Relations and Functions | Class 12 Maths | Aviate Learnings

1.2

Types of Relations

Reflexive, Symmetric, Transitive & Equivalence Relations

✈️ AAI ATC Exam Relevance: HIGH

What is a Relation?

A relation R from set A to set B is any subset of A × B. If (a, b) ∈ R, we write a R b (a is related to b). Relations generalise everyday connections between objects — two extreme cases are the empty relation (nothing is related) and the universal relation (everything is related).

📖 Key Definitions — Relations in Set A

Empty Relation: R = φ ⊂ A × A (no element related to any)
Universal Relation: R = A × A (every element related to every element)
Reflexive: (a, a) ∈ R for every a ∈ A
Symmetric: (a₁, a₂) ∈ R ⟹ (a₂, a₁) ∈ R, for all a₁, a₂ ∈ A
Transitive: (a₁, a₂) ∈ R and (a₂, a₃) ∈ R ⟹ (a₁, a₃) ∈ R
Equivalence Relation: R is reflexive AND symmetric AND transitive

📐 Quick Test Method for AAI ATC

To check Reflexive: Does (a, a) ∈ R for every a? Test one element at a time.
To check Symmetric: If (a, b) ∈ R, is (b, a) ∈ R? Swap and verify.
To check Transitive: If (a,b) ∈ R and (b,c) ∈ R, is (a,c) ∈ R? Chain check.
Equivalence ↔ All three hold simultaneously.

Equivalence Classes

An equivalence relation R on set X partitions X into mutually disjoint subsets called equivalence classes. The class containing element a is denoted [a], and consists of all elements b related to a. These classes satisfy: (i) all elements within a class are related, (ii) no element of one class is related to any element of another, (iii) their union is X.

🔗

NCERT Fig 1.1 — Lines showing perpendicularity relation (symmetric but not reflexive/transitive)

Insert NCERT figure showing three parallel lines L₁, L₂, L₃ demonstrating non-transitivity of ⊥

AAI ATC Tip: Equivalence relation is the most tested concept. Remember — R = {(a,b): a−b is divisible by n} is always an equivalence relation. "Congruent triangles", "parallel lines", "same birthday" are classic equivalence examples.

🎯 Practice MCQs — Types of Relations

The relation R in set A = {1, 2, 3} given by R = {(1,2), (2,1)} is:

Easy

Reflexive? (1,1) ∉ R → NOT reflexive.
Symmetric? (1,2) ∈ R and (2,1) ∈ R → YES symmetric.
Transitive? No pair (a,b),(b,c) both in R → vacuously transitive, but not reflexive so not equivalence.
Answer: Symmetric only

The relation R in ℤ defined by R = {(a,b): a−b is divisible by 5} is:

Moderate

Number of equivalence relations on set {1, 2, 3} containing (1, 2) and (2, 1) is:

Hard

Smallest: R₁ = {(1,1),(2,2),(3,3),(1,2),(2,1)}
Adding any pair forces symmetry and transitivity to yield the full universal relation R₂ = A×A.
Exactly 2 equivalence relations.

1.3

Types of Functions

One-One (Injective), Onto (Surjective) and Bijective

✈️ AAI ATC Exam Relevance: VERY HIGH

📖 Core Definitions

One-One (Injective): f: X→Y where f(x₁) = f(x₂) ⟹ x₁ = x₂. Distinct inputs give distinct outputs.
Onto (Surjective): f: X→Y where every y ∈ Y has some x ∈ X with f(x) = y. Range = Co-domain.
Bijective: f is both one-one AND onto. This is necessary for a function to be invertible.
Many-One: At least two elements of X map to the same element of Y.

📐 Quick Identification Rules

f(x) = 2x on ℝ→ℝ: One-one and onto (bijective)
f(x) = x² on ℝ→ℝ: Many-one and not onto (−ve numbers have no preimage)
f(x) = x² on ℕ→ℕ: One-one but not onto (e.g. 2 has no preimage)
f(x) = x² on ℝ→[0,∞): Many-one but onto
For finite set X: f: X→X is one-one ↔ onto (bijective)

🔀

NCERT Fig 1.2 (i) to (iv) — Arrow diagrams showing f₁, f₂, f₃, f₄

Insert NCERT figure showing four mapping diagrams: one-one not onto, many-one not onto, many-one onto, one-one onto (bijective)

Important Results for Finite Sets

A one-one function f: X→X (finite X) is necessarily onto
An onto function f: X→X (finite X) is necessarily one-one
For infinite sets, this property does NOT hold (e.g., f(x)=2x on ℕ is one-one but not onto)

AAI ATC Tip: To prove one-one: assume f(x₁)=f(x₂) and show x₁=x₂. To prove onto: take arbitrary y in co-domain, find x such that f(x)=y. f(x)=x² on ℝ is the most classic non-one-one example.

🎯 Practice MCQs — Types of Functions

The function f: ℝ→ℝ defined by f(x) = x² is:

Easy

Not one-one: f(−1) = f(1) = 1 but −1 ≠ 1
Not onto: −2 ∈ ℝ has no preimage (x² ≥ 0 always)
Answer: Neither one-one nor onto

The function f: ℝ→ℝ defined by f(x) = 3x is:

Moderate

One-one: f(x₁)=f(x₂) ⟹ 3x₁=3x₂ ⟹ x₁=x₂ ✓
Onto: For any y∈ℝ, x = y/3 ∈ ℝ and f(y/3)=y ✓
Answer: One-one and onto (Bijective)

Let f: ℕ→ℕ be defined by f(x) = x². Which of the following is correct?

Hard

On ℕ: f(x₁)=f(x₂) ⟹ x₁²=x₂² ⟹ x₁=x₂ (both positive) → One-one ✓
But 2∈ℕ has no preimage in ℕ (√2 ∉ ℕ) → Not onto ✗
Answer: One-one but not onto

1.4a

Composition of Functions

gof — applying one function after another

✈️ AAI ATC Exam Relevance: MODERATE

Definition of Composition

Let f: A→B and g: B→C. The composition gof: A→C is defined as:

gof(x) = g(f(x)) for all x ∈ A

First apply f to x, then apply g to that result. Note: gof ≠ fog in general (composition is NOT commutative).

📐 Key Properties of Composition

• If f and g are one-one, then gof is one-one
• If f and g are onto, then gof is onto
• If gof is one-one, then f is one-one (but g need not be)
• If gof is onto, then g is onto (but f need not be)
• Composition is associative: ho(gof) = (hog)of
• In general: gof ≠ fog

🔗

NCERT Fig 1.5 — Composition diagram showing A →f→ B →g→ C with gof: A→C

Insert NCERT arrow diagram showing the two-step mapping A to B to C via f then g

Example: f(x) = cos x, g(x) = 3x²
gof(x) = g(cos x) = 3cos²x
fog(x) = f(3x²) = cos(3x²)
At x=0: gof = 3, fog = 1 → gof ≠ fog

🎯 Practice MCQs — Composition of Functions

If f(x) = 2x + 1 and g(x) = x², then gof(2) equals:

Easy

f(2) = 2(2)+1 = 5
gof(2) = g(f(2)) = g(5) = 5² = 25

If f(x) = x + 2 and g(x) = 2x − 1, then fog(x) equals:

Moderate

fog(x) = f(g(x)) = f(2x−1) = (2x−1)+2 = 2x+1

If f: A→B and g: B→C are both one-one, then gof is:

Hard

Proof: Suppose gof(x₁) = gof(x₂)
→ g(f(x₁)) = g(f(x₂))
→ f(x₁) = f(x₂) (g is one-one)
→ x₁ = x₂ (f is one-one)
∴ gof is one-one

1.4b

Invertible Functions

Finding f⁻¹ — bijection is the key

✈️ AAI ATC Exam Relevance: HIGH

Definition of Invertible Function

A function f: X→Y is invertible if there exists g: Y→X such that:

gof = I_X (identity on X) and fog = I_Y (identity on Y)

The function g is called the inverse of f, written as f⁻¹. A function is invertible if and only if it is bijective (one-one and onto).

📐 Finding the Inverse

Step 1: Show f is bijective (one-one and onto)
Step 2: Let y = f(x), express x in terms of y
Step 3: That expression is f⁻¹(y)

Example: f(x) = 4x + 3
y = 4x + 3 → x = (y−3)/4
∴ f⁻¹(y) = (y−3)/4

Verify: gof(x) = g(4x+3) = (4x+3−3)/4 = x ✓
fog(y) = f((y−3)/4) = 4·(y−3)/4 + 3 = y ✓

↔️

NCERT Fig 1.3 — Graph of f(x) = 2x showing bijective function on ℝ

Insert NCERT graph showing f(x) = 2x as a straight line through origin — bijective on ℝ

Important: f ∘ f⁻¹ = I and f⁻¹ ∘ f = I

f⁻¹ exists ↔ f is bijective
If f and g are both bijective, then (gof)⁻¹ = f⁻¹og⁻¹
The inverse of a bijection is unique
f(x) = x² on ℝ→ℝ has NO inverse (not bijective)

AAI ATC Tip: Always first confirm the function is bijective before finding its inverse. The standard method: set y = f(x), solve for x, replace x with f⁻¹(y). Verify by composing.

🎯 Practice MCQs — Invertible Functions

If f: ℝ→ℝ is defined by f(x) = 2x − 3, then f⁻¹(x) equals:

Easy

y = 2x − 3 → 2x = y + 3 → x = (y+3)/2
∴ f⁻¹(x) = (x+3)/2

f: ℕ→Y is defined as f(x) = 4x + 3, where Y = {4x+3 : x∈ℕ}. The inverse f⁻¹(y) is:

Moderate

y = 4x + 3 → x = (y−3)/4
f is bijective on Y, so f⁻¹(y) = (y−3)/4
Verify: gof(x) = g(4x+3) = (4x+3−3)/4 = x ✓

Which of the following functions f: ℝ→ℝ is invertible?

Hard

f(x) = x²: not one-one (f(−1)=f(1)) → not invertible
f(x) = |x|: not one-one → not invertible
f(x) = 5x−7: one-one (5x₁−7=5x₂−7 ⟹ x₁=x₂) and onto ✓ → invertible
f(x) = sin x: not one-one on ℝ → not invertible (needs restricted domain)

✓

Chapter Summary

Key results at a glance for quick revision

Equivalence Relation

Reflexive + Symmetric + Transitive. Partitions the set into disjoint equivalence classes.

One-One (Injective)

f(x₁)=f(x₂) ⟹ x₁=x₂. Distinct inputs → distinct outputs. Test by contra.

Onto (Surjective)

Every y in co-domain has a preimage. Range = Co-domain. Test by solving f(x)=y.

Invertible ↔ Bijective

f⁻¹ exists iff f is one-one and onto. Find by solving y=f(x) for x.

Composition gof

gof(x)=g(f(x)). NOT commutative. Both bijective ⟹ gof bijective.

Finite Set Rule

f: X→X (finite X): one-one ↔ onto. Not true for infinite sets.

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