \text{}\text{}\text{}=\frac{4\left(\frac{4x+3}{6x–4}\right)+3}{6\left(\frac{4x+3}{6x–4}\right)-4}\\ \text{}\text{}\text{}=\frac{\left(\frac{16x+12+18x-12}{6x-4}\right)}{\left(\frac{24x+18-24x+16}{6x–4}\right)}\\ \text{}\text{}\text{}=\frac{34x}{34}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}fof\left(x\right)=x,\text{for all x}\ne \frac{2}{3}.\\ ⇒\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}fof=I\\ \text{Hence, the given function f is invertible and the inverse of f}\\ \text{is f itself}\text{.}\end{array}$

Q.5

$\begin{array}{l}\mathrm{State}\mathrm{with}\mathrm{reason}\mathrm{whether}\mathrm{following}\mathrm{functions} \mathrm{have}\mathrm{inverse}\\ \left(\mathrm{i}\right)\mathrm{f}:\left\{1,2,3,4\right\}\to \left\{10\right\}\mathrm{with}\\ \mathrm{ }\mathrm{f}=\left\{\left(1,10\right),\left(2,10\right),\left(3,10\right),\left(4,10\right)\right\}\\ \left(\mathrm{ii}\right) \mathrm{g}:\left\{5,6,7,8\right\}\to \left\{1,2,3,4\right\}\mathrm{with}\\ \mathrm{g}=\left\{\left(5,4\right),\left(6,3\right),\left(7,4\right),\left(8,2\right)\right\}\\ \left(\mathrm{iii}\right)\mathrm{h}:\left\{2,3,4,5\right\}\to \left\{7,9,11,13\right\}\mathrm{with}\\ \mathrm{ }\mathrm{h}=\left\{\left(2,7\right),\left(3,9\right),\left(4,11\right),\left(5,13\right)\right\}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{f}: \left\{1, 2, 3, 4\right\}\to \left\{10\right\}\mathrm{defined}\mathrm{as}:\\ \mathrm{ }\mathrm{f}= \left\{\left(1, 10\right), \left(2, 10\right), \left(3, 10\right), \left(4, 10\right)\right\}\\ \mathrm{From}\mathrm{the}\mathrm{given}\mathrm{definition}\mathrm{of}\mathrm{f},\mathrm{we}\mathrm{can}\mathrm{see}\mathrm{that}\mathrm{f}\mathrm{is}\mathrm{a}\mathrm{many}\\ \mathrm{one}\mathrm{function}\mathrm{as}:\mathrm{f}\left(1\right) =\mathrm{f}\left(2\right)=\mathrm{f}\left(3\right) =\mathrm{f}\left(4\right) = 10\\ \therefore \mathrm{f}\mathrm{is}\mathrm{not}\mathrm{one}–\mathrm{one}\mathrm{.}\\ \mathrm{Hence},\mathrm{function}\mathrm{f}\mathrm{does}\mathrm{not}\mathrm{have}\mathrm{an}\mathrm{inverse}\mathrm{.}\\ \left(\mathrm{ii}\right)\mathrm{ }\mathrm{g}: \left\{5, 6, 7, 8\right\}\to \left\{1, 2, 3, 4\right\}\mathrm{defined}\mathrm{as}:\\ \mathrm{g}= \left\{\left(5, 4\right), \left(6, 3\right), \left(7, 4\right), \left(8, 2\right)\right\}\\ \mathrm{From}\mathrm{the}\mathrm{given}\mathrm{definition}\mathrm{of}\mathrm{g},\mathrm{it}\mathrm{is}\mathrm{seen}\mathrm{that}\mathrm{g}\mathrm{is}\mathrm{a}\mathrm{many}\mathrm{one}\\ \mathrm{function}\mathrm{as}:\mathrm{g}\left(5\right) =\mathrm{g}\left(7\right) = 4\mathrm{.}\\ \therefore \mathrm{g}\mathrm{is}\mathrm{not}\mathrm{one}–\mathrm{one}\mathrm{.}\\ \mathrm{Hence},\mathrm{function}\mathrm{g}\mathrm{does}\mathrm{not}\mathrm{have}\mathrm{an}\mathrm{inverse}\mathrm{.}\\ \left(\mathrm{iii}\right)\mathrm{h}: \left\{2, 3, 4, 5\right\}\to \left\{7, 9, 11, 13\right\}\mathrm{defined}\mathrm{as}:\\ \mathrm{h}= \left\{\left(2, 7\right), \left(3, 9\right), \left(4, 11\right), \left(5, 13\right)\\ \mathrm{It}\mathrm{is}\mathrm{seen}\mathrm{that}\mathrm{all}\mathrm{distinct}\mathrm{elements}\mathrm{of}\mathrm{the}\mathrm{set}\left\{2, 3, 4, 5\right\}\\ \mathrm{have}\mathrm{distinct}\mathrm{images}\mathrm{under}\mathrm{h}\mathrm{.}\\ \therefore \mathrm{Function}\mathrm{h}\mathrm{is}\mathrm{one}–\mathrm{one}\mathrm{.}\\ \mathrm{Also},\mathrm{h}\mathrm{is}\mathrm{onto}\mathrm{since}\mathrm{for}\mathrm{every}\mathrm{element}\mathrm{y}\mathrm{of}\mathrm{the}\mathrm{set}\\ \left\{7, 9, 11, 13\right\},\mathrm{there}\mathrm{exists}\mathrm{an} \mathrm{element}\mathrm{x}\mathrm{in}\mathrm{the}\mathrm{set}\\ \left\{2, 3, 4, 5\right\} \mathrm{such}\mathrm{that}\mathrm{h}\left(\mathrm{x}\right) =\mathrm{y}\mathrm{.}\\ \mathrm{Thus},\mathrm{h}\mathrm{is}\mathrm{a}\mathrm{one}–\mathrm{one}\mathrm{and}\mathrm{onto}\mathrm{function}.\mathrm{Hence},\mathrm{h}\mathrm{has}\mathrm{an}\mathrm{inverse}\mathrm{.}\end{array}$

Q.6

$\begin{array}{l}Showthatf:\left[-1,1\right]\to R,givenbyf\left(x\right)=\frac{x}{x+2}is\text{\hspace{0.17em}}one–one.\\ Findtheinverseofthefunctionf:\left[-1,1\right]\to Range\end{array}$

Ans

$f: [−1, 1]→R is given as f x = x x+2 Let f x =f y ⇒ x x+2 = y y+2 ⇒ x y+2 =y x+2 ⇒ xy+2x=xy+2y ⇒ x=y ∴f is one-one function. It is clear that f: [−1, 1]→Range f is onto. ∴f: [−1, 1]→Range f is one-one and onto and therefore, the inverse of the function: f: [−1, 1]→ Range f exists. Let g: Range f→[−1, 1] be the inverse of f. Let y be an arbitrary element of range f. Since f: [−1, 1]→Range f is onto, we have: f x =y for some x∈ −1,1 ⇒ y= x x+2 ⇒xy+2y=x ⇒ 2y=x−xy =x 1−y ⇒ x= 2y 1−y , y≠1 Now, let us define g: Range f→[−1, 1] as g y = 2y 1−y , y≠1. Now, gof x =g f x =g x x+2 = 2 x x+2 1− x x+2 = 2x x+2−x = 2x 2 gof x =x fog y =f g y =f 2y 1−y = 2y 1−y 2y 1−y +2 = 2y 2y+2−2y = 2y 2 fog y =y ∴go f −1 = I −1,1 and fo g −1 = I Range f ∴ f −1 =g ⇒ f −1 y = 2y y−1 , y≠1. ⇒ f −1 x = 2x x−1 , x≠1. 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Q.7

$\begin{array}{l}\mathrm{Consider}\mathrm{f}:\mathrm{R}\to \mathrm{Rgiven}\mathrm{by}\mathrm{f}\left(\mathrm{x}\right)=4\mathrm{x}+3.\mathrm{Show}\mathrm{that}\mathrm{f}\mathrm{is}\\ \mathrm{invertible}.\mathrm{Find}\mathrm{the}\mathrm{inverse}\mathrm{off}.\end{array}$

Ans

$\begin{array}{l}\mathrm{f}:\mathrm{R}\to \mathrm{R}\mathrm{is}\mathrm{given}\mathrm{by},\\ \mathrm{f}\left(\mathrm{x}\right) = 4\mathrm{x}+ 3\\ \mathrm{One}–\mathrm{one}:\\ \mathrm{Let}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{y}\right)\mathrm{.}\\ ⇒4\mathrm{x}+ 3=4\mathrm{y}+ 3\\ ⇒ \mathrm{ }4\mathrm{x}=4\mathrm{y}\\ ⇒ \mathrm{x}=\mathrm{y}\\ \therefore \mathrm{f}\mathrm{is}\mathrm{a}\mathrm{one}–\mathrm{one}\mathrm{function}\mathrm{.}\\ \mathrm{Onto}:\\ \mathrm{For}\mathrm{y}\in \mathrm{R},\\ \mathrm{let}\mathrm{y}= 4\mathrm{x}+ 3\mathrm{.}\\ ⇒ \mathrm{ }\mathrm{x}=\frac{\mathrm{y}-3}{4}\in \mathrm{R}\\ \\ \mathrm{Therefore},\mathrm{for}\mathrm{any}\mathrm{y}\in \mathrm{R},\mathrm{there}\mathrm{exists},{\mathrm{f}}^{–1}\mathrm{exists}\mathrm{.}\\ \mathrm{Let}\mathrm{us}\mathrm{define}\mathrm{g}:\mathrm{R}\to \mathrm{R}\mathrm{by}\mathrm{g}\left(\mathrm{y}\right)=\frac{\mathrm{y}-3}{4}\\ \mathrm{Now},\mathrm{ }\left(\mathrm{gof}\right)\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\\ \mathrm{ }=\mathrm{g}\left(4\mathrm{x}+3\right)\\ \mathrm{ }=\frac{4\mathrm{x}+3-3}{4}\\ \left(\mathrm{gof}\right)\left(\mathrm{x}\right)=\mathrm{x}\\ \mathrm{ }\left(\mathrm{fog}\right)\left(\mathrm{y}\right)=\mathrm{f}\left(\mathrm{g}\left(\mathrm{y}\right)\right)\\ \mathrm{ }=\mathrm{f}\left(\frac{\mathrm{y}-3}{4}\right)\\ \mathrm{ }=4\left(\frac{\mathrm{y}-3}{4}\right)+3\\ \mathrm{ }\left(\mathrm{fog}\right)\left(\mathrm{y}\right)=\mathrm{y}\\ \therefore \mathrm{gof}=\mathrm{fog}={\mathrm{I}}_{\mathrm{R}}\\ \mathrm{Hence},\mathrm{f}\mathrm{is}\mathrm{invertible}\mathrm{and}\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{f}\mathrm{is}\mathrm{given}\mathrm{by}\\ {\mathrm{f}}^{-1}\left(\mathrm{y}\right)=\mathrm{g}\left(\mathrm{y}\right)\\ \mathrm{ }=\frac{\mathrm{y}-3}{4}.\\ ⇒ {\mathrm{f}}^{-1}\left(\mathrm{x}\right)=\frac{\mathrm{x}-3}{4}\end{array}$

Q.8

$\begin{array}{l}Considerf:{R}_{+}\to \left[4,\infty \right)givenbyf\left(x\right)={x}^{2}+4.Showthatfis\\ invertiblewiththeinverse{f}^{–1}ofgivenfby,where{R}^{+}istheset\\ ofallnon–negativereal\text{\hspace{0.17em}}\text{\hspace{0.17em}}numbers.\end{array}$

Ans

$f: R + → [4, ∞ ) is given as f(x) = x 2 + 4. One-one: Let f(x) = f(y) ⇒ x 2 + 4= y 2 + 4 ⇒ x 2 = y 2 ⇒ x=y as x=y∈ R + ∴ f is one−one function. Onto: For y∈ [4,∞), let y= x 2 + 4 x= y−4 ≥0 Therefore, for any y∈R, there exists x= y−4 ≥0, such that f x =f y−4 = y−4 2 +4 =y−4+4 f x =y ∴ f is onto. Thus, f is one-one and onto and therefore, f -1 exists. Let us define g: [4,∞)→ R + by, g y = y−4 Now, gof x =g f x =g x 2 +4 = x 2 +4 −4 gof x =x fog y =f g y =f y−4 = y−4 2 +4 =y−4+4 fog y =y ∴gof= fog= I R + . Hence, f is invertible and the inverse of f is given by f −1 y =g y = y−4 ⇒ f −1 x = x−4 MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaabAgacaqG6aGaaeiiaiaabkfadaWgaaWcbaGaae4kaaqa baGccqGHsgIRcaqGGaGaae4waiaabsdacaqGSaGaaeiiaiabg6HiLk aabMcacaqGGaGaaeyAaiaabohacaqGGaGaae4zaiaabMgacaqG2bGa aeyzaiaab6gacaqGGaGaaeyyaiaabohacaqGGaGaaeOzaiaabIcaca qG4bGaaeykaiaabccacaqG9aGaaeiiaiacycyG4bWaiGjGCaaaleqc ycyaiGjGcGaMagOmaaaakiacycyGGaGaiGjGbUcacGaMagiiaiacyc yG0aGaaeOlaaqaaiaab+eacaqGUbGaaeyzaiaab2cacaqGVbGaaeOB aiaabwgacaqG6aaabaGaaGPaVlaaykW7caaMc8UaaGPaVlaabYeaca qGLbGaaeiDaiaabccacaqGMbGaaeikaiaabIhacaqGPaGaaeiiaiaa b2dacaqGGaGaaeOzaiaabIcacaqG5bGaaeykaaqaaiabgkDiElaayk W7caaMc8UaaGPaVlaabIhadaahaaWcbeqaaiaabkdaaaGccaqGGaGa 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Q.9

$\begin{array}{l}Considerf:{R}_{+}\to \left[–5,\infty \right)givenbyf\left(x\right)=9{x}^{2}+6x–5.\\ Showthatfisinvertiblewith{f}^{-1}\left(y\right)=\left(\frac{\left(\sqrt{y+6}\right)-1}{3}\right)\end{array}$

Ans

$f: R + → [−5, ∞ ) is given as f(x) = 9x 2 + 6x−5. Let y be an arbitrary element of [−5,∞). Let y = 9x 2 + 6x−5 ⇒ 9x 2 + 6x− 5+y =0 ⇒ x= −6± 6 2 −4×9×− 5+y 2×9 = −6± 36+36 5+y 18 = −6±6 1+5+y 18 x= −1± 6+y 3 = 6+y −1 3 âˆµy≥−6⇒y+6≥0 ∴f is onto, thereby range f = [−5, ∞). Let us define g: [−5, ∞)→ R + as g y = y+6 −1 3 Now, gof x =g f x =g 9x 2 + 6x−5 = 9x 2 + 6x−5 +6 −1 3 = 9x 2 + 6x+1 −1 3 = 3x+1 2 −1 3 = 3x+1−1 3 =x And, fog y =f g y =f y+6 −1 3 =9 y+6 −1 3 2 + 6 y+6 −1 3 −5 =9 y+6−2 y+6 +1 9 +2 y+6 −2−5 =y+6−2 y+6 +1+2 y+6 −2−5 fog y =y ∴ gof= I R and fog= I −5,∞ Hence, f is invertible and the inverse of f is given by f −1 y =g y = y+6 −1 3 . 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Q.10

$\text{Let f :X→Y be an invertible function.Show that f has unique inverse.}$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{f}:\mathrm{X}\to \mathrm{Y}\mathrm{be}\mathrm{an}\mathrm{invertible}\mathrm{function}\mathrm{.}\\ \mathrm{Also},\mathrm{suppose}\mathrm{f}\mathrm{has}\mathrm{two}\mathrm{inverses}\left(\mathrm{say}{\mathrm{g}}_{\mathrm{1}}\mathrm{and}{\mathrm{g}}_{\mathrm{2}}\mathrm{\right)}\\ \mathrm{Then},\mathrm{for}\mathrm{all}\mathrm{y}\in \mathrm{Y},\mathrm{we}\mathrm{have}:\\ {\mathrm{fog}}_{\mathrm{1}}\left(\mathrm{y}\right)={\mathrm{I}}_{\mathrm{Y}}\left(\mathrm{y}\right)={\mathrm{fog}}_{2}\left(\mathrm{y}\right)\\ ⇒ \mathrm{f}\left({\mathrm{g}}_{1}\left(\mathrm{y}\right)\right)=\mathrm{f}\left({\mathrm{g}}_{2}\left(\mathrm{y}\right)\right)\\ ⇒ {\mathrm{g}}_{1}\left(\mathrm{y}\right)={\mathrm{g}}_{2}\left(\mathrm{y}\right)\left[\mathrm{f}\mathrm{is}\mathrm{invertible}⇒\mathrm{f}\mathrm{is}\mathrm{one}–\mathrm{one}\right]\\ ⇒ {\mathrm{g}}_{1}={\mathrm{g}}_{2}\left[\mathrm{g}\mathrm{is}\mathrm{one}–\mathrm{one}\right]\\ \mathrm{Hence},\mathrm{f}\mathrm{has}\mathrm{a}\mathrm{unique}\mathrm{inverse}\mathrm{.}\end{array}$

Q.11

$\begin{array}{l}\mathrm{Consider}\mathrm{f}:\left\{1,2,3\right\}\to \left\{\mathrm{a},\mathrm{b},\mathrm{c}\right\}\mathrm{given}\mathrm{by}\mathrm{f}\left(1\right)=\mathrm{a},\mathrm{f}\left(2\right)=\mathrm{b}\\ \mathrm{and}\mathrm{f}\left(3\right)=\mathrm{c}.\mathrm{Find}{\mathrm{f}}^{–1}\mathrm{and}\mathrm{show}\mathrm{that}{\left({\mathrm{f}}^{–1}\right)}^{–1}=\mathrm{f}.\end{array}$

Ans

$\begin{array}{l}\mathrm{Function}\mathrm{f}: \left\{1, 2, 3\right\}\to \left\{\mathrm{a},\mathrm{b},\mathrm{c}\right\}\mathrm{is}\mathrm{given}\mathrm{by},\\ \mathrm{f}\left(1\right) =\mathrm{a},\mathrm{f}\left(2\right) =\mathrm{b},\mathrm{and}\mathrm{f}\left(3\right) =\mathrm{c}\\ \mathrm{If}\mathrm{we}\mathrm{define}\mathrm{g}: \left\{\mathrm{a},\mathrm{b},\mathrm{c}\right\}\to \left\{1, 2, 3\right\}\mathrm{as}\mathrm{g}\left(\mathrm{a}\right) = 1,\mathrm{g}\left(\mathrm{b}\right) = 2,\\ \mathrm{g}\left(\mathrm{c}\right) = 3,\mathrm{then}\mathrm{we}\mathrm{have}:\\ \left(\mathrm{fog}\right)\left(\mathrm{a}\right)=\mathrm{f}\left(\mathrm{g}\left(\mathrm{a}\right)\right)=\mathrm{f}\left(1\right)=\mathrm{a}\\ \left(\mathrm{fog}\right)\left(\mathrm{b}\right)=\mathrm{f}\left(\mathrm{g}\left(\mathrm{b}\right)\right)=\mathrm{f}\left(2\right)=\mathrm{b}\\ \left(\mathrm{fog}\right)\left(\mathrm{c}\right)=\mathrm{f}\left(\mathrm{g}\left(\mathrm{c}\right)\right)=\mathrm{f}\left(3\right)=\mathrm{c}\\ \mathrm{and}\\ \left(\mathrm{gof}\right)\left(1\right)=\mathrm{g}\left(\mathrm{f}\left(1\right)\right)=\mathrm{g}\left(\mathrm{a}\right)=1\\ \left(\mathrm{gof}\right)\left(2\right)=\mathrm{g}\left(\mathrm{f}\left(2\right)\right)=\mathrm{g}\left(\mathrm{b}\right)=2\\ \left(\mathrm{gof}\right)\left(3\right)=\mathrm{g}\left(\mathrm{f}\left(3\right)\right)=\mathrm{g}\left(\mathrm{c}\right)=3\\ \therefore \mathrm{gof}={\mathrm{I}}_{\mathrm{X}}\mathrm{and}\mathrm{fog}={\mathrm{I}}_{\mathrm{Y}},\mathrm{where}\mathrm{X}=\left\{1,2,3\right\}\mathrm{and}\mathrm{Y}=\left\{\mathrm{a},\mathrm{b},\mathrm{c}\right\}.\\ \mathrm{Thus},\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{f}\mathrm{exists}\mathrm{and}{\mathrm{f}}^{–1}=\mathrm{g}\mathrm{.}\\ \therefore {\mathrm{f}}^{-1}:\left\{\mathrm{a},\mathrm{b},\mathrm{c}\right\}\to \left\{1,2,3\right\}\mathrm{is}\mathrm{given}\mathrm{by},\\ {\mathrm{f}}^{-\mathrm{1}}\left(\mathrm{a}\right)=1,{\mathrm{f}}^{-\mathrm{1}}\left(\mathrm{b}\right)=2,{\mathrm{f}}^{-\mathrm{1}}\left(\mathrm{c}\right)=3\\ \mathrm{The}\mathrm{inverse}\mathrm{of}{\mathrm{f}}^{–1}\mathrm{}⇒\mathrm{The}\mathrm{inverse}\mathrm{of}\mathrm{g}\\ \mathrm{We}\mathrm{define},\mathrm{h}:\left\{1,2,3\right\}\to \left\{\mathrm{a},\mathrm{b},\mathrm{c}\right\} \mathrm{as}\\ \mathrm{h}\left(1\right) =\mathrm{a},\mathrm{h}\left(2\right) =\mathrm{b},\mathrm{h}\left(3\right) =\mathrm{c},\mathrm{then}\mathrm{we}\mathrm{have}:\\ \left(\mathrm{goh}\right)\left(1\right)=\mathrm{g}\left(\mathrm{h}\left(1\right)\right)=\mathrm{g}\left(\mathrm{a}\right)=1\\ \left(\mathrm{goh}\right)\left(2\right)=\mathrm{g}\left(\mathrm{h}\left(2\right)\right)=\mathrm{g}\left(\mathrm{b}\right)=2\\ \left(\mathrm{goh}\right)\left(3\right)=\mathrm{g}\left(\mathrm{h}\left(3\right)\right)=\mathrm{g}\left(\mathrm{c}\right)=3\\ \mathrm{and}\\ \left(\mathrm{hog}\right)\left(\mathrm{a}\right)=\mathrm{h}\left(\mathrm{g}\left(\mathrm{a}\right)\right)=\mathrm{h}\left(1\right)=\mathrm{a}\\ \left(\mathrm{hog}\right)\left(\mathrm{b}\right)=\mathrm{h}\left(\mathrm{g}\left(\mathrm{b}\right)\right)=\mathrm{h}\left(2\right)=\mathrm{b}\\ \left(\mathrm{hog}\right)\left(\mathrm{c}\right)=\mathrm{h}\left(\mathrm{g}\left(\mathrm{c}\right)\right)=\mathrm{h}\left(3\right)=\mathrm{c}\\ \therefore \mathrm{goh}={\mathrm{I}}_{\mathrm{X}}\mathrm{ }\mathrm{and} \mathrm{hog}={\mathrm{I}}_{\mathrm{Y}},\mathrm{ }\mathrm{where}\mathrm{ }\mathrm{X}=\left\{1,2,3\right\}\mathrm{and}=\left\{\mathrm{a},\mathrm{b},\mathrm{c}\right\}.\\ \mathrm{Thus},\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{g}\mathrm{exists}\mathrm{and}\mathrm{ }{\mathrm{g}}^{-1}=\mathrm{h}⇒{\left({\mathrm{f}}^{-\mathrm{1}}\right)}^{-\mathrm{1}}=\mathrm{h}\mathrm{.}\\ \mathrm{It}\mathrm{can}\mathrm{be}\mathrm{noted}\mathrm{that}\mathrm{h}=\mathrm{f}\mathrm{.}\\ \mathrm{Hence},{\left({\mathrm{f}}^{–1}\right)}^{–1}=\mathrm{f}\mathrm{.}\end{array}$

Q.12

$\begin{array}{l}\mathrm{Let}\mathrm{f}:\mathrm{X}\to \mathrm{Y}\mathrm{be}\mathrm{aninvertible}\mathrm{function}.\mathrm{Show}\mathrm{that}\mathrm{the}\mathrm{inverse}\\ \mathrm{of}{\mathrm{f}}^{–1}\mathrm{is}\mathrm{f},\mathrm{i}.\mathrm{e}.,{\left({\mathrm{f}}^{–1}\right)}^{–1}=\mathrm{f}.\end{array}$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{f}:\mathrm{X}\to \mathrm{Y}\mathrm{be}\mathrm{an}\mathrm{invertible}\mathrm{function}\mathrm{.}\\ \mathrm{Then},\mathrm{there}\mathrm{exists}\mathrm{a}\mathrm{function}\mathrm{g}:\mathrm{Y}\to \mathrm{X}\mathrm{such}\mathrm{that}\mathrm{gof}={\mathrm{I}}_{\mathrm{X}}\\ \mathrm{and}\mathrm{fog}={\mathrm{I}}_{\mathrm{Y}}\mathrm{.}\\ \mathrm{Here},{\mathrm{f}}^{-1}=\mathrm{g}\mathrm{.}\\ \mathrm{Now},\mathrm{gof}={\mathrm{I}}_{\mathrm{X}}\mathrm{and}\mathrm{fog}={\mathrm{I}}_{\mathrm{Y}}\\ ⇒{\mathrm{f}}^{–1}\mathrm{of}={\mathrm{I}}_{\mathrm{X}}\mathrm{ }\mathrm{and}{\mathrm{fof}}^{–1}={\mathrm{I}}_{\mathrm{Y}}\\ \mathrm{Hence},{\mathrm{f}}^{-\mathrm{1}}:\mathrm{Y}\to \mathrm{X}\mathrm{is}\mathrm{invertible}\mathrm{and}\mathrm{f}\mathrm{is}\mathrm{the}\mathrm{inverse}\mathrm{of}{\mathrm{f}}^{-\mathrm{1}}\\ \mathrm{i}.\mathrm{e}., \left({\mathrm{f}}^{-\mathrm{1}}{\mathrm{\right)}}^{-1}=\mathrm{f}\mathrm{.}\end{array}$

Q.13

$\begin{array}{l}\mathrm{If}\mathrm{f}:\mathrm{R}\to \mathrm{R}\mathrm{be}\mathrm{given}\mathrm{by},\mathrm{f}\left(\mathrm{x}\right)={\left(3-{\mathrm{x}}^{3}\right)}^{\frac{1}{3}}\mathrm{ }\mathrm{then}\mathrm{f}\mathrm{of}\left(\mathrm{x}\right)\mathrm{is}\\ \left(\mathrm{A}\right){\mathrm{x}}^{\frac{1}{3}} \left(\mathrm{B}\right){\mathrm{x}}^{3}\left(\mathrm{C}\right)\mathrm{x}\left(\mathrm{D}\right)\left(3-{\mathrm{x}}^{3}\right).\end{array}$

Ans

$\begin{array}{l}\mathrm{f}:\mathrm{R}\to \mathrm{R}\mathrm{is}\mathrm{given}\mathrm{as} \mathrm{f}\left(\mathrm{x}\right)={\left(3-{\mathrm{x}}^{3}\right)}^{\frac{1}{3}}\\ \therefore \mathrm{fof}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\\ \mathrm{ }=\mathrm{f}\left\{{\left(3-{\mathrm{x}}^{3}\right)}^{\frac{1}{3}}\right\}\\ \mathrm{ }={\left[3-{\left\{{\left(3-{\mathrm{x}}^{3}\right)}^{\frac{1}{3}}\right\}}^{3}\right]}^{\frac{1}{3}}\\ \mathrm{ }={\left\{3-\left(3-{\mathrm{x}}^{3}\right)\right\}}^{\frac{1}{3}}\\ \mathrm{ }={\left(3-3+{\mathrm{x}}^{3}\right)}^{\frac{1}{3}}\\ \mathrm{ }\mathrm{fof}\left(\mathrm{x}\right)=\mathrm{x}\\ \mathrm{The}\mathrm{correct}\mathrm{answer}\mathrm{is}\left(\mathrm{C}\right)\mathrm{.}\\ \end{array}$

Q.14

$\begin{array}{l}\mathrm{Let} \mathrm{f}:\mathrm{R}-\left\{-\frac{4}{3}\right\}\to \mathrm{R}\mathrm{be}\mathrm{a}\mathrm{function}\mathrm{defined}\mathrm{as}\mathrm{f}\left(\mathrm{x}\right)=\frac{4\mathrm{x}}{3\mathrm{x}+4}.\\ \mathrm{The}\mathrm{inverse}\mathrm{of}\mathrm{f}\mathrm{is}\mathrm{map}\mathrm{g}:\mathrm{R}\mathrm{ange}\mathrm{f}\to \mathrm{R}-\left\{-\frac{4}{3}\right\}\mathrm{given}\mathrm{by}\\ \left(\mathrm{A}\right)\mathrm{ }\mathrm{g}\left(\mathrm{y}\right)=\frac{3\mathrm{y}}{3-4\mathrm{y}}\left(\mathrm{B}\right)\mathrm{ }\mathrm{g}\left(\mathrm{y}\right)=\frac{4\mathrm{y}}{4-3\mathrm{y}}\\ \left(\mathrm{C}\right)\mathrm{ }\mathrm{g}\left(\mathrm{y}\right)=\frac{4\mathrm{y}}{3-4\mathrm{y}}\left(\mathrm{D}\right)\mathrm{g}\left(\mathrm{y}\right)=\frac{3\mathrm{y}}{4-3\mathrm{y}}\end{array}$

Ans

$\begin{array}{l}\mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}\mathrm{ }\mathrm{f}:\mathrm{R}-\left\{-\frac{4}{3}\right\}\to \mathrm{R}\mathrm{ }\mathrm{is}\mathrm{defined}\mathrm{as}\mathrm{f}\left(\mathrm{x}\right)=\frac{4\mathrm{x}}{3\mathrm{x}+4}.\\ \mathrm{Let}\mathrm{ }\mathrm{y}\mathrm{be}\mathrm{an}\mathrm{arbitrary}\mathrm{element}\mathrm{of}\mathrm{Range}\mathrm{f}\mathrm{.}\\ \mathrm{Then},\mathrm{there}\mathrm{exists}\mathrm{x}\in \mathrm{R}-\left\{-\frac{4}{3}\right\}\mathrm{ }\mathrm{such}\mathrm{ }\mathrm{that}\mathrm{ }\mathrm{y}=\mathrm{f}\left(\mathrm{x}\right).\\ ⇒ \mathrm{ }\mathrm{y}=\frac{4\mathrm{x}}{3\mathrm{x}+4}\\ ⇒3\mathrm{xy}+4\mathrm{y}=4\mathrm{x}\\ ⇒-3\mathrm{xy}+4\mathrm{x}=4\mathrm{y}\\ ⇒\mathrm{ }\mathrm{x}\left(4-3\mathrm{y}\right)=4\mathrm{y}\\ ⇒ \mathrm{ }\mathrm{x}=\frac{4\mathrm{y}}{4-3\mathrm{y}}\\ \mathrm{Let}\mathrm{us}\mathrm{define}\mathrm{g}:\mathrm{Range} \mathrm{f}\to \mathrm{R}-\left\{-\frac{4}{3}\right\} \mathrm{as}\mathrm{ }\mathrm{g}\left(\mathrm{y}\right)=\frac{4\mathrm{y}}{4-3\mathrm{y}}\mathrm{ }\\ \mathrm{Now},\mathrm{ }\left(\mathrm{gof}\right)\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\\ \\ \mathrm{ }=\mathrm{g}\left(\frac{4\mathrm{x}}{3\mathrm{x}+4}\right)\\ \mathrm{ }=\frac{4\left(\frac{4\mathrm{x}}{3\mathrm{x}+4}\right)}{4-3\left(\frac{4\mathrm{x}}{3\mathrm{x}+4}\right)}\\ \mathrm{ }=\frac{16\mathrm{x}}{12\mathrm{x}+16-12\mathrm{x}}\\ \mathrm{ }=\mathrm{x}\\ \mathrm{And},\mathrm{ }\left(\mathrm{fog}\right)\left(\mathrm{y}\right)=\mathrm{f}\left(\mathrm{g}\left(\mathrm{y}\right)\right)\\ \mathrm{ }=\mathrm{f}\left(\frac{4\mathrm{y}}{4-3\mathrm{y}}\right)\\ \mathrm{ }=\frac{4\left(\frac{4\mathrm{y}}{4-3\mathrm{y}}\right)}{3\left(\frac{4\mathrm{y}}{4-3\mathrm{y}}\right)+4}\\ \mathrm{ }=\frac{16\mathrm{y}}{12\mathrm{y}+16-12\mathrm{y}}\\ \mathrm{ }=\frac{16\mathrm{y}}{16}\\ \mathrm{ }=\mathrm{y}\\ \therefore \mathrm{gof}={\mathrm{I}}_{\mathrm{R}-\left\{-\frac{4}{3}\right\}} \mathrm{and} \mathrm{fog}={\mathrm{I}}_{\mathrm{Range}\mathrm{ }\mathrm{f}}\\ \mathrm{Thus},\mathrm{g}\mathrm{is}\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{f}\mathrm{i}.\mathrm{e}.,{\mathrm{f}}^{–1}=\mathrm{g}\mathrm{.}\\ \mathrm{Hence},\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{f}\mathrm{is}\mathrm{the}\mathrm{map}\mathrm{g}:\mathrm{Range}, \mathrm{R}\to \mathrm{R}-\left\{-\frac{4}{3}\right\},\\ \mathrm{which}\mathrm{is}\mathrm{given}\mathrm{by} \mathrm{g}\left(\mathrm{y}\right)=\frac{4\mathrm{y}}{4-3\mathrm{y}}.\\ \mathrm{The}\mathrm{correct}\mathrm{answer}\mathrm{is}\mathrm{B}\mathrm{.}\end{array}$