# NCERT Solutions Class 11 Maths Chapter 2

## NCERT Solutions Class 11 Mathematics Chapter 2 – Relations and Functions

NCERT Solutions Class 11 Mathematics Chapter 2 is available on the Extramarks website. It covers all the fundamental and advanced concepts used in Algebra. Chapter 2 Solutions cover topics of a) relation – a set of ordered pairs and b) function – that expresses the correspondence between input and output. The Chapter covers all the essential concepts and is updated with the latest CBSE syllabus, 2022-2023. Further, the concepts explain the key points related to relations and functions with examples. With the help of NCERT Solutions Class 11 Mathematics Chapter 2, students can acquire detailed knowledge of the topic.

Beyond learning the concepts, it is essential to use the applications practically. Thus, the students can refer to NCERT Solutions Class 11 Mathematics Chapter 2 and NCERT Solutions for improving their performance. Relation and functions build-essential fundamentals for Mathematics, and it helps to understand the advanced concepts covered in other Chapters.

Stay tuned to the Extramarks website for the latest notification regarding syllabus updates. In addition, the students can refer to other solutions of both primary and secondary Classes including NCERT solutions Class 1, NCERT solutions Class 2, NCERT solutions Class 3, NCERT solutions Class 4, NCERT solutions Class 5, NCERT solutions Class 6, NCERT solutions Class 7 and others.

### Key Topics Covered In NCERT Solution for Class 11 Mathematics Chapter 2

Chapter 2 Class 11 Mathematics includes concepts based on Relations and Functions Class 11. NCERT Solutions students will be provided with examples of every concept covered throughout the Chapter. These examples make it  easier for students to grasp the implementation and application of the formulas that are hidden within this Chapter. They also adhere to the same format as in exam papers. This format makes it easier for students to become accustomed to it and  don’t  get nervous or anxious during exams.

NCERT Solutions Class 11 Maths Chapter 2 covers all the sums covered in the exercises and examples in this Chapter. This assists students move from basic to more complex problem-solving techniques and quickly builds  their understanding of this Chapter.

Every topic covered in NCERT Solutions Class 11 Maths Chapter 2 is thoroughly explained to help the students understand the functions and relationships.

The key topics covered in the solution are as follow:

 Chapter 2 Relations and Functions Exercise 2.1 Introduction and Cartesian Products of Sets Exercise 2.2 Understanding Relations, Properties, and Applications Exercise 2.3 Types and Functions Miscellaneous Exercise Miscellaneous

2.1 Introduction and Cartesian Products of Sets

This section will explain the basics of the Cartesian Product of Sets. They are given two sets that are not empty, P and Q. The cartesian product Q x P is the set of all ordered pairs from P and Q. These are just a few of the conclusions that can be drawn from this definition. First, two ordered teams can be equal if their first and second elements are equal.

P × Q = { (p,q) : p ∈ P, q ∈ Q }.

2.2 Understanding Relations, Properties, and Applications

A relation is a set which describes the mapping of elements from two sets. With the help of examples and problem samples, this exercise explains the concepts of codomain and domain. They are simple to intermediate in level and rely on an understanding of the attributes of a relationship.

• Relations: A relation R between a nonempty set X and a nonempty set Y can be called a subset of the Cartesian product sets X x Y. Their subset can be obtained by describing the relationship between each ordered pair in XxY. The domain of a relation R refers to the set of all the first elements. In comparison, the range of R includes all the second elements.
• Functions: A relation between a set X and a set Y can be called a function “f” if each element of set X contains one image from set Y. A function f, in other words, is a relationship where no two pairs of the relation have the same first element.

In NCERT Solutions Class 11 Maths Chapter 2, students can improve their knowledge of relations and functions.

2.3 Types and Functions

If every element in A has only one image from set B, a relation between A and B is called a function. This exercise is based on the real algebra functions and includes different formulae to obtain the results. The graphs for identity, polynomial rational, modulus and greatest integer are covered in detail. These solutions include many examples and sums to help the students understand the basic concepts of functions.

• Power Functions: A function of f(x), where x is a constant, is known as a power function.

p(x) = anx n + an−1x n−1 + … + a2x 2 + a1x + a0

• Rational Functions: We will now examine the last type of algebraic function, the rational function.

Two polynomials, p(x) and q(x): f(x) = p(x)/q(x)

• Exponential Functions are: These functions of the form F(x) = Ax are called exponential functions. The base a is a positive constant.
• Logarithmic Functions Logarithmic functions f(x = log ax), where base a is a positive constant, are functions that are the inverse exponential functions.

2.4 Miscellaneous

The questions in this NCERT Solutions Class 11 Maths Chapter 2 consolidate the topics covered in each activity, including representations of relations and functions and some practical problems based upon algebraic functions.

Some of the key concepts, formulas, and concepts that are explained within these answers are as follows:

• Ordered Pair: The pair of elements are put together in a certain pattern to form a pair. That is referred to as the ordered pair.
• Cartesian product of two sets: Let’s take A and B to constitute two sets that have finite values and the cartesian products of both groups are: A × B = {(a, b): a ∈ A, b ∈ B}.
• Relation: Relation “R” from an empty set ‘A’ to a nonempty set is a subset of their cartesian product, i.e. A B x A.
• Domain: The Domain of the Set R refers to the collection comprising all the first elements of ordered pairs of a relation set.
• The Range: A range in an R-related relation between set A and set B represents the collection of all elements that are second in the collection R.

Students can visit Extramarks website for getting access to NCERT Solutions Class 11 Mathematics Chapter 2 and various other study materials that will help students of Class 11 and Class 12 for their Mathematics preparation.

### NCERT Solutions Class 11 Mathematics Chapter 2: Exercise & Solutions

NCERT Solutions Class 11 Mathematics Chapter 2 explains key concepts based on the Chapter of relations and functions. Students can take advantage of the exercise and answer solutions and score more in the exam. To perform better in the exam, it is essential to improve problem-solving speed. There are 24 questions in 4 exercises, including 12 problems and various other  exercises.

The problems mentioned in the exercise and solutions are primarily based on representing relations and functions. The students will learn how to apply the basic operations of functions in problem-solving. Furthermore, the questions in the exercise are based on various sub-topics. It includes identity, constant, polynomial, and modulus functions.

Students can click on the  links below and refer to the exercise questions and solutions for NCERT Solutions Class 11 Mathematics Chapter 2:

• Class 11 Maths Chapter No. 2 Ex 2.1 – 10 Questions
• Class 11 Maths Chapter No. 2 Ex 2.2 – 9 Questions
• Class 11 Maths Chapter No. 2 Ex 2.3 – 5 Questions
• Class 11 Maths Chapter No. 2 Miscellaneous Ex – 12 Questions

Along with the Chapter 2 Mathematics Class 11 solutions, the students can also refer to other standards NCERT solutions on our website:

NCERT Solutions Class 6, NCERT Solutions Class 7, NCERT Solutions Class 8, NCERT Solutions for Class 9, Class 10, and Class 11

NCERT Exemplar Class 11 Mathematics

NCERT Exemplar helps the students to improve their speed in problem-solving. It contains long-form questions, short-form questions, and multiple-choice questions. Further, the Exemplar has various examples of rational functions, modulus functions, signum functions, and an integer function. The problems are majorly designed with high difficulty levels. It helps students advance their practice for competing in the board exams.

NCERT Solutions Class 11 Mathematics Chapter 2 helps students grasp the basics of relations and function. Further, they can attempt Exemplar questions that deal with various types of problems. The Chapter plays a vital role in laying a foundation for the upcoming Chapters. Thus, students can improve their knowledge of the cartesian product of sets, relations, signum functions, and integer functions.

### Key Features of NCERT Solutions Class 11 Mathematics Chapter 2

NCERT Solutions Class 11 Mathematics Chapter 2 promotes the learning of key concepts. With regular practice of the questions, the students can get the core fundamentals to have a firm grip over the subject. The key features in the solutions include:

• The solutions offer a step-by-step knowledge of the syllabus Class 11 Mathematics NCERT solutions Chapter 2.
• The format of the solutions is simple and makes it easy to learn all essential concepts.
• With the help of NCERT Solutions Class 11 Mathematics Chapter 2, students can attempt questions with a clear mind, maximising the scope of marks.
• It offers elaborative solutions with interactive examples with a brief and precise understanding of each term.
• The solutions are based on range, domain, functions, and algebra of relations.

## Chapter 2 – Relations and Functions

Q.1 Let A = {9, 10, 11, 12, 13} and let f: A → N be defined by f (n) = the highest prime factor of n. Find the range of f.

We have, A = {9, 10, 11, 12, 13}
and f(n) = the highest prime factor of n
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

The range of function = {Values of f(n)}, n ∈ N.
The range of f(n) = {3, 5, 11, 13}.

Q.2 Find the range of each of the following functions.

(i) f (x) = 2 – 3x, x ∈ R, x > 0.
(ii) f (x) = x2 + 2, x is a real number.
(iii) f (x) = x, x is a real number.

(i) Since, x > 0
So, – 3x < 0
2 – 3x < 2
⇒ f(x) < 2
Therefore, Range of f = (–∞, 2)

$\begin{array}{l}\left(\text{ii}\right)\text{f}\left(\text{x}\right)={\text{x}}^{2}+2,\text{ x is a real number.}\\ \text{Let y}={\text{x}}^{2}+2\\ ⇒{\text{x}}^{2}=\text{y}-2\\ ⇒\text{ x}=\sqrt{\text{y}-2}\\ \mathrm{Since},\text{x is a real number.}\\ \text{So, y}-\text{2}\ge 0\\ ⇒\text{ y}\ge 2\\ ⇒\text{ y}\in \left[2,0\right)\\ \mathrm{Therefore},\text{the range of function}\mathrm{is}\text{}\left[2,0\right).\\ \left(\mathrm{iii}\right)\text{f}\left(\text{x}\right)=\text{x},\text{x is a real number.}\\ \text{x is real number}⇒\text{f}\left(\text{x}\right)\text{is also a real number.}\\ \text{So, range of function}=\text{R}.\end{array}$

Q.3 Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that

(i) A × (B ∩ C) = (A × B) ∩ (A × C).
(ii) A × C is a subset of B × D.

Given, A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}

(i) B ∩ C = {}
= ∅
A × (B ∩ C)
= A × ∅
= ∅
A × B = {(1, 1), (1, 2),(1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}
A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}
(A × B) ∩ (A × C)
= ∅
L.H.S.= R.H.S. Hence proved.

(ii) Here,
A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}
B × D = {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5),
(2, 6), (2, 7), (2, 8), (3, 5), (3, 6),
(3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)}
Since,
(A × C) ∩ (B × D) = {(1, 5), (1, 6), (2, 5), (2, 6)}
= (A × C)
So, (A × C) is a subset of (B × D).

Q.4 State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly.

(i) If P = {m, n} and Q = {n, m}, then P × Q = {(m, n), (n, m)}.
(ii) If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (x, y) such that x ∈ A and y ∈ B.
(iii) If A = {1, 2}, B = {3, 4}, then A × (B ∩ ∅) = ∅ .

(i) False,
Since, P = {m, n} and Q = {n, m}
P × Q = {(m, n), (m, m), (n, n), (n, m)}

(ii) True.

(iii) True.

Q.5 Let f be the subset of Z × Z defined by f = {(ab, a + b): a, b ∈ Z}. Is f a function from Z to Z? Justify your answer.

$\begin{array}{l}\text{f}=\left\{\left(\mathrm{ab},\text{a}+\text{b}\right):\text{a},\text{b}\in \text{Z}\right\}\\ \mathrm{This}\text{can be written as:}\\ \text{f}\left(\mathrm{ab}\right)=\text{a}+\text{b},\text{ }\forall \text{a,b}\in \text{Z}\\ \mathrm{Substituting}\text{a}=2,\text{}-2\text{and b}=3,\text{ }-3\\ \text{ f}\left(2×3\right)=2+3\\ ⇒\text{ f}\left(6\right)=5\\ \text{ f}\left(-2×-3\right)=\left(-2\right)+\left(-3\right)\\ ⇒\text{ f}\left(6\right)=-5\\ \mathrm{Since},\text{element 6 has two different images 5 and}-\text{5.}\\ \text{So, relation f is not a function because elements have no}\\ \text{unique images.}\end{array}$

Q.6 Let A = {1,2,3,4}, B = {1,5,9,11,15,16} and f = {(1,5), (2,9), (3,1), (4,5), (2,11)} Are the following true?

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

Since, A x B ={(1, 1), (1, 5), (1, 9), (1, 11),
(1, 15), (1, 16), (2, 1), (2, 5), (2, 9), (2, 11),
(2, 15), (2, 16), (3, 1), (3, 5), (3, 9), (3, 11),
(3, 15), (3, 16), (4, 1), (4, 5), (4, 9), (4, 11),
(4, 15), (4, 16)}

(i) True.
Since, f is a subset of A x B, so, f is a relation.

(ii) False.
Since, image of 2 are 9 and 11 i.e., each element has not unique image. So, this relation is not a function.

Q.7

$\begin{array}{l}\text{Let R be a relation from N to N defined by}\\ \text{R = {(a, b): a, b}\in {\text{N and a = b}}^{\text{2}}\text{}.}\\ \text{Are the following true?}\\ \text{(i) (a,a)}\in \text{R, for all a}\in \text{N}\\ \text{(ii) (a,b)}\in \text{R, implies (b,a)}\in \text{R}\\ \text{(iii) (a,b)}\in \text{R, (b,c)}\in \text{R implies (a,c)}\in \text{R.}\end{array}$

$\begin{array}{l}\begin{array}{l}\mathrm{Since},\text{ R}=\left\{\left(\text{a},\text{b}\right):\text{a},\text{b}\in \mathrm{Nanda}={\text{b}}^{2}\right\}\\ \left(\text{i}\right)\text{ }\mathrm{Since},\text{ }3\in \text{N}⇒3\ne {3}^{2}=9\\ \mathrm{So},\text{}\left(\text{a},\text{a}\right)\in \text{R},\mathrm{foralla}\in \text{N is not true.}\\ \left(\mathrm{ii}\right)\mathrm{No},\text{ }\left(\text{a},\text{b}\right)\in \text{R},\mathrm{implies}\left(\text{b},\text{a}\right)\in \text{R not true.}\end{array}\\ \begin{array}{l}\text{Because, if }\left(4,2\right)\in \text{R}⇒4={2}^{2}\\ \mathrm{and}\text{ }\left(2,4\right)\in \text{R}⇒2\ne {4}^{2}=16\\ \mathrm{So},\text{ }\left(\text{a},\text{b}\right)\in \text{R},\mathrm{implies}\left(\text{b},\text{a}\right)\in \text{R not true.}\\ \left(\mathrm{iii}\right)\mathrm{No},\text{}\left(\text{a},\text{b}\right)\in \text{R},\left(\text{b},\text{c}\right)\in \mathrm{Rimplies}\text{ }\left(\text{a},\text{c}\right)\in \text{R is not true}.\end{array}\\ \begin{array}{l}\mathrm{Because},\\ \left(25,5\right)\in \text{R},\text{}\left(36,6\right)\in \text{R as}25,5,36,6\in \text{N}\end{array}\\ \begin{array}{l}\text{and 25}={5}^{2},\text{ }36={6}^{2}.\\ \text{ }25\ne 36⇒\left(25,36\right)\notin \text{R}\\ \mathrm{So},\text{ }\left(\text{a},\text{b}\right)\in \text{R},\left(\text{b},\text{c}\right)\in \mathrm{Rimplies}\text{ }\left(\text{a},\text{c}\right)\in \text{R is not true}.\end{array}\end{array}$

Q.8

$\begin{array}{l}\text{Let f = {(1,1), (2,3), (0,–1), (–1, –3)} be a function from}\\ \text{Z to Z defined by f(x) = ax + b, for some integers a, b.}\\ \text{Determine a, b.}\end{array}$

$\begin{array}{l}\mathrm{Given}:\\ \mathrm{f}\left(\mathrm{x}\right)=\mathrm{ax}+\mathrm{b}\\ \mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{f}=\left\{\left(1,1\right),\left(2,3\right),\left(0,-1\right),\left(-1,-3\right)\right\}\\ \mathrm{Then},\\ \mathrm{f}\left(1\right)=1⇒\mathrm{a}\left(1\right)+\mathrm{b}=1\\ ⇒\mathrm{a}+\mathrm{b}=1 ...\left(\mathrm{i}\right)\\ \mathrm{f}\left(0\right)=-1⇒\mathrm{a}\left(0\right)+\mathrm{b}=-1\\ ⇒\mathrm{b}=-1\\ \mathrm{Substituting}\text{value of b in equation}\left(\mathrm{i}\right),\text{we get}\\ \text{a}+\left(-1\right)=1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=1+1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=2\\ \mathrm{Thus},\text{a}=\text{2 and}\mathrm{b}=-1.\end{array}$

Q.9

$\begin{array}{l}\text{Let f, g : R}\mathrm{\to }\text{R be defined, respectively by f(x) = x + 1,}\\ \text{g(x) = 2x – 3. Find f + g, f – g and}\frac{\text{f}}{\text{g}}\text{.}\end{array}$

(f + g)(x) = f(x) + g(x)
= x + 1 + 2x – 3
= 3x – 2
(f g)(x) = f(x) g(x)
= (x + 1) (2x 3)
= x + 4

$\begin{array}{l}\left(\frac{\text{f}}{\text{g}}\right)\left(\text{x}\right)=\frac{\text{f}\left(\text{x}\right)}{\text{g}\left(\text{x}\right)}\\ \text{ }=\frac{\text{x}+1}{2\text{x}-3},\text{where x}\ne \frac{3}{2}\end{array}$

Q.10

$\begin{array}{l}\text{Let​ f =}\left\{\left(\text{x,\hspace{0.17em}}\frac{{\text{x}}^{\text{2}}}{{\text{1+x}}^{\text{2}}}\right)\text{: \hspace{0.17em}x}\in \text{R}\right\}\text{be a function from R to R.}\\ \text{Determine the range of f.}\end{array}$

$\begin{array}{l}\begin{array}{l}\text{f}=\left\{\left(\text{x},\text{ }\frac{{\text{x}}^{2}}{1+{\text{x}}^{2}}\right):\text{ x}\in \text{R}\right\}\mathrm{be}\text{ }\mathrm{a}\text{ }\mathrm{function}\text{ }\mathrm{from}\text{ }\mathrm{R}\text{ }\mathrm{to}\text{ }\mathrm{R}\\ ⇒\text{f}:\text{R}\to \text{R and f}\left(\text{x}\right)=\frac{{\text{x}}^{2}}{1+{\text{x}}^{2}},\text{ }1+{\text{x}}^{2}\ne 0\\ \mathrm{Range}\text{of f:}\\ \text{Let y}=\frac{{\text{x}}^{2}}{1+{\text{x}}^{2}}\\ ⇒\text{ y}\left(1+{\text{x}}^{2}\right)={\text{x}}^{2}\end{array}\\ \begin{array}{l}⇒\text{ y}+{\mathrm{yx}}^{2}={\text{x}}^{2}\\ ⇒\left(\text{y}-1\right){\text{x}}^{2}+\text{y}=0\end{array}\\ \begin{array}{l}⇒{\text{ x}}^{2}=\frac{\text{y}}{1-\text{y}}\\ ⇒\text{ x}=±\sqrt{\frac{\text{y}}{1-\text{y}}}\\ \mathrm{Since},\text{x is a real number. So,}\end{array}\\ \begin{array}{l}\text{ }\frac{\text{y}}{1-\text{y}}\ge 0\\ ⇒\frac{\text{y}}{\text{y}-1}\le 0\\ ⇒\frac{\text{y}-1}{\text{y}-1}\le 0\\ ⇒\text{ }0\le \text{y}<1\\ ⇒\text{y}\in \left[0,\text{ }1\right)\\ \mathrm{Thus},\text{range of function is [0,1).}\end{array}\end{array}$

Q.11

$\text{Find the domain and the range of the real function f defined by f (x) =}\left|\text{x –\hspace{0.17em}1}\right|\text{.}$

$\begin{array}{l}\mathrm{We}\text{have,}\\ \text{f}\left(\mathrm{x}\right)=|\mathrm{x}-1|\\ \mathrm{Here},\text{f}\left(\mathrm{x}\right)\text{is well defined for x}\in \text{R.}\\ \text{So, Domain of function is R.}\\ \mathrm{Now},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}|\mathrm{x}-1|\ge 0\text{for all x}\in \mathrm{R}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}0\le \text{\hspace{0.17em}}|\mathrm{x}-1|\le \mathrm{\infty }\text{\hspace{0.17em}\hspace{0.17em}for all x}\in \mathrm{R}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}0\le \text{\hspace{0.17em}f}\left(\mathrm{x}\right)\le \mathrm{\infty }\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}for all x}\in \mathrm{R}\\ ⇒\mathrm{f}\left(\mathrm{x}\right)\in \left[0,\mathrm{\infty }\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}for all x}\in \mathrm{R}\\ \mathrm{Hence},\text{range of function}=\left[0,\mathrm{\infty }\right).\end{array}$

Q.12

$\text{Find the domain and the range of the real function f defined by (x) =}\sqrt{\left(\text{x – 1}\right)}\text{.}$

$\begin{array}{l}\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\left(\mathrm{x}-1\right)}\\ \mathrm{Since},\text{f}\left(\mathrm{x}\right)\text{is a real number.}\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x}-1\ge 0\\ ⇒\mathrm{x}\ge 1\\ \mathrm{So},\text{domain of function is [1,}\mathrm{\infty }\text{).}\\ \text{Since,}\sqrt{\left(\mathrm{x}-1\right)}\ge 0\text{for x}\ge 1.\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{f}\left(\mathrm{x}\right)\ge 0\text{for x}\ge 1\\ \mathrm{Thus},\text{the range of function is [0,}\mathrm{\infty }\text{).}\end{array}$

Q.13

$\text{Find the domain of the function f(x) =}\frac{{\text{x}}^{\text{2}}\text{+2x+1}}{{\text{x}}^{\text{2}}\text{– 8x+12}}\text{.}$

$\begin{array}{l}\mathrm{The}\text{ }\mathrm{given}\text{ }\mathrm{function}\text{ }\mathrm{f}\left(\text{x}\right)=\frac{{\text{x}}^{2}+2\text{x}+1}{{\text{x}}^{2}-8\text{x}+12}\\ \mathrm{Here},{\text{​​​​​​ x}}^{2}-8\text{x}+12=\left(\text{x}-2\right)\left(\text{x}-6\right)\\ \mathrm{So},\text{ f}\left(\text{x}\right)=\frac{{\text{x}}^{2}+2\text{x}+1}{\left(\text{x}-2\right)\left(\text{x}-6\right)}\\ \mathrm{Here},\text{the function f}\left(\text{x}\right)\text{is defined for all the real values}\\ \text{except 2 and 6. So, the domain of f}\left(\text{x}\right)\text{is R}-\left\{2,6\right\}.\end{array}$

Q.14

$\mathrm{If}\text{}\mathrm{f}\left(\mathrm{x}\right)\text{}=\text{}{\mathrm{x}}^{2},\text{}\mathrm{find}\text{}\frac{\mathrm{f}\left(1.1\right)\text{}–\text{}\mathrm{f}\left(1\right)}{\left(1.1\text{}–\text{}1\right)}.$

$\begin{array}{l}\mathrm{Since},\text{f}\left(\text{x}\right)={\text{x}}^{2}\\ \frac{\text{f}\left(1.1\right)-\text{f}\left(1\right)}{\left(1.1-1\right)}=\frac{{\left(1.1\right)}^{2}-{\left(1\right)}^{2}}{0.1}\\ \text{ }=\frac{\left(1.1-1\right)\left(1.1+1\right)}{0.1}\\ \text{ }=\frac{0.1×2.1}{0.1}\\ \text{ }=2.1\\ \therefore \text{ }\frac{\text{f}\left(1.1\right)-\text{f}\left(1\right)}{\left(1.1-1\right)}=2.1\end{array}$

Q.15

$\begin{array}{l}\text{The function ‘t’ which maps temperature in degree Celsius }\mathrm{into}\\ \text{temperature in degree Fahrenheit is defined by t}\left(\text{C}\right)=\frac{\text{9C}}{\text{5}}\text{+32.}\\ \text{Find}\\ \text{(i) t(0)}\\ \text{(ii) t(28)}\\ \text{(iii) t(–10)}\\ \text{(iv) The value of C, when t(C) = 212.}\end{array}$

$\begin{array}{l}\begin{array}{l}\mathrm{The}\text{given function is:}\\ \text{ t}\left(\text{C}\right)=\frac{9\text{C}}{5}+32\\ \left(\text{i}\right)\text{ Putting C}=0\text{in t}\left(\text{C}\right),\text{we get}\\ \text{ t}\left(0\right)=\frac{9\left(0\right)}{5}+32\\ \text{ }=32\\ \left(\mathrm{ii}\right)\text{ Putting C}=28\text{in t}\left(\text{C}\right),\text{we get}\end{array}\\ \begin{array}{l}\text{ t}\left(28\right)=\frac{9\left(28\right)}{5}+32\\ \text{ }=82.4\\ \left(\mathrm{iii}\right)\text{ Putting C}=-10\text{in t}\left(\text{C}\right),\text{we get}\end{array}\\ \text{ t}\left(-10\right)=\frac{9\left(-10\right)}{5}+32\\ \begin{array}{l}\text{ }=14\\ \left(\mathrm{iv}\right)\text{ t}\left(\text{C}\right)=212\\ \text{ }\frac{9\text{C}}{5}+32=212\\ \text{ }\frac{9\text{C}}{5}=212-32\end{array}\\ \begin{array}{l}\text{ }=180\\ \text{ C}=180×\frac{5}{9}\\ \text{ }=100\\ \mathrm{Thus},\text{the value of t is 100 when t}\left(\text{C}\right)=212.\end{array}\end{array}$

Q.16 A function f is defined by f(x) = 2x –5. Write down the values of

(i) f (0),
(ii) f (7),
(iii) f (–3).

f(x) = 2x –5

(i) Substituting x = 0 in f(x), we get
f(0) = 2(0) – 5
= – 5

(ii) Substituting x = 7 in f(x), we get
f(7) = 2(7) – 5
= 9

(iii) Substituting x = – 3 in f(x), we get
f(– 3) = 2(– 3) – 5
= – 11

Q.17

$\begin{array}{l}\text{Find the domain and range of the following real functions:}\\ \begin{array}{l}\left(\text{i}\right)\text{ f}\left(\text{x}\right)\text{= –}\left|\text{x}\right|\\ \left(\text{ii}\right)\text{ f}\left(\text{x}\right)\text{=}\sqrt{\text{9}-{\text{x}}^{\text{2}}}\end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}\left(\text{i}\right)\text{ f}\left(\text{x}\right)=-\left|\text{x}\right|,\forall \text{x}\in \text{R}\\ \mathrm{Since},\text{}\left|\text{x}\right|=\left\{\begin{array}{l}\text{x},\text{ x}\ge 0\\ -\text{x},\text{ x}<0\end{array}\\ \mathrm{So},\text{ f}\left(\text{x}\right)=-\left|\text{x}\right|=\left\{\begin{array}{l}-\text{x},\text{ x}\ge 0\\ \text{ x},\text{ x}<0\end{array}\\ \mathrm{Since},\text{x}\in \text{R},\text{so domain of f}\left(\text{x}\right)\text{is R.}\\ \mathrm{Here},\text{}-\left|\text{x}\right|\text{is negative number for all real number, R.}\end{array}\\ \text{Therefore, Range of f}\left(\text{x}\right)\text{is (}-\infty ,\text{0].}\\ \begin{array}{l}\left(\mathrm{ii}\right)\mathrm{We}\text{have,}\\ \text{ f}\left(\text{x}\right)=\sqrt{9-{\text{x}}^{2}}\\ \text{f}\left(\text{x}\right)\text{is defined for all x satisfying}\end{array}\\ \begin{array}{l}\text{ 9}-{\text{x}}^{\text{2}}\ge 0\\ ⇒{\text{x}}^{2}-9\le 0\\ ⇒\left(\text{x}-3\right)\left(\text{x}+3\right)\le 0\end{array}\\ \begin{array}{l}⇒\text{ }-3\le \text{x}\le 3\\ ⇒\text{ x}\in \left[-3,3\right]\\ \mathrm{Hence},\text{Domain of f}\left(\text{x}\right)=\left[-3,3\right]\end{array}\\ \mathrm{Let}\text{y}=\text{f}\left(\text{x}\right).\mathrm{Then},\\ \begin{array}{l}\text{ y}=\sqrt{9-{\text{x}}^{2}}\\ ⇒{\text{ y}}^{2}=9-{\text{x}}^{2}\\ ⇒{\text{ x}}^{2}=9-{\text{y}}^{2}\\ ⇒\text{ x}=\sqrt{9-{\text{y}}^{2}}\end{array}\\ \mathrm{It}\text{is clear that x will take real values, if}\\ \begin{array}{l}\text{ }9-{\text{y}}^{2}\ge 0\\ ⇒{\text{ y}}^{2}-9\le 0\\ ⇒\text{ }\left(\text{y}-3\right)\left(\text{y}+3\right)\le 0\end{array}\\ ⇒\text{ }-3\le \text{y}\le 3\\ ⇒\text{ y}\in \left[-3,3\right]\\ \mathrm{Since},\text{y}=\sqrt{9-{\text{x}}^{2}}\ge 0\text{for all x}\in \left[-3,3\right]\\ \therefore \text{ y}\in \left[0,3\right]\text{for all x}\in \left[-3,3\right]\\ \mathrm{Hence},\text{range of f}\left(\text{x}\right)=\left[0,3\right]\end{array}$

Q.18 Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.

(i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}
(ii) {(2,1), (4,2), (6,3), (8,4), (10,5), (12,6), (14,7)}
(iii) {(1, 3), (1, 5), (2, 5)}.

(i) Given relation is a function, because there is no two different images of same elements.
Domain = {2, 5, 8, 11, 14, 17} and Range = {1}

(ii) Given relation is a function, because each element has unique image.
Domain = {2, 4, 6, 8, 10, 12, 14} and
Range = {1, 2, 3, 4, 5, 6, 7}

(iii) The given relation is not a function, because here one element has two different images,
i.e., 2 has two images 3 and 5.

Q.19 Let R be the relation on Z defined by R = {(a, b): a, b ∈ Z, a – b is an integer}. Find the domain and range of R

R = {(a, b): a, b ∈ Z, a – b is an integer}
Since, a and b both are integer Z and their difference is also an integer.
Therefore, domain of R = Z
and Range of R = Z.

Q.20 Let A = {x, y, z} and B = {1, 2}. Find the number of relations from A to B.

Since, A = {x, y, z} and B = {1, 2} So, A × B = {(x, 1), (x, 2), (y, 1), (y, 2), (z, 1), (z, 2)}
n(A × B) = 6
The number of subsets of A × B is 26. Therefore, number of relations from A into B will be 26.

Q.21 Write the relation R = {(x, x3): x is a prime number less than 10} in roster form.

R = {(x, x3): x is a prime number less than 10}
= {(x, x3): x = 2, 3, 5, 7}
= {(2, 8), (3, 27), (5, 125), (7, 343)}

Q.22 Determine the domain and range of the relation R defined by R = {(x, x + 5): x ∈ {0, 1, 2, 3, 4, 5}}.

R = {(x, x + 5): x ∈ {0, 1, 2, 3, 4, 5}} = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}
Domain of R = {0, 1, 2, 3, 4, 5}
Range of R = {5, 6, 7, 8, 9, 10}

Q.23 Let A = {1, 2, 3, 4, 6}. Let R be the relation on
A defined by {(a, b): a, b ∈ A, b is exactly divisible by a}.

(i) Write R in roster form
(ii) Find the domain of R
(iii) Find the range of R.

(i) R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3) , (3, 6), (4, 4), (6, 6)}
(ii) Domain of R = {1, 2, 3, 4 6}
(iii) Range of R = {1, 2, 3, 4, 6}

Q.24 The Fig 2.7 shows a relationship between the sets P and Q. Write this relation.
(i) in set-builder form (ii) roster form.
What is its domain and range?

Here, P = {5, 6, 7}, Q = {3, 4, 5}
(i) The set-builder form is R = {(x, y): y = x – 2; for x ∈P}
(ii) In roster form:
R = {(5, 3), (6, 4), (7, 5)}
Domain = {5, 6, 7}, Range = {3, 4, 5}

Q.25 A = {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B}. Write R in roster form.

Here, A = {1, 2, 3, 5} and B = {4, 6, 9}
R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B}
= {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}
Which is a roster form.

Q.26 Define a relation R on the set N of natural numbers by R = {(x, y): y = x + 5, x is a natural number less than 4; x, y ∈ N}. Depict this relationship using roster form. Write down the domain and the range.

R = {(x, y): y = x + 5, x is a natural number less than 4; x, y ∈N} = {(1, 6), (2, 7), (3, 8)}
Domain of R = {1, 2, 3}
Range of R = {6, 7, 8}

Q.27 Let A = {1, 2, 3,…, 14}. Define a relation R from A to A by R = {(x, y): 3x – y = 0, where x, y ∈ A}. Write down its domain, codomain and range.

We have:
A = {1, 2, 3,…, 14}
Relation R from A to A by R = {(x, y): 3x – y = 0, where x, y ∈ A}
= {(1, 3), (2, 6), (3, 9), (4, 12)}
Domain of Relation R = {1, 2, 3, 4}
Codomain = A
Range of relation R = {3, 6, 9, 12}

Q.28 The Cartesian product A × A has 9 elements among which are found (–1, 0) and (0, 1).
Find the set A and the remaining elements of A × A.

Number of elements in the cartesian product A × A = 9
Some given elements of A × A are (–1, 0) and (0, 1).
Let number of elements in set A, n(A) = 3
and number of elements in set A × A = 9
i.e., n(A × A) = n(A) × n(A)
9 = { n(A)}2
n(A) = 3
So, A = {–1, 0, 1}
Then, A × A = {(–1, –1), (–1, 0), (–1, 1), (0, –1,), (0, 0), (0,1), (1, –1,), (1, 0), (1, 1)}
Remaining elements of A × A are: (–1, –1), (–1, 1), (0, –1,), (0, 0), (1, –1,), (1, 0), (1, 1).

Q.29 Let A and B be two sets such that n(A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) are in A × B, find A and B, where x, y and z are distinct elements.

Number of elements in set A = 3
Number of elements in set B = 2,
Some elements of A × B are (x, 1), (y, 2), (z, 1).
Since, Cartesian product A × B of two sets A and B is given by A × B = {(a, b): a ∈ A, b ∈ B} So, A = {x, y, z} and B = {1, 2}.

Q.30 If A × B = {(a, x),(a , y), (b, x), (b, y)}. Find A and B.

Given, A × B = {(a, x),(a , y), (b, x), (b, y)}
Since, Cartesian product A × B of two sets A and B is given by A × B = {(a, b): a ∈ A, b ∈ B}.
Then, A = {a, b}, B = {x, y}.

Q.31 If A = {–1, 1}, find A × A × A.

Since, A = {–1, 1}
A × A = {(–1, –1), (–1, 1,), (1,–1), (1, 1)}
A × A × A = {–1, 1} × {(–1, –1), (–1, 1,), (1,–1), (1, 1)}
= {(1,–1, –1), (–1,–1, 1,), (–1,1,–1), (–1,1, 1), (1, –1, –1), (1,–1, 1,), (1, 1,–1), (1, 1, 1)}

Q.32 If G = {7, 8} and H = {5, 4, 2}, find G × H and H × G.

$\begin{array}{l}\text{Given: G}=\left\{\text{7},\text{8}\right\}\text{and H}=\left\{\text{5},\text{4},\text{2}\right\},\text{}\\ \text{ G}×\text{H}=\left\{\left(7,5\right),\left(7,4\right),\left(7,2\right),\left(8,5\right),\left(8,4\right),\left(8,2\right)\right\}\\ \text{ H}×\text{G}=\left\{\left(5,7\right),\left(5,8\right),\left(4,7\right),\left(4,8\right),\left(2,7\right),\left(2,8\right)\right\}\end{array}$

Q.33 If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in (A × B).

$\begin{array}{l}\mathrm{Number}\text{of elements in set A}=3\\ \text{i}.\text{e}.,\text{ n}\left(\text{A}\right)=3\\ \mathrm{Elements}\text{in set B}=\left\{3,4,5\right\}\\ ⇒\text{ n}\left(\text{B}\right)=3\\ \mathrm{So},\text{}\\ \text{Number of elements in A}×\text{B}=\text{n}\left(\text{A}\right)×\text{n}\left(\text{B}\right)\\ \text{ n}\left(\text{A}×\text{B}\right)=3×3\\ \text{ }=9\\ \mathrm{Thus},\text{ number of elements in A}×\text{B is 9.}\end{array}$

Q.34

$\text{If }\left(\frac{\text{x}}{\text{3}}\text{+1, y-}\frac{\text{2}}{\text{3}}\right)\text{=}\left(\frac{\text{5}}{\text{3}}\text{,}\frac{\text{1}}{\text{3}}\right)\text{, find the values of x and y.}$

$\begin{array}{l}\mathrm{Given}:\\ \text{\hspace{0.17em}}\left(\frac{\mathrm{x}}{3}+1,\mathrm{y}-\frac{2}{3}\right)=\left(\frac{5}{3},\frac{1}{3}\right)\\ \mathrm{Comparing}\text{both sides, we get}\\ \frac{\mathrm{x}}{3}+1=\frac{5}{3}⇒\frac{\mathrm{x}}{3}=\frac{5}{3}-1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{x}}{3}=\frac{2}{3}⇒\mathrm{x}=2\\ \mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{y}-\frac{2}{3}=\frac{1}{3}⇒\mathrm{y}=\frac{1}{3}+\frac{2}{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{y}=\frac{3}{3}=1\\ \mathrm{Thus},\text{x}=2\text{and y}=\text{1.}\end{array}$

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### 1. Which are the essential formulas in NCERT Solutions Class 11 Mathematics Chapter 2?

The Class 11 Mathematics Chapter 2 offers theoretical knowledge to solve a practical problem. However, the following are the formulas mentioned in NCERT Solutions Class 11 Mathematics Chapter 2:

• Cartesian Product of Two Sets: Let’s take A and B as two finite sets. Then the cartesian product of these two sets A × B = {(a, b): ∈ A, b ∈ B}.
• Relation: Relation ‘R’ of a nonempty set ‘A’ to a nonempty set ‘B’ is a subset of their cartesian product, i.e., A × B.
• Domain: The domain of a set R is the set of all first elements of the ordered pairs in a relation set.
• Range: The range of a relation ‘R’ from set A to set B is the set of all second elements in the set R.

### 2. Which are the important topics in NCERT Solutions Class 11 Mathematics Chapter 2?

The following are the essential topics in NCERT Solutions Class 11 Mathematics Chapter 2:

• Relations
• Functions
• Algebra of real functions
• Subtraction of real functions
• Multiplication of real functions
• The quotient of real functions

### 3. Should I practice all the questions in NCERT Solutions Class 11 Mathematics Chapter 2?

The students can attempt long-form and short-form questions. Further, they can also practice with multiple-choice questions to improve their speed and test their knowledge of the key concepts.