Chapter 1 — Relations and Functions — builds on your Class 11 knowledge and introduces types of relations, functions, composition, and inverse functions. Carries 6-8 marks in Board exams. Focus on identifying relation types and proving functions are bijective.
Key Concepts
Types of Relations
| Type | Definition | Example |
|---|---|---|
| Empty Relation | No element of A is related to any element | R = {} on A |
| Universal Relation | Every element is related to every element | R = A × A |
| Reflexive | (a, a) ∈ R for all a ∈ A | “is equal to” on integers |
| Symmetric | If (a, b) ∈ R, then (b, a) ∈ R | “is a sibling of” |
| Transitive | If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R | “is less than” |
| Equivalence Relation | Reflexive + Symmetric + Transitive | “is congruent to (mod n)” |
Types of Functions
| Type | Definition | Test |
|---|---|---|
| One-one (Injective) | f(a₁) = f(a₂) ⟹ a₁ = a₂ | Different inputs → different outputs |
| Onto (Surjective) | Range = Codomain | Every element in codomain has a pre-image |
| Bijective | Both one-one and onto | Invertible function |
Number of one-one functions: P(n(B), n(A)) if n(B) ≥ n(A), else 0
Number of onto functions: Uses inclusion-exclusion principle
Composition of Functions
Read right to left: first apply f, then apply g
g ∘ f ≠ f ∘ g in general (not commutative)
(g ∘ f) is one-one if both f and g are one-one
(g ∘ f) is onto if both f and g are onto
Inverse of a Function
f⁻¹(y) = x where f(x) = y
f ∘ f⁻¹ = I (identity function)
f⁻¹ ∘ f = I
Binary Operations
Commutative: a * b = b * a for all a, b
Associative: (a * b) * c = a * (b * c)
Identity element e: a * e = e * a = a for all a
Inverse of a: a * b = b * a = e
Solved Examples
Example 1
Q: Show that the relation R on Z defined by R = {(a,b) : a−b is divisible by 3} is an equivalence relation.
Solution:
Reflexive: a − a = 0, divisible by 3 ✓
Symmetric: If a−b is div by 3, then b−a = −(a−b) is also div by 3 ✓
Transitive: If a−b = 3k and b−c = 3m, then a−c = (a−b)+(b−c) = 3(k+m), div by 3 ✓
Hence R is an equivalence relation.
Example 2
Q: If f(x) = 2x + 3 and g(x) = x², find g ∘ f and f ∘ g.
Solution:
g ∘ f(x) = g(f(x)) = g(2x+3) = (2x+3)² = 4x² + 12x + 9
f ∘ g(x) = f(g(x)) = f(x²) = 2x² + 3
Clearly g ∘ f ≠ f ∘ g
Important Questions
1 Mark
- Is {(1,1), (2,2), (3,3)} reflexive on {1,2,3}?
- What is the identity element for addition on R?
3-5 Mark
- Prove that a relation is/is not an equivalence relation.
- Show that f is bijective and find f⁻¹.
- Find g ∘ f and f ∘ g for given functions.
Quick Revision Points
- Equivalence = reflexive + symmetric + transitive
- Bijective = one-one + onto = invertible
- Composition: g ∘ f means apply f first, then g
- f⁻¹ exists ⟺ f is bijective
- Binary operation: must be closed (result stays in the set)
Chapter Navigation
Previous: This is the first chapter
Next: Inverse Trigonometric Functions Class 12 Notes
Related Chapters in Class 12 Maths
Practice What You Learned
Test yourself with our JEE Main Mock Test Set 1 to see how well you’ve mastered the concepts.