What is the difference between relation and function?

A relation is any subset of cartesian product of two sets. A function is a special type of relation where each element of the domain is related to exactly one element in the codomain. In other words, all functions are relations, but not all relations are functions.

What are the types of functions in JEE Mathematics?

The main types of functions for JEE are: One-One (Injective), Onto (Surjective), Bijective (One-One and Onto), Identity Function, Constant Function, Polynomial Function, Rational Function, Modulus Function, Signum Function, Greatest Integer Function, and Exponential & Logarithmic Functions.

How to find domain and range of a function?

Domain is the set of all possible input values (x-values) for which the function is defined. To find domain: exclude values that make denominator zero, values that make expression under even root negative, and values outside the domain of logarithm. Range is the set of all possible output values (y-values). It can be found by solving y = f(x) for x and finding values of y for which x is real.

What is a composite function?

A composite function (fog)(x) = f(g(x)) is a function obtained by applying one function to the results of another. For fog to exist, the range of g must be a subset of the domain of f. Important: fog ≠ gof in general (composition is not commutative).

When does inverse function exist?

An inverse function f⁻¹ exists if and only if the function f is bijective (both one-one and onto). For f⁻¹ to exist: (1) Each y-value must correspond to exactly one x-value (one-one), (2) Every element in codomain must be mapped (onto). If f⁻¹ exists, then f(f⁻¹(x)) = x and f⁻¹(f(x)) = x.

What is an equivalence relation?

An equivalence relation is a relation that is reflexive (aRa for all a), symmetric (aRb implies bRa), and transitive (aRb and bRc implies aRc). Common examples include equality relation, congruence modulo n, and similarity relation. Equivalence relations partition a set into disjoint equivalence classes.

Relations and Functions for JEE 2025-26 | Complete Notes with Formulas & Solved Examples

Basic Concepts

Relations and Functions form the backbone of modern mathematics. Understanding these concepts is crucial not only for JEE but also for advanced mathematics, computer science, and engineering applications.

1.1 Cartesian Product

Definition

If A and B are two non-empty sets, the Cartesian product A × B is the set of all ordered pairs (a, b) where a ∈ A and b ∈ B.

A \times B = \{(a, b) : a \in A, b \in B\}

Example:

If A = {1, 2} and B = {3, 4}

A \times B = \{(1,3), (1,4), (2,3), (2,4)\}

💡 Important Properties

If n(A) = p and n(B) = q, then n(A × B) = pq
A × B ≠ B × A (not commutative)
A × ∅ = ∅ × A = ∅
(A × B) × C ≠ A × (B × C) but both have same number of elements

1.2 Relation

Definition

A relation R from set A to set B is a subset of the Cartesian product A × B.

R \subseteq A \times B

If (a, b) ∈ R, we say "a is related to b" and write aRb.

📝 Solved Example 1

Question: If A = {1, 2, 3} and B = {4, 5}, find the number of possible relations from A to B.

Solution:

Step 1: Find n(A × B)

n(A) = 3, n(B) = 2

n(A \times B) = 3 \times 2 = 6

Step 2: Number of subsets of A × B

Since a relation is any subset of A × B:

\text{Number of relations} = 2^{n(A \times B)} = 2^6 = 64

\text{Answer: 64 relations}

1.3 Function

Definition

A relation f from set A to set B is called a function if:

Every element of A has an image in B
No element of A has more than one image in B

f: A \to B

A is called the domain
B is called the codomain
Set of all images is called the range

⚠️ Key Difference: Relation vs Function

Aspect	Relation	Function
Definition	Any subset of A × B	Special type of relation
Mapping	Can be one-to-many	Must be one-to-one or many-to-one
Coverage	May not cover all elements of A	Must cover all elements of A

Types of Relations

Relations can be classified based on their properties. Understanding these types is crucial for solving JEE problems efficiently.

2.1 Empty Relation

A relation R in a set A is called an empty relation if no element of A is related to any element of A.

R = \phi \subset A \times A

2.2 Universal Relation

A relation R in a set A is called a universal relation if each element of A is related to every element of A.

R = A \times A

2.3 Reflexive Relation

A relation R in a set A is called reflexive if (a, a) ∈ R for every a ∈ A.

\forall a \in A, (a, a) \in R

Examples:

Equality relation (=) is reflexive: a = a
Less than or equal (≤) is reflexive: a ≤ a
Divisibility in integers: a | a

2.4 Symmetric Relation

A relation R in a set A is called symmetric if (a, b) ∈ R implies (b, a) ∈ R.

(a, b) \in R \Rightarrow (b, a) \in R

Examples:

Equality relation (=): if a = b then b = a
"Is married to" relation
"Is perpendicular to" in geometry

2.5 Transitive Relation

A relation R in a set A is called transitive if (a, b) ∈ R and (b, c) ∈ R implies (a, c) ∈ R.

(a, b) \in R \text{ and } (b, c) \in R \Rightarrow (a, c) \in R

Examples:

Less than (<): if a < b and b < c then a < c
Divisibility: if a | b and b | c then a | c
"Is ancestor of" relation

2.6 Equivalence Relation

Most Important for JEE!

A relation R in a set A is called an equivalence relation if it is:

Reflexive: (a, a) ∈ R for all a ∈ A
Symmetric: (a, b) ∈ R ⇒ (b, a) ∈ R
Transitive: (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R

Classic Examples:

Equality relation (=)
Congruence modulo n: a ≡ b (mod n)
Similarity of triangles

📝 Solved Example 2 (JEE Main Type)

Question: Let A = {1, 2, 3, 4} and R = {(1,1), (2,2), (3,3), (4,4), (1,2), (2,1), (2,3), (3,2)}. Check if R is an equivalence relation.

Solution:

Check 1: Reflexive?

All elements (1,1), (2,2), (3,3), (4,4) are present ✓

✓ R is reflexive

Check 2: Symmetric?

• (1,2) ∈ R and (2,1) ∈ R ✓

• (2,3) ∈ R and (3,2) ∈ R ✓

✓ R is symmetric

Check 3: Transitive?

• (1,2) ∈ R and (2,3) ∈ R but (1,3) ∉ R

✗ R is NOT transitive

\text{Answer: R is NOT an equivalence relation}

💡 Quick Memory Trick

Remember "RST" for Equivalence Relation:

Reflexive - relates to itself
Symmetric - if A→B then B→A
Transitive - if A→B and B→C then A→C

Types of Functions

Classification of functions is one of the highest-scoring topics in JEE. Master these definitions and tests to solve problems in under 1 minute.

3.1 One-One Function (Injective)

Definition

A function f: A → B is called one-one (injective) if different elements of A have different images in B.

f(x_1) = f(x_2) \Rightarrow x_1 = x_2

OR equivalently

x_1 \neq x_2 \Rightarrow f(x_1) \neq f(x_2)

Graphical Test:

Any horizontal line cuts the graph at most once.

3.2 Onto Function (Surjective)

Definition

A function f: A → B is called onto (surjective) if every element of B is the image of at least one element of A.

\text{Range of } f = \text{Codomain} = B

In other words: For every b ∈ B, there exists at least one a ∈ A such that f(a) = b.

3.3 Bijective Function (One-One and Onto)

Definition

A function f: A → B is called bijective if it is both one-one and onto.

Every element of A maps to a unique element in B (one-one)
Every element of B is mapped by some element in A (onto)

Important: Only bijective functions have inverse functions!

📝 Solved Example 3 (JEE Main 2023 Type)

Question: Check if f: ℝ → ℝ defined by f(x) = 3x + 5 is one-one and onto.

Solution:

Test for One-One:

Let f(x₁) = f(x₂)

\[3x_1 + 5 = 3x_2 + 5\]

\[3x_1 = 3x_2\]

\[x_1 = x_2\]

✓ f is one-one

Test for Onto:

Let y ∈ ℝ (codomain). We need to find x ∈ ℝ such that f(x) = y

\[3x + 5 = y\]

x = \frac{y-5}{3}

For any y ∈ ℝ, we can find x = (y-5)/3 ∈ ℝ

✓ f is onto

\text{Answer: f is bijective (one-one and onto)}

3.4 Other Important Functions

Identity Function

\[I(x) = x\]

Maps every element to itself

Bijective

Constant Function

\[f(x) = c\]

Maps all elements to constant c

Neither one-one nor onto

Polynomial Function

\[f(x) = a_nx^n + ... + a_1x + a_0\]

Degree n polynomial

Rational Function

f(x) = \frac{P(x)}{Q(x)}

Ratio of two polynomials

Function Type	Example	One-One?	Onto?
f(x) = x	Identity	✓ Yes	✓ Yes
f(x) = x²	ℝ → ℝ	✗ No	✗ No
f(x) = x³	ℝ → ℝ	✓ Yes	✓ Yes
f(x) = eˣ	ℝ → ℝ⁺	✓ Yes	✓ Yes
f(x) = \|x\|	ℝ → ℝ	✗ No	✗ No

Domain and Range

Finding domain and range is a guaranteed scoring topic in JEE. Learn the systematic approach to solve any domain-range problem in under 2 minutes.

4.1 Finding Domain

General Rules for Domain

1. Polynomial Functions

Domain = ℝ (all real numbers)

f(x) = x^n + ... + c \Rightarrow D_f = \mathbb{R}

2. Rational Functions

Denominator ≠ 0

f(x) = \frac{P(x)}{Q(x)} \Rightarrow Q(x) \neq 0

3. Square Root Functions

Expression under root ≥ 0

f(x) = \sqrt{g(x)} \Rightarrow g(x) \geq 0

4. Logarithmic Functions

Argument > 0

f(x) = \log g(x) \Rightarrow g(x) > 0

📝 Solved Example 4

Question: Find the domain of \(f(x) = \sqrt{x-3} + \frac{1}{x-5}\)

Solution:

Condition 1: For √(x-3) to be real

x - 3 \geq 0

x \geq 3

Condition 2: For 1/(x-5) to be defined

x - 5 \neq 0

x \neq 5

Combining both conditions:

x ≥ 3 AND x ≠ 5

Domain = [3, 5) \cup (5, \infty)

4.2 Finding Range

Methods for Finding Range

Method 1: Solve for x in terms of y

Put y = f(x), solve for x, find values of y for which x is real

Method 2: Use Calculus

Find minimum and maximum using derivatives

Method 3: Graphical Method

Sketch the graph and observe y-values covered

📝 Solved Example 5 (JEE Advanced Type)

Question: Find the range of \(f(x) = \frac{x-1}{x-2}\) where x ∈ ℝ - {2}

Solution:

Method: Solve for x in terms of y

Let y = f(x)

y = \frac{x-1}{x-2}

\[y(x-2) = x-1\]

\[yx - 2y = x - 1\]

\[yx - x = 2y - 1\]

\[x(y-1) = 2y - 1\]

x = \frac{2y-1}{y-1}

For x to be real and defined:

y - 1 ≠ 0

y ≠ 1

Range = \mathbb{R} - \{1\}

💡 Quick Domain-Range Table

Function	Domain	Range
f(x) = x²	ℝ	[0, ∞)
f(x) = √x	[0, ∞)	[0, ∞)
f(x) = \|x\|	ℝ	[0, ∞)
f(x) = 1/x	ℝ - {0}	ℝ - {0}
f(x) = eˣ	ℝ	(0, ∞)
f(x) = ln x	(0, ∞)	ℝ
f(x) = sin x	ℝ	[-1, 1]
f(x) = cos x	ℝ	[-1, 1]
f(x) = tan x	ℝ - {π/2 + nπ}	ℝ

Composite Functions

Composite functions form the basis of many advanced topics in calculus and algebra. This is a frequently asked topic in both JEE Main and Advanced.

5.1 Definition and Notation

Composite Function

Let f: A → B and g: B → C be two functions. The composite function gof (read as "g of f") is defined as:

(g \circ f)(x) = g(f(x))

Important: gof exists only if Range(f) ⊆ Domain(g)

Example:

If f(x) = x² and g(x) = x + 1, then:

(g \circ f)(x) = g(f(x)) = g(x^2) = x^2 + 1

(f \circ g)(x) = f(g(x)) = f(x+1) = (x+1)^2

⚠️ Key Point

Composition is NOT commutative: gof ≠ fog in general

Always check the order carefully in problems!

5.2 Properties of Composite Functions

Important Properties

1. Associativity

(h \circ g) \circ f = h \circ (g \circ f)

2. Identity Function Property

f \circ I = I \circ f = f

3. If f and g are one-one, then gof is one-one

4. If f and g are onto, then gof is onto

5. If f and g are bijective, then gof is bijective

📝 Solved Example 6

Question: If f(x) = 2x + 3 and g(x) = x² - 1, find (fog)(x) and (gof)(x).

Solution:

Finding (fog)(x) = f(g(x)):

\[f(g(x)) = f(x^2 - 1)\]

Replace x in f(x) with (x² - 1):

\[= 2(x^2 - 1) + 3\]

\[= 2x^2 - 2 + 3\]

(f \circ g)(x) = 2x^2 + 1

Finding (gof)(x) = g(f(x)):

\[g(f(x)) = g(2x + 3)\]

Replace x in g(x) with (2x + 3):

\[= (2x + 3)^2 - 1\]

\[= 4x^2 + 12x + 9 - 1\]

(g \circ f)(x) = 4x^2 + 12x + 8

Note: (fog)(x) ≠ (gof)(x) ✓ Composition is not commutative!

📝 Solved Example 7 (JEE Advanced Type)

Question: If f: ℝ → ℝ is defined by f(x) = 3x - 5, find fofof(x).

Solution:

Step 1: Find (fof)(x)

(f \circ f)(x) = f(f(x)) = f(3x - 5)

\[= 3(3x - 5) - 5 = 9x - 15 - 5\]

(f \circ f)(x) = 9x - 20

Step 2: Find (fofof)(x) = f((fof)(x))

f((f \circ f)(x)) = f(9x - 20)

\[= 3(9x - 20) - 5\]

\[= 27x - 60 - 5\]

(f \circ f \circ f)(x) = 27x - 65

Inverse Functions

Inverse functions are one of the most important concepts in JEE Mathematics. They connect directly to inverse trigonometric functions, which are heavily tested.

6.1 Definition of Inverse Function

Definition

Let f: A → B be a bijective function. The inverse function f⁻¹: B → A is defined by:

f^{-1}(y) = x \iff f(x) = y

Key Properties:

f(f^{-1}(x)) = x \text{ and } f^{-1}(f(x)) = x

⚠️ Condition for Existence

Inverse function exists if and only if f is bijective (both one-one and onto)

If f is not one-one → multiple x-values give same y → inverse not possible
If f is not onto → some y-values have no pre-image → inverse undefined for those y

6.2 How to Find Inverse Function

Step-by-Step Method

Check if f is bijective (necessary condition)
Write y = f(x)
Solve for x in terms of y to get x = g(y)
Replace y with x to get f⁻¹(x) = g(x)

📝 Solved Example 8

Question: Find the inverse of f(x) = 3x + 5.

Solution:

Step 1: Check if bijective

f(x) = 3x + 5 is a linear function with non-zero slope

✓ It is bijective (one-one and onto)

Step 2: Write y = f(x)

\[y = 3x + 5\]

Step 3: Solve for x in terms of y

\[y - 5 = 3x\]

x = \frac{y-5}{3}

Step 4: Replace y with x

f^{-1}(x) = \frac{x-5}{3}

Verification:

f(f^{-1}(x)) = f\left(\frac{x-5}{3}\right) = 3 \cdot \frac{x-5}{3} + 5 = x-5+5 = x

✓ Verified!

📝 Solved Example 9 (JEE Main Type)

Question: Find the inverse of \(f(x) = \frac{2x+3}{5x-2}\), x ≠ 2/5.

Solution:

Step 1: Write y = f(x)

y = \frac{2x+3}{5x-2}

Step 2: Solve for x

\[y(5x-2) = 2x+3\]

\[5xy - 2y = 2x + 3\]

\[5xy - 2x = 2y + 3\]

\[x(5y - 2) = 2y + 3\]

x = \frac{2y + 3}{5y - 2}

Step 3: Replace y with x

f^{-1}(x) = \frac{2x + 3}{5x - 2}, \quad x \neq \frac{2}{5}

Note: f(x) = f⁻¹(x) in this case! This is called a self-inverse function or involution.

6.3 Properties of Inverse Functions

Important Properties

1. Domain and Range Swap

Domain(f⁻¹) = Range(f)

Range(f⁻¹) = Domain(f)

2. Graph Symmetry

Graph of f⁻¹ is reflection of graph of f about the line y = x

3. Inverse of Inverse

(f^{-1})^{-1} = f

4. Inverse of Composite Function

(g \circ f)^{-1} = f^{-1} \circ g^{-1}

Note: Order reverses!

💡 JEE Quick Trick

To check if you found inverse correctly:

Verify: f(f⁻¹(x)) = x
Verify: f⁻¹(f(x)) = x
Check: Domain of f⁻¹ = Range of f

Special Functions

These special functions appear frequently in JEE problems and have unique properties that you must memorize for quick problem-solving.

7.1 Modulus Function (Absolute Value)

Definition

|x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}

Properties:

Domain: ℝ, Range: [0, ∞)
Even function: |−x| = |x|
|xy| = |x||y|
|x/y| = |x|/|y|, y ≠ 0
||x| − |y|| ≤ |x ± y| ≤ |x| + |y|

7.2 Signum Function

Definition

\text{sgn}(x) = \begin{cases} 1 & \text{if } x > 0 \\ 0 & \text{if } x = 0 \\ -1 & \text{if } x < 0 \end{cases}

Properties:

Domain: ℝ, Range: {−1, 0, 1}
sgn(x) = x/|x| for x ≠ 0
Odd function: sgn(−x) = −sgn(x)

7.3 Greatest Integer Function (Floor Function)

Definition

The greatest integer function [x] (or ⌊x⌋) gives the greatest integer less than or equal to x.

Examples:

[2.7] = 2
[−2.3] = −3 (not −2!)
[5] = 5
[−5] = −5

Properties:

Domain: ℝ, Range: ℤ (integers)
[x] ≤ x < [x] + 1
[x + n] = [x] + n, where n ∈ ℤ
[x] + [−x] = 0 if x ∈ ℤ, else −1

📝 Solved Example 10 (JEE Advanced Type)

Question: Solve: [x] + [2x] = 3

Solution:

Let x = n + f where n = [x] and 0 ≤ f < 1 (fractional part)

Case 1: 0 ≤ f < 0.5

Then 2x = 2n + 2f where 0 ≤ 2f < 1

So [2x] = 2n

\[[x] + [2x] = n + 2n = 3n = 3\]

Therefore n = 1

x ∈ [1, 1.5)

Case 2: 0.5 ≤ f < 1

Then 2x = 2n + 2f where 1 ≤ 2f < 2

So [2x] = 2n + 1

\[[x] + [2x] = n + 2n + 1 = 3n + 1 = 3\]

Therefore 3n = 2, which gives n = 2/3 (not an integer!)

No solution in this case

\text{Answer: } x \in [1, 1.5)

7.4 Fractional Part Function

Definition

\{x\} = x - [x]

Properties:

Domain: ℝ, Range: [0, 1)
0 ≤ {x} < 1 for all x
{x} = 0 if and only if x is an integer
{x + n} = {x} for any integer n

⚠️ Common Mistakes with Special Functions

[−2.3] = −3, NOT −2 (take greatest integer ≤ x)
|x| = 3 has TWO solutions: x = 3 and x = −3
[x] + [−x] = 0 only if x is an integer
Greatest integer function is NOT continuous

Advanced Topics

8.1 Even and Odd Functions

Even Function

\[f(-x) = f(x)\]

Graph symmetric about y-axis

Examples:

f(x) = x²
f(x) = cos x
f(x) = |x|

Odd Function

\[f(-x) = -f(x)\]

Graph symmetric about origin

Examples:

f(x) = x³
f(x) = sin x
f(x) = x/(1+x²)

Important Properties

• Even + Even = Even

• Odd + Odd = Odd

• Even × Even = Even

• Odd × Odd = Even

• Even × Odd = Odd

• Any function f(x) can be written as: f(x) = [f(x) + f(−x)]/2 + [f(x) − f(−x)]/2

where first term is even and second is odd

8.2 Periodic Functions

Definition

A function f(x) is called periodic if there exists a positive number T such that:

f(x + T) = f(x) \text{ for all } x

The smallest such T is called the fundamental period.

Standard Periods:

sin x, cos x: Period = 2π
tan x, cot x: Period = π
|sin x|, |cos x|: Period = π
sin²x, cos²x: Period = π

8.3 Important Theorems

Theorem 1: Uniqueness of Inverse

If f is bijective, then f⁻¹ is unique and bijective

Theorem 2: Composition of Bijections

If f and g are bijective, then gof is bijective and (gof)⁻¹ = f⁻¹og⁻¹

Theorem 3: Intermediate Value Property

If f is continuous on [a,b] and k is between f(a) and f(b), then there exists c ∈ (a,b) such that f(c) = k

🎯 JEE Strategy for Relations and Functions

Time allocation: Spend 3-4 days mastering this chapter (15-20% JEE weightage)
Focus areas: Domain-range (30%), Composite & Inverse (40%), Types of functions (30%)
Practice: Minimum 100 problems covering all function types
Shortcuts: Memorize all standard domains and ranges
Graph skills: Learn to sketch modulus, signum, and greatest integer quickly

📝 Previous Year Questions Analysis

JEE Main (Last 5 Years)

✓ Domain and Range: 35%
✓ Composite Functions: 25%
✓ Inverse Functions: 20%
✓ Types of Functions: 15%
✓ Relations: 5%

JEE Advanced (Last 5 Years)

✓ Composite & Inverse: 40%
✓ Special Functions: 30%
✓ Domain & Range: 20%
✓ Functional Equations: 10%

Weightage Analysis

JEE Main: 12-16 marks (3-4 questions)

JEE Advanced: 15-20 marks (4-5 questions)

Difficulty Level: Easy to Difficult

Time Required: 4-5 hours practice

Download Complete PYQ PDF (2015-2024)

🎯 Practice Problem Set

Level 1: Basic (JEE Main Standard)

If A = {1, 2, 3} and B = {4, 5}, find the number of relations from A to B.
Check if R = {(1,1), (2,2), (3,3), (1,2), (2,1)} is an equivalence relation in {1, 2, 3}.
Find domain and range of f(x) = √(9 - x²).
If f(x) = 2x + 3 and g(x) = x², find (fog)(x) and (gof)(x).
Check if f(x) = x³ is one-one and onto from ℝ to ℝ.
Find the inverse of f(x) = (x - 1)/(x + 2).
Evaluate [2.7] + [−2.7].
Solve |x - 3| = 5.

Level 2: Intermediate (JEE Main/Advanced)

Find domain of f(x) = 1/√(x² - 5x + 6).
Find range of f(x) = (x² + 2)/(x² + 1).
If f(x) = x/(1 + |x|), show that f is bijective from ℝ to (−1, 1).
Find fofof(x) if f(x) = 1/(1 - x).
Solve [x]² - 5[x] + 6 = 0.
If f: ℝ → ℝ is defined by f(x) = 2x³ - 5, find f⁻¹.
Express f(x) = x³ + x as sum of even and odd functions.
Find the period of f(x) = sin(2x) + cos(3x).

Level 3: Advanced (JEE Advanced/Olympiad)

Find domain of f(x) = log(x² - 4x + 3) + √(16 - x²).
If f: (0, ∞) → ℝ is defined by f(x) = ln x, show that f(fog) = f + fog for any g.
Solve the functional equation: f(x + y) = f(x) + f(y) for all x, y ∈ ℝ.
Find all functions f: ℝ → ℝ such that f(x + 1) = f(x) + 1 and f(1/x) = f(x)/x² for x ≠ 0.
If f(x) = {x²} (fractional part), find discontinuities in [0, 2].
Prove that if f and g are periodic with periods T₁ and T₂, then f + g is periodic if T₁/T₂ is rational.
Find the number of bijective functions from {1,2,3,4,5} to itself that satisfy f(f(x)) = x.
Solve ||x - 1| - 2| = 3.

Get Solutions PDF with Step-by-Step Working

Year	JEE Main	JEE Advanced	Topic Focus
2024	4 Questions (16 marks)	5 Questions (18 marks)	Domain-range, Composite functions
2023	3 Questions (12 marks)	4 Questions (15 marks)	Inverse functions, Special functions
2022	3 Questions (12 marks)	4 Questions (16 marks)	One-one onto, Greatest integer

📑 Quick Navigation - Relations and Functions

Relations & Functions JEE Main & Advanced 2025-26

Basic Concepts

1.1 Cartesian Product

Definition

💡 Important Properties

1.2 Relation

Definition

📝 Solved Example 1

1.3 Function

Definition

⚠️ Key Difference: Relation vs Function

Types of Relations

2.1 Empty Relation

2.2 Universal Relation

2.3 Reflexive Relation

2.4 Symmetric Relation

2.5 Transitive Relation

2.6 Equivalence Relation

Most Important for JEE!

📝 Solved Example 2 (JEE Main Type)

💡 Quick Memory Trick

Types of Functions

3.1 One-One Function (Injective)

Definition

3.2 Onto Function (Surjective)

Definition

3.3 Bijective Function (One-One and Onto)

Definition

📝 Solved Example 3 (JEE Main 2023 Type)

3.4 Other Important Functions

Identity Function

Constant Function

Polynomial Function

Rational Function

Domain and Range

4.1 Finding Domain

General Rules for Domain

📝 Solved Example 4

4.2 Finding Range

Methods for Finding Range

📝 Solved Example 5 (JEE Advanced Type)

💡 Quick Domain-Range Table

Composite Functions

5.1 Definition and Notation

Composite Function

⚠️ Key Point

5.2 Properties of Composite Functions

Important Properties

📝 Solved Example 6

📝 Solved Example 7 (JEE Advanced Type)

Inverse Functions

6.1 Definition of Inverse Function

Definition

⚠️ Condition for Existence

6.2 How to Find Inverse Function

Step-by-Step Method

📝 Solved Example 8

📝 Solved Example 9 (JEE Main Type)

6.3 Properties of Inverse Functions

Important Properties

💡 JEE Quick Trick

Special Functions

7.1 Modulus Function (Absolute Value)

Definition

7.2 Signum Function

Definition

7.3 Greatest Integer Function (Floor Function)

Definition

📝 Solved Example 10 (JEE Advanced Type)

7.4 Fractional Part Function

Definition

⚠️ Common Mistakes with Special Functions

Advanced Topics

8.1 Even and Odd Functions

Even Function

Odd Function

Important Properties

8.2 Periodic Functions

Definition