CBSE Class 12 Maths Revision Notes Chapter 1

Mathematics is regarded to be one of the most demanding subjects for students of all classes. The students of higher classes face difficulty preparing for their examinations since high-level Mathematics concepts are introduced. The Class 12 Mathematics Chapter 1 notes- Relations and Functions is an important chapter and defines different concepts along with their graphs.

Quick Links

Extramarks provide one of the best academic notes to help students ace their examinations. These notes are prepared by a team of experts as per the latest guidelines issued by the CBSE board. With all the important formulae and definitions, the notes act as a helping guide for students. They aid in quick revision on the days before the exam.

Students may visit the Extramarks web portal to get all the latest news and information about the CBSE examinations, paper patterns, marking schemes and more. Students may also get access to various academic study materials such as Class 12 Mathematics Chapter 1 notes for free.

Key Topics Covered In Class 12 Mathematics Chapter 1 Notes

The key topics covered under class 12 Mathematics chapter 1 notes include the following.

Relation:

Definition: Relation is the connection between two or more sets of values. If (a,b) ∈ R, then a R b means a is related to b under R.

For example: if A = {9, 16, 25} and B = {5, 4, 3, -3, -4, -5} then set A is related to set B under R, where R = {(x,y), x=y2 / x A and y B}

Types of Relations:

The types of relation mentioned in class 12 Mathematics chapter 1 notes are

Empty Relation

The relation R in any set A is said to be empty if no element in A is related to any element of the same set, i.e., R = φ A × A

Consider set A= {2, 3, 5} and R= {(x,y), x + y > 9}, then R is an empty relation or void relation.

Universal Relation

The relation R in any set A is said to be universal if every element in A is related to every element of the same set, R = A × A

For set A= {2, 3, 5} and R= {(x,y), x + y > 0}, then R is a universal relation.

NOTE: Both Empty and Universal Relations are sometimes called trivial relations.

Reflexive Relation

The relation R in any set A is said to be reflexive; if every element in A is related to itself, i.e., (a, a) ∈ R for every a ∈ A, then a R a is Reflexive.

Symmetric Relation

A relation R in a set A is said to be symmetric if the elements in R can be swapped.

R is symmetric relation if (a, b) ∈ R (b, a) ∈ R, i.e., aRb = bRa a, b, c A.

If S={a, b, c} then relation R6={(a, b), (b, a), (c, c)}

Transitive Relation

A relation R in a set A is said to be transitive if (a, b) ∈ R and (b, c) ∈ R (a, c) ∈ R, which means that if (a,b) belongs to R and (b,c) belongs to R then (a,c) will also belong to R a, b, c R.

Equivalence Relation

If relation R is reflexive, symmetric and transitive, then it is said to be an equivalence relation.

Notation: a b.

If R={(x, y)/ len (x) = len (y)}

Then R is reflexive: len (a) =len (a)

R is symmetric: len (a)=len (b) and len (b)=len (a)

R is transitive: len (a) = len (b) and len (b) = len (c) then len (a) = len (c)

Therefore, R is an Equivalence Relation.

Equivalence Classes:

Notation: {a} or [a]

Let A be a non-empty set. a ∈ A

An equivalence class is the set of all the points that are in relation to a set A under R.

Functions:

A Function f is a special kind of relation. Let f be the function from set A to set B, i.e., f: A B if every element in A is associated or is mapped with only one element in B.

Set A is called the domain, and Set B is called the Co-domain of the function f.

The range is defined as the set of all possible values of function f.

For example: if x3is a function, all values of x are the domain, and the values of x3 are the range.

Types of Function:

Types of function under class 12 Mathematics chapter 1 notes include

One-One or Injective function:

Function f is injective if each element of set A is mapped to the distinct element of set B.

Mathematically, f: X → Y is an one-one or injective function, if f(x1) = f(x2) ⇒ x1 = x2 ∀ x1 , x2 ∈ X.

For example: f(x)=3x is a one-one function.

Onto or Surjective Function:

Function f is said to be Onto if at least one element in the domain is mapped to every element in the codomain. It means for a surjective function, the range, codomain, and the image are equal.

Mathematically, f: X → Y is onto or surjective function if for the given y ∈ Y, ∃ x ∈ X such that f(x) = y.

For example: f(x)=x-2 is a surjective function.

Bijective Function:

Function f is said to be bijective if it is both injective and surjective.

Mathematically, f: X → Y is one-one and onto or bijective, if f is one-one and onto.

For example: f(x)=3x is a one-one and onto function.

Composition of a Function:

Notation: fog or gof

If Function f: A → B and g: B → C, then the composition of function gof: A → C is given as gof(x) = g(f(x)) for all X in set A

NOTE: fog is used to represent f(g(x))

The function gof is the set of all values of x in the domain of f, where f(x) is in the domain of g.

Properties:

Let f: X → Y, g: Y → Z and h: X → Z, then the functions show the following properties:

Associativity: f(gh)=(fg)h
If functions f and g are injective, then gof is also injective
If functions f and g are surjective, then gof is also surjective

Invertible Function:

A function is said to be invertible if f: X → Y then ∃ g: Y → X such that gof = Ix and fog = Iy.

Here g is known as the inverse of f. It is denoted by f-1.

Condition- If the function f: X → Y is bijective (one-one and onto), then it is said to be invertible.

Therefore, let f: X → Y be a bijective function and f(x) =y, then f-1: X → Y defined as f-1(y)=x, is known as the invertible function of f and is always unique.

Binary Operation

Notation: *

A binary operation is used to associate two elements of a set. They are mathematical operations (addition, subtraction, multiplication and division) performed between two elements of a set.

If S is a non-empty set, let * be the binary operation on S, then a*b is defined for every element belonging to the set S, i.e. for all a,b ∈ S.

We can say that * is a function such that *(x): A × A → A

Example: let there be two real numbers a,b ∈ R then a+b ∈ R.

Properties of Binary Operations:

The properties of binary operations include

1. Commutative property:

The binary operation *: A × A → A is commutative on set A, i.e., for all a, b ∈ A a*b = b*a

2. Associative property

The binary operation *: A × A → A is associative on set A, i.e., for all x, y, z ∈ A x*(y*z) = (x*y)*z

3. Existence of Identity

For the binary operation *: A × A → A, there exists an element e such that a*e = e*a= a for all a, b ∈ A, where e is the identity element of Set A

4. Existence of inverse

For the binary operation *: A × A → A, there exists an element b such that a*b = b*a= a for all a, b ∈ A, where b is the inverse of a denoted by a-1.

Theorems:

Prove the associativity of the composition of a function. i.e., if f: X → Y, g: Y → Z and h: X → Z, then

show that fo(goh) =(fog)oh

Prove that two invertible functions f: X → Y, g: Y → Z, then (gof)-1= f-1o g-1

Students may refer to these Class 12 Mathematics Chapter 1 notes while preparing for their examination, as they include all concepts that are a part of the syllabus.

Class 12 Mathematics Chapter 1 Notes: Exercises & Answer Solutions

Chapter 1 Mathematics class 12 notes begin with basic notations and definitions of Relations and Functions. To understand this chapter easily, students need to revise the concepts learnt in Class 11. Students will know about different types of relations, function and their properties, such as associativity, commutativity, the existence of identity and inverse element, etc.

To help students understand this chapter better, Extramarks provides detailed Class 12 Mathematics Chapter 1 notes. It includes all the important definitions, formulae, theorems and properties of this chapter. With the help of the CBSE solutions, students can solve and understand questions using step-by-step and well-explained answers. These Class 12 Mathematics Chapter 1 notes will allow students to retain their learning and reduce simple mistakes.

For quick revision, refer to the Extramarks Exercises & Answer Solutions and MCQ questions under the CBSE sample papers of this chapter from the links below.

Extramarks, an online learning platform, focuses on providing students with a wonderful learning experience. To attain high scores, students are advised to refer to the CBSE revision notes and CBSE previous year question papers and practice as many CBSE extra questions as possible.

NCERT Exemplar Class 12 Mathematics

With the help of the NCERT Exemplar, students can gain an in-depth knowledge and develop a strong foundation in Mathematics. Students can use the Class 12 Mathematics Chapter 1 notes, NCERT Exemplar and other NCERT books to perform well in the exams and pursue their dreams. NCERT Exemplar includes many practice questions of all difficulty levels, which will help in developing analytical and problem-solving abilities. Using the NCERT Exemplar and Class 12 Mathematics Chapter 1 notes, students will learn several tips, tricks and shortcut methods to tackle any question in the examination quickly.

Key Features of Class 12 Mathematics Chapter 1 notes

The key features of Class 12 Mathematics Chapter 1 notes provided by Extramarks are

These notes are prepared by the subject elites at Extramarks.
The notes follow the CBSE syllabus and are based on the latest guidelines issued by the board.
The class 12 Mathematics chapter 1 notes offer the student a wide variety of questions to practice to get acquainted with the concepts in the Class 12 Mathematics Chapter 1.
Studying with the help of notes will help to build confidence and reduce exam stress and anxiety.
It helps to develop advanced skills and prepares the students for the CBSE Class 10 board exams as well as the competitive examinations like JEE.
The class 12 chapter 1 Mathematics notes help to develop time management skills in students and reduce the chances of making simple mistakes.

For more guidance on CBSE exams, keep visiting Extramarks. Students are provided with the latest news on the CBSE paper pattern, marking system and important preparation tips. Students can also access the Class 12 Mathematics Chapter 1 notes for free.

Q.1 If f: [1, 2, 3] [a, b, c] and g: [a, b, c] [apple, ball, cat] defined as f(1) = a, f(2) = b, f(3) = c, g(a) = apple, g(b) = ball and g(c) = cat. Show that f, g and gof are invertible. Find out f ^-1, g^-1 and (gof)^-1 and show that (gof)^-1 = f ^-1og^-1.

Ans

$\begin{array}{l} By definition, f and gare bijective functions . \\ Let f^{- 1} : [a, b, c] \to [1, 2, 3] and g^{- 1} : [apple, ball, cat] \to [a, b, c] \\ be defined as \\ f^{- 1} (a) = 1, f^{- 1} (b) = 2, f^{- 1} (c) = 3 \\ g^{- 1} (apple) = a, g^{- 1} (ball) = b, g^{- 1} (cat) = c \\ \Rightarrow f^{- 1} of = I_{(123)} \\ {fof}^{- 1} = I_{(a, b, c)} \\ and g^{- 1} = I_{(a, b, c)} \\ {gog}^{- 1} = I_{(apple, ball, cat)} \\ Now, gof : [1, 2, 3] \to [apple, ball, cat] is given by \\ gof (1) = apple, gof (2) = ball, gof (3) = cat . We can define \\ {gof}^{- 1} [apple, ball, cat] \to [1, 2, 3] by \\ {gof}^{- 1} (apple) = 1 \\ {gof}^{- 1} (ball) = 2 \\ {gof}^{- 1} (cat) = 3 \\ We can see that {gof}^{- 1} o (gof) = I_{\{1, 2, 3\}} and \\ (gof) o {(gof)}^{- 1} = I_{\{apple, ball, cat\}} \\ Thus, f, g and gof are invertible . \\ Now, \\ f^{- 1} {og}^{- 1} (apple) = f^{- 1} (g^{- 1} (apple)) = f^{- 1} (a) = 1 = {gof}^{- 1} (apple) \\ f^{- 1} {og}^{- 1} (ball) = f^{- 1} (g^{- 1} (ball)) = f^{- 1} (b) = 2 = {gof}^{- 1} (ball) \\ f^{- 1} {og}^{- 1} (cat) = f^{- 1} (g^{- 1} (cat)) = f^{- 1} (c) = 3 = {gof}^{- 1} (cat) \\ Hence, {(gof)}^{- 1} = f^{- 1} {og}^{- 1} \end{array}$

Q.2

Ans

$\begin{array}{l} Define a binary operation * on the set \{0, 1, 2, 3, 4, 5\} \\ as a * b = \{\begin{cases} a + b if a + b < 6 \\ a + b - 6 if a + b \geq 6 \end{cases} \\ Show that 0 is the identify for this operation and each \\ element of the set is invertible with 6 - a being the inverse of a . \\ (i) e is the identify element if a * e = e * a = a \\ a * o = a + o = a \\ o * a = o + a = a \\ \Rightarrow a * o = o * a \\ Hence, 0 is the identify of the operation . \\ (ii) b is the inverse of a if \\ a * b = b * a = a \\ Now, a * (6 - a) = a + (6 - a) - 6 = 0 \\ (6 - a) * a = (6 - a) + a - 6 = 0 \\ Hence, each element of the given set is invertible \\ with 6 - a being the inverse of a . \end{array}$

Q.3 Let * be the binary operation on the set Q of rational numbers which is defined as follows

(i) a * b = a – b
(ii) a * b = a² + b²
(iii) a * b = a + ab
Find which of the binary operations are commutative and which are associative.

Ans

(i) Here a * b = a – b
Now b *a = b – a, but a – b ≠ b – a
⇒ a * b ≠ b * a
⇒ * is not commutative

a*(b*c) = a*(b – c) = a – (b – c) = a – b + c
(a*b)*c = (a – b)*c = (a – b) – c
Thus, a*(b*c) ≠ (a * b)*c
⇒ * is not associative

(ii) Here a*b = a² + b²
b*a = b² + a² = a² + b²
a*b = b*a
⇒ * is commutative

a*(b*c) = a*(b² + c²) = a² + (b² + c²)²
(a*b)*c = (a² + b²)*c = (a²+ b²)² + c²
Thus, a*(b*c) ≠ (a*b)*c
⇒ * is not associative

(iii) Here a * b = a + ab
Now b *a = b + ba
⇒ a * b ≠ b * a
⇒ * is not commutative

a*(b*c) = a*(b +bc) = a + a(b + bc) = a + ab + abc
(a*b)*c = (a + ab)*c = a + ab + (a + ab)c = a + ab + ac + abc
Thus, a*(b*c) ≠ (a * b)*c
⇒ * is not associative

Q.4

$\begin{array}{l} If f (x) = \frac{(4 x + 3)}{(6 x - 4)}, show that fof (x) = x_{1} \forall x \neq \frac{2}{3} . \\ What is the inverse of f ? \end{array}$

Ans

$\begin{array}{l} Here, f (x) = \frac{4 x + 3}{6 x - 4}, x \neq \frac{2}{3} \\ (a) fof (x) = f (f (x)) \\ = f (\frac{4 x + 3}{6 x - 4}) \\ = \frac{4 (\frac{4 x + 3}{6 x - 4}) - 3}{6 (\frac{4 x + 3}{6 x - 4}) - 4} \\ = \frac{\frac{4 (4 x + 3) + 3 (6 x - 4)}{6 x - 4}}{\frac{6 (4 x + 3) + 4 (6 x - 4)}{6 x - 4}} \\ = \frac{4 (4 x + 3) + 3 (6 x - 4)}{6 (4 x + 3) - 4 (6 x - 4)} \\ = \frac{16 x + 12 + 18 x - 12}{24 x + 18 - 24 x + 16} = \frac{34 x}{34} = x \end{array}$ $\begin{array}{l} (b) Put f (x) = y \\ \Rightarrow y = \frac{4 x + 3}{6 x - 4} \\ \Rightarrow y (6 x - 4) = 4 x + 3 \\ \Rightarrow 6 xy - 4 y = 4 x + 3 \\ \Rightarrow 6 xy - 4 x = 4 y + 3 \\ \Rightarrow x (6 y - 4) = 4 y + 3 \\ \Rightarrow x = \frac{4 y + 3}{6 y - 4} \end{array}$

Q.5 Consider the binary operation ∧ on the set {1, 2, 3, 4, 5} defined by a ∧ b = min{a, b}. Write the composition table of the operation ∧ .

Ans

Operation ∧ table on the set {1, 2, 3, 4, 5} is as follows.

∧	1	2	3	4	5
1	1	1	1	1	1
2	1	2	2	2	2
3	1	2	3	3	3
4	1	2	3	4	4
5	1	2	3	4	5

Q.6 Let the * binary operation on N be defined by a*b = HCF of a and b. Is * commutative? Is * associative? Does there exist identity for this operation on N?

Ans

Here a*b = HCF of a and b.

(a) We know HCF of a, b = HCF of b, a

⇒ a*b = b*a

Hence, * is commutative.

(b) a*(b*c) = a*(HCF of b, c)

= HCF of a and HCF of b, c

= HCF of a, b and c

Now (a*b)*c = (HCF of a, b)*c

= HCF of a, b and HCF of c

= HCF of a, b and c

⇒ a*(b*c) = (a*b)*c

Hence, * is associative.

i.e., 1*a = a*1 = 1
1 ≠ a

∴ There does not exist any identity for this operation

Q.7 Let f: N → R be a function defined as f(x) = 4x² + 12x + 15. Show that f: N → S where, S is the range of f(x) is invertible. Find the inverse of f.

Ans

$\begin{array}{l} f is invertible if f : N \to S is a bijection . \\ f is one - one : For any x, y \in N \\ f (x) = f (y) \\ \Rightarrow 4 x^{2} + 12 x + 15 = 4 y^{2} + 12 y + 15 \\ \Rightarrow 4 x^{2} + 12 x = 4 y^{2} + 12 y \\ \Rightarrow 4 (x^{2} - y^{2}) + 12 (x - y) = 0 \\ \Rightarrow (x^{2} - y^{2}) + 3 (x - y) = 0 \\ \Rightarrow (x - y) (x + y) + 3 (x - y) = 0 \\ \Rightarrow (x - y) (x + y + 3) = 0 [x + y + 30 for anyx, yN] \\ \Rightarrow x - y = 0 \\ \Rightarrow x = y \\ So, f : N \to S is one - one . \\ Obviously f : N \to S is onto . \\ Hence f : N \to S is invertible . \\ Let f^{- 1} denote the inverse of f . Then, \\ {fof}^{- 1} (x) = x for all x \in S \\ \Rightarrow f (f^{- 1} (x)) = x for all x \in S \\ \Rightarrow 4 {(f^{- 1} (x))}^{2} +12 f^{- 1} (x) +15 = x for all x \in S \\ \Rightarrow 4 {\{f^{- 1} (x)\}}^{2} + 12 f^{- 1} (x) + 15 = 0 \\ \Rightarrow f^{- 1} (x) = \frac{- 12 \pm \sqrt{144 - 16 (15 - x)}}{8} = \frac{- 3 \pm \sqrt{x - 6}}{2} \\ \Rightarrow f^{- 1} (x) = \frac{- 3 + \sqrt{x - 6}}{2} [f^{- 1} (x) \in N, ∴ f^{- 1} (x) > 0] \end{array}$

Q.8

$\begin{array}{l} Let A = {- 1, 0, 1, 2}, B = {- 4, 2, 0, 2} and f, g: A \to B be \\ function defined by f (x) = x^{2} - x, x \in A and g (x) = 2 | x - \frac{1}{2} | - 1, x \in A. \\ Are f and g equal? Justify your answer. \end{array}$

Ans

$\begin{array}{l} At x = -1, \\ f (x) = 1^{2} + 1 = 2 \\ g (x) = 2 | - 1 - \frac{1}{2} | - 1 \\ = 2 \times \frac{3}{2} - 1 = 2 \\ At x = 0 \\ f (0) = 0 \\ g (0) = 2 | \frac{- 1}{2} | - 1 = 2 \times \frac{1}{2} - 1 = 0 \\ At x =1 \\ f (1) = 1^{2} - 1 = 0 \\ g (1) = 2 | 1 - \frac{1}{2} | - 1 = 2 \times \frac{1}{2} - 1 = 0 \\ At x = 2 \\ f (2) = 2^{2} - 2 = 2 \\ g (2) = 2 | 2 - \frac{1}{2} | - 1 = 3 - 1 = 2 \\ ∴ f (a) = g (a) for all a \in A \\ \Rightarrow f and g are equal functions . \end{array}$

Q.9 Let f, g and h be functions from R →R. Then show that
(i) (f + g)oh = foh + goh
(ii) (f.g)oh = (foh).(goh)

Ans

Here f: R →R, g: R →R, h: R →R(i) ((f + g)oh)(x) = (f + g)[h(x)]
= f[h(x)] + g[h(x)]
= foh(x) + goh(x)
⇒ (f + g)oh = foh + goh

(ii) ((f.g)oh)(x) = (f.g)[h(x)]
= f[h(x)].g[h(x)]
=(foh)(x).(goh)(x)
⇒ (f.g)oh = (foh).(goh)

Q.10

$\begin{array}{l} Let A = R – {3} and B =R – {- 1} . Consider the function \\ f: A \to B defined by f (x) = \frac{x - 2}{x - 3} . Is f one-one onto ? \\ Justify your answer. \end{array}$

Ans

$\begin{array}{l} f : A \to B where A = R - {3} and B = R - {1} . f is defined by f (x) = \frac{x - 2}{x - 3} . \\ (a) f (x_{1}) = \frac{x_{1} - 2}{x_{1} - 3} f (x_{2}) = \frac{x_{2} - 2}{x_{2} - 3} \\ f (x_{1}) = f (x_{2}) \\ \Rightarrow \frac{x_{1} - 2}{x_{1} - 3} = \frac{x_{2} - 2}{x_{2} - 3} \\ \Rightarrow (x_{1} - 2) (x_{2} - 3) = (x_{2} - 2) (x_{1} - 3) \\ \Rightarrow x_{1} x_{2} - 3 x_{1} - 2 x_{2} + 6 = x_{1} x_{2} - 2 x_{1} - 3 x_{2} + 6 \\ \Rightarrow - 3 x_{1} - 2 x_{2} = - 2 x_{1} - 3 x_{2} \\ \Rightarrow x_{1} = x_{2} \\ ∴ f is one - one \\ (b) Let y = \frac{x - 2}{x - 3} \\ \Rightarrow xy - 3 y = x - 2 \\ or xy - x = 3 y - 2 \\ or x (y - 1) = 3 y - 2 \\ \Rightarrow x = \frac{3 y - 2}{y - 1} \\ \Rightarrow For every value of y \in B except y =1 there is a pre - image, \\ x = \frac{3 y - 2}{y - 1} \\ \Rightarrow f is onto \\ This f is one - one and onto . \end{array}$

Q.11

$\begin{array}{l} Show that the functionf : \to R given by f (x) = \{\begin{matrix} 1 if x > 0 \\ 0 if x = 0 \\ - 1 if x < 0 \end{matrix}\} \\ is neither one - one nor onto . \end{array}$

Ans

(a) f(1) = f(2) = 1

1 and 2 have the same images.

⇒ f(x₁) = f(x₂) = 1 for x > 0
But x₁ ≠ x₂

Similarly -1 and -2 have the same images.

Hence f is not one-one.

(b) Except -1, 0, 1 no other element of co-domain of f has any pre-image in its domain.

⇒ f is not onto.

Q.12 Check the injectivity and surjectivity of the following:
(i) f: N → N given by f(x) = x²

(ii) f: R → R given by f(x) = x²

Ans

(i) f: N → N given by f(x) = x²
(a) f(x₁) = f(x₂) ⇒ x₁²= x₂²⇒ x₁= x₂

⇒f is one – one, i.e., it is injective

(b) There are many such elements belonging to co-domain that have no pre-image in their domain N e.g. 3 ∈ co domain N. But there is no pre- image in domain of f.

⇒ f is not onto, i.e., not surjective.

(ii) f: R R given by f(x) = x²
(a) f is not one – one because f(-1) = f (1) = 1
-1 and 1 have the same image
⇒ f is not injective.
(b) -2 ∈ co domain R of f. But √-2 does not belong to domain R of f.
⇒ f is not onto, i.e., f is not surjective.

Q.13 Show that the Relation R on the set A = {x ∈ Z : 0 ≤ x ≤ 12}, given by R = {(a, b) : |a – b| is a multiple of 4} is an equivalence relation.

Ans

R = {(a, b) : |a – b| is a multiple of 4}, where a, b ∈ A ={x Z : 0 ≤ x ≤ 12} = {0, 1, 2, …, 12}.

Reflexivity: For any a ∈ A, |a – a| = 0, which is a multiple of 4.

⇒ (a, a) ∈ R, for all a ∈ A.

So, R is reflexive.

Symmetry: Let (a, b) ∈ R. Then,

(a, b) ∈ R

⇒ |a – b| is a multiple of 4
⇒ |a – b| = 4k for some k ∈ N
⇒ |b – a| = 4k for some k ∈ N
⇒ (b, a) ∈ R

So, R is symmetric.

Transitive: Let (a, b) ∈ R and (b, c) ∈ R. Then,
⇒ |a – b| is a multiple of 4 and |b – c| is a multiple of 4
⇒ |a – b| = 4k and |b – c| = 4m for some k, m ∈ N
⇒ a – b = ± 4k and b – c = ± 4m
⇒ a – c = ± 4k ± 4m
⇒ a – c is a multiple of 4
⇒ |a – c| is a multiple of 4
⇒ (a, c) ∈ R

So, R is transitive.

Hence, R is an equivalence relation.

Q.14 Determine whether the following relation is reflexive, symmetric and transitive: Relation R on the set N of all natural numbers is defined as
R = {(x, y): y = x + 5 and x < 4}

Ans

R = {(x, y): y = x + 5 and x < 4}
∴ R = {(1, 6), (2, 7), (3, 8)}

Reflexivity: (1, 1), (2, 2) etc. are not in R. So, R is not reflexive.

Symmetry: (1, 6) ∈ R but (6, 1) ∉ R. So, R is not symmetric.

Transitivity: Since (1, 6) ∈ R and there is no ordered pair in R which has 6 as the first element. Same is the case for (2, 7) and (3, 8). So, R is not transitive.

Q.15 Show that 1/x is the inverse of x ≠ 0 for the multiplication operation on R.

Ans

We know that x X (1/x) = 1 (Identity for ‘x’). So, 1/x is the multiplicative inverse of x.

Q.16 Let f = {(3, 1), (2, 3), (1, 2)}, find f ^-1if f is one-one and onto.

Ans f is invertible. So, f ^-1 = {(1, 3), (3, 2), (2, 1)}.

Q.17

$\begin{array}{l} If a function is defined as f (x) = \frac{x}{x - 1}, x \neq 1, then \\ find of f (x) \end{array}$

Ans

$\begin{array}{l} If a function is defined as f (x) = \frac{x}{x - 1}, x \neq 1, then \\ find of f (x) \\ f (x) = \frac{x}{x - 1} \\ \Rightarrow f of (x) = \frac{\frac{x}{x - 1}}{\frac{x}{x - 1} - 1} \\ = \frac{\frac{x}{x - 1}}{\frac{x - x + 1}{x - 1}} \\ = \frac{\frac{x}{x - 1}}{\frac{1}{x - 1}} \\ Now, fof (x) = x \\ \Rightarrow fofof (x) = \frac{x}{x - 1} [same as given function] \\ on repeating the above step, we get fofofof (x) = x . \end{array}$

Q.18 A relation R in a set A is called …, if no element of A is elated to any element of A.

AnsEmpty Relation

Q.19 A relation R in a set A is called …, if each element of A is elated to every element of A.

AnsUniversal Relation

Q.20 Name the relation R in a set A satisfying the condition (a, a) ∈ R, for every a ∈ A.

Ans

The relation R is reflexive.

Q.21 Name the relation R in a set A satisfying the condition (a₁, a₂) ∈ R ⇒ (a₂, a₁) ∈ R for all (a₁, a₂) ∈ A.

Ans

The relation R is symmetric.

Q.22 R is a relation in a set {1, 2, 4} given by R = {(1, 1), (2, 2), (4, 4)}. State whether R is reflexive or symmetric.

Ans

R is reflexive. [ (a, a) ε R, for every a ε {1, 2, 4}]

Q.23 Let f: R R; f(x) = sinx and g: R R; g(x) = x², find fog and gof.

AnsClearly, fog and gof both exist.
Now, (gof)(x) = g(f(x)) = g(sinx) = (sinx)² = sin²x.
And, (fog)(x) = f(g(x)) = f(x²) = sinx².

Q.24 If f is one-one and onto, then what is f ^-1o f ?

Ans

f ^-1of = I_f

Q.25 Let * be a binary operation on set Q of rational numbers as follows:

a*b = (a-b)² . Is this operation commutative?

Ans

a*b = (a-b)²
b*a = (b-a)²
(a-b)²= (b-a)²
Therefore, a*b = b*a
So, the given operation is commutative.

Q.26

$If f (x) = x+8 and g (x) = x-8, \forall x \in R, find fog (8)$

Ans

$\begin{array}{l} f (x) = x + 8, g (x) = x - 8 \\ ∴ fog (x) = f [g (x)] \\ \Rightarrow fog (x) = f (x - 8) \\ \Rightarrow fog (x) = (x - 8) + 8 \\ \Rightarrow fog (x) = x \\ \Rightarrow fog (8) = 8 \end{array}$

Q.27

$If f (x) {= x}^{2} +3 and g (x) = \frac{x}{x+2}, \forall x \in R, find gof (2)$

Ans

$\begin{array}{l} f (x) {= x}^{2} + 3, g (x) = \frac{x}{x+2} \\ ∴ gof (x) = g [f (x)] \\ = g [x^{2} + 3] \\ = \frac{x^{2} + 3}{(x^{2} + 3) + 2} \\ \Rightarrow gof (x) = \frac{x^{2} + 3}{x^{2} + 3 + 2} \\ \Rightarrow gof (x) = \frac{x^{2} + 3}{x^{2} + 5} \\ \Rightarrow gof (2) = \frac{4 + 3}{4 + 5} = \frac{7}{9} \end{array}$

Q.28 Let * be a binary operation defined by a*b=2a+b+ab+1. Find 3*5.

Ans

Given that

a*b = 2a+b+ab+1

⇒ 3*5 = 2×3+5+3×5+1

= 6+5+15+1

= 27

Q.29

$\begin{array}{l} If a function is defined as f (x) = \frac{x}{x - 1}; x \neq 1, then \\ find fof (x) . \end{array}$

Ans

$\begin{array}{l} f (x) = \frac{x}{x - 1} \\ \Rightarrow fof (x) = \frac{\frac{x}{x - 1}}{\frac{x}{x - 1} - 1} \\ = \frac{\frac{x}{x - 1}}{\frac{x - x + 1}{x - 1}} \\ = \frac{\frac{x}{x - 1}}{\frac{1}{x - 1}} = x \\ ∴ fof (x) = x \end{array}$

Q.30

$\begin{array}{l} If a function is defined as f (x) = \frac{x}{\sqrt{1 + x^{2}}}, then \\ find fof (x) . \end{array}$

Ans

$\begin{array}{l} f (x) = \frac{x}{\sqrt{1 + x^{2}}} \\ \Rightarrow fof (x) = \frac{\frac{x}{\sqrt{1 + x^{2}}}}{\sqrt{1 + {(\frac{x}{1 + x^{2}})}^{2}}} \\ = \frac{\frac{x}{\sqrt{1 + x^{2}}}}{\sqrt{1 + \frac{x^{2}}{1 + x^{2}}}} \\ = \frac{\frac{x}{\sqrt{1 + x^{2}}}}{\sqrt{\frac{1 + 2 x^{2}}{1 + x^{2}}}} \\ = \frac{x}{\sqrt{1 + x^{2}}} \times \frac{\sqrt{1 + x^{2}}}{\sqrt{1 + x^{2}}} \\ = \frac{x}{\sqrt{1 + 2 x^{2}}} \end{array}$

Q.31 Let A = {1, 2, 3, 8}, B = {4, 5, 6} and let f = {(1, 4), (2, 5), (3, 6)} be defined from A to B. Show that f is not a function from A to B.

AnsSince there is no image in set B for every element of set A (that is, element 8 of set A has no image in set B), therefore, f is not a function from set A to set B.

Q.32 Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show that f is one-one.

Ans

Since every element of set A has only one image in set B, that is 1→ 4, 2→ 5, 3→ 6, therefore, f is one-one function.

Q.33 Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 4), (3, 6)} be a function from A to B. Show that f is many one.

Ans

Since elements 1 and 2 have same image 4 in set B, (1→ 4, 2→ 4, 3→ 6 ), therefore, f is many one function.

Q.34 Let A = {1, 2}, B = {4, 5, 6} and let f = {(1, 4), (2, 5), (2, 6)} be a function from A to B. Show that f is not a function.

Ans

Since every element of set A has no unique image in set B, that is, element 2 has two images 5 and 6 in set B (1→ 4, 2→ 5, 2→ 6), therefore, f is not a function.

Q.35 If Set A has 2 elements and Set B has 3 elements, find the number of relations defined from Set A to Set B.

Ans

Since number of elements in set A = 2

Number of elements in set B = 3

∴ Number of elements in set (AXB) = 23=6

The number of all possible subsets of Set (AXB) = 2⁶

⇒ The number of relations defined from Set A to Set B is 2⁶ = 64.

Q.36

$\begin{array}{l} Show that the function f:R \to R defined by f (x) = \frac{2 x - 1}{3}, \\ x \in R is one-one and on to function. Also, find the \\ inverse of the function f. \end{array}$

Ans

$\begin{array}{l} {Let x}_{1,} x_{2} \in R such that f (x_{1}) = f (x_{2}) . \\ f (x_{1}) = f (x_{2}) \\ \Rightarrow \frac{2 x_{1} - 1}{3} = \frac{2 x_{1} - 1}{3} \\ \Rightarrow 2 x_{1} - 1 = 2 x_{2} - 1 \\ \Rightarrow 2 x_{1} = 2 x_{1} \\ \Rightarrow x_{1} = x_{2} \\ ∴ f is one-one. \\ Again, y = \frac{2 x - 1}{3} \\ \Rightarrow 2 x - 1 = 3 y \\ \Rightarrow 2 x = 3 y + 1 \\ \Rightarrow x = \frac{3 y + 1}{2} \\ Now, f (x) = f (\frac{3 y + 1}{2}) \forall x, y \in R \\ = \frac{2 (\frac{3 y + 1}{2}) - 1}{3} \\ = \frac{3 y -}{3} \\ = \frac{3 y}{3} \\ = y \\ ∴ f is onto function. \\ Since f is one-one and onto function, therefore, f is invertible. \\ Again, y = \frac{2 x - 1}{3} \\ \Rightarrow 2 x - 1 = 3 y \\ \Rightarrow 2 x = 3 y + 1 \\ x = \frac{3 y + 1}{2} \\ \Rightarrow f^{- 1} (y) = \frac{3 y + 1}{2} [f (x) = y \Rightarrow x = f^{- 1} (y)] \\ \Rightarrow f^{- 1} (x) = \frac{3 x + 1}{2} [Replacing y by x] \end{array}$

Q.37

$Find the domain of the function f (x) = \sqrt{36 - x^{2} .}$

Ans

$\begin{array}{l} We know that all values of x for which \\ function is defined is called domain. \\ f (x) = \sqrt{36 - x^{2}} \\ For function to be defined \\ 36 - x^{2} \geq 0 \\ \Rightarrow x^{2} - 36 \leq 0 \\ \Rightarrow x^{2} - 6^{2} \leq 0 \\ \Rightarrow (x - 6) (x + 6) \leq 0 \\ \Rightarrow x \geq - 6 and x \leq 6 \\ \Rightarrow - 6 \leq x \leq 6 \\ \Rightarrow x \in [- 6, 6] \\ Hence, domain of the given function is [- 6, 6] . \end{array}$

Q.38

$Find the domain of the function f (x) = \sqrt{x^{2} - 7 x + 10 .}$

Ans

$\begin{array}{l} We know that all values of x for which function \\ is defined is called domain. \\ f (x) = \sqrt{x^{2} - 7 x + 10} \\ For function to be defined, \\ x^{2} - 7 x + 10 \geq 0 \\ \Rightarrow (x - 2) (x - 5) \geq 0 \\ \Rightarrow (x - 2) \geq 0 and (x - 5) \geq 0 \\ \Rightarrow x \geq 2 and x \geq 5 \Rightarrow x \geq 5 . .. (1) \\ \Rightarrow Also (x - 2) \leq 0 and (x - 5) \leq 0 \\ \Rightarrow x \leq 2 and x \leq 5 \Rightarrow x \leq 2 . \dots (2) \\ From (1) and (2) we get, \\ - \infty < \times \leq 2 and 5 \leq \times \infty \\ Hence, domain of the given function is \\ (- \infty, 2] \cup [5, \infty) . \end{array}$

Q.39

$\begin{array}{l} Find the domain of the function \\ f (x) = \sqrt{x^{3} - 6 x^{2} + 11 x} - 6 . \end{array}$

Ans

$\begin{array}{l} We know that all values of x for which function \\ is defined is called domain. \\ ∴ for function to be defined \\ x^{3} - 6 x^{2} + 11 x – 6 \geq . \dots (1) \\ Let P (x) = x^{3} - 6 x^{2} + 11 x - 6 \\ Now for getting its factor, put x = - 1, 1, 0 . \dots \\ {(1)}^{3} - 6 {(1)}^{2} + 11 (1) - 6 \\ = 1 - 6 +11 - 6 \\ = 12 - 12 \\ 0 \\ Hence, (x - 1) is one of the factors of P (x) . \\ Now on dividing P (x) by (x - 1), we get \\ P (x) = (x - 1) (x^{2} - 5 x + 6) \\ P (x) = (x - 1) (x - 2) (x - 3) \end{array}$

$\begin{array}{l} The expression is +ve in the interval \\ [1, 2] and [3, + \infty] . \\ Hence, the domain of the given function is \\ [1, 2] \cup [3, + \infty] . \end{array}$

Q.40

$\begin{array}{l} Show that the function f:R \to R defined by f (x) = \frac{5 x + 2}{3}, \\ x \in R is one-one and on to function. Also find the \\ inverse of the function f. \end{array}$

Ans

$\begin{array}{l} {Let x}_{1,} x_{2} \in R such that f (x_{1}) = f (x_{2}) . \\ f (x_{1}) = f (x_{2}) \\ \Rightarrow \frac{5 x_{1} + 2}{3} = \frac{5 x_{2} + 2}{3} \\ \Rightarrow 5 {fx}_{1} + 2 = 5 {fx}_{2} + 2 \\ \Rightarrow 5 {fx}_{1} = 5 {fx}_{2} \\ \Rightarrow x_{1} = x_{2} \\ ∴ f is one-one \\ Again, y = \frac{fx+2}{3} \\ \Rightarrow 5 x + 2 = 3 y \\ \Rightarrow 5 x = 3 y - 2 \\ \Rightarrow x = \frac{3 y - 2}{5} \end{array}$ $\begin{array}{l} Now, f (x) = f (\frac{3 y - 2}{5}) \forall x, y \in R \\ \Rightarrow = \frac{5 (\frac{3 y - 2}{5}) + 2}{3} \\ = \frac{3 y - 2 + 2}{3} \\ = \frac{3 y}{3} \\ = y \\ ∴ f is onto function. \\ Since f is one-one and onto function. therefore. it is invertible. \\ Again, y = \frac{5 x + 2}{3} \\ \Rightarrow 5 x + 2 = 3 y \\ \Rightarrow 5 x = 3 y - 2 \\ \Rightarrow x = \frac{3 y - 2}{5} \\ \Rightarrow f^{- 1} (y) = \frac{3 y - 2}{5} [f (x) = y \Rightarrow {x = f}^{- 1} (y)] \\ \Rightarrow f^{- 1} (x) = \frac{3 x - 2}{5} \end{array}$

Q.41

$Find the range of the function f (x) = \frac{x^{2} + 2}{x^{2} + 1} \forall x \in R.$

Ans

$\begin{array}{l} Let y = \frac{x^{2} + 2}{x^{2} + 1} \forall \times \in R \\ \Rightarrow {yx}^{2} + y = x^{2} + 2 \\ \Rightarrow (y - 1) x^{2} = 2 - y \\ \Rightarrow x^{2} = \frac{(2 - y)}{y - 1} \\ \Rightarrow x = \sqrt{\frac{(2 - y)}{y - 1}} . \dots\dots. (1) \\ {Since x}^{2} \geq 0 \\ So, \frac{(2 - y)}{y - 1} \geq 0 \\ \Rightarrow \frac{}{} \frac{(y - 2)}{y - 1} \leq 0 \\ Now, using number line, we get \end{array}$

$\begin{array}{l} \Rightarrow 1 \leq y \leq 2 \\ but from (1) we get, y \neq 1 \\ \Rightarrow 1< y \leq 2 \\ \Rightarrow y \in (1, 2) \\ Range of the given function is (1, 2) . \end{array}$

Q.42

$\begin{array}{l} Show that the function f: [- 1, 1] \to R defined \\ by f (x) = \frac{x}{x + 2} is one-one. Find the inverse \\ of the function f. \end{array}$

Ans

$\begin{array}{l} {Let x}_{1}, x_{2} \in [- 1, 1] such that f (x_{1}) = f (x_{2}) . \\ f (x_{1}) = f (x_{2}) \\ \Rightarrow \frac{x_{1}}{x_{1} + 2} = \frac{x_{2}}{x_{2} + 2} \\ \Rightarrow x_{1} x_{2} + 2 x_{1} x_{2} + 2 x_{2} \\ \Rightarrow 2 x_{1} = 2 x_{2} \\ \Rightarrow x_{1} = x_{2} \\ ∴ f is one-one \\ Again, y = \frac{x}{x + 2} \\ \Rightarrow yx + 2 y = x \\ \Rightarrow 2 y = x - xy \\ \Rightarrow x (1 - y) = 2 y \\ \Rightarrow x = \frac{2 y}{1 - y} . \dots.. (1) \\ f (x) = f (\frac{2 y}{1 - y}) \\ = \frac{\frac{2 y}{1 - y}}{\frac{2 y}{1 - y} + 2} \\ = \frac{2 y}{2 y + 2 - 2 y} \\ = \frac{2 y}{2} \\ = y \\ ∴ f (x) = y [so, f (x) is on to] \\ \Rightarrow {x = f}^{- 1} (y) \\ Using this (1) become \\ \Rightarrow f^{- 1} (y) = \frac{2 y}{1 - y} \\ \Rightarrow f^{- 1} (x) = \frac{2 x}{1 - x} [Replacing y by x] \end{array}$

Q.43

$\begin{array}{l} Let f: R \to R be defined as f (x) = 11 \times -9. Find the \\ function g:R \to {such that gof = fog I}_{R} . \end{array}$

Ans

$\begin{array}{l} {We have, gof =I}_{R} \\ \Rightarrow gof (x) = I_{R} (x) for all \times \in R \\ \Rightarrow g [f (x)] = x for all \times \in R [I_{R} (x) = x] \\ \Rightarrow g (11 \times 9) = x for all \times \in R [f (x) = 11 x - 9] \\ \Rightarrow g (y) = x \\ Where y = 11x -9 \\ \Rightarrow 11x = y+9 \\ \Rightarrow x = \frac{y + 9}{11} \\ \Rightarrow g (y) = x \\ \Rightarrow g (y) = \frac{y + 9}{11} \\ \Rightarrow g (x) = \frac{x + 9}{11} \\ Thus, g:R \to R defined by g (x) = \frac{x + 9}{11} is the \\ required function. \end{array}$

Q.44

$\begin{array}{l} Let f : R \to R be defined as f (x) = ax+b. \\ Show that f (x) is one-one. Also find the function \\ g : R \to R such that gof = fog I. \end{array}$

Ans

$\begin{array}{l} Let x 1, x_{2} \in R such that f (x_{1}) = f (x_{2}) . \\ f (x_{1}) = f (x_{2}) \\ \Rightarrow {ax}_{1}_{2} \\ \Rightarrow {ax}_{1} = {ax}_{2} \\ \Rightarrow x_{1} = x_{2} \\ ∴ f is one - one . \\ We have, gof = I_{R} \\ \Rightarrow gof (x) = I_{R} (x) for all \times \in R \\ \Rightarrow g [f (x)] = x for all x \in R [I_{R} (x) = x] \\ \Rightarrow g (ax + b) = x for all x \in R [f (x) = ax + b] \\ \Rightarrow g (y) = x \\ Where y = ax + b \\ \Rightarrow ax = y - b \\ \Rightarrow x = \frac{y - b}{a} . \dots.. (1) \\ \Rightarrow g (y) = x \\ \Rightarrow g (y) = \frac{y - b}{a} using (1) \\ \Rightarrow g (x) = \frac{x - b}{a} \\ Thus, g : R \to R defined by g (x) = \frac{x - b}{a} is the required function . \end{array}$

Q.45 Consider the binary operation ∧ on the set {1, 2, 3, 4, 5} defined by a∧b = min {a, b}. Write the operation table of the operation ∧.

Ans

Since a*b = min {a, b}

⇒1*1 = min {1,1} = 1

⇒ 1*2 = min {1,2} = 1

⇒ 1*3 = min {1,3} = 1

⇒ 1*4 = min {1,4} = 1

⇒ 1*5 = min {1,5} = 1

⇒ 2*1 = min {2,1} = 1

⇒ 2*2 = min {2,2} = 2

⇒ 2*3 = min {2,3} = 2

⇒ 2*4 = min {2,4} = 2

⇒ 2*5 = min {2,5} = 2

⇒3*1 = min {3,1} = 1

⇒3*2 = min {3,2} = 2

⇒3*3 = min {3,3} = 3

⇒3*4 = min {3,4} = 3

⇒3*5 = min {3,5} = 3

⇒ 4*1 = min {4,1} = 1

⇒ 4*2 = min {4,2} = 2

⇒ 4*3 = min {4,3} = 3

⇒ 4*4 = min {4,4} = 4

⇒ 4*5 = min {4,5} = 4

⇒ 5*1 = min {5,1} = 1

⇒ 5*2 = min {5,2} = 2

⇒ 5*3 = min {5,3} = 3

⇒ 5*4 = min {5,4} = 4

⇒ 5*5 = min {5,5} = 5

Thus, operation table of operation (*) calculated above is

*	1	2	3	4	5
1	1	1	1	1	1
2	1	2	2	2	2
3	1	2	3	3	3
4	1	2	3	4	4
5	1	2	3	4	5

Q.46 Consider the binary operation * on the set {1, 2, 3, 4, 5} given by the following multiplication table. Using this table compute

*	1	2	3	4	5
1	1	1	1	1	1
2	1	2	1	2	1
3	1	1	3	1	1
4	1	2	1	4	1
5	1	1	1	1	5

(i) (3*4)*5.
(ii) 3*(4*5).
(iii) (2*3)*(4*5).
(iv) if * is commutative.

Ans

We have
(i) (3*4)*5 = 1*5 [Since, 3*4=1] =1
(ii) 3*(4*5) = 3*1 [Since, 4*5=1] =1
(iii) (2*3)*(4*5) = 1*1 [∴2*3=1 and 4*5=1] =1
(iv) Since (2*5)=1 and (5*2)=1
⇒(2*5) = (5*2)
⇒ Operation * is commutative.

Q.47 Show that addition, subtraction and multiplication are binary operations on R, but division is not a binary operation on R. Further, show that division is a binary operation on the set R of non-zero real numbers.

Ans

$\begin{array}{l} + : R \times R \to R is given by (a, b) \to a + b \\ If 2 \in R, 3 \in R then 2+3 = 5 \in R \\ \Rightarrow Operation ‘+’ is binary operation on R \\ - : R \times R \to R is given by (a, b) \to a - b \\ If 2 \in R, 3 \in R then 2 - 3 = - 1 \in R \\ \Rightarrow Operation ‘ - ‘ is binary operation on R \\ \times : R \times R \to R is given by (a,b) \to ab \\ If 2 \in R, O \in R then 2 \times 0 = 0 \in R \\ \Rightarrow Operation ‘ \times ‘ is binary operation on R \\ \div : R \times R \to R is given by (a, b) \to \frac{a}{b} \\ If 2 \in R, 0 \in R then \frac{2}{0} = \infty \in R \\ \Rightarrow Operation ‘ \div ‘ is not a binary operation on R \\ \div : R \times R \to is given by (a, b) \to \frac{a}{b}, where R is the set of \\ non-zero real numbers (given) . \\ If 2 \in R, \in R then \frac{2}{3} \in R . \\ \Rightarrow Operation ‘ \div ‘ is a binary operation on R. \end{array}$

Q.48

$If a function is defined as f (x) = \frac{x}{\sqrt{{1+x}^{2}}}, then find fofofofof (x) .$

Ans

$\begin{array}{l} f (x) = \frac{x}{1 + x^{2}} \\ \Rightarrow fof (x) = \frac{\frac{x}{\sqrt{1 + x^{2}}}}{\sqrt{1 + \frac{x}{1 + x^{2}}}} \\ = \frac{\frac{x}{\sqrt{1 + x^{2}}}}{\sqrt{1 + \frac{x^{2}}{1 + x^{2}}}} \\ = \frac{\frac{x}{\sqrt{1 + x^{2}}}}{\sqrt{\frac{1 + 2 x^{2}}{1 + x^{2}}}} \\ = \frac{\frac{x}{\sqrt{1 + x^{2}}}}{} x \frac{\sqrt{1 + x^{2}}}{\sqrt{1 + 2 x^{2}}} \\ = \frac{x}{\sqrt{1 + 2 x^{2}}} \\ = \frac{x}{\sqrt{1 + x^{2}}} \\ Now, fofof (x) = \frac{\frac{x}{\sqrt{1 + x^{2}}}}{\sqrt{1 + \frac{x^{2}}{1 + 2 x^{2}}}} \\ = \frac{\frac{x}{\sqrt{1 + 2 x^{2}}}}{\sqrt{\frac{1 + 2 x^{2} + x^{2}}{1 + 2 x^{2}}}} = \frac{\frac{x}{\sqrt{1 + 2 x^{2}}}}{\sqrt{\frac{1 + 3 x^{2}}{1 + 2 x^{2}}}} \\ = \frac{x}{\sqrt{1 + x^{2}}} x \frac{\sqrt{1 + 2 x^{2}}}{1 + 3 x^{2}} \\ = \frac{x}{\sqrt{1 + 3 x^{2}}} \\ on observing, we get \\ for fof there is 2, fofof there is 3, fofofof there will be 4 \\ and therefore, fofofofof (x) = \frac{x}{\sqrt{1 + 5 x^{2}}} \end{array}$

Q.49

$\begin{array}{l} Let : A \to B and g:B \to {A be two functions such that fog = I}_{B} . \\ Then, show that f is a surjection and g is an injection. \end{array}$

Ans

$\begin{array}{l} f is surjection: we have to prove here that every element in B \\ has its pre – image in A. \\ Let b be an arbitrary element of B. \\ Since, g:B \to A. Therefore, g (b) \in A \\ Let g (b) = a \\ Now, f (a) = f (g (b)) [a = g (b)] \\ \Rightarrow f (a) = fog (b) \\ \Rightarrow f (a) = I_{B} (b) [fog = I_{B}] \\ \Rightarrow f (a) = b \\ Thus, \forall b \in B there exists a \in A such that f (a) = b. \\ So, f is a surjection. \\ {g is an injection : let x}_{1} {, x}_{2} be any two elements of B that \\ g (x_{1}) = g (x_{2}) . Then, \\ g (x_{1}) = g (x_{2}) \\ \Rightarrow f (g (x_{1})) = f (g (x_{2}^{})) \\ \Rightarrow fog (x_{1}) = fog (x_{2}^{}) \\ \Rightarrow I_{B} (x_{1}) = I_{B} (x_{2}) \\ \Rightarrow x_{1} {= x}_{2} \\ Thus, g (x_{1}) = g (x_{2}) \Rightarrow x_{1} = x_{2} \forall x_{1} {, x}_{2}, \in B. \\ So, g is an injection. \end{array}$

Q.50

$\begin{array}{l} If f, g : R \to R are defined respectively by f (x) = x^{2} + 3 x + 1, \\ g (x) = 2 x - 3, find \\ (i) fog (ii) gof (iii) fof (iv) gog . \end{array}$

Ans

$\begin{array}{l} Since the range of f = doamin of g and range of g = domain of f, \\ therefore, fog, gof, gog and fof exists . \\ (i) For any x \in R, we have \\ fog (x) = f (g (x)) = f (2 x - 3) \\ = {(2 x - 3)}^{2} + 3 (2 x - 3) + 1 \\ = 4 x^{2} - 6 x + 1 \\ Therefore, \\ fog : R \to R is defined by (fog) (x) = 4 x^{2} - 6 x + 1 for all x \in R . \\ (ii) For any x \in R, we have \\ (gof) (x) = g (f (x)) \\ = g (x^{2} + 3 x + 1) \\ = 2 (x^{2} + 3 x + 1) - 3 \\ = 2 x^{2} + 6 x - 1 \\ Therefore, \\ gof : R \to R is defined by (gof) (x) = 2 x^{2} + 6 x - 1 for all x \in R . \\ (ii) For any x \in R_{1} we have \\ (gof) (x) = g (f (x)) \\ = g (x^{2} + 3 x + 1) \\ = 2 (x^{2} + 3 x + 1) - 3 \\ = 2 x^{2} + 6 x - 1 \\ Therefore, \\ gof : R \to R is defined by (gof) (x) = 2 x^{2} + 6 x - 1 for all x \in R . \\ (iii) For any x \in R, we have \\ (fof) (x) = f (f (x)) \\ = f (x^{2} + 3 x + 1) \\ = {(x^{2} + 3 x + 1)}^{2} + 3 (x^{2} + 3 x + 1) + 1 \\ = x^{4} + 9 x^{2} + 1 + 6 x^{3} + 6 x + 2 x^{2} + 3 x^{2} + 9 x + 3 + 1 \\ = x^{4} + 6 x^{3} + 14 x^{2} + 15 x + 5 \\ Therefore, \\ fof : R \to R is defined by \\ (fof) (x) = x^{4} + 6 x^{3} + 14 x^{2} + 15 x + 5 for all x \in R . \\ (iv) for any x \in R, we have \\ (gog) (x) = g (g (x)) \\ = g (2 x - 3) \\ = 2 (2 x - 3) - 3 \\ = 4 x - 9 \\ Therefore, \\ gog : R \to R is defined by (gog) (x) = 4 x - 9 for all x \in R . \end{array}$

Q.51

$\begin{array}{l} Let A = R – {3} and B = R – {1} . Consider the finction f:A \to B \\ defined by f (x) = \frac{x - 2}{x - 3} . Is f one – one and onto? \\ Justify your answer. \end{array}$

Ans

$\begin{array}{l} f : A \to B where A = R – {3} and B = R – {1}, f is defined by \\ f (x) = \frac{x - 2}{x - 3} . \\ (a) f (x_{1}) = \frac{x_{1} - 2}{x_{1} - 3}; f (x_{2}) = \frac{x_{2} - 2}{x_{2} - 3} \\ f (x_{1}) = f (x_{2}) \\ \Rightarrow \frac{x_{1} - 2}{x_{1} - 3} = \frac{x_{2} - 2}{x_{2} - 3} \\ \Rightarrow (x_{1} - 2) (x_{2} - 3) = (x_{2} - 2) (x_{1} - 3) \\ \Rightarrow x_{1} x_{2} - 3 x_{1} - 2 x_{2} + 6 = x_{1} x_{2} - 2 x_{1} - 3 x_{2} + 6 \\ \Rightarrow - 3 x_{1} - 2 x_{2} = - 2 x_{1} - 3 x_{2} \\ \Rightarrow x_{1} = x_{2} \\ ∴ f is one-one. \\ (b) y \in B=R- {1} . Then, y \neq 1 \\ The function f is onto if there exists x \in A such that \\ \Rightarrow f (x) = y \\ \Rightarrow \frac{x - 2}{x - 3} = y \\ \Rightarrow x - 2 = y (x - 3) \\ \Rightarrow x - 2 = yx - 3 y \\ \Rightarrow x - yx = 2 - 3 y \\ \Rightarrow x (1 - y) = 2 - 3 y \\ \Rightarrow x = \frac{2 - 3 y}{1 - y} \in A [y \neq 1] \\ Thus, for any y \in B, there exists \frac{2 - 3 y}{1 - y} \in A such that \\ f (\frac{2 - 3 y}{1 - y}) = \frac{\frac{2 - 3 y}{1 - y} - 2}{\frac{2 - 3 y}{1 - y} - 3} = \frac{\frac{2 - 3 y - 2 (1 - y)}{1 - y}}{\frac{2 - 3 y (1 - y)}{1 - y}} \\ = \frac{2 - 3 y - 2 + 2 y}{2 - 3 y - 3 + 3 y} = \frac{- y}{- 1} \\ f (\frac{2 - 3 y}{1 - y}) = y \\ ∴ f is onto. \\ Hence, function f is one-one and onto. \end{array}$

Q.52

$\begin{array}{l} Three relations R_{1}, R_{2}, R_{3} are defined on set A = (a, b, c) as follows : \\ (i) R_{1} = \{(a, a), (a, b), (a, c), (b, b), (b, c), (c, a) (c, b), (c, c)\} \\ (ii) R_{2} = \{(a, b), (b, a), (a, c) (c, a)\} \\ (iii) R_{3} = \{(a, b), (b, c), (c, a)\} \\ {Find whether each of the relations R}_{1}, R_{2}, R_{3} arereflexive, \\ symmetric and transitive. \end{array}$

Ans

$\begin{array}{l} (i) Reflexive : Clearly, (a, a), (b, b), (c, c) \in R_{1} . \\ So, R_{1} is reflexive on A. \\ Symmetric: We observe that (a, b) \in R_{1} but (b, a) \notin R_{1} . \\ {So, R}_{1} is not symmetric on A. \\ Transitive : We find that (b, c) \in R_{1} and (c, a) \in R_{1} but (b, a) \in R_{1} . \\ S {o, R}_{1} is not transitive on A. \\ (ii) Reflexive: Clearly, (a, a), (b . b), (c, c) \notin R_{2} . \\ So, R_{2} is not a reflexive on A. \\ Symmetric: We find that the ordered pairs obtained by interchanging \\ {the components of ordered pairs in R}_{2} {are also in R}_{2} . \\ So, R_{2} is a symmetric relation on A. \\ Transitive: We find that (a, b) \in R_{2} and (b, a) \in R_{2} . \\ So, R_{2} is not transitive on A. \\ (iii) Reflexive: Since none of (a, a), (b, b), (c, c) is {an element of R}_{3} . \\ {So, R}_{3} is not a reflexive relation on A. \\ Symmetric: We observe that (b, c) \in R_{3} but (c, b) \notin R_{3} . \\ So, R_{3} is not symmetric on A. \\ Transitive:We find that (a, b) \in R_{3} and (b, c) \in R_{3} but (a, c) \notin R_{3} . \\ So, R_{3} is not transitive on A. \end{array}$

Q.53

$\begin{array}{l} Show that f: [01, 1] \to Range f, given by f (x) = \frac{x}{x + 2} is one-one. \\ Find the inverse of the function f. \end{array}$

Ans

$\begin{array}{l} f : [01, 1] \to Range is given as f (x) = \frac{x}{x + 2} . \\ Let f (x) = f (y) . \\ \Rightarrow \frac{x}{x + 2} = \frac{y}{y + 2} \\ \Rightarrow xy + 2 x = xy + 2 y \\ \Rightarrow 2 x = 2 y \\ \Rightarrow x = y \\ ∴ f is a one - one function . \\ It is clear that f : [- 1, 1] \to Range f is onto . \\ ∴ f : [- 1, 1] \to Range f is one - one and onto and therefore, \\ the inverse of the function \\ f : [- 1, 1] \to Range f exists . \\ Let g : Range f \to [- 1, 1] be the inverse of f . \\ Let y be an arbitrary element of range f . \\ Since f : [- 1, 1] \to Range f is onto, we have : \\ y = f (x) for some x \in [- 1, 1] \\ \Rightarrow y = \frac{x}{x + 2} \\ \Rightarrow xy +2 y = x \\ \Rightarrow x (1 - y) = 2 y \\ \Rightarrow x = \frac{2 y}{1 - y}, y \neq 1 \\ Now, let us define g : Range f \to [- 1, 1] as \\ g (y) = \frac{2 y}{1 - y}, y \neq 1 \\ Now, (gof) (x) = g (f (x)) = g (\frac{x}{x + 2}) \\ = \frac{2 (\frac{x}{x + 2})}{1 - \frac{x}{x + 2}} = \frac{2 x}{x + 2 - x} \\ = \frac{2 x}{2} = x \\ (fog) (y) = f (g (y)) = f (\frac{2 y}{1 - y}) \\ = (\frac{\frac{2 y}{1 - y}}{\frac{2 y}{1 - y} + 2}) = \frac{2 y}{2 y + 2 - 2 y} \\ = \frac{2 y}{2} = y \\ ∴ gof = I_{[- 1, 1]} and fog = I_{Range f} \\ ∴ f^{- 1} = g \\ \Rightarrow f^{- 1} (y) = \frac{2 y}{1 - y}, y - \neq 1 \end{array}$

Q.54

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

Ans

$\begin{array}{l} Function f: {1, 2, 3} \to {a, b, c} is given by f (1) = a, f (2) = b, f (3) = c. \\ If (1) = a, f (2) = b, f (3) = c \\ If we define g : {a, b, c} \to {1, 2, 3} as g (a) = 1, g (b) = 2, g (c) = 3, \\ then we have : \\ (fog) (a) = f (g (a)) = f (1) a \\ (fog) (b) = f (g (b)) = f (2) b \\ (fog) (c) = f (g (c)) = f (3) c \\ And \\ (gof) (1) = g (f (1)) = g (a) 1 \\ (gof) (2) = g (f (2)) = g (a) 2 \\ (gof) (3) = g (f (3)) = g (a) 3 \\ ∴ gof = Ix and fog = Iy, where x = {1, 2, 3} and y = {a, b, c} \\ {Thus , the inverse of exists and f}^{- 1} = g . \\ ∴ f^{-1} : {a, b, c} \to {1, 2, 3} is given by, \\ f^{- 1} (a) = 1, f^{- 1} (b) = 2, f^{- 1} (c) = 3 \\ {Let us now find the inverse of f}^{- 1} i . e ., find the inverse of g. \\ If we define h : {1, 2, 3} \to {a, b, c} as \\ h (1) = a, h (2) = b, h (3) = c, then we have : \\ (goh) (1) = g (h (1)) = g (a) 1 \\ (goh) (2) = g (h (2)) = g (b) 2 \\ (goh) (3) = g (h (3)) = g (c) 3 \\ and \\ (hog) (a) = h (g (a)) = h (1) a \\ (hog) (b) = h (g (b)) = h (2) b \\ (hog) (c) = h (g (c)) = h (3) c \\ ∴ {goh = I}_{x} {and hog = I}_{y}, where x = {1, 2, 3} and y = {a, b, c} . \\ {Thus, the inverse of g exists and g}^{- 1} = h \Rightarrow (f^{- 1}) = h . \\ It can be noted that h = f. \\ Hence, {(f^{- 1})}^{- 1} = f. \end{array}$

Q.55 Let A = {2, 3, 4}, B = {5, 6, 7, 9} and let f = {(1, 4), (3, 6), (4, 7)} be a function from A to B, then function is ______________.

Ans

Since, every element of set A has only one image in set B i.e., 2→5, 3→6 and 4→7. Therefore, f is one-one function.

Please register to view this section

FAQs (Frequently Asked Questions)

1. How does Extramarks help me in my Mathematics exam preparation?

Extramarks, an online learning platform, provides informative and detailed notes that students must use to prepare for their exams. The Class 12 Mathematics Chapter 1 notes are prepared by subject matter experts in an easy language and stepwise manner. The chapter’s important questions and key points are included in these CBSE Solutions. Students can quickly glance through important formulas, properties, theorems and derivations with some examples. The class 12 Mathematics notes Chapter 1 through all chapters will be students’ helping guide in the preparation. Visit Extramarks and make learning fun!

2. Is Chapter 1- Relations and Functions difficult?

No. The first chapter of Mathematics Class 12 is quite easy. There are several problems bifurcated into four exercises that help students gain a deeper understanding and ace the chapter. The answers to these exercise problems are available on the Extramarks platform. Students must also include the Class 12 Mathematics Chapter 1 notes in their preparation for a better understanding

CBSE Class 12 Maths Revision Notes Chapter 1

Class 12 Mathematics Chapter 1 Notes

Key Topics Covered In Class 12 Mathematics Chapter 1 Notes

Relation:

Types of Relations:

Functions:

Types of Function:

Composition of a Function:

Invertible Function:

Binary Operation

Theorems:

Class 12 Mathematics Chapter 1 Notes: Exercises & Answer Solutions

NCERT Exemplar Class 12 Mathematics

Key Features of Class 12 Mathematics Chapter 1 notes

Please register to view this section

FAQs (Frequently Asked Questions)

1. How does Extramarks help me in my Mathematics exam preparation?

2. Is Chapter 1- Relations and Functions difficult?

CBSE Related Links

NCERT Solutions

Solved Board Papers

CBSE Class

ICSE Class

Sample Paper

Exam Weightage

Board Paper Solution 2020

Test Preparation

Other Services

About Us

Terms & Conditions

CBSE Class 12 Maths Revision Notes Chapter 1

Class 12 Mathematics Chapter 1 Notes

Key Topics Covered In Class 12 Mathematics Chapter 1 Notes

Relation:

Types of Relations:

Functions:

Types of Function:

Composition of a Function:

Invertible Function:

Binary Operation

Theorems:

Class 12 Mathematics Chapter 1 Notes: Exercises & Answer Solutions

NCERT Exemplar Class 12 Mathematics

Key Features of Class 12 Mathematics Chapter 1 notes

Share

Please register to view this section

FAQs (Frequently Asked Questions)

1. How does Extramarks help me in my Mathematics exam preparation?

2. Is Chapter 1- Relations and Functions difficult?

Share

CBSE Related Links

NCERT Solutions

Solved Board Papers

CBSE Class

ICSE Class

Sample Paper

Exam Weightage

Board Paper Solution 2020

Test Preparation

Other Services

About Us

Terms & Conditions