Class 12 Maths Notes: Relations and Functions (Chapter 1)
Master Chapter 1 of Class 12 Mathematics. Easy explanation of Types of Relations, Equivalence Relations, Functions, and exam PYQs.
1. Introduction to Relations and Functions
Relations and Functions are the foundational building blocks of calculus and advanced algebra. Before you can differentiate, integrate, or work with limits — you need to understand what a function truly is and how sets relate to each other.
A relation maps elements from one set to another. A function takes that concept further — it enforces a strict rule: every input must map to exactly one output.
2. Types of Relations
A relation R in a set A is defined by specific properties. All three are tested heavily in board exams, usually as "prove that R is an equivalence relation" questions.
- Reflexive: Every element is related to itself. For all a ∈ A, (a, a) ∈ R.
Example: "is equal to" — every number equals itself. - Symmetric: If (a, b) ∈ R, then (b, a) ∈ R.
Example: "is a sibling of" — if A is B's sibling, B is A's sibling. - Transitive: If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R.
Example: "is less than" — if a < b and b < c, then a < c.
3. Equivalence Relation
A relation that satisfies all three — Reflexive, Symmetric, and Transitive — is called an Equivalence Relation. This is the most common 5-mark proof question in board exams.
Standard example: The relation R = {(a, b) : a and b have the same remainder when divided by n} is an equivalence relation on Z.
4. Types of Functions
- One-One (Injective): If f(x₁) = f(x₂) → x₁ = x₂. No two different inputs give the same output. Every element in the domain has a distinct image in the co-domain.
- Onto (Surjective): For every y ∈ Y (co-domain), there exists some x ∈ X (domain) such that f(x) = y. In other words, the Range = Co-domain. Nothing in the co-domain is left unmapped.
- Bijective: A function that is simultaneously One-One and Onto. Only bijective functions have a well-defined inverse function f⁻¹.
Proof Structure: Proving a Function is One-One
Step 1: Assume f(x₁) = f(x₂) for some x₁, x₂ in the domain.
Step 2: Algebraically simplify both sides of the equation.
Step 3: Arrive at x₁ = x₂. This proves the function is injective (One-One).
Example: f(x) = 3x − 4. Assume f(x₁) = f(x₂) → 3x₁ − 4 = 3x₂ − 4 → x₁ = x₂. ✓ One-One.
5. Composition of Functions
If f : A → B and g : B → C, then the composite function (g ∘ f) : A → C is defined as:
(g ∘ f)(x) = g(f(x))
Important: f ∘ g and g ∘ f are generally not equal. The order matters.
6. Inverse of a Function
A function f : X → Y has an inverse f⁻¹ : Y → X if and only if f is bijective.
To find f⁻¹(x): Write y = f(x), then express x in terms of y. That expression gives you f⁻¹(y).
Example: f(x) = 3x − 4. Let y = 3x − 4 → x = (y + 4)/3. So f⁻¹(x) = (x + 4)/3.
⚠️ Watch Out! — Domain vs. Co-domain Trap
A function's "onto" status depends entirely on its declared co-domain. The function f(x) = x² is onto if it maps ℝ → [0, ∞), but not onto if declared as f : ℝ → ℝ, because negative numbers in ℝ are never a square of any real input.
Always check the domain and co-domain carefully before deciding if a function is onto.
Summary: Key Rules at a Glance
| Concept | Key Rule |
|---|---|
| Equivalence Relation | Reflexive + Symmetric + Transitive — all three must hold |
| One-One (Injective) | f(x₁) = f(x₂) ⟹ x₁ = x₂ |
| Onto (Surjective) | Range = Co-domain |
| Bijective | One-One AND Onto → inverse exists |
| Composite Function | (g ∘ f)(x) = g(f(x)) |
| Inverse Function | Exists if and only if f is Bijective |
Practice Questions (PYQs)
- Show that the relation R in ℝ defined as R = {(a, b) : a ≤ b²} is neither reflexive, symmetric, nor transitive.
- Check whether f : ℝ → ℝ defined by f(x) = 3 − 4x is one-one, onto, or bijective.
- Let A = {1, 2, 3}. Give an example of a relation on A which is symmetric but neither reflexive nor transitive.
- If f(x) = (4x + 3)/(6x − 4), show that f(f(x)) = x for all x ≠ 2/3. What does this tell you about the function?
- Prove that the relation R in the set of integers Z defined by R = {(a, b) : 2 divides (a − b)} is an equivalence relation.