Last Updated: April 2026 | Prepared by Ready For Boards
CBSE Class 12 Mathematics Chapter 1 Relations and Functions — Complete Notes, Examples and MCQs
Relations and Functions is the opening chapter of CBSE Class 12 Mathematics and forms a critical foundation for the entire course. This chapter carries approximately 10 marks in the board examination and tests students on types of relations, types of functions, composition, invertibility, and binary operations. This comprehensive guide covers all concepts with definitions, solved examples, and board exam tips.
Chapter Overview
- Marks Weightage: ~10 marks in board exam (1-mark MCQs + 2-mark + 5-mark questions)
- Unit: Relations and Functions (Unit I — Relations and Functions)
- Key Topics: Types of Relations, Types of Functions, Composition, Inverse Functions, Binary Operations
- Prerequisites: Basic set theory (Class 11), domain and range concepts
Types of Relations
A relation R from set A to set B is a subset of the Cartesian product A × B. When the relation is on set A itself (from A to A), it is called a relation on A.
| Type of Relation | Definition | Example |
|---|---|---|
| Reflexive | (a, a) ∈ R for every a ∈ A | R = {(1,1), (2,2), (3,3)} on A = {1,2,3} |
| Symmetric | If (a, b) ∈ R, then (b, a) ∈ R for all a, b ∈ A | R = {(1,2), (2,1)} — symmetric because both directions present |
| Transitive | If (a,b) ∈ R and (b,c) ∈ R, then (a,c) ∈ R | R = {(1,2), (2,3), (1,3)} — transitive because (1,2),(2,3) implies (1,3) |
| Equivalence | Reflexive + Symmetric + Transitive | “is equal to” relation on integers is an equivalence relation |
| Empty Relation | No element of A is related to any element: R = φ | R = {} — no pairs at all |
| Universal Relation | Every element is related to every element: R = A × A | All possible pairs in A × A |
Solved Example 1 — Identifying Relation Type
Q: Let A = {1, 2, 3} and R = {(1,1), (2,2), (3,3), (1,2), (2,1)}. Is R reflexive, symmetric, transitive?
Solution:
- Reflexive: (1,1), (2,2), (3,3) all present — YES
- Symmetric: (1,2) present and (2,1) also present — YES
- Transitive: (1,2) ∈ R and (2,1) ∈ R — do we have (1,1)? YES. (2,1) ∈ R and (1,2) ∈ R — do we have (2,2)? YES. So transitive — YES
- Conclusion: R is an Equivalence Relation
Types of Functions
A function f: A → B assigns exactly one element of B to each element of A.
| Type | Definition | Example |
|---|---|---|
| One-One (Injective) | Different elements in domain have different images in codomain: f(a) = f(b) implies a = b | f(x) = 2x from R to R — each x gives unique 2x |
| Onto (Surjective) | Every element of codomain has at least one preimage: for every b ∈ B, there exists a ∈ A such that f(a) = b | f(x) = x from R to R — every real number is covered |
| Bijective | Both one-one (injective) and onto (surjective) | f(x) = x + 1 from R to R — each x gives unique x+1 and every real is covered |
| Into | At least one element in codomain has NO preimage (not onto) | f: {1,2} → {1,2,3} with f(1)=1, f(2)=2 — 3 has no preimage |
Solved Example 2 — One-One Test
Q: Is f: R → R defined by f(x) = x² one-one?
Solution: f(2) = 4 and f(-2) = 4. So f(2) = f(-2) but 2 ≠ -2. Therefore, f is NOT one-one.
Solved Example 3 — Onto Test
Q: Is f: R → R defined by f(x) = x² onto?
Solution: f(x) = x² ≥ 0 for all x ∈ R. So negative real numbers (e.g., -1) have no preimage. Therefore, f is NOT onto.
Composition of Functions
If f: A → B and g: B → C are two functions, then the composition (g∘f): A → C is defined by:
(g∘f)(x) = g(f(x)) for all x ∈ A
Properties of Composition
- Composition is associative: h∘(g∘f) = (h∘g)∘f
- Composition is generally NOT commutative: f∘g ≠ g∘f in general
- If f and g are both one-one, then g∘f is one-one
- If f and g are both onto, then g∘f is onto
- If g∘f is one-one, then f is one-one (but g need not be)
Solved Example 4 — Composition
Q: If f(x) = 2x and g(x) = x + 1, find (f∘g)(x) and (g∘f)(x).
Solution:
- (f∘g)(x) = f(g(x)) = f(x+1) = 2(x+1) = 2x + 2
- (g∘f)(x) = g(f(x)) = g(2x) = 2x + 1 = 2x + 1
- Since (f∘g)(x) ≠ (g∘f)(x), composition is not commutative here.
Invertible Functions
A function f: A → B is invertible if and only if it is bijective (one-one and onto). The inverse function f⁻¹: B → A satisfies:
f⁻¹(f(x)) = x for all x ∈ A, and f(f⁻¹(y)) = y for all y ∈ B
Solved Example 5 — Finding Inverse
Q: Find the inverse of f: R → R defined by f(x) = 3x + 5.
Solution: Let y = 3x + 5. Solving for x: x = (y – 5)/3. Therefore, f⁻¹(y) = (y – 5)/3, or written as f⁻¹(x) = (x – 5)/3.
Binary Operations
A binary operation * on a set S is a function *: S × S → S, i.e., it takes two elements of S and gives one element of S.
Properties of Binary Operations
| Property | Definition | Example |
|---|---|---|
| Commutative | a * b = b * a for all a, b ∈ S | Addition: a + b = b + a ✓ |
| Associative | a * (b * c) = (a * b) * c for all a, b, c ∈ S | Multiplication: a(bc) = (ab)c ✓ |
| Identity Element | e ∈ S such that a * e = e * a = a for all a ∈ S | For addition: e = 0. For multiplication: e = 1 |
| Inverse Element | For each a ∈ S, there exists b ∈ S such that a * b = b * a = e | For addition: inverse of a is -a. For multiplication: inverse of a is 1/a |
Solved Example 6 — Binary Operation
Q: Let * be defined on Z (integers) as a * b = a + b – 2. Find the identity element.
Solution: We need a * e = a, so a + e – 2 = a, which gives e = 2. Verify: a * 2 = a + 2 – 2 = a ✓. The identity element is e = 2.
Board Exam Pattern for This Chapter
- 1-mark MCQ: Typically tests definition-based identification of relation type or function property
- 2-mark question: Verifying if a given relation is reflexive/symmetric/transitive, or finding fog/gof
- 5-mark question: Proving a function is bijective and finding its inverse, OR proving a relation is an equivalence relation
For complete CBSE Class 12 Mathematics preparation including video lectures, solved NCERT exercises, and previous 10 years board papers, visit Ready For Boards courses.
Practice Quiz — Relations and Functions
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Frequently Asked Questions — Relations and Functions Class 12
Q1. What is the difference between a relation and a function?
A relation from set A to set B is any subset of A × B — one element of A can be related to multiple elements of B. A function is a special relation where each element of A (domain) is related to exactly one element of B (codomain). All functions are relations, but not all relations are functions.
Q2. How do you prove a function is bijective in board exams?
To prove bijectivity, show two things separately: (1) One-one: Assume f(a) = f(b) and show this implies a = b. (2) Onto: Take any arbitrary element y in the codomain and find a preimage x in the domain such that f(x) = y. If both are proved, the function is bijective. Always write both proofs clearly for full marks.
Q3. Can the composition fog be defined when gof is not?
Yes. For fog(x) = f(g(x)), we need: range of g ⊆ domain of f. For gof(x) = g(f(x)), we need: range of f ⊆ domain of g. These are independent conditions, so fog may be defined even when gof is not, depending on the domains and codomains involved.
Q4. What is an equivalence class and is it important for boards?
An equivalence class of an element a under an equivalence relation R is the set of all elements related to a: [a] = {x ∈ A : (x, a) ∈ R}. Equivalence classes partition the set A into disjoint subsets. For CBSE Class 12 boards, being able to find equivalence classes given a relation is a common 5-mark question and must be practised thoroughly.
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