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# NCERT Solutions for Class 11 Maths Miscellaneous Exercise Chapter 2 - Relations and Functions

Last updated date: 12th Aug 2024
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## NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Miscellaneous Exercise

NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions includes solutions to all Miscellaneous Exercise problems. The NCERT Solutions for Maths Class 11 Miscellaneous Exercises are based on the ideas presented in Maths Chapter 2. This activity is crucial for both the CBSE Board examinations and competitive tests.

Table of Content
1. NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Miscellaneous Exercise
2. Access NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions
2.1Miscellaneous Exercise
3. Class 11 Maths Chapter 2: Exercises Breakdown
4. CBSE Class 11 Maths Chapter 2 Other Study Materials
5. Chapter-Specific NCERT Solutions for Class 11 Maths
FAQs

Focus on the different types of relations and functions covered in the chapter. This understanding is important for solving problems effectively and improving your math skills. To do well in exams, download the CBSE Class 11 Maths Syllabus in PDF format and practice them offline regularly.

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## Access NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions

### Miscellaneous Exercise

1. The relation $f$  is defined by $f(x)=\left\{ \begin{matrix} {{x}^{2}},0\le x\le 3 \\ 3x,3\le x\le 10 \\ \end{matrix} \right.$

And the relation $g$ is defined by $g(x)=\left\{ \begin{matrix} {{x}^{2}},0\le x\le 2 \\ 3x,2\le x\le 10 \\ \end{matrix} \right.$

Show that $f$ is a function and $g$ is not a function.

Ans: According to the problem, we have the function $f$ as,

$f(x)=\left\{ \begin{matrix} {{x}^{2}},0\le x\le 3 \\ 3x,3\le x\le 10 \\ \end{matrix} \right.$

We can see that, for $x=3$ ,

$f(x)={{3}^{2}}\,=9$ from the first given condition.

And again, $f(x)=3\times 3=9$ from the second condition.

But now,

$g(x)=\left\{ \begin{matrix} {{x}^{2}},0\le x\le 2 \\ 3x,2\le x\le 10 \\ \end{matrix} \right.$

We can see that, for $x=2$ ,

$f(x)={{2}^{2}}\,=4$ from the first given condition.

And again, $f(x)=3\times 2=6$ from the second condition.

Thus, the domain of the relation $g$ is having two different images from a single element.

So, it can be concluded that the relation is not a function.

2. If $f(x)={{x}^{2}}$ , find $\dfrac{f(1.1)-f(1)}{(1.1-1)}$ .

Ans: We have the function, $f(x)={{x}^{2}}$ .

So, we will have,

$\dfrac{f(1.1)-f(1)}{(1.1-1)}$ equalling to,

$\dfrac{{{(1.1)}^{2}}-{{1}^{2}}}{1.1-1}$ , putting the values.

After further simplification,

$\dfrac{1.21-1}{0.1}$

$=\dfrac{0.21}{0.1}$

$=2.1$

3. Find the domain of the function $f(x)=\dfrac{{{x}^{2}}+2x+1}{{{x}^{2}}-8x+12}$

Ans: According to the problem, we have the given function as,

$f(x)=\dfrac{{{x}^{2}}+2x+1}{{{x}^{2}}-8x+12}$

Let us try to simplify the given function and bring it to a form where we can analyze the problem.

The denominator can be factorized as,

${{x}^{2}}-8x+12$

$={{x}^{2}}-6x-2x+12$

$=x(x-6)-2(x-6)$

$=(x-2)(x-6)$

So, we see that the function is defined for every real numbers except $6,2$ .

Thus, the domain of the function will be, $R-\{2,6\}$

4. Find the domain and the range of the real function $f$ defined by $f(x)=\sqrt{(x-1)}$.

Ans: We have the given function as, $f(x)=\sqrt{(x-1)}$ .

Clearly, the term inside the root sign must be non-negative.

So, the function is valid for all values of $x\ge 1$ .

Thus, the domain of the function will be, $[1,\infty )$ .

Now, again, for $x\ge 1$, the value of the function will always be greater than or equal to zero.

So, the range of the function is, $[0,\infty )$ .

5. Find the domain and the range of the real function $f$ defined by $f(x)=\left| x-1 \right|$.

Ans: The function which is given is, $f(x)=\left| x-1 \right|$ .

We can clearly see that, the function is well defined for all the real numbers.

Thus, it can be concluded that, the domain of the function is $R$ .

And for every $x\in R$ , the function gives all non-negative real numbers.

So, the range of the function is the set of all non-negative real numbers. i.e, $[0,\infty )$ .

6. Let $f=\left\{ \left( x,\dfrac{{{x}^{2}}}{1+{{x}^{2}}} \right):x\in R \right\}$ be a function from $R$ to $R$. Determine the range of $f$ .

Ans: We have our given function as,$f=\left\{ \left( x,\dfrac{{{x}^{2}}}{1+{{x}^{2}}} \right):x\in R \right\}$

Expressing it by term to term, we are getting,

$f=\left\{ \left( 0,0 \right),\left( \pm 0.5,\dfrac{1}{5} \right),\left( \pm 1,\dfrac{1}{2} \right),\left( \pm 1.5,\dfrac{9}{13} \right),\left( \pm 2,\dfrac{4}{5} \right),\left( 3,\dfrac{9}{10} \right),\left( 4,\dfrac{16}{17} \right),.... \right\}$

And we also know, the range of $f$ is the set of all the second elements. We can also see that the terms are greater than or equal to $0$ but less than $1$ .

So, the range of the function is, $[0,1)$ .

7. Let $f,g:R\to R$ be defined, respectively by $f(x)=x+1,g(x)=2x-3$ . Find $f+g,f-g$ and $\dfrac{f}{g}$ .

Ans: We have the functions defined as, $f,g:R\to R$is defined as, $f(x)=x+1,g(x)=2x-3$ .

Thus, the function

$(f+g)(x)=f(x)+g(x)$

$=(x+1)+(2x-3)$

$=3x-2$

So, the function $(f+g)(x)=3x-2$ .

Again, the function,

$(f-g)(x)=f(x)-g(x)$

$=(x+1)-(2x-3)$

$=-x+4$

So, the function $(f-g)(x)=-x+4$.

Similarly,

$\left( \dfrac{f}{g} \right)(x)=\dfrac{f(x)}{g(x)}$ where $g\left( x \right)\ne 0$ and also $x\in R$ .

Now, putting the values,

$\left( \dfrac{f}{g} \right)(x)=\dfrac{x+1}{2x-3}$

where,

$2x-3\ne 0$

$\Rightarrow x\ne \dfrac{3}{2}$

8. Let $f=\left\{ \left( 1,1 \right),\left( 2,3 \right),\left( 0,-1 \right),\left( -1,-3 \right) \right\}$ be a function from $Z$ to $Z$ defined by $f(x)=ax+b$ , for some integers $a,b$ . Determine $a,b$

Ans: We have the given function as,  $f=\left\{ \left( 1,1 \right),\left( 2,3 \right),\left( 0,-1 \right),\left( -1,-3 \right) \right\}$ and also $f(x)=ax+b$ .

As, $(1,1)\in f$ , we get,

$a\times 1+b=1$

$\Rightarrow a+b=1$

And again, $(0,-1)\in f$ , from this we can get,

$a\times 0+b=-1$

$\Rightarrow b=-1$

Putting this value in the first equation, we have,

$a-1=1$

$\Rightarrow a=2$

So, the value of $a$ and $b$ are respectively, $2,-1$ .

9. Let $R$ be a relation from $N$ to $N$ defined by $R = \left\{ {\left( {a,b} \right):\,a,b \in \,N\,and\,a = {b^2}} \right\}$ . Are the following true? Justify your answer in each case.

(i) $\left( a,a \right)\in R$ , for all $a\in N$ .

Ans: We are given our relation as, $R=\left\{ (a,b):a,b\in N\,and\,a={{b}^{2}} \right\}$

Let us take, $2\in N$ .

But we have, $2\ne {{2}^{2}}=4$

So, the statement that $\left( a,a \right)\in R$ , for all $a\in N$is not true.

(ii) $\left( a,b \right)\in R$ , implies $\left( b,a \right)\in R$

Ans: We are given our relation as, $R=\left\{ (a,b):a,b\in N\,and\,a={{b}^{2}} \right\}$

Let us take, $(9,3)\in N$ .

We have to check if, $\left( 3,9 \right)\in N$or not.

But, the condition of the relation says,  $R=\left\{ (a,b):a,b\in N\,and\,a={{b}^{2}} \right\}$and ${{9}^{2}}\ne 3$ .

So, the statement $\left( a,b \right)\in R$ , implies $\left( b,a \right)\in R$ is not true.

(iii) $\left( a,b \right)\in R,\left( b,c \right)\in R$ implies $\left( a,c \right)\in R$ .

Ans: We are given our relation as, $R=\left\{ (a,b):a,b\in N\,and\,a={{b}^{2}} \right\}$

Now, let us take, $\left( 9,3 \right)\in R,\left( 16,4 \right)\in R$ .

We have to check if, $\left( 9,4 \right)\in N$or not.

Thus can also easily see, $9\ne {{4}^{2}}=16$ .

So, the given statement $\left( a,b \right)\in R,\left( b,c \right)\in R$ implies $\left( a,c \right)\in R$ .

10. Let $A=\{1,2,3,4\},B=\{1,5,9,11,15,16\}$ and $f=\left\{ \left( 1,5 \right),\left( 2,9 \right),\left( 3,1 \right),\left( 4,5 \right),(2,11) \right\}$ . Are the following true? Justify your answer in each case.

(i) $f$ is a relation from $A$ to $B$ .

Ans: We are provided with two sets, $A=\{1,2,3,4\},B=\{1,5,9,11,15,16\}$

Thus, the Cartesian product of these two sets will be,

$A\times B=${ $\left( 1,1 \right),\left( 1,5 \right),\left( 1,9 \right),\left( 1,11 \right),\left( 1,15 \right),\left( 1,16 \right),$

$\left( 2,1 \right),\left( 2,5 \right),\left( 2,9 \right),\left( 2,11 \right),\left( 2,15 \right),\left( 2,16 \right),$

$\left( 3,1 \right),\left( 3,5 \right),\left( 3,9 \right),\left( 3,11 \right),\left( 3,15 \right),\left( 3,16 \right),$

$\left( 4,1 \right),\left( 4,5 \right),\left( 4,9 \right),\left( 4,11 \right),\left( 4,15 \right),\left( 4,16 \right)$

And it is also given that,

$f=\left\{ \left( 1,5 \right),\left( 2,9 \right),\left( 3,1 \right),\left( 4,5 \right),(2,11) \right\}$

A relation from a non-empty set $A$ to a non-empty set $B$is a subset of the Cartesian product $A\times B$ .

Thus, it can be easily checked that $f$ is a relation from $A$ to B.

(ii) f is a function from $A$ to $B$ .

Ans: We are provided with two sets, $A=\{1,2,3,4\},B=\{1,5,9,11,15,16\}$

Thus, the Cartesian product of these two sets will be,

$A\times B=\{\left( 1,1 \right),\left( 1,5 \right),\left( 1,9 \right),\left( 1,11 \right),\left( 1,15 \right),\left( 1,16 \right),$

$\left( 2,1 \right),\left( 2,5 \right),\left( 2,9 \right),\left( 2,11 \right),\left( 2,15 \right),\left( 2,16 \right),$

$\left( 3,1 \right),\left( 3,5 \right),\left( 3,9 \right),\left( 3,11 \right),\left( 3,15 \right),\left( 3,16 \right),$

$\left( 4,1 \right),\left( 4,5 \right),\left( 4,9 \right),\left( 4,11 \right),\left( 4,15 \right),\left( 4,16 \right)$

And it is also given that,

$f=\left\{ \left( 1,5 \right),\left( 2,9 \right),\left( 3,1 \right),\left( 4,5 \right),(2,11) \right\}$

If we check carefully, we see that the first element $2$ is providing us two different value of the image $9,11$ .

So, it can be concluded that $f$ is not a function from $A$ to B.

11. Let $f$ be the subset of $Z\times Z$ defined by $f=\left\{ \left( ab,a+b \right):a,b\in Z \right\}$ . If $f$ a function from $Z$ to $Z$. Justify your answer.

Ans: Our given relation $f$ is defined as $f=\left\{ \left( ab,a+b \right):a,b\in Z \right\}$.

We also know that a relation will be called a function from $A$ to $B$ if every element of the set $A$ has unique images in set $B$ .

Let us take 4 elements, $2,6,-2,-6\in Z$ .

So, for the first two elements,

$\left( 2\times 6,2+6 \right)\in f$

$\Rightarrow \left( 12,8 \right)\in f$

And for the last two elements,

$\left( -2\times -6,-2+-6 \right)\in f$

$\Rightarrow \left( 12,-8 \right)\in f$

So, it is clearly visible that one single element $12$ having two different images $8,-8$. Thus, the relation is not a function.

12. Let $A=\left\{ 9,10,11,12,13 \right\}$ and let $f:A\to N$ be defined by $f(n)=$ the highest prime factor of $n$ . Find the range of $f$ .

Ans: We have our given set as, $A=\left\{ 9,10,11,12,13 \right\}$ and the relation is given as $f(n)=$ the highest prime factor of $n$.

The prime factor of $9$ is $3$ .

The prime factors of 10 is $2,5$ .

The prime factor of $11$ is $11$ .

The prime factor of 12 is $2,3$ .

The prime factor of $13$ is $13$ .

Thus, it can be said,

$f(9)=$ the highest prime factor of $9=3$ .

$f(10)=$ the highest prime factor of $10=5$ .

$f(11)=$ the highest prime factor of $11=11$ .

$f(12)=$ the highest prime factor of $12=3$ .

$f(13)=$ the highest prime factor of $13=13$ .

Now, the range of the function will be, $\left\{ 3,5,11,13 \right\}$.

## Conclusion

NCERT Solutions for Maths Miscellaneous Exercise Class 11 Chapter 2 Relations and Functions, by Vedantu, provides clear explanations for every problem. Expert teachers have prepared these solutions, following CBSE guidelines. Focus on understanding different types of relations and functions, as these are important for the chapter. Practising these solutions will help you understand the key concepts and improve your problem-solving skills, which is important for doing well in exams. Download the PDF and practice regularly for the best results.

## Class 11 Maths Chapter 2: Exercises Breakdown

 Exercise Number of Questions Exercise 2.1 10 Questions & Solutions Exercise 2.2 9 Questions & Solutions Exercise 2.3 5 Questions & Solutions

## Chapter-Specific NCERT Solutions for Class 11 Maths

Given below are the chapter-wise NCERT Solutions for Class 11 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 11 Maths Miscellaneous Exercise Chapter 2 - Relations and Functions

1. What is the key focus of the NCERT Solutions of Miscellaneous Exercise Class 11 Chapter 2?

The NCERT Solutions of Maths Miscellaneous Exercise Class 11 Chapter 2 focuses on applying the concepts of relations and functions. It covers different types of relations, functions, and their properties. Understanding these concepts is crucial. This exercise helps in practising how to relate sets and functions comprehensively. It tests your ability to apply theory to various problems.

2. How many questions are there in the NCERT Solutions of Miscellaneous Exercise Class 11 Chapter 2?

There are 12 questions in the Miscellaneous Exercise Class 11 Chapter 2. These questions cover a wide range of topics within relations and functions. They include different types of problems to test your understanding. Practising all these questions is important for thorough preparation. The variety helps in covering all possible exam scenarios.

3. What types of questions are frequently asked in exams from this NCERT Class 11 Maths Chapter 2 Miscellaneous Solutions?

Exams often ask questions that prove the properties of functions. You might need to identify different types of relations. Solving complex problems involving sets and mappings is common. These questions test your comprehensive understanding of the topic. They are designed to assess how well you can apply theoretical concepts.

4. How important is NCERT Class 11 Maths Chapter 2 Miscellaneous Solutions for board exams?

NCERT Class 11 Maths Chapter 2 Miscellaneous Solutions is very important for board exams. It covers a wide range of topics and problem types. These questions are often seen in board exams. Understanding and solving these problems can help you score well. They give a complete review of the chapter’s concepts.

5. Are there any specific strategies to solve the questions in NCERT Class 11 Maths Chapter 2 Miscellaneous Solutions?

Practice is key to solving these questions effectively. Focus on identifying and applying properties of functions and relations. Review previous examples to understand different problem types. Understanding the logic behind each problem is crucial. Regular practice will build confidence and improve accuracy.

6. What are the common mistakes to avoid in NCERT Class 11 Maths Ch 2 Miscellaneous Exercise Solutions?

In NCERT Relation and Function Class 11 Miscellaneous Exercise, one common mistake is confusing different types of functions and relations. It’s important to be clear about definitions and properties. Misunderstanding these can lead to errors. Pay close attention to the details in each problem. Practice helps in avoiding these common pitfalls.

7. What topics are covered in the Relation and Function Class 11 Miscellaneous Exercise?

The Relation and Function Class 11 Miscellaneous Exercise Solutions covers a mix of questions from all the topics in the chapter, including relations and functions, types of relations, and types of functions.

8. How can the Relation and Function Class 11 Miscellaneous Exercise Solutions help in exam preparation?

The Class 11 Ch 2 Miscellaneous Exercise provides a comprehensive review of the entire chapter, helping students practice and reinforce their understanding of key concepts. It includes a variety of questions that can aid in better exam preparation.

9. Are the solutions for the Class 11 Ch 2 Miscellaneous Exercise available for free?

Yes, the NCERT Solutions for the Class 11 Ch 2 Miscellaneous Exercise are available for free download in PDF format, making it easy for students to access and practice offline.

10. Who prepares the NCERT Solutions for the Class 11 Ch 2 Miscellaneous Exercise?

The solutions are prepared by expert teachers who follow the latest CBSE guidelines, ensuring that students get accurate and reliable answers to all the questions in the exercise.