When the elements of two or more non-empty sets are related to each other in some way, it is called a relation. There are eight types of relations in mathematics:

Empty Relation: An empty relation exists if the elements of two sets M and N are not connected to each other in any way. It is also called a void relation, and can be written as R = φ. For example, whenever set M = {4, 5, 6} at that point, one of the empty relations can be R = {m, n} where, |m – n| = 7.

Empty Relation: R = φ ⊂ M × M

Universal Relation: When every element in set G is mapped to every element in set H, it is known as a universal relation. For example, if set G = {g, h, i}, one of the universal relations will be R = {g, h} where, |g – h| ≥ 0.

Universal Relation: R = G × G

Identity Relation: An identity relation exists if every element of a set is mapped to only itself. For example, in a set L = {j, k, l}, the identity relation will be I = {j, j}, {k, k}, {l, l}.

Identity Relation: I = {(j, j), j ∈ L}

Inverse Relation: When every element in set P is the inverse of every element in set Q, it is known as an inverse relation. For example if set P = {(p, q), (r, s)}, then the inverse relation will be R-1 = {(q, p), (s, r)}.

Inverse Relation = R-1 = {(q, p): (p, q) ∈ R}

Reflexive Relation: When every element of a set is mapped to itself, it is known as a reflexive relation. For example, in a set Y = {45, 46}, the reflexive relation will be R = {(45, 45), (46, 46), (45, 46), (46, 45)}.

Reflexive Relation: (y, y) ∈ R

Symmetric Relation: If p=q is true and q=p is also true, then a symmetric relation exists between the elements of a set. In other words, a relation R is symmetric only if (q, p) ∈ R is true when (p, q) ∈ R. For example, in a set P = {5, 8}, the symmetric relation will be R = {(5, 8), (8, 5)}.

Symmetric Relation: pRq ⇒ qRp, ∀ p, q ∈ P

Transitive Relation: A relation is transitive if (j, k) ∈ R and (k, l) ∈ R, then (j, l) ∈ R.

Transitive Relation: jRk and kRl ⇒ jRl ∀ j, k, l ∈ R

Equivalence Relation: A relation that is simultaneously reflexive, symmetric and transitive is known as an equivalence relation.

A function between two sets exists if the relation is such that each element of one set is related to exactly one element in the other set. Let X and Y be any two non-empty sets. Mapping from X to Y will be a function only when every element in set X has only one image in set Y. Additionally, a function cannot have two pairs of the same first element. A function from set X to set Y is represented by F: X→Y. There are four types of functions in mathematics.

• One-To-One Function: A function f: X → Y is one-to-one if each element of set X corresponds to a unique element in set Y. It is also known as an injective function. For example, if x1 ∈ X and x2 ∈ Y, then f is defined as f: X → Y such that f (x1) = f (x2).
• Many-To-One Function: If two or more elements of set X are mapped to the same element of set Y, it is known as a many-to-one function.
• Onto Function: If for every element of set Y, there is at least one or more than one element mapped in set X, then the function is known as onto function. Onto is also referred to as Surjective Function.
• One-To-One Correspondence Function: A one-to-one correspondence function is when the function satisfies both the injective and surjective properties. In other words, the function f maps each element of X with a unique element of Y, and every element of Y has a pre-image in X.

Composition of Functions

The composition of a function is an operation where two functions, m and n, generate a new function, o, in such a way that o(x) = m(n(x))[1] [2] . Here, function n is applied to the function of x. In other words, one function is applied to the result of another function.

Let m : P → Q and n : Q → R be two functions. Then the composition of m and n, denoted by nom, is defined as the function nom : P → R given by mΟn(x) = m(n(x)), ∀ x ∈ P.

For example, if m(x) = 3x + 6, and n(x) = x – 4, then m(n(x)) [3] [4] = 3(x – 4) + 6.

Properties of Composite Functions:

• Composite functions are associative, that is (m[5] [6] Οn)[7] [8] Οh = m[9] [10] Ο(n[11] [12] Οh)[13] [14] .
• Composite functions are not commutative, that is m[15] [16] Οn ≠ n[17] [18] Οm.[19] [20] 
• The function composition of one-to-one function is always one to one.
• The function composition of two onto functions is always onto.

Invertible Function

An inverse function is when a function reverses into another function. In other words, if a function f takes x to y then, the inverse of f will take y to x. If the function is denoted by f or F, then the inverse function is denoted by f-1 or F-1.

For example, if f(x) = x + 3 = y, then g(y) = y – 3 = x, which is f-1(x).

A binary operation ∗ on a set P is a function ∗ : P × P → P.

∗ (p, q) is denoted by p ∗ q.

Some properties of binary operations:

• Commutative Binary Operation: A binary operation * on set P is commutative if p * q = q * p, ∀ p, q ∈ P.
• Associative Binary Operation: A binary operation * on set M is associative, if m * (n * o) = (m * n) * o, ∀ m, n, o ∈ M.

Note: For a binary operation, the bracket in an associative property can be ignored. But in the absence of associative property, the bracket cannot be ignored.

• Identity Element: An element e ∈ M is the identity element of a binary operation * on set M, if m * e = e * m = m, ∀ m ∈ M. The identity element is unique.

Note: Zero is an identity for the addition operation on R, and one is an identity for the multiplication operation on R.

• Invertible Element or Inverse: Let * : S × S → S be a binary operation and let e ∈ S be its identity element. An element s ∈ S is invertible with respect to the operation *, if there exists an element t ∈ S such that s * t = t * s = e, ∀ t ∈ S. Element t is called the inverse of element s and is denoted by s-1.

Note: Inverse of an element, if it exists, is unique.

Example 1

Let M be the set of all students of a girls school. Show that the relation R in M given by R = {(m, n) : m is brother [1] [2] of n} is the empty relation and R′ = {(m, n) : the difference between heights of m and n is less than 4 meters} is the universal relation.

Solution

Since the school is a girls school, no student of the school can be brother [3] [4] of any student of the school. Hence, R = φ, showing that R is the empty relation. It is also obvious that the difference between heights of any two students of the school has to be less than 4 meters. This shows that R′ = M × M is the universal relation.

Example 2

Let P be the set of all triangles in a plane with R a relation in P given by R = {(P1 , P2 ) : P1 is congruent to P2}. Show that R is an equivalence relation.

Solution

Because every triangle is congruent to itself, R is reflexive. Moreover, (P1 , P2) ∈ R ⇒ P1 is congruent to P2 ⇒ P2 is congruent to P1 ⇒ (P2, P1) ∈ R. Therefore, R is symmetric. Furthermore, (P1, P2), (P2, P3) ∈ R ⇒ P1 is congruent to P2 and P2 is congruent to P3 ⇒ P1 is congruent to P3 ⇒ (P1, P3) ∈ R. Hence, R is an equivalence relation.

Example 3

Show that the relation R in the set {14, 15, 16} given by R = {(14, 14), (15, 15), (16, 16), (14, 15), (15, 16)} is reflexive but neither symmetric nor transitive.

Solution

R is reflexive since (14, 14), (15, 15) and (16, 16) lie in R. Additionally, R is not symmetric, as (14, 15) ∈ R but (15, 14) ∉ R. Similarly, R is not transitive, as (14, 15) ∈ R and (15, 16) ∈ R but (14, 16) ∉ R.

Example 4

Show that an onto function g : {21, 22, 23} → {21, 22, 23} is always one-one.

Solution

Suppose g is not one-one. Then there exists two elements, say 21 and 22 in the domain whose image in the co-domain is the same. Also, the image of 23 under g can be only one element. Therefore, a maximum of two elements of the co-domain {21, 22, 23} can be contained within the range set. This shows that g is not onto, a contradiction. Hence, g must be one-one.

Example 5

Show that a one-one function h : {6, 7, 8} → {6, 7, 8} must be onto.

Solution

Since h is one-one, three elements of {6, 7, 8} must be taken to three different elements of the co-domain {6, 7, 8} under h. Hence, h has to be onto.

Example 6

Consider f : P → N, g : Q → N and h : R → N defined as f(p) = 2p, g(q) = 3q + 4 and h(r) = sin r, ∀ p, q and r in N. Show that h[5] [6] Ο(g[7] [8] Οf) = (h[9] [10] Οg) of.[11] [12] 

Solution

We have hΟ(gΟf) (p) = h(gΟf (p)) = h(g(f(p))) = h(g(2p)) = h(3(2p) + 4) = h(6p + 4) = sin (6p+4), ∀ p ∈ P.

Also, ((hΟg)Ο f ) (p) =(hΟg) (f(p)) = (hΟg) (2p) = h (g(2p)) = h(3(2p) + 4) = h(6p + 4) = sin (6p + 4), ∀ p ∈ N.

This proves that hΟ(gΟf) = (hΟg) Ο f. This result is true in general situations as well.

Example 7

Show that subtraction, addition and multiplication are binary operations on R, but division is not a binary operation on R. Also, show that division is a binary operation on the set R∗ of nonzero real numbers.

Solution

 + : R × R → R is given by

 (m, n) → m + n

– : R × R → R is given by

 (m, n) → m – n

× : R × R → R is given by

 (m, n) → mn

Since ‘+’, ‘–’ and ‘×’ are functions, they are binary operations on R.

 But ÷: R × R → R, given by (m, n) → m/n, is not a function, therefore not a binary operation, as for n = 0, m/n is not defined.

However, ÷ : R∗ × R∗ → R∗, given by (m, n) → m/n is a function, therefore a binary operation on R∗.

Example 8

Show that –q is not the inverse of q ∈ N for the addition operation + on N and 1/q is not the inverse of q ∈ N for multiplication operation × on N, for q ≠ 1.[13] [14] 

Solution

Since –q ∉ N, –q can not be inverse of q for addition operation on N, although –q satisfies q + (– q) = 0 = (–q) + q.

Similarly, for q ≠ 1 in N, 1/q ∉ N, which implies that other than 1 no element of N has inverse for multiplication operation on N.

Example 9

Show that the number of binary operations on {1, 2} having 1 as identity and having 2 as the inverse of 2 is exactly one.

Solution

A binary operation ∗ on {1, 2} is a function from {1, 2} × {1, 2} to {1, 2}, i.e., a function from {(1, 1), (1, 2), (2, 1), (2, 2)} → {2, 1}. Since 1 is the identity for the desired binary operation ∗, ∗ (1, 1) = 1, ∗ (1, 2) = 2, ∗ (2, 1) = 2 and the only choice left is for the pair (2, 2). Since 2 is the inverse of 1, i.e., ∗ (2, 2) must be equal to 1. Thus, the number of desired binary operations is only one.

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• Practising the Relations and Functions Class 12 problems ensures the student has strong retention of knowledge of the lesson.

Can you give a summary of the topics covered in CBSE Class 12 Maths Chapter 1?

The topic covered in the CBSE Maths Chapter 1 is Relations and Functions Class 12. The sub-topics covered are Types of Relations, Types of Functions, Composition of Functions and Invertible Functions, and Binary Operations. The solutions to the exercise problems are solved by highly experienced professionals. By providing the NCERT Solutions, the aim is to help students understand the lesson and score high marks in the Class 12 CBSE exams.

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• Watch videos explaining the Relations and Functions Class 12 chapter.
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• Problems and exercises given in the NCERT Solutions are in accordance with the CBSE board. Practice these exercises to get used to answering board exam-style questions, preparing you well for the exam.

What videos should I refer to for CBSE class 12 maths chapter 1?

The videos listed below can be accessed on the MSVgo app. These engaging videos explain the chapter in detail, and will help give a clear understanding of the lessons. Download the app and start watching these videos:

1. Equivalence Relations and Classes
2. Composition of Functions
3. Invertible Functions
4. Types of Relations
5. Introduction to Functions
6. Types of Functions
7. Binary Operations

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