Chapter 1 Relations and Functions

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NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions

Class 12 Maths Chapter 1, Relations and Functions, introduces students to relations, different types of functions, domain and range, and special functions that are used throughout higher mathematics. A strong understanding of these basics is essential for later chapters like calculus and inverse trigonometric functions.

ALLEN provides you with the Chapter 1 Relations and Functions NCERT Solutions prepared by subject experts and are fully aligned with the current syllabus prescribed by CBSE. These solutions explain each step clearly, helping Class 12 students understand how and why a method is used. Practising these questions improves accuracy, builds confidence, and prepares students for board exams as well as competitive exams. NCERT Solutions Class 12 Maths Chapter 1 are drafted in simple language to make revision easier and improve the efficiency of preparation before the exams.

1.0Key Concepts of Class 12 Maths Chapter 1 Relations and Functions

Class 12 Maths Chapter 1, Relations and Functions, introduces the basic concepts of relations and functions and explains how elements of sets are connected with each other. The chapter covers the following important concepts:

Relations: Learn the definition of relations, ordered pairs, and different ways to represent relations using sets.
Types of Relations: Understand reflexive, symmetric, transitive, equivalence, and empty relations.
Functions: Learn about functions as special types of relations in which each input has a unique output.
Domain, Codomain, and Range: Understand the domain, codomain, and range of a function and how input and output values are related.
Types of Functions: Study one-one, many-one, onto, into, and constant functions along with their properties.
Special Functions: Learn about identity, polynomial, rational, and modulus functions.

2.0NCERT Class 12 Maths Chapter 1 Relations and Functions : Detailed Solutions

EXERCISE - 1.1

Determine whether each of the following relations are reflexive, symmetric and transitive : (i) Relation R in the set $A = {1, 2, 3 \dots 13, 14}$ defined as $R = {(x, y) : 3 x - y = 0}$ (ii) Relation R in the set N of natural numbers defined as $R = {(x, y) : y = x + 5$ and $x < 4}$ (iii) Relation R in the set $A = {1, 2, 3, 4, 5, 6}$ as $R = {(x, y) : y$ is divisible by x $}$ (iv) Relation R in the set Z of all integers defined as $R = {(x, y) : x - y$ is an integer $}$ (v) Relation R in the set A of human beings in a town at a particular time given by (a) $R = {(x, y) : x$ and y work at the same place} (b) $R = {(x, y) : x$ and y live in the same locality} (c) $R = {(x, y) : x$ is exactly 7 cm taller than y $}$ (d) $R = {(x, y) : x$ is wife of y $}$ (e) $R = {(x, y) : x$ is father of y $}$

Sol.

(i) $A = {1, 2, 3 \dots 13, 14}; R = {(x, y) : 3 x - y = 0}$ Reflexive: $R = {(1, 3), (2, 6), (3, 9), (4, 12)}$ $∵ (1, 1) \in / R$ ,

So, $R$ is not reflexive. Symmetric: $(1, 3) \in R$ , but $(3, 1) \in / R$ . So, R is not symmetric Transitive: $(1, 3), (3, 9) \in R$ , but $(1, 9) \in / R$ . So, R is not transitive Hence, R is neither reflexive, nor symmetric, nor transitive.

(ii) $R = {(x, y) : y = x + 5$ and $x < 4}$ $\Rightarrow R = {(1, 6), (2, 7), (3, 8)}$ Reflexive: $∵ (1, 1) \in / R$ . So, $R$ is not reflexive. Symmetric : $(1, 6) \in R$ , but $(6, 1) \in / R$ . $∴ R$ is not symmetric.

Transitive : Since there are no three elements x, $y, z \in N$ such that $(x, y) \in R, (y, z) \in R$ but $(x, z) \in / R$

$∴ R$ is transitive. Hence, R is neither reflexive, nor symmetric but it is transitive. (iii) $A = {1, 2, 3, 4, 5, 6}$ ; $R = {(x, y) : y$ is divisible by x $}$

Reflexive: Let $x \in A$ such that if $(x, x) \in R$

⇒ x is divisible by x, which is true, $\forall x \in A$ ∴ R is reflexive.

Symmetric: Let $x, y \in A$ such that if

$(x, y) \in R$

⇒ y is divisible by x ⇒ x is not divisible by y $\Rightarrow (y, x) \in / R$ For example: $(2, 4) \in R$ but $(4, 2) \in / R$ . ∴ R is not symmetric. Transitive: Let $x, y, z \in A$ such that if $(x, y) \in R$ and $(y, z) \in R$ ⇒ y is divisible by x and z is divisible by y.

$\Rightarrow \frac{y}{x} = k_{1} \in I (y, z) \in R \Rightarrow \frac{z}{y} = k_{2} \in I$

Equation (1) × (2) gives

$\Rightarrow \Rightarrow \frac{y}{x} \times \frac{z}{y} = k_{1} \times k_{2} \frac{z}{x} = (k_{1} k_{2}) \in I (x, z) \in R .$

$∴ R$ is transitive. Hence, R is reflexive and transitive but not symmetric.

(iv) $R = {(x, y) : x - y$ is an integer $}$ Reflexive: If $(x, x) \in R$ $\Rightarrow x - x = 0$ , which is an integer, $\forall x \in Z$ . $∴ R$ is reflexive.

Symmetric: If $(x, y) \in R$

$\Rightarrow (x - y)$ is an integer. $\Rightarrow (y - x)$ is also an integer. $\Rightarrow (y, x) \in R$ ⇒ R is symmetric.

Transitive: If $(x, y) \in R$

$\Rightarrow x - y = k_{1} \in I$

and $(y, z) \in R$

$\Rightarrow y - z = k_{2} \in I$

Adding equation (1) and (2), we get

$(x - y) + (y - z) = (k_{1} + k_{2}) \in I x - z = (k_{1} + k_{2}) \in I$

$∴ (x, z) \in R$

⇒ R is transitive. Hence, R is reflexive, symmetric and transitive.

(v) Given $A = {x : x$ is a human being in a town} (a) $R = {(x, y) : x$ and y work at the same place $}$

Reflexive: If $(x, x) \in R \Rightarrow x$ and x work at the same place, which is true, $\forall x \in A$

$∴ R$ is reflexive.

Symmetric: If $(x, y) \in R$

⇒ x and y work at the same place. ⇒ y and x also work at the same place. $\Rightarrow (y, x) \in R$ . $∴ R$ is symmetric.

Transitive: If $(x, y) \in R$ and $(y, z) \in R$

⇒ x and y work at the same place and y and z work at the same place. ⇒ x and z work at the same place. $\Rightarrow (x, z) \in R$ $∴ R$ is transitive.

Hence, R is reflexive, symmetric, and transitive.

(b) $R = {(x, y) : x$ and y live in the same locality $}$ Reflexive : If $(x, x) \in R \Rightarrow x$ and x live in the same locality, which is true, $\forall x \in A$ $∴ R$ is reflexive.

Symmetric: If $(x, y) \in R \Rightarrow x$ and y live in the same locality.

⇒ y and x also live in the same locality $\Rightarrow (y, x) \in R$ $∴ R$ is symmetric.

Transitive: Let $(x, y) \in R$ and $(y, z) \in R$ .

⇒ x and y live in the same locality and y and z live in the same locality. ⇒ x and z live in the same locality. $\Rightarrow (x, z) \in R$ $∴ R$ is transitive.

Hence, R is reflexive, symmetric and transitive.

Reflexive: If $(x, x) \in R \Rightarrow x$ is exactly 7 cm taller than x, which is not true, for any $x \in A$ . Since human being x cannot be taller than himself. So, $R$ is not reflexive. Symmetric: If $(x, y) \in R$

⇒ x is exactly 7 cm taller than y ⇒ y is exactly 7 cm smaller $∴ (y, x) \in / R$ $∴ R$ is not symmetric.

Transitive: If $(x, y), (y, z) \in R$

⇒ x is exactly 7 cm taller than y and y is exactly 7 cm taller than z. ⇒ x is exactly 14 cm taller than z. $∴ (x, z) \in / R ∴ R$ is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.

(d) $R = {(x, y) : x$ is wife of y $}$

Reflexive: If $(x, x) \in R$

⇒ x is wife of x, which is not true, for any $x \in A$ $∴ R$ is not reflexive.

Symmetric: If $(x, y) \in R$

⇒ x is wife of y $\Rightarrow y$ is husband of x. $∴ (y, x) \in / R$ $∴ R$ is not symmetric.

Transitive: There are no three element x, y, $z \in A$ such that $(x, y) \in R$ and $(y, z) \in R$ but $(x, z) \in / R$ So, R is transitive. Hence, R is neither reflexive, nor symmetric, but R is transitive.

(e) $R = {(x, y) : x$ is father of y $}$ Reflexive: If $(x, x) \in R \Rightarrow x$ is the father of x, which is not true, for any $x \in A$ ∴ R is not reflexive. Symmetric : If $(x, y) \in R \Rightarrow x$ is the father of y ⇒ y is son or daughter of x. $∴ (y, x) \in / R ∴ R$ is not symmetric. Transitive: If $(x, y) \in R$ and $(y, z) \in R$ , ⇒ x is the father of y and y is the father of z $\Rightarrow x$ is grand father of z. $∴ (x, z) \in / R$ $∴ R$ is not transitive. Hence, R is neither reflexive, nor symmetric, nor transitive.

Show that the relation R in the set R of real numbers, defined as $R = {(a, b) : a \leq b^{2}}$ is neither reflexive nor symmetric nor transitive. Sol. Reflexive: $R = {(a, b) : a \leq b^{2}}$ Let $a \in R$ such that if $(a, a) \in R \Rightarrow a \leq a^{2}$ is not true for all $a \in R$ i.e. (a, a) $\in / R, \forall a \in R$ Let $a = \frac{1}{2} \in R$ $∵ \frac{1}{2} \in / \frac{1}{2 ^{2}} \Rightarrow \frac{1}{2} \in / \frac{1}{4} \Rightarrow (\frac{1}{2}, \frac{1}{2}) \in / R$ $∴ R$ is not reflexive Symmetric: Let $a, b \in R$ such that if $(a, b) \in R \Rightarrow a \leq b^{2} \Rightarrow b \neq \leq a^{2}$ For eg. $1 \leq 2^{2} \Rightarrow (1, 2) \in R$ but $2 ≮ 1^{2}$ $\Rightarrow (2, 1) \in / R$ $∴ R$ is not symmetric Transitive: Let $a, b, c \in R$ such that if $(a, b) \in R$ and $(b, c) \in R$ $\Rightarrow a \leq b^{2}$ and $b \leq c^{2} \Rightarrow a \neq \leq c^{2}$ For example: $2 \leq (- 3)^{2}$ and $(- 3) \leq 1^{2}$ but $2 ≮ 1^{2}$ $\Rightarrow (2, - 3) \in R$ and $(- 3, 1) \in R$ but $(2, 1) \in / R$ $∴ R$ is not transitive Hence, R is neither reflexive nor symmetric nor transitive.
Check whether the relation R defined in the set ${1, 2, 3, 4, 5, 6}$ as $R = {(a, b) : b = a + 1}$ is reflexive, symmetric or transitive. Sol. Reflexive: Given $A = {1, 2, 3, 4, 5, 6}$ . $R = {(a, b) : b = a + 1}$ $\Rightarrow R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}$ $∵ (1, 1) \in / R$ , where $1 \in A$ $∴ R$ is not reflexive

Symmetric: $(1, 2) \in R$ but $(2, 1) \in / R$ , where $1, 2 \in A$ . So R is not symmetric Transitive: $(1, 2) \in R$ and $(2, 3) \in R$ but $(1, 3) \in / R$ , where $1, 2, 3 \in A$ .

∴ R is not transitive Hence R is neither reflexive, nor symmetric, nor transitive.

Show that the relation R in $R$ defined as $R = {(a, b) : a \leq b}$ , is reflexive and transitive but not symmetric. Sol. Reflexive: Given $R = {(a, b) : a \leq b}$ Let $a \in R$ such that if $(a, a) \in R$ $\Rightarrow a \leq a$ , which is true, $\forall a \in R$ ∴ R is reflexive

Symmetric: Let $a, b \in R$ such that if $(a, b) \in R \Rightarrow a \leq b \Rightarrow b \geq a \Rightarrow (b, a) \in / R$ For eg.: $1 \leq 2 \Rightarrow (1, 2) \in R$ But $2 \neq \leq 1 \Rightarrow (2, 1) \in / R$

$∴ R$ is not symmetric

Transitive: Let $a, b, c \in R$ such that if $(a, b) \in R$ and $(b, c) \in R$

$\Rightarrow a \leq b$ and $b \leq c \Rightarrow a \leq c \Rightarrow (a, c) \in R$ $∴ R$ is transitive Hence R is reflexive and transitive but not symmetric.

Check whether the relation R in R defined as $R = {(a, b) : a \leq b^{3}}$ is reflexive, symmetric or transitive. Sol. Reflexive: Given $R = {(a, b) : a \leq b^{3}}$ If $(a, a) \in R \Rightarrow a \leq a^{3}$ . Let $a = \frac{1}{2} \in R$ If $\frac{1}{2} R \frac{1}{2} \Rightarrow \frac{1}{2} \leq \frac{1}{2 ^{3}} \Rightarrow \frac{1}{2} \leq \frac{1}{8}$ (Not true) $∴ R$ is not reflexive

Symmetric: Let $a, b \in R$ such that if $(a, b) \in R \Rightarrow a \leq b^{3} \Rightarrow b \leq a^{3}$ For example: $1 \leq 2^{3}$ but $2 \neq \leq 1^{3}$

$\Rightarrow (1, 2) \in R$ but $(2, 1) \in / R$ ∴ R is not symmetric

Transitive: Let $a, b, c \in R$ such that if $(a, b) \in R$ and $(b, c) \in R$

$\Rightarrow a \leq b^{3}$ and $b \leq c^{3} \Rightarrow a \leq c^{3}$ (Not true) Let $a = 100, b = 5, c = 2$

$(a, b) \in R (b, c) \in R \Rightarrow (100, 5) \in R \Rightarrow 100 \leq (5)^{3} \Rightarrow (5, 2) \in R \Rightarrow 5 \leq (2)^{3}$

But $(a, c) \in R \Rightarrow (100, 2) \in R$ $\Rightarrow 100 \leq (2)^{3}$ (Not true) $∴ R$ is not transitive Hence R is neither reflexive nor symmetric nor transitive.

Show that the relation R in the set {1, 2, 3} given by $R = {(1, 2), (2, 1)}$ is symmetric but neither reflexive nor transitive. Sol. Reflexive: Let $A = {1, 2, 3}$ . A relation R on A is defined as $R = {(1, 2), (2, 1)}$ . $∵ (1, 1) \in / R$ , where $1 \in A$ $∴ R$ is not reflexive

Symmetric: $(1, 2) \in R$ and $(2, 1) \in R$

$∴ R$ is symmetric

Transitive: $(1, 2) \in R$ and $(2, 1) \in R$ but $(1, 1) \in / R$

$∴ R$ is not transitive. Hence, R is symmetric but neither reflexive nor transitive.

Show that the relation R in the set A of all the books in a library of a college, given by $R = {(x, y) : x$ and y have same number of pages} is an equivalence relation. Sol. Reflexive: Set A is the set of all books in the library of a college. $R = {(x, y) : x$ and y have the same number of pages}Let $x \in A$ such that if $(x, x) \in R \Rightarrow x$ and x has the same number of pages. Which is the true, $\forall x \in A$ ∴ R is reflexive

Symmetric: Let $x, y \in A$ such that if $(x, y) \in R$

$\Rightarrow x$ and y have the same number of pages. ⇒ y and x have the same number of pages. $\Rightarrow (y, x) \in R$ $∴ R$ is symmetric.

Transitive: Let $x, y, z \in A$ such that if $(x, y) \in R$ and $(y, z) \in R$ .

⇒ x and y have the same number of pages and y and z have the same number of pages. ⇒ x and z have the same number of pages $\Rightarrow (x, z) \in R$ $∴ R$ is transitive.

Since, R is reflexive, symmetric and transitive on A Hence, R is an equivalence relation on A .

Show that the relation R in the set $A = {1, 2, 3, 4, 5}$ given by $R = {(a, b)$ : $∣ a - b ∣$ is even}, is an equivalence relation. Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of ${1, 3, 5}$ is related to any element of {2, 4}. Sol. Reflexive: $A = {1, 2, 3, 4, 5}$ , $R = [{a, b) : ∣ a - b ∣$ is even $}$ Let $a \in A$ such that if $(a, a) \in R$ $\Rightarrow ∣ a - a ∣ = 0$ , is even which is true, $\forall a \in A$ $∴ R$ is reflexive.

Symmetric: Let $a, b \in A$ such that if $(a, b) \in R \Rightarrow ∣ a - b ∣$ is even

$\Rightarrow ∣ - (b - a) ∣ = ∣ b - a ∣$ is also even $\Rightarrow (b, a) \in R$ $∴ R$ is symmetric.

Transitive: Let $a, b, c \in R$ such that if $(a, b) \in R$ and $(b, c) \in R$ . $\Rightarrow ∣ a - b ∣$ and $∣ b - c ∣$ both are even $\Rightarrow ∣ a - b ∣ = 2 k_{1}$ and $∣ b - c ∣ = 2 k_{2}, (k_{1}, k_{2} \in Z)$ $\Rightarrow (a - b) = \pm 2 k_{1}$ and $(b - c) = \pm 2 k_{2}$ On adding equations (1) & (2), we get : $(a - c) = 2 (\pm k_{1} \pm k_{2})$ $\Rightarrow ∣ a - c ∣ = 2 ∣ (\pm k_{1} \pm k_{2}) ∣$ $\Rightarrow ∣ a - c ∣$ is even $\Rightarrow (a, c) \in R$ $∴ R$ is transitive Since R is reflexive, symmetric and transitive on A. Hence, R is an equivalence relation on A . Since, the difference of two even (or odd) be always even. So, every element of set ${1, 3, 5}$ is related to each other and also every element of set {2, 4} is related to each other but no element of {1, 3, 5} is related to any element of {2, 4}.

Show that each of the relation R in the set $A = {x \in Z : 0 \leq x \leq 12}$ , given by (i) $R = {(a, b) : ∣ a - b ∣$ is a multiple of 4 $}$ (ii) $R = {(a, b) : a = b}$ is an equivalence relation. Find the set of all elements related to 1 in each case. Sol. Given that : $A = {x \in Z : 0 \leq x \leq 12}$ $= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}$ (i) $R = {(a, b) : ∣ a - b ∣}$ is a multiple of 4 $}$ Reflexive: Let $a \in A$ such that if $(a, a) \in R$ $\Rightarrow ∣ a - a ∣ = 0$ is a multiple of 4 which is true, $\forall a \in A$ $∴ R$ is reflexive. Symmetric: Let $a, b \in A$ such that if $(a, b) \in R \Rightarrow ∣ a - b ∣$ is a multiple of 4 . $\Rightarrow ∣ b - a ∣$ is also multiple of 4 . $\Rightarrow (b, a) \in R ∴ R$ is symmetric. Transitive: Let $a, b, c \in A$ such that if $(a, b) \in R$ and $(b, c) \in R$ $\Rightarrow ∣ a - b ∣$ and $∣ b - c ∣$ both are multiple of 4 $\Rightarrow ∣ a - b ∣ = 4 k_{1}$ and $∣ b - c ∣ = 4 k_{2}, (k_{1}, k_{2} \in Z)$ $\Rightarrow a - b = \pm 4 k_{1}$ and $b - c = \pm 4 k_{2}$ On adding equations (1) & (2), we get : $a - c = 4 (\pm k_{1} \pm k_{2})$ $\Rightarrow ∣ a - c ∣ = 4 ∣ (\pm k_{1} \pm k_{2}) ∣$ $\Rightarrow ∣ a - c ∣$ is multiple of 4 $\Rightarrow (a, c) \in R$ $∴ R$ is transitive Since R is reflexive, symmetric and transitive on A Hence, R is an equivalence relation on A . Further, let $x \in A$ such that $(x, 1) \in R$ $\Rightarrow ∣ x - 1∣$ is multiple of 4 $\Rightarrow ∣ x - 1∣ = 0, 4, 8, \dots\dots$ $\Rightarrow x - 1 = 0, 4, 8 \Rightarrow x = 1, 5, 9$ Hence, required set $= {1, 5, 9}$ (ii) $R = {(a, b) : a = b}$ Reflexive: Let $a \in A$ , such that $(a, a) \in R$ , $\Rightarrow a = a$ . which is true, $\forall a \in A$ . So, R is reflexive. Symmetric: Let $a, b \in A$ such that if $(a, b) \in R$ . $\Rightarrow a = b \Rightarrow b = a \Rightarrow (b, a) \in R$ $∴ R$ is symmetric. Transitive: Let $a, b, c \in A$ such that if $(a, b) \in R$ and $(b, c) \in R$ . $\Rightarrow a = b$ and $b = c \Rightarrow a = c$ $\Rightarrow (a, c) \in R$ $∴ R$ is transitive. Since R is reflexive, symmetric and transitive on A Hence, R is an equivalence relation on A Let $x \in A$ such that $(x, 1) \in R \Rightarrow x = 1$ Hence, required set $= {1}$
Give an example of a relation. Which is (i) Symmetric but neither reflexive nor transitive. (ii) Transitive but neither reflexive nor symmetric. (iii) Reflexive and symmetric but not transitive. (iv) Reflexive and transitive but not symmetric. (v) Symmetric and transitive but not reflexive.

Sol．

（i） Let $A = {5, 6, 7}$ ． Reflexive：Define a relation $R$ on $A$ as $R = {(5, 6), (6, 5)}$ ． Since $(5, 5) \in / R$ ，where $5 \in A$ ． So，R is not reflexive Symmetric：Since $(5, 6) \in R$ and $(6, 5) \in R$ ． So， R is symmetric Transitive：Since $(5, 6) \in R$ and $(6, 5) \in R$ but $(5, 5) \in / R$ ．So， R is not transitive Hence，relation R is symmetric but neither reflexive nor transitive．（ii） Consider a relation R in N defined as： $R = {(a, b) : a < b}$ Reflexive：Let $a \in N$ such that if $(a, a) \in R$ $\Rightarrow a < a$ ，which is not true for any $a \in N$ $∴ R$ is not reflexive

Symmetric：Let $a, b \in N$ such that if $(a, b) \in R \Rightarrow a < b$

$\Rightarrow b > a \Rightarrow (b, a) \in / R$ ∴ R is not symmetric

Transitive：Let $a, b, c \in N$ such that if $(a, b) \in R$ and $(b, c) \in R$

$\Rightarrow a < b$ and $b < c \Rightarrow a < c$ $\Rightarrow (a, c) \in R$ ．So， R is transitive

Hence，relation R is transitive but neither reflexive nor symmetric．

（iii） Let $A = {4, 6, 8}$ ． Define a relation R on A as： $R = {(4, 4)$ ，（6，6），（8，8），（4，6），（6，4），（6，8），（8，6）} Reflexive：Since， $(a, a) \in R, \forall a \in A$ ．So， R is reflexive Symmetric：Since， $(a, b) \in R \Rightarrow (b, a) \in R$ ， for all $a, b \in A$ ．So， R is symmetric． Transitive：Since， $(4, 6) \in R$ and $(6, 8) \in R$ but $(4, 8) \in / R$ ．So， R is not transitive． Hence， R is reflexive symmetric but not transitive．（iv） Define a relation $R_{1}$ in R as： $R_{1} = {(a, b) : a^{3} \geq b^{3}}$ Reflexive：Let $a \in R$ such that if $(a, a) \in R_{1}$ $\Rightarrow a^{3} \geq a^{3}$ ，which is true $\forall a \in R$ $∴ R_{1}$ is reflexive．

Symmetric：Let $a, b \in R$ such that if $(a, b) \in R_{1} \Rightarrow a^{3} \geq b^{3} \Rightarrow b^{3} \leq a^{3}$ For example： $2^{3} \geq 1^{3}$ but $1^{3} \leq 2^{3}$

$\Rightarrow (2, 1) \in R_{1}$ but $(1, 2) \in / R_{1}$ $∴ R_{1}$ is not symmetric．

Transitive：Let $a, b, c \in R$ such that if $(a, b) \in R_{1}$ and $(b, c) \in R_{1}$

$\Rightarrow a^{3} \geq b^{3}$ and $b^{3} \geq c^{3} \Rightarrow a^{3} \geq c^{3} \Rightarrow (a, c) \in R_{1}$ $∴ R$ is transitive． Hence，relation $R_{1}$ is reflexive and transitive but not symmetric．（v） Let $A = {1, 2, 3}$ ． Define a relation R on A as： $R = {(1, 1) (2, 2)}$ Reflexive：Since $(3, 3) \in / R$ ，where $3 \in A$ ． So， $R$ is not reflexive Symmetric：If $(a, b) \in R$ Here， $a = b \Rightarrow b = a \Rightarrow (b, a) \in R$ $∴ R$ is symmetric Transitive：If $(a, b) \in R$ and $(b, c) \in R$ Here $a = b$ and $b = c \Rightarrow a = c \Rightarrow (a, c) \in R$ $∴ R$ is transitive Hence，relation $R$ is symmetric and transitive but not reflexive．

Show that the relation R in the set A of points in a plane given by $R = {(P, Q)$ ：distance of the point P from the origin is same as the distance of the point Q from the origin $}$ ，is an equivalence relation．Further，show that the set of all points related to a point $P \neq = (0, 0)$ is the circle passing through P with origin as centre．

Sol. Given the set A of points in a plane and a relation R on A selection as $R = {(P, Q)$ : distance of point P from the origin is the same as the distance of point Q from the origin} Reflexive: Let $P \in A$ such that if $(P, P) \in R$ $\Rightarrow OP = OP$ , which is true, $\forall P \in A$ $∴ R$ is reflexive.

Symmetric : Let $P, Q \in A$ such that if $(P, Q) \in R \Rightarrow OP = OQ$

$\Rightarrow OQ = OP \Rightarrow (Q, P) \in R$ $∴ R$ is symmetric.

Transitive: Let $P, Q, S \in A$ such that if $(P, Q) \in R$ and $(Q, S) \in R$

$\Rightarrow OP = OQ$ and $OQ = OS$ $\Rightarrow OP = OS \Rightarrow (P, S) \in R$ $∴ R$ is transitive. Since R is reflexive symmetric and transitive. Hence, R is an equivalence relation. The set of all points related to $P \neq = (0, 0)$ will be those points whose distance from the origin is the same as the distance of point P from the origin. In other words, If $O (0, 0)$ is the origin and $OP = k$ , then the set of all points related to P is at a distance of k from the origin. Hence, this set of points form a circle with the centre as the origin and this circle passes through point P.

Show that the relation R defined in the set A of all triangles as $R = {(T_{1}, T_{2}) : T_{1}$ is similar to $T_{2}$ }, is equivalence relation. Consider three right angle triangles $T_{1}$ with sides $3, 4, 5, T_{2}$ with sides 5, 12, 13 and $T_{3}$ with sides 6, 8, 10. Which triangles among $T_{1}, T_{2}$ and $T_{3}$ are related? Sol. Given, set A of all triangles and a relation R on A defined by $R = {(T_{1}, T_{2}) : T_{1}$ is similar to $T_{2}$ } Reflexive: Since every triangle is similar to itself, so R is reflexiveSymmetric: Let $T_{1}, T_{2} \in A$ such that if $(T_{1}, T_{2}) \in R \Rightarrow T_{1}$ is similar to $T_{2} \Rightarrow T_{2}$ is similar to $T_{1}$ . $\Rightarrow (T_{2}, T_{1}) \in R$ ∴ R is symmetric.

Transitive: Let $T_{1}, T_{2}, T_{3} \in A$ such that if $(T_{1}, T_{2}), (T_{2}, T_{3}) \in R$ .

$\Rightarrow T_{1}$ is similar to $T_{2}$ and $T_{2}$ is similar to $T_{3}$ . $\Rightarrow T_{1}$ is similar to $T_{3} . \Rightarrow (T_{1}, T_{3}) \in R$ $∴ R$ is transitive.

Science, R is reflexive, symmetric and transitive. Hence, R is an equivalence relation. Given three right angle triangle $T_{1}$ with sides 3, 4, 5, $T_{2}$ with sides 5, 12, 13 and $T_{3}$ with sides 6, 8, 10

$∵ \frac{3}{6} = \frac{4}{8} = \frac{5}{10} = \frac{1}{2}$ Since, the corresponding sides of triangles $T_{1}$ and $T_{3}$ are in the same ratio. Then, triangle $T_{1}$ is similar to triangle $T_{3}$ . Hence, $T_{1}$ is related to $T_{3}$ .

Show that the relation R defined in the set A of all polygons as $R = {(P_{1}, P_{2}) : P_{1}$ and $P_{2}$ have same number of sides}, is an equivalence relation. What is the set of all elements in A related to the right angled triangle T with sides 3, 4 and 5? Sol. Given a relation R defined in the set A of all polygons as : $R = {(P_{1}, P_{2}) : P_{1}$ and $P_{2}$ have same number of sides} Reflexive: Let $P_{1} \in A$ such that if $(P_{1}, P_{1}) \in R$ as the same polygon has the same number of sides with itself. $∴ R$ is reflexive.

Symmetric: Let $P_{1}, P_{2} \in A$ such that if $(P_{1}, P_{2}) \in R \Rightarrow P_{1}$ and $P_{2}$ have same number of sides

$\Rightarrow P_{2}$ and $P_{1}$ have same number of sides $\Rightarrow (P_{2}, P_{1}) \in R$ ∴ R is symmetric.

Transitive: Let $P_{1}, P_{2}, P_{3} \in A$ such that If $(P_{1}, P_{2}), (P_{2}, P_{3}) \in R$ .

$\Rightarrow P_{1}$ and $P_{2}$ have the same number of sides and $P_{2}$ and $P_{3}$ have the same number of sides. $\Rightarrow P_{1}$ and $P_{3}$ have the same number of sides. $\Rightarrow (P_{1}, P_{3}) \in R ∴ R$ is transitive.

Since R is reflexive, symmetric and transitive Hence, R is an equivalence relation. The elements in A related to the right-angled triangle (T) with sides 3, 4 and 5 are those polygons which have 3 sides (since T is a polygon with 3 sides). Hence, the required set is the set of all triangles of A which are related to T.

Let L be the set of all lines in XY plane and R be the relation in L defined as $R = {(L_{1}, L_{2}) : L_{1}$ is parallel to $L_{2}}$ . Show that R is an equivalence relation. Find the set of all lines related to the line $y = 2 x + 4$ . Sol. Given, L be the set of all lines in XY plane and relation R on set L defined as $R = {(L_{1}, L_{2}) : L_{1}$ is parallel to $L_{2}}$ Reflexive: Let $L_{1} \in L$ such that if $(L_{1}, L_{1}) \in R$ $\Rightarrow L_{1} ∥ L_{1}$ which is true, $\forall L_{1} \in L$ $∴ R$ is reflexive

Symmetric: Let $L_{1}, L_{2} \in L$ such that if $(L_{1}, L_{2}) \in R$ .

$\Rightarrow L_{1}$ is parallel to $L_{2}$ $\Rightarrow L_{2}$ is parallel to $L_{1}$ $\Rightarrow (L_{2}, L_{1}) \in R$ $∴ R$ is symmetric.

Transitive: Let $L_{1}, L_{2}, L_{3} \in L$ such that if $(L_{1}, L_{2}), (L_{2}, L_{3}) \in R$ .

$\Rightarrow L_{1}$ is parallel to $L_{2}$ and $L_{2}$ is parallel to $L_{3}$ . $\Rightarrow L_{1}$ is parallel to $L_{3}$ . $∴ R$ is transitive.

Since R is reflexive, symmetric and transitive Hence, R is an equivalence relation.

The set of all lines related to the line $y = 2 x + 4$ is the set of all lines that are parallel to the line $y = 2 x + 4$ . Slope of line $y = 2 x + 4$ is $m = 2$ . It is known that parallel lines have the same slope. The line parallel to the given line is of the form $y = 2 x + c$ , where $c \in R$ . Hence, the set of all lines related to the given line is given by $y = 2 x + c$ , where $c \in R$ .

Choose the correct answer in the following questions from 15 to 16.

Let R be the relation in the set ${1, 2, 3, 4}$ given by $R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3)$ , $(3, 3), (3, 2)}$ . Choose the correct answer. (A) R is reflexive and symmetric but not transitive. (B) R is reflexive and transitive but not symmetric. (C) R is symmetric and transitive but not reflexive. (D) $R$ is an equivalence relation.

Sol. (B) Let $A = {1, 2, 3, 4}$ , given a relation R on A defined as $R = {(1, 2), (2, 2), (1, 1) ((4, 4), (1, 3)$ , (3, 3), $(3, 2)$ } Reflexive: Since $(a, a) \in R, \forall a \in A$ . So, R is reflexive. Symmetric: Since, $(1, 2) \in R$ , but $(2, 1) \in / R$ . So, R is not symmetric. Transitive: Since at least three elements $a, b$ , $c \in A$ does not exists Such that $(a, b) \in R$ and $(b, c) \in R$

$\Rightarrow (a, c) \in / R$ $∴ R$ is transitive.

Hence, R is reflexive and transitive but not symmetric.

Let R be the relation in the set N given by $R = {(a, b) : a = b - 2, b > 6}$ . Choose the correct answer. (A) $(2, 4) \in R$ (B) $(3, 8) \in R$ (C) $(6, 8) \in R$ (D) $(8, 7) \in R$ Sol. (C)

$R = {(a, b) : a = b - 2, b > 6}$

$∵ b > 6$ and $a = b - 2$ $∴ R = {(5, 7), (6, 8), (7, 9) \dots .$ .

Here $(6, 8) \in R$

EXERCISE - 1.2

Show that the function $f : R_{*} \to R_{*}$ defined by $f (x) = \frac{1}{x}$ is one-one and onto, where $R_{*}$ is the set of all non-zero real numbers. Is the result true, if the domain $R_{*}$ is replaced by N with codomain being same as $R_{*}$ ? Sol. Given that $f : R_{*} \to R_{*}$ is defined by $f (x) = \frac{1}{x}$ Let $x_{1}, x_{2} \in R_{*}$ (domain) such that if $f (x_{1}) = f (x_{2}) \Rightarrow \frac{1}{x _{1}} = \frac{1}{x _{2}} \Rightarrow x_{1} = x_{2}$ $∴ f$ is one-one. $y \in R_{*}$ (codomain) and $f (x) = y$ $\Rightarrow \frac{1}{x} = y \Rightarrow x = \frac{1}{y} \forall y \in R_{*}$ , there exists $x = \frac{1}{y} \in R$ such that $f (x) = \frac{1}{( \frac{1}{y} )} = y$ . $∴ f$ is onto. Thus, the given function f is one-one and onto. Now, consider function $g : N \to R_{*}$ defined by $g (x) = \frac{1}{x}$ . Let $x_{1}, x_{2} \in N$ . Let $y \in R_{*}$ and if $g (x_{1}) = g (x_{2})$ $\Rightarrow \frac{1}{x _{1}} = \frac{1}{x _{2}} \Rightarrow x_{1} = x_{2}$ $∴ g$ is one-one.Let $y = g (x) = \frac{1}{x}$ $\Rightarrow x = \frac{1}{y} \in / N, \forall y \in R_{*}$ For example: $y = \frac{2}{3} \in R_{*}$ then $x = \frac{3}{2} \in / N$ $∴ g$ is not onto Function g is one-one but not onto. Hence result is not true.
Check the injectivity and surjectivity of the following functions: (i) $f : N \to N$ given by $f (x) = x^{2}$ (ii) $f : Z \to Z$ given by $f (x) = x^{2}$ (iii) $f : R \to R$ given by $f (x) = x^{2}$ (iv) $f : N \to N$ given by $f (x) = x^{3}$ (v) $f : Z \to Z$ given by $f (x) = x^{3}$ Sol. (i) $f : N \to N$ is given by, $f (x) = x^{2}$ Let $x_{1}, x_{2} \in N$ , such that if $f (x_{1}) = f (x_{2})$ $\Rightarrow x_{1}^{2} = x_{2}^{2}$ $\Rightarrow (x_{1} - x_{2}) (x_{1} + x_{2}) = 0$ $\Rightarrow x_{1} - x_{2} = 0$ $\Rightarrow x_{1} = x_{2} (∵ x_{1} + x_{2} \neq = 0)$ $∴ f$ is injective. Let $y \in N$ (codomain) Put $y = f (x) = x^{2}$ $\Rightarrow x = y \in / N, \forall y \in N$ Every element of codomain does not have preimages in domain. Range $\neq =$ Co-domain For eg. range of $f = {1, 4, 9 \dots$ . $} \neq = co - d o main$ of f $∴ f$ is not surjective. Hence, function f is injective but not surjective. (ii) $f : Z \to Z$ is given by, $f (x) = x^{2}$ Let $x_{1}, x_{2} \in Z$ , such that If $f (x_{1}) = f (x_{2}) \Rightarrow x_{1}^{2} = x_{2}^{2}$ $\Rightarrow (x_{1} - x_{2}) (x_{1} + x_{2}) = 0$

$\Rightarrow x_{1} - x_{2} = 0$ or $x_{1} = x_{2}$ $\Rightarrow x_{1} + x_{2} = 0$ or $x_{1} = - x_{2}$ ∵ Function does not have unique solution $∴ f$ is not injective

Let $y \in Z$ (co-domain) Put $y = f (x) = x^{2} \Rightarrow x = y \in / Z, \forall y \in Z$ Every element of co-domain does not have pre images in domain. Range $\neq =$ Co-domain For example: $f (- 1) = f (1) = 1$ , but $- 1 \neq = 1$ ,

∴ f is not injective.

Since the range of $f = {0, 1, 4, 9 \dots\dots} \neq =$ co-domain of f.

$∴ f$ is not surjective.

Hence, function f is neither injective nor surjective.

(iii) $f : R \to R$ is given by $f (x) = x^{2}$ Let $x_{1}, x_{2} \in R$ , such that if $f (x_{1}) = f (x_{2})$ $\Rightarrow x_{1}^{2} = x_{2}^{2} \Rightarrow (x_{1} - x_{2}) (x_{1} + x_{2}) = 0$ $\Rightarrow x_{1} - x_{2} = 0$ and $x_{1} = x_{2}$ $\Rightarrow x_{1} + x_{2} = 0$ and $x_{1} = - x_{2}$

Function does not have unique solution

$∴ f$ is not injective.

Let $y \in R$ (co-domain) Put $y = f (x) = x^{2} \Rightarrow x = y \in / R, \forall y \in R$ Every element of co-domain do not have pre image in domain Range $\neq =$ co-domain For example : $f (- 1) = f (1) = 1$ , but $- 1 \neq = 1$ ,

∴ f is not injective.

Since, range of $f \neq = co$ -domain of f

$∴ f$ is not surjective.

Hence, function f is neither injective nor surjective.

(iv) $f : N \to N$ given by, $f (x) = x^{3}$ Let $x_{1}, x_{2} \in N$ (domain) such that if $f (x_{1}) = f (x_{2})$ $\Rightarrow x_{1}^{3} = x_{2}^{3}$ $\Rightarrow (x_{1} - x_{2}) (x_{1}^{2} + x_{1} x_{2} + x_{2}^{2}) = 0$ $\Rightarrow x_{1} - x_{2} = 0$ $\Rightarrow x_{1} = x_{2} (∵ x_{1}^{2} + x_{1} x_{2} + x_{2}^{2} \neq = 0)$ $∴ f$ is injective. Let $y \in N$ (co-domain) Put $y = f (x) = x^{3} \Rightarrow x = (y)^{\frac{1}{3}} \in / N, \forall y \in N$ Every element of co-domain do not have pre-images in domain. Range $\neq =$ co-domain Range of $f = {1, 8, 27 \dots\dots} \neq =$ co-domain of f $∴ f$ is not surjective. Hence, function f is injective but not surjective. (v) $f : Z \to Z$ is given by, $f (x) = x^{3}$ Let $x_{1}, x_{2} \in Z$ (domain) such that if $f (x_{1}) = f (x_{2})$ $\Rightarrow x_{1}^{3} = x_{2}^{3}$ $\Rightarrow (x_{1} - x_{2}) (x_{1}^{2} + x_{1} x_{2} + x_{2}^{2}) = 0$ $\Rightarrow x_{1} = x_{2} (∵ x_{1}^{2} + x_{1} x_{2} + x_{2}^{2} > 0)$ $∴ f$ is injective. Let $y \in Z$ (co-domain) Put $y = f (x) = x^{3} \Rightarrow x = (y)^{\frac{1}{3}} \in / Z, \forall y \in Z$ Every element of co-domain does not have pre image in domain Range $\neq =$ co-domain Range of $f = {0, \pm 1, \pm 8, \pm 27, \dots} \neq =$ co-domain of f. $∴ f$ is not surjective. Hence, function $f$ is injective but not surjective.

Prove that the Greatest Integer Function $f : R \to R$ given by $f (x) = [x]$ , is neither oneone nor onto, where [x] denotes the greatest integer less than or equal to x. Sol. $f : R \to R$ is given by, $f (x) = [x]$ It is seen that $f (1.2) = [1.2] = 1$ , $f (1.9) = [1.9] = 1$ . $∴ f (1.2) = f (1.9)$ , but $1.2 \neq = 1.9$ . $∴ f$ is not one - one. Range of $f = I$ ⇒ Range of $f \neq = co$ -domain of f. $∴ f$ is not onto. Hence, the greatest integer function is neither one-one nor onto.
Show that the Modulus Function $f : R \to R$ given by $f (x) = ∣ x ∣$ , is neither one-one nor onto, where |x| is x, if x is positive or 0 and $∣ x ∣$ is -x , if x is negative. Sol. $f : R \to R$ is given by,

$f (x) = ∣ x ∣ = {x, - x, if x \geq 0 if x < 0$

It is seen that $f (- 1) = ∣ - 1∣ = 1, f (1) = ∣1∣ = 1$ . $∴ f (- 1) = f (1)$ , but $- 1 \neq = 1$ ∴ f is not one-one. Range of $f = R^{+} \cup {0}$ ⇒ range of $f \neq = co$ -domain of f. ∴ f is not onto. Hence, the modulus function is neither one-one nor onto.

Show that the Signum Function $f : R \to R$ , given by $f (x) = ⎩ ⎨ ⎧ 1, 0, - 1 if x > 0 if x = 0 if x < 0$ is neither one-one nor onto. Sol. $f : R \to R$ is given by, $f (x) = ⎩ ⎨ ⎧ 1, 0, - 1, if x > 0 if x = 0 if x < 0$ $∵ f (x) = 1, \forall x \in (0, \infty)$ and $f (x) = - 1, \forall x \in (- \infty, 0)$ $∴ f$ is not one-one. $∵$ Range of $f = {0, \pm 1}$ ∵ Range of $f \neq =$ co-domain of f. So, f is not onto. Hence, the signum function is neither one-one nor onto.
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. Sol. It is given that $A = {1, 2, 3}, B = {4, 5, 6, 7}$ . $f : A \to B$ is defined as $f = {(1, 4), (2, 5), (3, 6)}$ . $∴ f (1) = 4, f (2) = 5, f (3) = 6$

Since, different elements of A have different images in B Hence, function f is one-one.

In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer. (i) $f : R \to R$ defined by $f (x) = 3 - 4 x$ (ii) $f : R \to R$ defined by $f (x) = 1 + x^{2}$

Sol.

(i) $f : R \to R$ defined by $f (x) = 3 - 4 x$ . Let $x_{1}, x_{2} \in R$ such that

$f (x_{1}) = f (x_{2}) \Rightarrow 3 - 4 x_{1} = 3 - 4 x_{2}$

$\Rightarrow - 4 x_{1} = - 4 x_{2} \Rightarrow x_{1} = x_{2}$ $∴ f$ is one-one. Let $y \in R$ (co-domain) put $y = f (x) \Rightarrow y = 3 - 4 x$ $\Rightarrow x = \frac{3 - y}{4} \in R$ (domain), $f (\frac{3 - y}{4}) = 3 - 4 (\frac{3 - y}{4}) = y, \forall y \in R$ (co-domain) ∴ f is onto. Since, f is one-one and onto. Hence, f is bijective. (ii) $f : R \to R$ is defined as, $f (x) = 1 + x^{2}$ . Let $x_{1}, x_{2} \in R$ such that $f (x_{1}) = f (x_{2})$ $1 + x_{1}^{2} = 1 + x_{2}^{2} \Rightarrow x_{1}^{2} = x_{2}^{2} \Rightarrow x_{1} = \pm x_{2}$ $∵ f (1) = 2, f (- 1) = 2 \Rightarrow f (1) = f (- 1)$ but $1 \neq = - 1$ $∴ f$ is not one-one. Here range of $f = [1, \infty)$ ∵ Range of $f \neq = co$ -domain of f. ∴ f is not onto . Hence, f is neither one-one nor onto.

Let A and B be sets. Show that $f : A \times B \to B \times A$ such that $f (a, b) = (b, a)$ is bijective function. Sol. $f : A \times B \to B \times A$ is defined as $f (a, b) = (b, a)$ . Let $(a_{1}, b_{1}), (a_{2}, b_{2}) \in A \times B$ such that if $f (a_{1}$ , $b_{1}) = f (a_{2}, b_{2})$ $\Rightarrow (b_{1}, a_{1}) = (b_{2}, a_{2})$ $\Rightarrow b_{1} = b_{2}$ and $a_{1} = a_{2}$ $\Rightarrow (a_{1}, b_{1}) = (a_{2}, b_{2})$ . So, f is one-one.

Now, let $(b, a) \in B \times A$

$\Rightarrow (a, b) \in A \times B$ i.e. $\forall (b, a) \in B \times A, \exists (a, b) \in A \times B$ such that $f (a, b) = (b, a)$ . (Definition of f) So, f is onto. Hence, f is bijective.

Let f : N → N be defined by

$f (n) = {\frac{n + 1}{2}, \frac{n}{2}, if n is odd if n is even, for all n \in N .$

State whether the function f is bijective. Justify your answer Sol. $f (1) = \frac{1 + 1}{2} = 1, f (2) = \frac{2}{2} = 1$ $\Rightarrow f (1) = f (2)$ but $1 \neq = 2$ $∴ f$ is not one-one.

Hence, f is not bijective.

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 and onto? Justify your answer. Sol. $A = R - {3}, B = R - {1}$ $f : A \to B$ is defined as $f (x) = \frac{x - 2}{x - 3}$ let $x_{1}, x_{2} \in A$ such that $f (x_{1}) = f (x_{2})$ $\Rightarrow \frac{x _{1} - 2}{x _{1} - 3} = \frac{x _{2} - 2}{x _{2} - 3}$ $\Rightarrow (x_{1} - 2) (x_{2} - 3) = (x_{2} - 2) (x_{1} - 3)$ $\Rightarrow x_{1} x_{2} - 3 x_{1} - 2 x_{2} + 6 = x_{1} x_{2} - 3 x_{2} - 2 x_{1} + 6$ $\Rightarrow - 3 x_{1} - 2 x_{2} = - 3 x_{2} - 2 x_{1}$ $\Rightarrow 3 x_{1} - 2 x_{1} = 3 x_{2} - 2 x_{2}$ $\Rightarrow x_{1} = x_{2}$ $∴ f$ is one - one. Let $y \in B = R - {1}$ put $f (x) = y$ $\Rightarrow \frac{x - 2}{x - 3} = y \Rightarrow x - 2 = xy - 3 y$ $\Rightarrow x (1 - y) = - 3 y + 2$ $\Rightarrow x = \frac{2 - 3 y}{1 - y} \in A, \forall y \in B$ i.e. every element of B has pre-image(s) in A Range = Co-domain ∴ f is onto. Hence, function f is one-one and onto.

Choose the correct answer in the following questions from 11 &12.

Let $f : R \to R$ be defined as $f (x) = x^{4}$ . Choose the correct answer. (A) f is one - one onto (B) f is many-one onto (C) f is one-one but not onto (D) f is neither one-one nor onto Sol. (D) $f : R \to R$ is defined as $f (x) = x^{4}$ $∵ f (1) = 1, f (- 1) = 1$ $\Rightarrow f (1) = f (- 1)$ but $1 \neq = - 1$ . So, f is not one-one Again, Range of $f = [0, \infty)$ ⇒ Range of f $\subset$ co-domain of f . So, f is not onto. Hence, function f is neither one-one nor onto.
Let $f : R \to R$ be defined as $f (x) = 3 x$ . Choose the correct answer. (A) f is one-one onto (B) f is many-one onto (C) f is one-one but not onto (D) f is neither one-one nor onto

Sol. (A) Let $x_{1}, x_{2} \in R$ such that $f (x_{1}) = f (x_{2})$ $\Rightarrow 3 x_{1} = 3 x_{2} \Rightarrow x_{1} = x_{2}$ . So $f$ is one - one. Let $y = f (x) \forall y \in R$

$y = 3 x \Rightarrow x = \frac{y}{3} \in R$

Range = Co-domain . So f is onto. Hence, function f is one-one and onto.

MISCELLANEOUS EXERCISE

Show that function $f : R \to {x \in R : - 1 < x < 1}$ defined by $f (x) = \frac{x}{1 + ∣ x ∣}, x \in R$ is one-one and onto function. Sol. Given that $f : R \to {x \in R : - 1 < x < 1}$ is defined as $f (x) = \frac{x}{1 + ∣ x ∣}, x \in R$

$\Rightarrow f (x) = {\frac{x}{1 + x} \frac{x}{1 - x};; x \geq 0 x < 0$

For one-one function : Case-I: when $x_{1}, x_{2} > 0$ Let $f (x_{1}) = f (x_{2})$

$\Rightarrow \Rightarrow \Rightarrow \frac{x _{1}}{1 + x _{1}} = \frac{x _{2}}{1 + x _{2}} x_{1} + x_{1} x_{2} = x_{2} + x_{1} x_{2} x_{1} = x_{2}$

Case-II: when $x_{1}, x_{2} < 0$ Let $f (x_{1}) = f (x_{2})$

$\Rightarrow \frac{x _{1}}{1 - x _{1}} = \frac{x _{2}}{1 - x _{2}} \Rightarrow x_{1} - x_{1} x_{2} = x_{2} - x_{1} x_{2} \Rightarrow x_{1} = x_{2}$

Case-III: when $x_{1} > 0, x_{2} < 0$ Let $f (x_{1}) = f (x_{2})$

$\Rightarrow \frac{x _{1}}{1 + x _{1}} = \frac{x _{2}}{1 - x _{2}} \Rightarrow x_{1} - x_{1} x_{2} = x_{2} + x_{1} x_{2} \Rightarrow x_{1} - x_{2} - 2 x_{1} x_{2} = 0$

which is not possible $(∵ x_{1} > 0, x_{2} < 0)$

$∴ x_{1} \neq = x_{2} \Rightarrow f (x_{1}) \neq = f (x_{2})$

∴ f is one-one For onto function :

Case-I When $x \geq 0$ Let $y = f (x)$

$∴ y = \frac{x}{1 + x}$

$\Rightarrow y + xy = x$ or $y = x (1 - y)$

$\Rightarrow x = \frac{y}{1 - y}$

or $x = - \frac{y}{( y - 1 )} \geq 0$

$\Rightarrow \frac{y}{y - 1} \leq 0 + 0 1 - +$

$y \in [0, 1)$

Case-II When $x < 0$ Let $y = f (x)$

$∴ y = \frac{x}{1 - x}$

$\Rightarrow y - xy = x$ or $y = x + xy$

$\Rightarrow x = \frac{y}{1 + y}$

$\Rightarrow x = \frac{y}{1 + y} < 0$

$c_{0}^{+} 0 - 0_{0}^{+}$

$y \in (- 1, 0)$

From Case I and case II, $y \in [0, 1) \cup (- 1, 0)$ Range $= (- 1, 1) =$ Co-domain So, f is onto. Hence, f is one-one and onto function.

Show that the function $f : R \to R$ given by $f (x) = x^{3}$ is injective. Sol. $f : R \to R$ is given as $f (x) = x^{3}$ . Let $x_{1}, x_{2} \in R$ such that if $f (x_{1}) = f (x_{2})$

$\Rightarrow x_{1}^{3} = x_{2}^{3}$

$\Rightarrow (x_{1} - x_{2}) (x_{1}^{2} + x_{2}^{2} + x_{1} x_{2}) = 0$ $\Rightarrow x_{1} - x_{2} = 0$ $\Rightarrow x_{1} = x_{2} (∵ x_{1}^{2} + x_{1} x_{2} + x_{2}^{2} > 0)$ ∴ f is injective

Given a non empty set X, consider P(X) which is the set of all subsets of X. Define the relation R in P (X) as follows: For subsets A, B in P(X), A R B if and only if $A \subset B$ . Is R an equivalence relation on $P (X)$ ? Justify you answer: Sol. Since every set is a subset of itself, ARA, $\forall A \in P (X)$ . $∴ R$ is reflexive. Let $ARB \Rightarrow A \subset B . \Rightarrow B \neq \subset A$ . For $Ex :$ , if $A = {1, 2}$ and $B = {1, 2, 3}$ , then $A \subset B$ but $B \neq \subset A$ R is not symmetric. Since, R is not symmetric. Hence, R is not an equivalence relation on P(X).
Find the number of all onto functions from the set ${1, 2, 3, \dots, n$ ) to itself. Sol. Since f is onto function, so all elements of set ${1, 2, \dots n}$ have unique pre-image in set {1, 2,...n} So, the total number of onto function $= n \times (n - 1) \times (n - 2) \times \dots \times 2 \times 1 = n!$
Let $A = {- 1, 0, 1, 2}, B = {- 4, - 2, 0, 2}$ and $f, g : A \to B$ be functions defined by $f (x) = x^{2} - x$ , $x \in A$ and $g (x) = x - \frac{1}{2} - 1, x \in A$ . Are f and g equal? Justify your answer. Sol. Given $A = {- 1, 0, 1, 2}, B = {- 4, - 2, 0, 2}$ Also, $f : A \to B, f (x) = x^{2} - x, \forall x \in A$ $g : A \to B$ , $g (x) = 2 x - \frac{1}{2} - 1, \forall x \in A$ It is observed that :

$f (- 1) = (- 1)^{2} - (- 1) = 1 + 1 = 2 g (- 1) = 2 (- 1) - \frac{1}{2} - 1 = 2 (\frac{3}{2}) - 1 = 3 - 1 = 2$

$\Rightarrow f (- 1) = g (- 1)$ $f (0) = (0)^{2} - 0 = 0$ ; $g (0) = 2 0 - \frac{1}{2} - 1 = 2 (\frac{1}{2}) - 1 = 1 - 1 = 0$ $\Rightarrow f (0) = g (0)$ $f (1) = (1)^{2} - 1 = 1 - 1 = 0$ ; $g (1) = 2 1 - \frac{1}{2} - 1 = 2 (\frac{1}{2}) - 1 = 0$ $\Rightarrow f (1) = g (1)$ $f (2) = 2^{2} - 2 = 4 - 2 = 2$ ; $g (2) = 2 2 - \frac{1}{2} - 1 = 2 (\frac{3}{2}) - 1 = 3 - 1 = 2$ $\Rightarrow f (2) = g (2)$ Thus $\forall a \in A, f (a) = g (a)$ $∴ f$ and g are equal functions.

Let $A = {1, 2, 3}$ . Then number of relations containing $(1, 2)$ and $(1, 3)$ which are reflexive and symmetric but not transitive is (A) 1 (B) 2 (C) 3 (D) 4 Sol. (A) The given set is $A = {1, 2, 3}$ . The smallest relation containing $(1, 2)$ and $(1, 3)$ which is reflexive and symmetric, but not transitive is given by: $R = {(1, 1), (2, 2), (3, 3), (1, 2), (1, 3), (2, 1)$ , $(3, 1)$ } Since $(1, 1), (2, 2), (3, 3) \in R$ . So R is reflexive Since $(1, 2), (2, 1) \in R$ and $(1, 3), (3, 1) \in R$ . So R is symmetric Since $(3, 1), (1, 2) \in R$ , but $(3, 2) \in / R$ . So R is not transitive Now, if we add ordered pairs $(3, 2)$ and $(2, 3)$ to relation R , then relation R will become transitive. Hence, the total number of desired relations is one.
Let $A = {1, 2, 3}$ . Then number of equivalence relations containing $(1, 2)$ is (A) 1 (B) 2 (C) 3 (D) 4 Sol. (B) It is given that $A = {1, 2, 3}$ . The smallest equivalence relation containing $(1, 2)$ is given by, $R_{1} = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}$ Now, we are left with only four pairs i.e., (2, 3), (3, 2), (1, 3), and $(3, 1)$ . If we add any one pair [say (2, 3)] to $R_{1}$ , then for symmetry we must add $(3, 2)$ . Also, for transitivity we are required to add $(1, 3)$ and $(3, 1)$ . Hence, another equivalence relation $R_{2} = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)$ , (2, 3), (3, 2), (1, 3), (3, 1)} This shows that the total number of equivalence relations containing $(1, 2)$ is two i.e. $R_{1}$ and $R_{2}$

3.0Class 12 Maths NCERT Solutions – Chapter-wise Links

Access complete NCERT Solutions for Class 12 Maths chapter-wise, with clear exercise answers, important formulas, and step-by-step solutions to help students understand and solve Maths questions.

Chapter Number	Chapterwise NCERT Solutions
Chapter 2	Inverse Trigonometric Functions
Chapter 3	Matrices
Chapter 4	Determinants
Chapter 5	Continuity and Differentiability
Chapter 6	Application of Derivatives
Chapter 7	Integrals
Chapter 8	Application of Integrals
Chapter 9	Differential Equations
Chapter 10	Vector Algebra
Chapter 11	Three Dimensional Geometry
Chapter 12	Linear Programming
Chapter 13	Probability

4.0Class 12 Maths Chapter 1 Relations and Functions Exercise-wise Solutions

Exercise	Number of Questions	Important Topics
Exercise 1.1 Solutions	16 Questions	Types of relations, reflexive, symmetric, transitive and equivalence relations
Exercise 1.2 Solutions	12 Questions	One-one and onto functions, injective and surjective functions, domain and range, special functions
Miscellaneous Exercise Solutions	7 Questions	Relations and functions, one-one and onto functions, composition and inverse functions, mixed-concept questions

5.0Key Features and Benefits of Class 12 Maths Chapter 1 Relations and Functions

Solutions are prepared according to the latest NCERT syllabus for Class 12 Maths followed by CBSE schools.
Step-by-step explanations help students understand the method used to solve each question.
Exercises include questions based on important concepts of relations and functions covered in the chapter.
Practising NCERT questions with solutions helps students improve their accuracy and problem-solving skills.
A clear understanding of relations, functions, domain, codomain, range, and different types of functions helps strengthen the basics of the chapter.
Strong fundamentals in this chapter can help students understand higher-level concepts in mathematics and prepare for different entrance and competitive examinations.

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Frequently Asked Questions

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What topics are mainly covered in NCERT Class 12 Maths Chapter 1 Relations and Functions?

2

Why is Relations and Functions important for Class 12 board exams?

3

How do Class 12 Maths Chapter 1 NCERT Solutions help in understanding functions?

5

How does class 12 maths chapter 1 NCERT Solutions improve overall problem-solving skills?

Where can I get free NCERT Solutions for Relations and Functions Class 12 PDF?

What are the most commonly asked questions from Relations and Functions in NCERT Class 12 Maths?

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NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions

1.0Key Concepts of Class 12 Maths Chapter 1 Relations and Functions

2.0NCERT Class 12 Maths Chapter 1 Relations and Functions : Detailed Solutions

EXERCISE - 1.1

EXERCISE - 1.2

MISCELLANEOUS EXERCISE

3.0Class 12 Maths NCERT Solutions – Chapter-wise Links

4.0Class 12 Maths Chapter 1 Relations and Functions Exercise-wise Solutions

5.0Key Features and Benefits of Class 12 Maths Chapter 1 Relations and Functions

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