Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# NCERT Solutions for Class 12 Maths Miscellaneous Exercise Chapter 1 - Relation and Functions

Last updated date: 13th Aug 2024
Total views: 10.5k
Views today: 0.10k

## NCERT for Class 12 Maths Chapter 1 Miscellaneous Exercise Solutions - Free PDF Download

NCERT Solutions for Class 12 Maths Chapter 1 Relation and Functions includes solutions to all Miscellaneous Exercise problems. Relation and Functions Class 12 NCERT Solutions Miscellaneous Exercises are based on the concepts presented in Maths Chapter 1. This activity is crucial for both the CBSE Board examinations and competitive tests. To perform well on the board exam, download the NCERT for Class 12 Maths Chapter 1 Miscellaneous Exercise Solutions in PDF format and practice them offline. Students can download the revised Class 12 Maths NCERT Solutions from our page, which is prepared so that you can understand it easily.

Table of Content
1. NCERT for Class 12 Maths Chapter 1 Miscellaneous Exercise Solutions - Free PDF Download
2. Access NCERT Solutions for Class 12 Maths Chapter 1 Relation and Functions
2.1Miscellaneous Exercise
3. Class 12 Maths Chapter 1: Exercises Breakdown
4. CBSE Class 12 Maths Chapter 1 Other Study Materials
5. Chapter-Specific NCERT Solutions for Class 12 Maths
FAQs

Class 12 Chapter 1 Maths Miscellaneous Exercise Solutions are aligned with the updated CBSE guidelines for Class 12, ensuring students are well-prepared for exams. Access the latest Class 12 Maths Syllabus here.

Competitive Exams after 12th Science

## Access NCERT Solutions for Class 12 Maths Chapter 1 Relation and Functions

### Miscellaneous Exercise

1. Show that function $\text{f: R }\to \text{ }\!\!\{\!\!\text{ x}\in \text{R:-1 x 1 }\!\!\}\!\!\text{ }$ defined by $\text{f}\left( \text{x} \right)\text{=}\dfrac{\text{x}}{\text{1+ }\!\!|\!\!\text{ x }\!\!|\!\!\text{ }}\text{,x}\in \text{R}$ is one – one and onto function.

Ans: The function $\text{f: R }\to \text{ }\!\!\{\!\!\text{ x}\in \text{R:-1 x 1 }\!\!\}\!\!\text{ }$ is defined as $\text{f}\left( \text{x} \right)\text{=}\dfrac{\text{x}}{\text{1+ }\!\!|\!\!\text{ x }\!\!|\!\!\text{ }}\text{,x}\in \text{R}$.

For the function $\text{f}$ to be one – one:

$\text{f}\left( \text{x} \right)\text{ = f}\left( \text{y} \right)$, where $\text{x, y}\in \text{R}$.

$\Rightarrow \dfrac{\text{x}}{\text{1+ }\!\!|\!\!\text{ x }\!\!|\!\!\text{ }}\text{=}\dfrac{\text{y}}{\text{1+ }\!\!|\!\!\text{ y }\!\!|\!\!\text{ }}$

Assuming that $\text{x}$ is positive and $\text{y}$ is negative:

$\dfrac{\text{x}}{\text{1+x}}\text{=}\dfrac{\text{y}}{\text{1+y}}$

$\Rightarrow \text{2xy=x-y}$

Since, $\text{x y}\Rightarrow \text{x-y 0}$.

But $\text{2xy}$ is negative.

Therefore, $\text{2xy }\ne \text{ x - y}$.

Hence, $\text{x}$ being positive and $\text{y}$ being negative is not possible. Similarly $\text{x}$ being negative and $\text{y}$ being positive can also be ruled out.

So, $\text{x}$ and $\text{y}$ have to be either positive or negative.

Assuming that both $\text{x}$ and $\text{y}$ are positive:

$\text{f(x)=f(y)}$

$\Rightarrow \dfrac{\text{x}}{\text{1+x}}\text{=}\dfrac{\text{y}}{\text{1+y}}$

$\Rightarrow \text{x+xy=y+xy}$

$\Rightarrow \text{x=y}$

Assuming that both $\text{x}$ and $\text{y}$ are negative:

$\text{f(x)=f(y)}$

$\Rightarrow \dfrac{\text{x}}{\text{1+x}}\text{=}\dfrac{\text{y}}{\text{1+y}}$

$\Rightarrow \text{x+xy=y+xy}$

$\Rightarrow \text{x=y}$

Therefore, the function $\text{f}$ is one – one.

For onto:

$\text{y}\in \text{R}$ such that $\text{-1 y 1}$.

If $\text{y}$ is negative, then, there exists $\text{x = }\dfrac{\text{y}}{\text{1+y}}\in \text{R}$ such that

$\text{f}\left( \dfrac{\text{y}}{\text{1+y}} \right)\text{=}\dfrac{\left( \dfrac{\text{y}}{\text{1+y}} \right)}{\text{1+}\left| \dfrac{\text{y}}{\text{1+y}} \right|}$

$\text{=}\dfrac{\dfrac{\text{y}}{\text{1+y}}}{\text{1+}\left( \dfrac{\text{-y}}{\text{1+y}} \right)}$

$\text{=}\dfrac{\text{y}}{\text{1+y-y}}$

$\text{=y}$

If $\text{y}$ is positive, then, there exists $\text{x = }\dfrac{\text{y}}{\text{1-y}}\in \text{R}$ such that

$\text{f}\left( \dfrac{\text{y}}{\text{1-y}} \right)\text{=}\dfrac{\left( \dfrac{\text{y}}{\text{1-y}} \right)}{\text{1+}\left| \dfrac{\text{y}}{\text{1-y}} \right|}$

$\text{=}\dfrac{\dfrac{\text{y}}{\text{1-y}}}{\text{1+}\left( \dfrac{\text{-y}}{\text{1-y}} \right)}$

$\text{=}\dfrac{\text{y}}{\text{1-y+y}}$

$\text{=y}$

Therefore, the function $\text{f}$ is onto.

Hence the given function $\text{f}$ is both one–one and onto.

2. Show that the function $\text{f: R }\to \text{ R}$ given by $\text{f}\left( \text{x} \right)\text{ = }{{\text{x}}^{\text{3}}}$ is injective.

Ans: The given function $\text{f: R }\to \text{ R}$ is given as $\text{f}\left( \text{x} \right)\text{ = }{{\text{x}}^{\text{3}}}$.

For the function $\text{f}$ to be one – one:

$\text{f}\left( \text{x} \right)\text{ = f}\left( \text{y} \right)$ where $\text{x, y}\in \text{R}$.

$\Rightarrow {{\text{x}}^{\text{3}}}\text{=}{{\text{y}}^{\text{3}}}$           …… (1)

We need to show that $\text{x=y}$.

Assuming that $\text{x}\ne \text{y}$, then,

$\Rightarrow {{\text{x}}^{\text{3}}}\ne {{\text{y}}^{\text{3}}}$

Since this is a contradiction to (1), therefore, $\text{x=y}$.

Hence, the function $\text{f}$ is injective.

3. Given a non-empty set ${X}$, consider ${P}\left( {X} \right)$ which is the set of all subsets of ${X}$. Define the relation ${R}$ in ${P}\left( {X} \right)$ as follows:

For subsets ${A,B}$ in ${P}\left( {X} \right)$, ${ARB}$ if and only if ${A}\subset {B}$. Is ${R}$ an equivalence relation on ${P}\left( {X} \right)$? Justify your answer.

Ans: We know that every set is a subset of itself, ${ARA}$ for all ${A}\in {P}\left( {X} \right)$

Therefore ${R}$ is reflexive.

Let ${ARB}\Rightarrow {A}\subset {B}$.

This does not mean that ${B}\subset {A}$.

If ${A = }\left\{ {1, 2} \right\}$ and ${B = }\left\{ {1, 2, 3} \right\}$, then it cannot be implied that ${B}$ is related to ${A}$.

Therefore ${R}$ is not symmetric.

If ${ARB}$ and ${BRC}$ , then;

${A}\subset {B}$ and ${B}\subset {C}$

$\Rightarrow {A}\subset {C}$

$\Rightarrow {ARC}$

Therefore ${R}$ is transitive.

Hence, ${R}$ is not an equivalence relation as it is not symmetric.

4. Find the number of all onto functions from the set ${ }\!\!\{\!\!{ 1, 2, 3, }...{, n }\!\!\}\!\!{ }$ to itself.

Ans: The total number of onto maps from ${ }\!\!\{\!\!{ 1, 2, 3, }...{ , n }\!\!\}\!\!{ }$ to itself will be same as the total number of permutations on ${n}$ symbols ${1, 2, 3, }...{ , n}$.

Since the total number of permutations on ${n}$ symbols ${1, 2, 3, }...{ , n}$ is ${n}$, thus total number of onto maps from ${ }\!\!\{\!\!{ 1, 2, 3, }...{ , n }\!\!\}\!\!{ }$ to itself are ${n}$.

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

Ans: Let ${A=}\left\{ {-1, 0, 1, 2} \right\}{,B=}\left\{ {-4, -2, 0, 2} \right\}$ and ${f,g:A }\to { B}$ are defined by ${f}\left( {x} \right){=}{{{x}}^{{2}}}{-x,}\,{x}\in {A}$ and ${g}\left( {x} \right){=2}\left| {x-}\dfrac{{1}}{{2}} \right|{-1,x}\in {A}$.

${f}\left( {-1} \right){=}{{\left( {-1} \right)}^{{2}}}{-}\left( {-1} \right)$

${=1+1}$

${=2}$

And,

${g}\left( {-1} \right){=2}\left| \left( {-1} \right){-}\dfrac{{1}}{{2}} \right|{-1}$

${=2}\left( \dfrac{{3}}{{2}} \right){-1}$

${=3-1}$

${=2}$

$\Rightarrow {f}\left( {-1} \right){=g}\left( {-1} \right)$

${f}\left( {0} \right){=}{{\left( {0} \right)}^{{2}}}{-}\left( {0} \right)$

${=0}$

And,

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

${=1-1}$

${=0}$

$\Rightarrow {f}\left( {0} \right){=g}\left( {0} \right)$

${f}\left( {1} \right){=}{{\left( {1} \right)}^{{2}}}{-}\left( {1} \right)$

${=1-1}$

${=0}$

And,

${g}\left( {1} \right){=2}\left| \left( {1} \right){-}\dfrac{{1}}{{2}} \right|{-1}$

${=2}\left( \dfrac{1}{{2}} \right){-1}$

${=1-1}$

${=0}$

$\Rightarrow {f}\left( {1} \right){=g}\left( {1} \right)$

${f}\left( {2} \right){=}{{\left( {2} \right)}^{{2}}}{-}\left( {2} \right)$

${=4-2}$

${=2}$

And,

${g}\left( {2} \right){=2}\left| \left( {2} \right){-}\dfrac{{1}}{{2}} \right|{-1}$

${=2}\left( \dfrac{{3}}{{2}} \right){-1}$

${=3-1}$

${=2}$

$\Rightarrow {f}\left( {2} \right){=g}\left( {2} \right)$

Therefore, ${f}\left( {a} \right){=g}\left( {a} \right)\forall {a}\in {A}$. Hence the functions ${f}$ and ${g}$ are equal.

6. Let $\mathbf{\text{A=}\left\{ \text{1, 2, 3} \right\}}$ Then number of relations containing $\mathbf{\left( \text{1, 2} \right)}$ and $\mathbf{\left( \text{1, 3} \right)}$  which are reflexive and symmetric but not transitive is

(A) $\mathbf{\text{1}}$

(B) $\mathbf{\text{2}}$

(C) $\mathbf{\text{3}}$

(D) $\mathbf{\text{4}}$

Ans: We are given a set $\text{A=}\left\{ \text{1, 2, 3} \right\}$.

Let us take the relation $\text{R}$, containing $\left( \text{1, 2} \right)$ and $\left( \text{1, 3} \right)$, as $\text{R=}\left\{ \left( \text{1, 1} \right)\text{, }\left( \text{2, 2} \right)\text{, }\left( \text{3, 3} \right)\text{, }\left( \text{1, 2} \right)\text{, }\left( \text{1, 3} \right)\text{, }\left( \text{2, 1} \right)\text{, }\left( \text{3, 1} \right) \right\}$.

As we can see that $\left( \text{1, 1} \right)\text{, }\left( \text{2, 2} \right)\text{, }\left( \text{3, 3} \right)\in \text{R}$, therefore relation $\text{R}$ is reflexive.

Since $\left( \text{1, 2} \right)\text{, }\left( \text{1, 3} \right)\text{, }\left( \text{2, 1} \right)\in \text{R}$, the relation $\text{R}$ is symmetric.

The relation Relation $\text{R}$ is not transitive because $\left( \text{1, 2} \right)\text{,}\,\left( \text{3, 1} \right)\in \text{R}$, but $\left( \text{3, 2} \right)\notin \text{R}$.

The relation Relation $\text{R}$ will become transitive on adding and two pairs $\left( \text{3, 2} \right)\text{, }\left( \text{2, 3} \right)$.

Therefore the total number of desired relations is one.

The correct answer is option $(A)$ $\text{1}$.

7. Let $\mathbf{\text{A=}\left\{ \text{1, 2, 3} \right\}}$ Then number of equivalence relations containing $\mathbf{\left( \text{1, 2} \right)}$ is

(A) $\mathbf{\text{1}}$

(B) $\mathbf{\text{2}}$

(C) $\mathbf{\text{3}}$

(D) $\mathbf{\text{4}}$

Ans: We are given a set $\text{A=}\left\{ \text{1, 2, 3} \right\}$.

Let us take the relation $\text{R}$, containing $\left( \text{1, 2} \right)$ as $\text{R=}\left\{ \left( \text{1, 1} \right)\text{, }\left( \text{2, 2} \right)\text{, }\left( \text{3, 3} \right)\text{, }\left( \text{1, 2} \right)\text{, }\left( \text{2, 1} \right) \right\}$.

Now the pairs left are $\left( \text{2, 3} \right)\text{, }\left( \text{3, 2} \right)\text{, }\left( \text{1, 3} \right)\text{, }\left( \text{3, 1} \right)$

In order to add one pair, say $\left( \text{2, 3} \right)$, we must add $\left( \text{3, 2} \right)$ for symmetry. And we are required to add $\left( \text{1, 3} \right)\text{, }\left( \text{3, 1} \right)$ for transitivity.

So, only the equivalence relation (bigger than $\text{R}$) is the universal relation.

Therefore, the total number of equivalence relations containing $\left( \text{1, 2} \right)$ are two.

Hence, the correct answer is (B) $\text{2}$.

## Conclusion

Miscellaneous Exercise Class 12 Chapter 1 of Maths is important for understanding various concepts thoroughly. Relations and Functions Miscellaneous Exercise Class 12 covers diverse problems that require the application of multiple formulas and techniques. It's important to focus on understanding the underlying principles behind each question rather than just memorizing solutions.

## Class 12 Maths Chapter 1: Exercises Breakdown

 Exercise Number of Questions Exercise 1.1 16 Questions & Solutions Exercise 1.2 12 Questions & Solutions

## Chapter-Specific NCERT Solutions for Class 12 Maths

Given below are the chapter-wise NCERT Solutions for Class 12 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 12 Maths Miscellaneous Exercise Chapter 1 - Relation and Functions

1. What topics are covered in the Miscellaneous Exercise of Chapter 1?

The Miscellaneous Exercise in Chapter 1 covers a variety of problems related to relations and functions, including types of relations, types of functions, composition of functions, and inverse functions.

2. How do I identify different types of relations?

Different types of relations include reflexive, symmetric, transitive, and equivalence relations. You can identify them by checking the properties they satisfy.

3. What are the main types of functions discussed?

The main types of functions discussed are one-one (injective), onto (surjective), and bijective functions. Understanding their definitions and properties is crucial.

4. How is the composition of functions performed?

The composition of functions involves applying one function to the result of another function. If ff and gg are functions, then the composition (g∘f)(x)is g(f(x)).

5. What is an inverse function?

An inverse function reverses the effect of the original function. If ff is a function, then its inverse $f^{-1}$ satisfies $(f^{-1}(x))$=x for all x in the domain of $f^{-1}$.

6. Are there any specific formulas I need to remember?

Yes, remember the definitions and properties of different types of relations and functions, the formulas for the composition of functions, and how to find the inverse of a function.

7. How can I solve the problems in the Relations and Functions Miscellaneous Exercise Class 12?

To solve the problems, carefully read each question, understand the concepts involved, and apply the appropriate formulas and methods learned in the chapter.

8. Why are relations and functions important in math?

Relations and functions are fundamental concepts in mathematics that are used in various fields such as algebra, calculus, and computer science. They help in understanding and modelling real-world situations.

9. How can the NCERT solutions help me with miscellaneous exercise?

NCERT solutions provide step-by-step explanations for each problem, helping you understand the methods and concepts required to solve them. They are a valuable resource for exam preparation.

10. Are there any tricky questions in the Miscellaneous Exercise Class 12 Chapter 1?

Yes, the Miscellaneous Exercise includes some challenging questions designed to test your understanding of the entire chapter. Practice them thoroughly to improve your problem-solving skills.