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Chapter 1  ·  Class XII Mathematics

Refining the Map Between Sets

Relations and Functions

From "Is It a Function?" to "Is It Invertible?" — The Foundation for All of Class XII

📖 Read Full Notes ⚡ Formula Capsules

Chapter Snapshot

9Concepts
8Formulae
4–6%Exam Weight
1–2Avg Q's
ModerateDifficulty

Why This Chapter Matters for Entrance Exams

CBSEJEE Main

CBSE Boards regularly ask a full question on proving a relation is an equivalence relation, or checking whether a function is one-one/onto/invertible. JEE Main tests this more sparingly, but the bijectivity concept underlies inverse trigonometric functions (Ch 2) and is quietly assumed in every later chapter.

Key Concept Highlights

Empty & Universal Relations
Reflexive, Symmetric, Transitive Relations
Equivalence Relations
Equivalence Classes
One-One (Injective) Functions
Onto (Surjective) Functions
Bijective Functions
Composition of Functions
Invertible Functions

Important Formula Capsules

$\mathrm{Reflexive:\ (a,a) \in R\ \forall\ a \in A}$
$\mathrm{Symmetric:\ (a,b)\in R \Rightarrow (b,a)\in R}$
$\mathrm{Transitive:\ (a,b),(b,c)\in R \Rightarrow (a,c)\in R}$
$f\ \text{is one-one} \iff f(x_1)=f(x_2) \Rightarrow x_1=x_2$
$f\ \text{is onto} \iff \mathrm{Range}(f) = \mathrm{Codomain}$
$f\ \text{is invertible} \iff f\ \text{is bijective}$
$(g \circ f)^{-1} = f^{-1} \circ g^{-1}$
$(f^{-1})^{-1} = f$

What You Will Learn

Navigate to Chapter Resources

📖 Full Notes Complete NCERT Coverage 🧠 MCQ Practice Topic-wise Questions 🎯 PYQ Bank JEE · CBSE · BITSAT 📘✔️ Exercises Step by step Solutions to textbook exercises ✔️❌ True / False Concept-based True & False questions

🏆 Exam Strategy & Preparation Tips

CBSE loves "check if R is an equivalence relation" and "show f is a bijection" as full 5-mark proofs — practise the standard proof templates until they are automatic. Do not skip this chapter even though its own JEE weightage is small: sloppy function fundamentals cost marks in every later calculus chapter. Time investment: 2 days.

Chapter 1 · CBSE · Class XII
🔗

Relation

Relations and Functions NCERT Class 12 Mathematics Class 12 Maths Chapter 1 Relation Function Types of Relations Reflexive Relation Symmetric Relation Transitive Relation Equivalence Relation Types of Functions One-One Function Many-One Function Onto Function Into Function Bijective Function Injective Function Surjective Function Composition of Functions Composite Function Identity Function Invertible Function Inverse Function Domain Codomain Range NCERT Solutions CBSE Class 12 Maths JEE Main Mathematics CUET Mathematics Board Exam Preparation
📋
Table of Contents
16 sections
🗺️ Overview
The concept of a relation in Mathematics is derived from the everyday meaning of the word "relation", where two objects are said to be related if there exists some identifiable connection between them. In mathematics, this idea is represented rigorously using the language of sets and ordered pairs.

Relations form the foundation of higher mathematics. They are extensively used in Algebra, Coordinate Geometry, Calculus, Graph Theory, Computer Science, Artificial Intelligence, Database Management Systems, Logic, Statistics and Discrete Mathematics. Understanding relations is essential before studying functions because every function is a relation, but every relation is not a function.
📘 Definition
📌 Important Note
💡 Cartesian Product
🗒️ Components Of A Relation
Term Meaning
Domain Set of all first components appearing in the relation
Codomain Entire destination set
Range Set of second components actually related
🔢 Formula
✏️ Illustrative Example
Let $A=\{1,2,3\}$ and $B=\{4,5,6\}$ Consider
\[R=\{(1,4),(2,5),(2,6)\}\]
Then
  • Domain = {1,2}
  • Codomain = {4,5,6}
  • Range = {4,5,6}
🗂️ Types of Relations
1. Empty Relation (Void Relation)
Definition
A relation \(R\) on a set \(A\) is called an empty relation if no element of \(A\) is related to any element of \(A\).
\[R=\phi\subseteq A\times A\]
Key Characteristics
  • No ordered pair exists.
  • Contains zero elements.
  • Smallest possible relation.
  • It is a subset of every Cartesian product.
Example 1
Let
\[A=\{1,2,3\}\]
Define
\[R=\{(a,b):a>b\text{ and }a < b\}\]
Since no ordered pair can satisfy both conditions simultaneously,
\[R=\phi\]
Example 2
Let
\[A=\{2,4,6\}\]
Define relation
\[aRb\iff a+b=1\]
No pair satisfies this condition.

Therefore,
\[R=\phi\]
Quick Observation
Every empty relation is a subset of every relation.
2. Universal Relation
Definition
A relation $R$ on a set $A$ is called a universal relation if every element of $A$ is related to every element of $A$.
\[R=A\times A\]
Key Characteristics
  • Contains every possible ordered pair.
  • Largest possible relation on the set.
  • Total number of ordered pairs equals $n(A)^2$.
Formula
If $n(A)=n$ then
\[n(R)=n^2\]
Example
Let $A=\{1,2\}$ Then
\[ A\times A= \{ (1,1), (1,2), (2,1), (2,2) \} \]
Hence $R=A\times A$ is the universal relation.
⚖️ Comparison of Empty and Universal Relation
Property Empty Relation Universal Relation
Definition No ordered pair All ordered pairs
Notation $\phi$ $A\times A$
Number of pairs 0 $n^2$
Smallest/Largest Smallest Largest
🔢 Important Formulae
✏️ Example
Solved Example
1
Question
Let $A=\{1,2,3\}$ Find the number of ordered pairs in the universal relation.
Universal relation contains every ordered pair of the Cartesian product.
  1. 1
    Find number of elements.
  2. 2

    Use \(n^2\)

\[n=3\]
Therefore,
\[n(R)=3^2=9\]
9 ordered pairs.
2
Question
How many relations are possible on a set containing three elements?
Total relations \(=2^{n^2}\)
\[n=3\]
\[2^{3^2}=2^9=512\]
512 relations.
📐 Derivation of Number of Relations
Suppose
\[n(A)=m,\qquad n(B)=n\]
Then Cartesian product contains \(mn\) ordered pairs.
Every ordered pair has only two possibilities:
  • Included in the relation
  • Not included in the relation
Therefore,
\[\underbrace{2\times2\times2\times\cdots\times2}_{mn\text{ times}} =2^{mn}\]
Hence,
\[\boxed{\text{Number of relations}=2^{mn}}\]
⚡ Exam Tip
❌ Common Mistakes
  • Writing relation as an element instead of a subset.
  • Using curly braces incorrectly while writing ordered pairs.
  • Confusing universal relation with universal set.
  • Writing range equal to codomain without verification.
  • Using \(2^n\) instead of \(2^{n^2}\) for relations on a set.
📋 CBSE Competency-Based (Case Study/HOTS)

In a school, every student is connected to every other student through an online discussion platform. Let the set of students be

\[ A=\{S_1,S_2,S_3,S_4\} \]

  1. What type of relation does this represent?
  2. How many ordered pairs are present?
  3. How many different relations can be formed on this set?
Solution

Since every student is related to every student,

\[ R=A\times A \]

Therefore, it is a Universal Relation.

Number of ordered pairs:

\[ 4^2=16 \]

Number of possible relations:

\[ 2^{16}=65536 \]

🌟 Significance
  • Fundamental concept for Functions, Binary Operations and Mapping.
  • Frequently asked in CBSE one-mark, competency-based and case-study questions.
  • Very important for JEE Main, NDA, CUET, BITSAT and State Engineering Entrance Examinations.
  • Questions based on counting the number of relations appear regularly in competitive examinations.
  • Forms the theoretical base for Reflexive, Symmetric, Transitive and Equivalence Relations studied later in the chapter.
🔗

Example 1

📋
Table of Contents
9 sections
❓ Question
Let A be the set of all students of a boys' school. Show that the relation
\[R=\{(a,b):a\text{ is sister of }b\}\]
is the empty relation, and the relation
\[R'=\{(a,b):\text{the difference between the heights of }a\text{ and }b\text{ is less than }3\text{ metres}\}\]
is the universal relation.
💡 Concept used
🗺️ Roadmap

  1. Identify the underlying set.

  2. Examine the condition defining the first relation.

  3. Determine whether any ordered pair satisfies the condition.

  4. Repeat the process for the second relation.

  5. Conclude whether each relation is empty or universal.

🧩 Solution
Given: The given set \(A=\{\text{all students of a boys' school}\}\) contains only male students.
Solution to Part (1)
  1. Consider the relation
    \[ R=\{(a,b):a\text{ is sister of }b\}. \]
  2. Since every student in the school is a boy, no student can be the sister of another student. Therefore, there does not exist even a single ordered pair satisfying the given condition.
    Hence,
    \[R=\phi.\]
Therefore, R is an Empty Relation.
Solution to Part (2)
  1. Now consider the relation
    \[ R'=\{(a,b):\text{difference between the heights of }a\text{ and }b\text{ is less than }3\text{ metres}\}.\]
  2. In a school, the height difference between any two students can never be as large as 3 metres. Thus, every possible ordered pair satisfies the given condition.
  3. Hence,
    \[R'=A\times A\]
Therefore, R' is the Universal Relation.
🌟 Significance
Why This Question is Important?
  • Illustrates how real-life situations can be translated into mathematical relations.
  • Tests conceptual understanding instead of calculations.
  • Frequently asked as a CBSE competency-based or one-mark conceptual question.
  • Develops the ability to distinguish between Empty and Universal Relations using logical reasoning.
⚡ Exam Tip
❌ Common Mistakes
  • Assuming a relation is empty without checking whether at least one ordered pair satisfies the condition.
  • Confusing Universal Relation with Universal Set.
  • Writing \(R=A\) instead of the correct notation \(R=A\times A\).
📋 Higher Order Thinking (HOTS)
  • Assuming a relation is empty without checking whether at least one ordered pair satisfies the condition.
  • Confusing Universal Relation with Universal Set.
  • Writing \(R=A\) instead of the correct notation \(R=A\times A\).
Higher Order Thinking (HOTS)

Question:

Let

\[ A=\{\text{all employees of a company}\}. \]

Consider the relation

\[ R=\{(a,b):a\text{ is the father of }b\}. \]

Is this relation empty, universal or neither?

Answer:

The relation is neither empty nor universal. Some employees may be father and son, but certainly not every pair satisfies the condition, and the relation is not empty because such pairs may exist.

🔑 Key Takeaway
🔗

Properties of Relations

📋
Table of Contents
8 sections
🗺️ Overview
After understanding different types of relations, the next important step is to study their properties. These properties help us classify relations and form the basis of advanced mathematical concepts such as equivalence relations, partitions of sets, quotient sets and many applications in Computer Science, Artificial Intelligence, Graph Theory and Database Management Systems.

In CBSE Board examinations and competitive entrance examinations like JEE Main, NDA, CUET, BITSAT, conceptual questions based on these properties are frequently asked. Students should therefore understand the meaning of each property rather than memorizing the definitions.
🗂️ Types / Category
  • Reflexive Relation
    Definition
    A relation R on a set A is said to be Reflexive if every element of the set is related to itself.
    Mathematically,
    \[(a,a)\in R,\qquad \forall a\in A\]
    or equivalently,
    \[\forall\; a\in A,\; a\;R\;a.\]
    Meaning,
    Every element must have a self-loop. If even one element is not related to itself, then the relation is not reflexive.
    Example,
    Let $A=\{1,2,3\}$ and
    \[R=\{(1,1),(2,2),(3,3),(1,2),(2,1)\}.\]
    Since every element appears with itself,
    \[(1,1),(2,2),(3,3)\in R,\]
    therefore R is reflexive.
    Counter Example,
    Consider
    \[R=\{(1,1),(2,2)\}\]
    on
    \[A=\{1,2,3\}.\]
    Since $(3,3)\notin\; R,$ the relation is not reflexive.
    Shortcut Test
    • Count the elements of the set.
    • Check whether every diagonal pair \((a,a)\) is present.
    • If all diagonal pairs are present, the relation is reflexive.


    REFLEXIVE PROPERTY OF A RELATION Every element a ∈ A is related to itself: (a, a) ∈ R Set A (a, a) a (b, b) b (c, c) c Key Takeaway: For a relation to be reflexive, EVERY node must have a self-loop. ∀ x ∈ A, x R x
  • Symmetric Relation
    Definition
    A relation R on a set A is called Symmetric if whenever one element is related to another, the reverse relation also exists.
    Mathematically,
    \[(a_1,a_2)\in\; R\Rightarrow (a_2,a_1)\in\; R,\qquad\forall\; a_1,a_2\in\; A.\]
    or simply,
    \[a\;R\;b\Rightarrow b\;R\;a.\]
    Meaning,
    The direction of every ordered pair can be reversed.
    Example,
    Let $A=\{1,2,3\}$ and
    \[R=\{(1,2),(2,1),(2,3),(3,2)\}.\]
    Every ordered pair has its reverse.

    Therefore,
    R is symmetric.
    Counter Example,
    If $(2,3)\in\; R$ but $(3,2)\notin\; R,$ then the relation is not symmetric.
    Shortcut Test
    • Pick any ordered pair.
    • Immediately search for its reverse.
    • If every reverse pair exists, the relation is symmetric.


    SYMMETRIC PROPERTY OF A RELATION If (a, b) ∈ R, then (b, a) ∈ R must also exist Set A c d (c, c) (a, b) (b, a) a b ⇔ Key Takeaway: Relations are two-way streets. Every directed edge has an opposite arrow return path. a R b ⇒ b R a
  • Transitive Relation
    Definition
    A relation R on a set A is called Transitive if \((a_1,a_2)\in R\) and \((a_2,a_3)\in R\) together imply
    \[(a_1,a_3)\in\; R,\qquad\forall\; a_1,a_2,a_3\in\; A.\]
    or simply,
    \[a\;R\;b,\;b\;R\;c\Rightarrow a\;R\;c.\]
    Meaning
    A transitive relation allows us to "skip the middle element." If
    \[a\rightarrow b,\qquad b\rightarrow c,\]
    then we can conclude
    \[a\rightarrow c.\]
    Example
    Let $A=\{1,2,3\}$ and
    \[R=\{(1,2),(2,3),(1,3)\}.\]
    Since \((1,2)\in R\) and \((2,3)\in R\), we also have \((1,3)\in R\).

    Therefore, R is transitive.
    Counter Example
    If \((1,2)\in R\) and \((2,3)\in R\) but \((1,3)\notin R\), then the relation is not transitive.

    TRANSITIVE PROPERTY OF A RELATION If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R must also exist Set A (a, b) (b, c) (a, c) a b c Key Takeaway: Transitivity creates a direct shortcut anytime a two-step path exists. (a R b ∧ b R c) ⇒ a R c
📌 Equivalence Relation
📘 Definition
A relation R on a set A is called an Equivalence Relation if it satisfies all three properties simultaneously:
  • Reflexive
  • Symmetric
  • Transitive
Mathematically,
\(R\text{ is an equivalence relation}\) if and only if $R$ is reflexive, symmetric, and transitive.
EQUIVALENCE RELATION A relation R is an Equivalence Relation if and only if it satisfies all three properties: Set A (Equivalence Class) a b c 1. Reflexive ∀a, (a, a) ∈ R (Self-loops on all nodes) 2. Symmetric (a, b) ∈ R ⇒ (b, a) ∈ R (All connections two-way) 3. Transitive (a, b) & (b, c) ∈ R ⇒ (a, c) ∈ R (Direct shortcuts exist)
1
Example
Let
\[R=\{(a,b):a\equiv b\pmod 5\}\]
This relation is reflexive, symmetric, and transitive. Therefore, R is an equivalence relation.
🗒️ Comparison of the Three Properties
Property Condition Easy Memory Trick
Reflexive \((a,a)\in R\) Self relation
Symmetric \((a,b)\Rightarrow(b,a)\) Reverse relation
Transitive \((a,b),(b,c)\Rightarrow(a,c)\) Skip the middle
✏️ Example
Solved Example
2
Question
Let \(A=\{1,2,3\}\) and \(R=\{(1,1),(2,2),(3,3),(1,2),(2,1)\}\). Determine whether the relation is reflexive, symmetric and transitive.
To determine whether the relation is reflexive, symmetric, and transitive, we will check each property one by one.
  1. 1
    Check all self-pairs.
  2. 2
    Check reverse pairs.
  3. 3
    Check transitive chains.
  1. Check Reflexivity
    The relation is reflexive if it contains all self-pairs.
    In this case, the self-pairs are (1,1), (2,2), and (3,3).
    Since all self-pairs are present in R, the relation is reflexive.
  2. Check Symmetry
    The relation is symmetric if for every pair (a,b) in R, the pair (b,a) is also in R.
    In this case, we have (1,2) and (2,1) in R.
    Since both pairs are present, the relation is symmetric.
  3. Check Transitivity
    The relation is transitive if for every pair (a,b) and (b,c) in R, the pair (a,c) is also in R.
    In this case, we have (1,2) and (2,1) in R, but (1,1) is also in R.
    Therefore, the relation is transitive.
Since the relation is reflexive, symmetric, and transitive, we conclude that R is an equivalence relation.
⚡ Exam Tip
❌ Common Mistakes
  • Checking only one ordered pair for symmetry.
  • Ignoring one element while testing reflexivity.
  • Testing transitivity incompletely.
  • Assuming every symmetric relation is transitive.
  • Confusing reflexive with symmetric.
📋 CBSE Competency-Based (HOTS)

Question

Let

\[ A=\{1,2,3\} \]

and

\[ R=\{(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2),(1,3),(3,1)\}. \]

Determine whether R is an Equivalence Relation.

Answer

The relation contains all self-pairs, every ordered pair has its reverse, and every possible transitive requirement is satisfied. Hence, the relation is Reflexive, Symmetric and Transitive.

Therefore,

\[ \boxed{R\text{ is an Equivalence Relation}.} \]

🌟 Significance
  • One of the most important topics in the Relations and Functions chapter.
  • Frequently appears in CBSE competency-based and case-study questions.
  • Regularly tested in JEE Main, NDA, CUET, BITSAT and State Engineering Entrance examinations.
  • Forms the foundation for Partition of Sets and Equivalence Classes in higher mathematics.
🔗

Example 2

📋
Table of Contents
12 sections
❓ Question
Let T be the set of all triangles in a plane. A relation R is defined on T by
\[R=\{(T_1,T_2):T_1\text{ is congruent to }T_2\}.\]
Show that R is an Equivalence Relation.
💡 Concept
📖 Theory
🗺️ Roadmap
  1. Check whether every triangle is related to itself.
  2. Check whether congruence works in both directions.
  3. Check whether congruence can be transferred from one triangle to another.
  4. Conclude whether all three properties are satisfied.
🧩 Solution
Reflexive Property
  1. Every triangle is congruent to itself because all its corresponding sides and angles are equal.
    Therefore,
    \[(T_1,T_1)\in\; R,\qquad \forall\; T_1\in T.\]
Hence, R is Reflexive.
Symmetric Property
  1. Suppose
    \[(T_1,T_2)\in\; R.\]
    then
    \[T_1\cong T_2.\]
    Since congruence is reversible,
    \[T_2\cong T_1.\]
    Therefore,
    \[(T_2,T_1)\in\; R.\]
Hence, R is Symmetric.
Transitive Property
  1. Suppose
    \[(T_1,T_2)\in R\]
    and
    \[(T_2,T_3)\in R.\]
    Then
    \[T_1\cong T_2\]
    and
    \[T_2\cong T_3.\]
    Since two triangles congruent to the same triangle are congruent to each other,
    \[T_1\cong T_3.\]
    Therefore,
    \[(T_1,T_3)\in\; R.\]
Hence, R is Transitive.
Since the relation is
  • Reflexive
  • Symmetric
  • Transitive
it follows that
\[\boxed{R\text{ is an Equivalence Relation}.}\]
🔬 Alternative Proof (One-Line Method for Boards)

Reflexive:

\[T_1\cong T_1\]

Symmetric:

\[T_1\cong T_2\Rightarrow T_2\cong T_1\]

Transitive:

\[T_1\cong T_2,\;T_2\cong T_3\Rightarrow T_1\cong T_3\]

Therefore,

\[\boxed{R\text{ is an Equivalence Relation}.}\]

🎨 SVG Diagram
Geometrical Illustration
EQUIVALENCE RELATION: CONGRUENCE (≅) Two triangles are related if they are congruent ($T_i \cong T_j$ via SSS, SAS, ASA, AAS, RHS) Set of Triangles (S) T₁ ≅ T₁ T₂ ≅ T₂ T₃ ≅ T₃ T₁ ≅ T₂ T₂ ≅ T₁ T₂ ≅ T₃ T₃ ≅ T₂ Transitive Link: T₁ ≅ T₃ T₁ T₂ T₃ 1. Reflexive T₁ ≅ T₁ Any triangle is congruent to itself 2. Symmetric T₁ ≅ T₂ ⇒ T₂ ≅ T₁ Congruence works both ways 3. Transitive (T₁ ≅ T₂ ∧ T₂ ≅ T₃) ⇒ T₁ ≅ T₃ Chain reaction via SSS/SAS/etc.
👁️ Observation
⚡ Exam Tip
❌ Common Mistakes
  • Proving only one or two properties and concluding that the relation is an equivalence relation.
  • Using the symbol "=" instead of the congruence symbol "\(\cong\)".
  • Skipping the logical implication while proving symmetry and transitivity.
  • Not writing the final conclusion explicitly.
📋 CBSE Competency-Based (HOTS)

Question

Let S be the set of all squares in a plane. Define the relation

\[R=\{(S_1,S_2):S_1\text{ is congruent to }S_2\}.\]

Is the relation an equivalence relation?

Answer

Yes.

  • Every square is congruent to itself (Reflexive).
  • If one square is congruent to another, then the converse is also true (Symmetric).
  • If the first square is congruent to the second and the second to the third, then the first is congruent to the third (Transitive).

Hence,

\[\boxed{R\text{ is an Equivalence Relation}.}\]

🌟 Significance
  • This is one of the standard proofs asked in CBSE Board examinations.
  • Frequently appears in competency-based and case-study questions.
  • Understanding this example makes it easier to prove equivalence relations involving equality, similarity, parallelism and modular arithmetic.
  • The concept is also useful in higher mathematics, computer science, geometry and abstract algebra.
🔑 Key Takeaway
🔗

Example 3

📋
Table of Contents
11 sections
❓ Question
Let L be the set of all lines in a plane. A relation R is defined on L by
\[R=\{(L_1,L_2):L_1\text{ is perpendicular to }L_2\}.\]
Show that R is Symmetric but neither Reflexive nor Transitive.
💡 Concept
📖 Theory Behind the Question
🗺️ Roadmap
  1. Test whether every line is perpendicular to itself.
  2. Verify whether the order of perpendicular lines can be reversed.
  3. Check whether perpendicularity is preserved through a chain of three lines.
  4. State the final conclusion.
🧩 Solution
Reflexive Property
  1. For a relation to be reflexive, every line must be perpendicular to itself.
    However, a line cannot intersect itself at an angle of $90^\circ.$
    Therefore,
    \[(L_1,L_1)\notin\; R.\]
Hence, R is not Reflexive.
Symmetric Property
  1. Suppose
    \[(L_1,L_2)\in\; R.\]
    Then
    \[L_1\perp L_2.\]
    Since perpendicularity is mutual,
    \[L_2\perp L_1.\]
    Therefore,
    \[(L_2,L_1)\in\; R.\]
Hence, R is Symmetric.
Transitive Property
  1. Suppose
    \[(L_1,L_2)\in\; R\]
    and
    \[(L_2,L_3)\in\; R.\]
    Then
    \[L_1\perp L_2\]
    and
    \[L_2\perp L_3.\]
    Since both L1 and L3 are perpendicular to the same line L2, they are parallel to each other. Therefore,
    \[L_1\parallel L_3.\]
    Hence,
    \[(L_1,L_3)\notin\; R\]
    Thus,
    \[(L_1,L_2)\in R,\qquad(L_2,L_3)\in R,\]
    but
    \[(L_1,L_3)\notin\ R.\]
Therefore, R is not Transitive.
Property Result Reason
Reflexive ✗ No A line cannot be perpendicular to itself.
Symmetric ✓ Yes Perpendicularity works in both directions.
Transitive ✗ No Two lines perpendicular to the same line become parallel.

Hence,

\[ \boxed{\text{The relation is Symmetric but neither Reflexive nor Transitive.}} \]

🎨 SVG Diagram
Geometrical Illustration
LINES PERPENDICULAR TO A COMMON LINE Demonstrates that if L₁ ⊥ L₂ and L₃ ⊥ L₂, then L₁ ∥ L₃ L₁ L₃ L₂ L₁ ∥ L₃
🔬 Alternative Proof (Board Examination)

Reflexive:

\[(L_1,L_1)\notin R.\]

Therefore, the relation is not reflexive.

Symmetric:

\[L_1\perp L_2\Rightarrow L_2\perp L_1.\]

Hence, the relation is symmetric.

Transitive:

\[L_1\perp L_2,\quad L_2\perp L_3\]

implies

\[L_1\parallel L_3.\]

Therefore,

\[(L_1,L_3)\notin R.\]

Hence, the relation is not transitive.

⚡ Exam Tip
❌ Common Mistakes
  • Thinking that every symmetric relation is also transitive.
  • Writing that a line is perpendicular to itself.
  • Ignoring that two lines perpendicular to the same line are parallel.
  • Concluding that the relation is an equivalence relation simply because it is symmetric.
📋 CBSE Competency-Based (HOTS)

Question

Let L be the set of all lines in a plane and define the relation

\[R=\{(L_1,L_2):L_1\parallel L_2\}.\]

Determine whether the relation is reflexive, symmetric and transitive.

Answer
  • A line is parallel to itself, so the relation is Reflexive.
  • If \(L_1\parallel L_2\), then \(L_2\parallel L_1\), so it is Symmetric.
  • If \(L_1\parallel L_2\) and \(L_2\parallel L_3\), then \(L_1\parallel L_3\), so it is Transitive.

Hence,

\[\boxed{\text{The relation "is parallel to" is an Equivalence Relation.}}\]

🌟 Significance
  • A classic proof-based question in the CBSE Class 12 Relations and Functions chapter.
  • Frequently asked to test conceptual understanding of reflexive, symmetric and transitive properties.
  • Important for JEE Main, CUET, NDA, BITSAT and other entrance examinations.
  • Helps students understand that each property must be verified independently.
🔑 Key Takeaway
🔗

Example 4

📋
Table of Contents
11 sections
❓ Question
Show that the relation
\[R=\{(1,1),(2,2),(3,3),(1,2),(2,3)\}\]
defined on the set
\[A=\{1,2,3\}\]
is Reflexive but neither Symmetric nor Transitive.
💡 Concept
📖 Theory Behind the Question
🗺️ Roadmap
  1. Check whether every diagonal pair is present.
  2. Check whether every ordered pair has its reverse.
  3. Check every possible transitive chain.
  4. Draw the final conclusion.
🧩 Solution
Reflexive Property
  1. The set is
    \[A=\{1,2,3\}.\]
    For reflexivity, the following ordered pairs must belong to the relation:
    \[(1,1),\quad (2,2),\quad (3,3).\]
    All these ordered pairs are present in
    \[R=\{(1,1),(2,2),(3,3),(1,2),(2,3)\}.\]
    Therefore,
    \[(a,a)\in\; R,\qquad \forall\; a\in\; A.\]
Hence, R is Reflexive.
Symmetric Property
  1. Observe that
    \[(1,2)\in\; R.\]
    If the relation were symmetric, then
    \[(2,1)\in\; R\]
    must also belong to the relation.
    However,
    \[(2,1)\notin\ R.\]
    Therefore,
    \[(1,2)\in\; R\quad\text{but}\quad(2,1)\notin\; R.\]
Hence, R is not Symmetric.
Transitive Property
  1. Notice that
    \[(1,2)\in\; R\]
    and
    \[(2,3)\in R.\]
    For transitivity, $(1,3)$ must also belong to the relation.
    However,
    \[(1,3)\notin\; R.\]
Hence, R is not Transitive.
Property Status Reason
Reflexive ✓ Yes All self-pairs are present.
Symmetric ✗ No \((1,2)\in R\) but \((2,1)\notin R\).
Transitive ✗ No \((1,2),(2,3)\in R\) but \((1,3)\notin R\).

Hence,

\[\boxed{\text{The relation is Reflexive but neither Symmetric nor Transitive.}}\]
🎨 SVG Diagram
CLASSIFYING A RELATION: INDEPENDENT VERIFICATION To fully classify a relation, each of the three fundamental properties must be verified independently. 1. REFLEXIVE a b Self-Loop Check Every element must be related to itself. Rule: ∀a ∈ A (a, a) ∈ R 2. SYMMETRIC a b Reverse-Pair Check Every ordered pair must have its reverse pair. Rule: (a, b) ∈ R ⇒ (b, a) ∈ R 3. TRANSITIVE a b c Shortcut Check If (a,b) and (b,c) exist, (a,c) must also exist. Rule: (a,b) ∧ (b,c) ∈ R ⇒ (a,c) ∈ R SUMMARY FOR RELATION CLASSIFICATION Satisfies all 3 = Equivalence Relation | Fails any 1 = Non-Equivalence Relation
🔬 Alternative Board Examination Solution
Reflexive:
\[(1,1),(2,2),(3,3)\in\; R.\]
Hence, the relation is reflexive.

Symmetric:
\[(1,2)\in\; R\]

but
\[(2,1)\notin\; R.\]
Hence, the relation is not symmetric.

Transitive:
\[(1,2),(2,3)\in\; R\]
but
\[(1,3)\notin\; R.\]
Hence, the relation is not transitive.
⚡ Exam Tip
❌ Common Mistakes
  • Checking only one diagonal pair while testing reflexivity.
  • Assuming that the presence of self-pairs guarantees symmetry.
  • Looking for \((3,1)\) instead of the required pair \((1,3)\) while testing transitivity.
  • Concluding that a relation is an equivalence relation because it is reflexive.
📋 CBSE Competency-Based (HOTS)

Question

Let

\[A=\{1,2,3,4\}\]

and

\[R=\{(1,1),(2,2),(3,3),(4,4),(1,2),(2,1),(2,3)\}.\]

Determine whether the relation is reflexive, symmetric and transitive.

Hint
  • All self-pairs are present, so the relation is reflexive.
  • The pair \((2,3)\) exists but \((3,2)\) does not, so it is not symmetric.
  • Since \((1,2)\) and \((2,3)\) belong to the relation but \((1,3)\) does not, it is not transitive.
🌟 Significance
  • One of the most frequently asked proof-based questions in CBSE Class 12 Relations and Functions.
  • Commonly appears as a competency-based or short-answer question.
  • Important for JEE Main, CUET, NDA, BITSAT and various state entrance examinations.
  • Develops the ability to classify relations systematically using counterexamples.
🔑 Key Takeaway
🔗

Example 5

📋
Table of Contents
11 sections
❓ Question
Show that the relation
\[R=\{(a,b):2\text{ divides }(a-b)\}\]
defined on the set of integers $\mathbb{Z}$ is an Equivalence Relation.
💡 Concept
📖 Theory Behind the Question
🗺️ Roadmap
  1. Verify Reflexive property.
  2. Verify Symmetric property.
  3. Verify Transitive property.
  4. Conclude whether the relation is an Equivalence Relation.
🧩 Solution
Reflexive Property
  1. For every integer
    \[a\in\mathbb Z,\]
    we have
    \[a-a=0\]
    Since
    \[2\mid0,\]
    it follows that
    \[(a,a)\in R,\qquad \forall\; a\in\mathbb Z.\]
Hence, R is Reflexive.
Symmetric Property
  1. Suppose
    \[(a,b)\in\;R.\]
    Then
    \[2\mid(a-b).\]
    Therefore,
    \[a-b=2k\]
    for some integer $k$
    Multiplying both sides by $-1,$
    we get
    \[b-a=-2k=2(-k).\]
    Since
    \[-k\in\;\mathbb Z,\]

    it follows that
    \[2\mid(b-a).\]
    Therefore,
    \[(b,a)\in\; R.\]
Hence, R is Symmetric.
Transitive Property
  1. Suppose $(a,b)\in\; R$ and $(b,c)\in\; R.$ Then
    \[2\mid(a-b)\]
    and
    \[2\mid(b-c).\]
    Therefore, there exist integers
    \[m,\;n\]
    such that
    \[a-b=2m \tag{1}\]
    and
    \[b-c=2n.\tag{2}\]
    Adding equations (1) and (2),
    \[(a-b)+(b-c)=2m+2n.\]
    Hence,
    \[a-c=2(m+n).\]
    Since
    \[m+n\in\;\mathbb Z,\]
    it follows that
    \[2\mid(a-c).\]
    Therefore,
    \[(a,c)\in\; R.\]
Hence, R is Transitive.
Property Status Reason
Reflexive ✓ Yes \(2\mid(a-a)\)
Symmetric ✓ Yes \(2\mid(a-b)\Rightarrow2\mid(b-a)\)
Transitive ✓ Yes Sum of two even integers is even.

Therefore,

\[ \boxed{R\text{ is an Equivalence Relation on }\mathbb Z.} \]

🔬 Alternative Proof (Board Examination)

Reflexive:

\[2\mid(a-a).\]

Hence, the relation is reflexive.

Symmetric:

\[2\mid(a-b)\Rightarrow 2\mid(b-a).\]

Hence, the relation is symmetric.

Transitive:

\[2\mid(a-b),\quad 2\mid(b-c)\]

imply

\[2\mid(a-c).\]

Hence, the relation is transitive.

Therefore,

\[\boxed{R\text{ is an Equivalence Relation}.}\]

⚡ Exam Tip
❌ Common Mistakes
  • Writing \(2\mid(a+b)\) instead of \(2\mid(a-b)\).
  • Skipping the proof of symmetry.
  • Not introducing integers \(m\) and \(n\) while proving transitivity.
  • Concluding without verifying all three properties.
📋 CBSE Competency-Based (HOTS)

Question

Define a relation on

\[ \mathbb Z \]

by

\[ (a,b)\in R \iff 3\mid(a-b). \]

Show that the relation is an equivalence relation.

Hint
  • Prove Reflexive using \(a-a=0\).
  • Prove Symmetric by multiplying the difference by \(-1\).
  • Prove Transitive by adding the two divisibility equations.
  • This proof is identical to the modulo 2 case.
🌟 Significance
  • One of the most important proof-based examples in the Relations and Functions chapter.
  • Frequently asked in CBSE Board examinations and competency-based assessments.
  • Forms the foundation of Congruence Modulo Arithmetic studied in higher mathematics.
  • Very important for JEE Main, CUET, NDA, BITSAT and Olympiad examinations.
🔑 Key Takeaway
🔗

Example 6

📋
Table of Contents
12 sections
❓ Question
Let \(A=\{1,2,3,4,5,6,7\}\) and let the relation \(R=\{(a,b):\text{both }a\text{ and }b\text{ are either odd or even}\}.\) Show that
  • R is an Equivalence Relation.
  • Every element of the subset
    \[\{1,3,5,7\}\]
    is related to every other element of the same subset.
  • Every element of the subset
    \[\{2,4,6\}\]
    is related to every other element of the same subset.
  • No element of
    \[\{1,3,5,7\}\]
    is related to any element of
    \[\{2,4,6\}.\]
💡 Concept
📖 Theory Behind the Question
🗺️ Roadmap
  1. Verify Reflexive property.
  2. Verify Symmetric property.
  3. Verify Transitive property.
  4. Identify the equivalence classes.
  5. Show that elements of different classes are never related.
🧩 Solution
Reflexive Property
  1. Let
    \[a\in A.\]
    Every element is either odd or even.
    Therefore, $a$ and $a$ have the same parity.
    Hence,
    \[(a,a)\in R,\quad\forall\; a\in A.\]
Therefore, R is Reflexive.
Symmetric Property
  1. Suppose $(a,b)\in\; R.$ Then both $a$ and $b$ are either odd or even.
    Clearly, $b$ and $a$ also have the same parity.
    Therefore,
    \[(b,a)\in\; R.\]
Hence, R is Symmetric.
Transitive Property
  1. Suppose $(a,b)\in\; R$ and $(b,c)\in\; R.$
    Then
    • a and b have the same parity.
    • b and c have the same parity.
    Therefore, $a,\;b,\;c$ must all be either odd or all be even.
    Hence,
    \[(a,c)\in\; R.\]
Therefore, R is Transitive.
Property Status
Reflexive ✓ Yes
Symmetric ✓ Yes
Transitive ✓ Yes

Hence,

\[\boxed{R\text{ is an Equivalence Relation}.}\]

📌 Equivalence Classes
🔬 Alternative Board Examination Solution

Reflexive:

Every element has the same parity as itself.Hence,

\[(a,a)\in R.\]

Symmetric:

If

\[(a,b)\in R,\]

then both have the same parity.Therefore,

\[(b,a)\in R.\]

Transitive:

If

\[(a,b)\in R\]

and

\[(b,c)\in R,\]

then all three numbers have the same parity.Hence,

\[(a,c)\in R.\]

Therefore,

\[\boxed{R\text{ is an Equivalence Relation}.}\]

⚡ Exam Tip
❌ Common Mistakes
  • Confusing "odd or even" with "odd and even".
  • Assuming every odd number is related to every even number.
  • Not mentioning the equivalence classes after proving the relation.
  • Ignoring that each element belongs to exactly one equivalence class.
📋 CBSE Competency-Based (HOTS)

Question

Let

\[A=\{1,2,3,4,5,6,7,8\}\]

Define the relation

\[ (a,b)\in R\iff a\text{ and }b\text{ have the same remainder when divided by }3.\]

Show that this relation is an equivalence relation and determine its equivalence classes.

Hint:

The equivalence classes are

\[\{3,6\},\qquad\{1,4,7\},\qquad\{2,5,8\}.\]

🌟 Significance
  • One of the most important examples illustrating the concept of equivalence classes.
  • Frequently asked in CBSE competency-based and case-study questions.
  • Forms the foundation for the theorem: "Every Equivalence Relation partitions a set."
  • Very important for JEE Main, CUET, NDA, BITSAT and higher mathematics.
🔑 Key Takeaway
🔗

Types of Functions

📋
Table of Contents
12 sections
🗺️ Overview
Functions are one of the most fundamental concepts in mathematics. Depending upon the relationship between the domain and the codomain, functions are classified into different types. The three most important types are:
  • Injective (One-One) Function
  • Surjective (Onto) Function
  • Bijective (One-One and Onto) Function
Understanding these functions is essential not only for CBSE Board examinations but also for competitive examinations such as JEE Main, CUET, NDA, BITSAT and higher studies in Linear Algebra, Calculus, Abstract Algebra and Computer Science.
💡 Prerequisite Concepts
🗂️ Types / Category
  • Injective Function (One-One Function)
    Definition
    A function
    \[f:X\rightarrow Y\]
    is called an Injective Function or One-One Function if distinct elements of the domain have distinct images.
    Mathematically,
    \[f(x_1)=f(x_2)\Rightarrow x_1=x_2,\quad\forall\; x_1,x_2\in X.\]
    The logically equivalent (and often more intuitive) statement is
    \[x_1\neq x_2\Rightarrow f(x_1)\neq f(x_2)\]
    Meaning
    No two different inputs can have the same output.
    Each output has at most one pre-image.
    Example 1
    Consider
    \[f:\{1,2,3\}\rightarrow\{2,4,6,8\}\]
    defined by
    \[f(x)=2x\]
    x f(x)
    1 2
    2 4
    3 6
    Since different inputs produce different outputs, the function is Injective.
    Example 2
    \[f(x)=3x+5,\qquad x\in\;\mathbb R\]
    Suppose
    \[f(x_1)=f(x_2)\]
    Then
    \[3x_1+5=3x_2+5\]
    Therefore,
    \[x_1=x_2\]
    Hence,
    \[f(x)=3x+5\]
    is one-one.
    Graphical Interpretation
    An injective function passes the Horizontal Line Test. Every horizontal line intersects its graph at most once. Non-Example (Many-One Function) Consider
    \[f(x)=x^2,\qquad x\in\mathbb R\]
    Here,
    \[f(2)=4=f(-2)\]
    Since
    \[2\neq-2,\]
    the function is Many-One, not Injective.

    INJECTIVE FUNCTION (ONE-TO-ONE) A function f: X → Y is injective if distinct elements in X map to distinct elements in Y Domain (X) x₁ x₂ x₃ Codomain (Y) y₁ y₂ y₃ y₄ f(x) Key Condition: No two elements in X map to the same Y. Unmapped codomain elements (like y₄) are permitted in Injective functions. f(x₁) = f(x₂) ⇒ x₁ = x₂
  • Surjective Function (Onto Function)
    Definition
    A function
    \[f:X\rightarrow Y\]
    is called Surjective (or Onto) if every element of the codomain has at least one pre-image.
    Mathematically,
    \[\forall\; y\in\; Y,\;\exists x\in\; X\text{ such that }f(x)=y\]
    Equivalent Condition,
    A function is onto if
    \[\boxed{\text{Range}=\text{Codomain}}\]
    Meaning,
    Every element of the codomain is "covered" by the function.
    No element of the codomain remains unused.
    Example 1
    Let
    \[X=\{1,2,3\},\quad Y=\{a,b,c\}\]
    and define
    \[f(1)=a,\quad f(2)=b,\quad f(3)=c.\]
    Every element of $Y$ is obtained as an image.
    Hence, $f$ is onto.
    Example 2
    Consider
    \[f:\mathbb R\rightarrow\mathbb R,\quad f(x)=x^3\]
    Given any
    \[y\in\mathbb R,\]
    choose
    \[x=\sqrt[3]{y}\]
    Then
    \[f(x)=y\]
    Therefore,
    \[f(x)=x^3\]
    is onto.
    Non-Example
    Consider
    \[f:\mathbb R\rightarrow\mathbb R,\quad f(x)=x^2.\]
    Negative numbers are never images of the function.
    Hence,
    \[\text{Range}=[0,\infty)\neq\mathbb R.\]
    Therefore, the function is not onto.

    SURJECTIVE FUNCTION (ONTO) A function f: X → Y is surjective if every element in Y has at least one preimage in X Domain (X) x₁ x₂ x₃ x₄ Codomain (Y) y₁ y₂ y₃ f(x) Key Condition: Every element in Y is mapped to. Multiple elements in X may map to the same Y (e.g., x₂, x₃ → y₂), as long as no Y is left out. Range(f) = Codomain(Y)
  • Bijective Function (One-One and Onto)
    Definition
    A function
    \[f:X\rightarrow Y\]
    is called Bijective if it is both
    • Injective (One-One)
    • Surjective (Onto)
    That is, $f$ is one-one and onto simultaneously.
    Meaning,
    Every element of the codomain has exactly one pre-image.
    There is a perfect one-to-one correspondence between the two sets.
    Example
    Let
    \[f:\mathbb R\rightarrow\mathbb R,\quad f(x)=2x+3\]
    Injective:
    \[2x_1+3=2x_2+3\Rightarrow x_1=x_2\]
    Surjective:
    Given
    \[y\in\mathbb R,\]
    choose
    \[x=\frac{y-3}{2}\]
    Then
    \[f(x)=y\]
    Hence,
    \[f(x)=2x+3\]
    is Bijective.

    BIJECTIVE FUNCTION (ONE-TO-ONE & ONTO) A function f: X → Y is bijective if it is both Injective and Surjective (Perfect One-to-One Correspondence) Domain (X) x₁ x₂ x₃ Codomain (Y) y₁ y₂ y₃ f(x) f⁻¹(y) 1. Injective: Every X maps to a unique Y. 2. Surjective: Every Y is mapped to (Range = Codomain). Bijective ⇔ Invertible (f⁻¹ exists)
🔗 Relationship among Different Types of Functions
Type Different Inputs Give Different Outputs? Every Codomain Element Has a Pre-image?
Injective ✓ Yes May or may not
Surjective May or may not ✓ Yes
Bijective ✓ Yes ✓ Yes
⚖️ Comparison of Injective, Surjective and Bijective Functions
Property Injective Surjective Bijective
Distinct images ✓ May or may not ✓
Range = Codomain May or may not ✓ ✓
Inverse Exists Not always Not always ✓ Always
🧮 Theorem
🧮 Theorem
  • Every bijective function is both injective and surjective.
  • Every injective function need not be surjective.
  • Every surjective function need not be injective.
  • Every bijective function possesses a unique inverse function.
  • A function has an inverse if and only if it is bijective.
🧩 Solution
1
Question
Determine whether
\[f:\mathbb R\rightarrow\mathbb R,\quad f(x)=2x+5\]
is injective, surjective or bijective.
  • Use algebra for injectivity.
  • Find pre-image for surjectivity.
  1. 1

    Assume \(f(x_1)=f(x_2)\).

  2. 2

    Find \(x\) in terms of \(y\).

  3. 3

    Conclude the type.

Injective:
  1. \[2x_1+5=2x_2+5\Rightarrow x_1=x_2.\]
Surjective:
  1. Given
    \[y\in\mathbb R,\]
  2. choose
    \[x=\frac{y-5}{2}\]
  3. Then
    \[f(x)=y\]
Hence,
\[\boxed{f\text{ is Bijective}}\]
2
Question
Determine the type of
\[f(x)=x^2,\quad f:\mathbb R\rightarrow\mathbb R.\]
Injective
  1. Since
    \[f(2)=f(-2)=4,\]
The function is not injective.
Surjective
  1. Also
    Negative real numbers are not images.
Therefore, The function is not Surjective.
\[\boxed{\text{Function is neither Injective nor Surjective}.}\]
⚡ Exam Tip
❌ Common Mistakes
  • Ignoring the codomain while checking surjectivity.
  • Concluding that every linear function is bijective without checking the domain and codomain.
  • Confusing one-one with onto.
  • Writing "Range = Domain" instead of "Range = Codomain".
  • Forgetting that \(x^2\) is one-one only on restricted domains such as \([0,\infty)\).
📋 CBSE Competency-Based (HOTS)

Question

Let

\[f:\mathbb R\rightarrow\mathbb R\]

be defined by

\[f(x)=x^3-1.\]

Determine whether the function is injective, surjective or bijective.

Hint
  • The function is strictly increasing on \(\mathbb R\), so it is injective.
  • Given any \(y\in\mathbb R\), choose
    \[ x=\sqrt[3]{y+1}. \]
    Then \(f(x)=y\).
  • Hence, the function is bijective.
🌟 Significance
  • One of the highest-weightage topics in the Relations and Functions chapter.
  • Frequently asked in CBSE Board examinations as proof-based, competency-based and case-study questions.
  • Forms the basis of Inverse Functions studied later in the chapter.
  • Very important for JEE Main, CUET, NDA, BITSAT and higher mathematics.
  • Questions involving graphs, domain restrictions and inverse functions are directly based on these concepts.
🔑 Key Takeaway
🔗

Example 7

📋
Table of Contents
11 sections
❓ Question
Let $A$ be the set of all 50 students of Class X in a school. Define a function
\[f:A\rightarrow\mathbb N\]
by
\[f(x)=\text{Roll number of student }x.\]
Show that the function f is One-One (Injective) but not Onto (Surjective).
💡 Concept
📖 Theory
🗺️ Roadmap
  1. Check whether two different students can have the same roll number.
  2. Check whether every natural number is assigned to a student.
  3. Conclude the type of function.
🧩 Solution
Prove that the Function is One-One
  1. Suppose
    \[x_1,x_2\in A\]
    and
    \[f(x_1)=f(x_2)\]
    This means that both students have the same roll number.
    However, according to school rules, every student is assigned a unique roll number.
    Therefore,
    \[x_1=x_2\]
    Hence,
    \[f(x_1)=f(x_2)\Rightarrow x_1=x_2\]
Thus, \(\boxed{f\text{ is One-One (Injective).}}\)
Prove that the Function is Not Onto
  1. Assume that the roll numbers assigned to the students are
    \[1,2,3,\ldots,50.\]
  2. The codomain is
    \[\mathbb N=\{1,2,3,\ldots\}.\]
  3. Consider the natural number 51
    Since there are only 50 students, no student has roll number 51
  4. Hence
    \(51\) is not the image of any element of \(A\}
  5. Therefore,
    \[\exists\,y\in\mathbb N\]
  6. such that there is no
    \[x\in A\]
  7. for which
    \[f(x)=y\]
Hence,
\[\boxed{f\text{ is not Onto (Surjective).}}\]
🔬 Alternative Board Examination Solution

Injective:

If

\[f(x_1)=f(x_2),\]

then both students have the same roll number. Since roll numbers are unique,

\[x_1=x_2.\]

Therefore,

\[f\]

is one-one.

Not Onto:

The codomain is

\[\mathbb N.\]

Natural numbers such as

\[51,52,\ldots\]

are not roll numbers of any student.

Hence,

\[f\]

is not onto.

⚡ Exam Tip
❌ Common Mistakes
  • Assuming that one-one automatically implies onto.
  • Ignoring the codomain \(\mathbb N\) while checking surjectivity.
  • Confusing roll numbers with the number of students.
  • Concluding that the function is bijective without verifying surjectivity.
📋 CBSE Competency-Based (HOTS)

Question

A company has 200 employees. Define the function

\[f:E\rightarrow\mathbb N\]

where \(f(x)\) is the employee ID assigned to employee \(x\). Determine whether the function is injective, surjective or bijective.

Hint
  • Every employee has a unique employee ID, so the function is injective.
  • The codomain is the set of all natural numbers, so infinitely many natural numbers are not employee IDs.
  • Therefore, the function is not surjective and hence not bijective.
🌟 Significance
  • A standard CBSE example illustrating the difference between injective and surjective functions.
  • Frequently appears in competency-based and application-oriented questions.
  • Strengthens the understanding that the codomain is essential when testing surjectivity.
  • Forms the foundation for inverse functions and bijections studied later in the chapter.
🔑 Key Takeaway
🔗

Example 8

📋
Table of Contents
10 sections
❓ Question
Show that the function
\[f:\mathbb N\rightarrow\mathbb N\]
defined by
\[f(x)=2x\]
is One-One (Injective) but not Onto (Surjective).
💡 Concept
📖 Theory
🗺️ Roadmap
  1. Assume two images are equal and prove the corresponding inputs are equal.
  2. Find an element of the codomain that has no pre-image.
  3. Conclude the type of the function.
🧩 Solution
Prove that the Function is One-One
  1. Suppose
    \[f(x_1)=f(x_2)\]
  2. Then
    \[2x_1=2x_2\]
  3. Dividing both sides by 2
    \[x_1=x_2\]
  4. Therefore,
    \[f(x_1)=f(x_2)\Rightarrow x_1=x_2\]
Hence,
\[\boxed{f\text{ is One-One (Injective).}}\]
Prove that the Function is Not Onto
  1. The codomain is
    \[\mathbb N=\{1,2,3,4,5,\ldots\}\]
  2. Consider the element
    \[1\in\mathbb N.\]
  3. If the function were onto, there would exist some
    \[x\in\mathbb N\]
  4. such that
    \[f(x)=1\]
  5. That is,
    \[2x=1\]
  6. Solving for $x$
  7. we get
    \[x=\frac{1}{2},\]
    which is not a natural number
  8. Hence, there is no
    \[x\in\mathbb N\]
  9. such that
    \[f(x)=1\]
Therefore,
\[\boxed{f\text{ is not Onto (Surjective).}}\]
Alternative Explanation
  1. Since
    \[f(x)=2x\]
  2. every image is an even natural number:
    \[2,4,6,8,\ldots\]
  3. The odd natural numbers
    \[1,3,5,7,\ldots\]
    never appear as outputs.
  4. Therefore,
    \[\text{Range}=\{2,4,6,8,\ldots\}\neq\mathbb N.\]
Hence, the function cannot be onto.
Property Status Reason
Injective (One-One) ✓ Yes \(2x_1=2x_2\Rightarrow x_1=x_2\).
Surjective (Onto) ✗ No Odd natural numbers have no pre-image.
Bijective ✗ No A bijection must be both one-one and onto.
🎨 SVG Diagram
Mapping Diagram
f: ℕ → ℕ where f(x) = 2x Function is One-One (Injective) but NOT Onto (Not Surjective) Domain (ℕ) 1 2 3 4 ⋮ Codomain (ℕ) 1 2 3 4 5 6 ⋮ f(x)=2x Odd Numbers Left Out! (No x gives 2x = 1, 3, 5...) 1. ONE-ONE (INJECTIVE): YES ✓ Let f(x₁) = f(x₂) ⇒ 2x₁ = 2x₂ ⇒ x₁ = x₂ Distinct inputs always map to distinct outputs. 2. ONTO (SURJECTIVE): NO ✗ For odd y = 1 ∈ ℕ, 2x = 1 ⇒ x = 1/2 ∉ ℕ Range (Evens) ≠ Codomain ℕ (Range is a proper subset).
⚡ Exam Tip
❌ Common Mistakes
  • Writing that every natural number is an image under \(f(x)=2x\).
  • Forgetting that odd numbers are part of the codomain.
  • Claiming the function is bijective because it is one-one.
  • Ignoring the specified domain and codomain while checking surjectivity.
📋 CBSE Competency-Based (HOTS)

Question

Consider the function

\[ f:\mathbb N\rightarrow\mathbb N, \qquad f(x)=3x. \]

Determine whether the function is injective, surjective or bijective.

Hint
  • Use \(3x_1=3x_2\) to prove injectivity.
  • Observe that numbers such as \(1,2,4,5,\ldots\) are not multiples of 3, so they have no pre-image.
  • Hence, the function is injective but not surjective.
🌟 Significance
  • A classic CBSE proof-based example for identifying one-one but not onto functions.
  • Frequently asked in Board examinations, competency-based questions and entrance tests.
  • Builds the foundation for understanding inverse functions and bijections.
  • Illustrates the importance of comparing the range with the codomain.
🔑 Key Takeaway
🔗

Exercise 9

📋
Table of Contents
10 sections
❓ Question
Prove that the function
\[f:\mathbb{R}\rightarrow\mathbb{R}\]
defined by is One-One (Injective) as well as Onto (Surjective).
💡 Concept
📖 Theory Behind the Question
🗺️ Roadmap
  1. Assume two images are equal and prove that the corresponding inputs are equal.
  2. Take an arbitrary real number in the codomain.
  3. Find its pre-image.
  4. Conclude that the function is bijective.
🧩 Solution
Prove that the Function is One-One
  1. Assume
    \[f(x_1)=f(x_2)\]
  2. then
    \[2x_1=2x_2\]
  3. Dividing both sides by 2 we obtain
    \[x_1=x_2\]
  4. Hence,
    \[f(x_1)=f(x_2)\Rightarrow x_1=x_2\]
Therefore,
\[\boxed{f\text{ is One-One (Injective).}}\]
Prove that the Function is Onto
  1. let
    \[y \in\; \mathbb{R}\]
    be an arbitrary real number.
  2. Choose
    \[x=\frac{y}{2}\]
  3. Since
    \[\frac{y}{2}\in\mathbb R,\]
    it belongs to the domain.
  4. Now,
    \[f\left(\frac{y}{2}\right)=2\left(\frac{y}{2}\right)=y\]
  5. Thus, every real number $y$ is the image of some real number
    \[x=\frac{y}{2}\]
Therefore,
\[\boxed{f\text{ is Onto (Surjective).}}\]
🎨 SVG Diagram
f: ℝ → ℝ where f(x) = 2x Function is BOTH One-One (Injective) AND Onto (Surjective) ⇒ Bijective Domain (ℝ) -1 0 1/2 1 ⋮ Codomain (ℝ) -2 0 1 2 ⋮ f(x)=2x 1. ONE-ONE (INJECTIVE): YES ✓ Let f(x₁) = f(x₂) ⇒ 2x₁ = 2x₂ ⇒ x₁ = x₂ Distinct real inputs yield distinct real outputs. 2. ONTO (SURJECTIVE): YES ✓ For any y ∈ ℝ, choose x = y/2 ∈ ℝ Then f(x) = 2(y/2) = y. Every real number is mapped to!
⚡ Exam Tip
❌ Common Mistakes
  • Assuming that every one-one function is automatically onto.
  • Choosing a pre-image without verifying that it belongs to the domain.
  • Ignoring the codomain while checking surjectivity.
  • Not concluding explicitly that the function is bijective after proving both properties.
📋 CBSE Competency-Based (HOTS)

Show that the function

\[f:\mathbb R\rightarrow\mathbb R,\quad f(x)=5x-7\]

is bijective. Also determine its inverse.

Hint
  • Use \(5x_1-7=5x_2-7\) to prove injectivity.
  • For any \(y\in\mathbb R\), choose \(x=\dfrac{y+7}{5}\).
  • The inverse function is
    \[f^{-1}(x)=\frac{x+7}{5}.\]
🌟 Significance
  • A standard CBSE proof-based question on bijective functions.
  • Frequently appears in Board examinations, competency-based questions and entrance tests such as JEE Main and CUET.
  • Provides the foundation for the concept of inverse functions.
  • Illustrates that every non-zero linear function on \(\mathbb R\) is bijective.
🔑 Key Takeaway
🔗

Example 10

📋
Table of Contents
10 sections
❓ Question
Show that the function \(f : N → N\), given by \(f(1) = f(2) = 1\) and \(f(x) = x – 1\), for every x > 2, is onto but not one-one.
💡 Concept
📖 Theory Behind the Question
🗺️ Roadmap
  1. Find two different inputs having the same image.
  2. Take an arbitrary element of the codomain.
  3. Find a pre-image for every element.
  4. Conclude the type of function.
🧩 Solution
Prove that the Function is Not One-One
  1. Observe that
    \[f(1)=1\]
    and
    \[f(2)=1\]
  2. Therefore,
    \[f(1)=f(2)\]
  3. but,
    \[1\neq2\]
  4. Hence,
    \[f(1)=f(2)\Rightarrow 1\neq2\]
Hence,
\[\boxed{f\text{ is not One-One (Injective).}}\]
Prove that the Function is Onto
  1. Let
    \[y \in \mathbb{N}\]
  2. We consider two cases.<br>
  3. Case 1 : \(\(y=1)\)<br> Since
    \[f(1)=1\]
    the element 1 has a pre-image.
  4. Case 2 : \((y>1)\)<br>Choose
    \[x=y+1.\]
  5. Since
    \[y+1>2\]
  6. the function rule becomes
    \[f(x)=x-1\]
  7. Therefore,
    \[f(y+1)=(y+1)-1=y\]
  8. Thus, every natural number has at least one pre-image.
Hence,
\[\boxed{f\text{ is Onto (Surjective).}}\]
Property Status Reason
Injective (One-One) ✗ No \(f(1)=f(2)=1\).
Surjective (Onto) ✓ Yes Every natural number has at least one pre-image.
Bijective ✗ No A bijection must be both one-one and onto.
Hence,
\[\boxed{\text{The function is Onto but not One-One.}}\]
🎨 SVG Diagram
Mapping Diagram
f: ℕ → ℕ where f(1)=f(2)=1 & f(x)=x–1 (x > 2) Function is NOT One-One (Many-to-One) but Onto (Surjective) Domain (ℕ) 1 2 3 4 5 ⋮ Codomain (ℕ) 1 2 3 4 ⋮ f(1) = f(2) = 1 1. ONE-ONE (INJECTIVE): NO ✗ f(1) = 1 and f(2) = 1 ⇒ f(1) = f(2) However, 1 ≠ 2. Multiple inputs yield the same output. 2. ONTO (SURJECTIVE): YES ✓ For y = 1, f(1) = 1. For y > 1, choose x = y + 1 ∈ ℕ Then f(y + 1) = (y + 1) - 1 = y. Every element in ℕ is mapped!
⚡ Exam Tip
❌ Common Mistakes
  • Checking only one part of a piecewise-defined function.
  • Claiming the function is injective because most outputs have unique pre-images.
  • Forgetting to verify the special case \(y=1\).
  • Concluding that an onto function must also be one-one.
📋 CBSE Competency-Based (HOTS)

Question

Define

\[f:\mathbb N\rightarrow\mathbb N\]

by

\[f(1)=1,\quad f(n)=n-1,\quad n>1.\]

Determine whether the function is injective, surjective or bijective.

Hint
  • The outputs of 1 and 2 are both equal to 1, so the function is not injective.
  • Every natural number has a pre-image: \(1\) is the image of \(1\), and every \(y>1\) is the image of \(y+1\).
  • Hence, the function is surjective but not injective.
🌟 Significance
  • A standard CBSE example illustrating an onto function that is not one-one.
  • Frequently asked in Board examinations and competency-based questions.
  • Helps students distinguish between injective and surjective functions.
  • Demonstrates how to analyse piecewise-defined functions.
🔑 Key Takeaway
🔗

Example 12

📋
Table of Contents
10 sections
❓ Question
Show that the function \(f : R → R\), defined as \(f(x) = x^2\), is neither one-one nor onto.
💡 Concept
📖 Theory Behind the Question
🗺️ Roadmap
  1. Find two distinct inputs with the same image.
  2. Find an element of the codomain having no pre-image.
  3. Conclude the type of the function.
🧩 Solution
Prove that the Function is Not One-One
  1. Consider the two distinct real numbers
    \[1\quad\text{and}\quad1.\]
  2. Their images are
    \[f(-1)=(-1)^2=1\]
    and
    \[f(1)=1^2=1\]
  3. Thus,
    \[f(-1)=f(1)\]
  4. but
    \[-1\neq1\]
\[\text{Hence, }\boxed{f\text{ is not One-One (Injective).}}\]
Prove that the Function is Not Onto
  1. The codomain is
    \[\mathbb R\]
  2. Consider the real number
    \[-2\]
  3. If the function were onto, there would exist some
    \[x\in\mathbb R\]
  4. such that
    \[x^2=-2\]
  5. However,
    \[x^2\ge0\]
    for every real number $x$
  6. Therefore,
    \[x^2=-2\]
    has no real solution.
  7. Hence, $-2$
    has no pre-image.
\[\text{Hence, }\boxed{f\text{ is not Onto (Surjective).}}\]
Property Status Reason
Injective (One-One) ✗ No \(f(-1)=f(1)=1\).
Surjective (Onto) ✗ No Negative real numbers have no pre-image.
Bijective ✗ No The function is neither one-one nor onto.
Hence,
\[ \boxed{f(x)=x^2\text{ is neither Injective nor Surjective}.} \]
🎨 SVG Diagram
f: ℝ → ℝ where f(x) = x² Function is NOT One-One (Many-to-One) and NOT Onto (Not Surjective) Domain (ℝ) -2 -1 0 1 2 ⋮ Codomain (ℝ) -1 0 1 4 ⋮ f(x)=x² Negatives Left Out! (No x ∈ ℝ gives x² < 0) 1. ONE-ONE (INJECTIVE): NO ✗ f(-1) = 1 and f(1) = 1 ⇒ f(-1) = f(1) However, -1 ≠ 1. Multiple inputs share the same output. 2. ONTO (SURJECTIVE): NO ✗ For y = -1 ∈ ℝ, x² = -1 ⇒ x = √(-1) ∉ ℝ Range [0, ∞) ≠ Codomain ℝ (Negative numbers have no pre-image).
⚡ Exam Tip
❌ Common Mistakes
  • Assuming \(x^2\) is one-one because positive numbers have unique squares.
  • Ignoring the negative values of the domain.
  • Confusing the range \([0,\infty)\) with the codomain \(\mathbb R\).
  • Forgetting that surjectivity depends on the codomain, not just the formula.
📋 CBSE Competency-Based (HOTS)

Question

Consider the function

\[f:[0,\infty)\rightarrow[0,\infty),\quad f(x)=x^2.\]

Determine whether the function is injective, surjective or bijective.

Hint6
  • On the restricted domain \([0,\infty)\), different inputs have different squares.
  • Every non-negative real number \(y\) has the pre-image \(x=\sqrt{y}\).
  • Hence, the function becomes bijective.
🌟 Significance
  • A classical example used to distinguish between injective and surjective functions.
  • Frequently asked in CBSE Board examinations, competency-based questions and JEE Main.
  • Demonstrates how changing the domain or codomain can change the type of a function.
  • Forms the basis for understanding inverse functions of restricted domains.
🔑 Key Takeaway
🔗

Exaple 12

📋
Table of Contents
11 sections
❓ Question
Show that \(f : N → N\), given by
\[f(x) = \begin{cases} x + 1, & \text{if } x \text{ is odd} \\ x - 1, & \text{if } x \text{ is even} \end{cases}\]
is both one-one and onto.
💡 Concept
📖 Theory Behind the Question
🗺️ Roadmap

  1. Assume

    \[f(x_1)=f(x_2).\]

  2. Consider all possible parity cases.

  3. Show that equal images imply equal inputs.

  4. Take an arbitrary natural number.

  5. Find its pre-image.

  6. Conclude that the function is bijective.

🧩 Solution
Prove that the Function is One-One
  1. Assume
    \[f(x_1)=f(x_2)\]
  2. We consider different cases.
  3. Case I : One Number is Odd and the Other is Even; Suppose \(x_1\) is odd and \(x_2\) is even. Then
    \[f(x_1)=x_1+1\]
    and
    \[f(x_2)=x_2-1\]
  4. Since
    f(x_1)=f(x_2)
  5. we get
    x_1+1=x_2-1
  6. However,
    \[x_2=x_1+2\]
  7. which is also odd because adding 2 preserves parity. This contradicts the assumption that
    \(x_2\) is even. Hence, this case is impossible.
  8. Similarly, the possibility that
    \(x_1\) is even and \(x_2\) is odd can be shown to lead to a contradiction.
  9. Case II : Both Numbers are Odd, then
    \[f(x_1)=x_1+1\]
    and
    \[f(x_2)=x_2+1\]
  10. Therefore,
    \[x_1+1=x_2+1\]
  11. Hence,
    \[x_1=x_2\]
  12. Case III : Both Numbers are Even, then
    \[f(x_1)=x_1-1\]
    and
    \[f(x_2)=x_2-1\]
  13. Thus,
    \[x_1-1=x_2-1\]
  14. Therefore,
    \[x_1=x_2\]
  15. Hence, in every possible case,
    \[f(x_1)=f(x_2) \Rightarrow x_1=x_2\]
\[\text{Therefore, }\boxed{f\text{ is One-One (Injective).}}\]
Prove that the Function is Onto
  1. Let
    \[y\in\mathbb N.\]
  2. Again, we consider two cases. Case I : y is Odd. Suppose
    \[y=2r+1\]
  3. Choose
    \[x=2r\]
    which is even.
  4. then
    \[f(x)=x-1=(2r+2)-1=2r+1=y\]
  5. Hence
    Every odd natural number has a pre-image.
  6. Case II : y is Even. Suppose
    \[y=2r\]
  7. Choose
    \[x=2r+1\]
    which is odd.
  8. then
    \[f(x)=x+1=(2r+1)+1=2r+2=y\]
  9. Hence
    Every even natural number has a pre-image.
\[\text{Therfore, }\boxed{f\text{ is Onto (Surjective).}}\]
Property Status Reason
Injective (One-One) ✓ Yes Equal images imply equal inputs.
Surjective (Onto) ✓ Yes Every natural number has exactly one pre-image.
Bijective ✓ Yes The function is both one-one and onto.

Hence,

\[ \boxed{f:\mathbb N\rightarrow\mathbb N\text{ is Bijective}.} \]
🎨 SVG Diagram
Mapping Diagram
f: ℕ → ℕ where f(x) = x+1 (if x odd) & x–1 (if x even) Function is BOTH One-One (Injective) AND Onto (Surjective) ⇒ Bijective Domain (ℕ) 1 2 3 4 ⋮ Codomain (ℕ) 1 2 3 4 ⋮ Pairs Swap Places! 1. ONE-ONE (INJECTIVE): YES ✓ If f(x₁) = f(x₂), both x₁, x₂ have same parity ⇒ x₁ = x₂ Distinct inputs map 1-to-1 via pair-swapping (Self-Inverse). 2. ONTO (SURJECTIVE): YES ✓ For odd y, f(y+1) = y. For even y, f(y-1) = y Every natural number y has a valid pre-image in ℕ.
🧩 Alternative Board Examination Solution

Injective:

If

\[f(x_1)=f(x_2),\]

then \(x_1\) and \(x_2\) cannot have different parity. If both are odd or both are even, simple algebra gives

\[x_1=x_2.\]

Hence, the function is one-one.

Surjective:

  • For every odd number
    \[2r+1,\]
    its pre-image is
    \[2r+2.\]
  • For every even number
    \[2r,\]
    its pre-image is
    \[2r-1.\]

Therefore,

\[\boxed{f\text{ is Onto}.}\]

⚡ Exam Tip
❌ Common Mistakes
  • Not considering odd and even cases separately.
  • Assuming \(x_2=x_1+2\) is even when it is actually odd if \(x_1\) is odd.
  • Forgetting to find pre-images for both odd and even outputs.
  • Concluding bijectivity before proving both injectivity and surjectivity.
📋 CBSE Competency-Based (HOTS)

Question

Define

\[g:\mathbb N\rightarrow\mathbb N\]

by

\[g(x)=\begin{cases}x+2,&x\text{ is odd}\\x-2,&x\text{ is even and }x>2\\2,&x=2.\end{cases}\]

Determine whether the function is injective, surjective or bijective.

Hint: Analyse the odd and even cases separately and check whether every natural number has a unique pre-image.

🌟 Significance
  • A classical CBSE example of a bijective piecewise-defined function.
  • Frequently asked in Board examinations and competency-based questions.
  • Illustrates how parity arguments simplify proofs of injectivity.
  • Builds a strong foundation for inverse functions and permutations.
🔑 Key Takeaway
🔗

Example 13

📋
Table of Contents
11 sections
❓ Question
Show that an onto function \(f : {1, 2, 3} → {1, 2, 3}\) is always one-one.
💡 Concept
📖 Theory Behind the Question
🗺️ Roadmap
  1. Assume that the function is onto.
  2. Assume, for contradiction, that it is not one-one.
  3. Show that two different inputs having the same image reduce the number of distinct outputs.
  4. Conclude that some element of the codomain has no pre-image.
  5. Obtain a contradiction.
🧩 Solution
  1. Suppose
    \[f:\{1,2,3\}\rightarrow\{1,2,3\}\]
    is onto.
  2. Assume, contrary to our claim, that $f$ is not one-one.
    Then there exist two distinct elements of the domain having the same image.
  3. Without loss of generality, suppose
    \[f(1)=f(2)\]
  4. Since $1$ and $2$ have the same image,
    these two elements contribute only one distinct output
  5. The remaining element, $3$
    can contribute at most one more distinct output.
  6. Therefore, the range of the function can contain at most $2$ distinct elements. That is,
    \[|\text{Range}(f)|\le2\]
  7. However, the codomain
    \[\{1,2,3\}\]
  8. contains
    3 elements.
  9. Therefore,
    $f$ cannot be onto.
    This contradicts the given condition.
    Hence, our assumption is false.
\[\text{Therefore, }\boxed{f\text{ is One-One (Injective).}}\]
Property Reason
Onto Given in the question.
One-One Otherwise, the range would contain fewer than three elements.
Bijective Since the function is both onto and one-one.

Hence,

\[ \boxed{\text{Every onto function from }\{1,2,3\}\text{ to }\{1,2,3\}\text{ is bijective}.}\]

🧩 Alternative Proof Using the Pigeonhole Principle

The domain and codomain each contain exactly three elements.

If two elements of the domain are mapped to the same element of the codomain, then only two distinct images can be produced.

Consequently, one element of the codomain is left without a pre-image, contradicting the assumption that the function is onto.

Hence, every onto function from

\[\{1,2,3\}\]

to itself must also be one-one.

🎨 SVG Diagram
f : {1, 2, 3} → {1, 2, 3} (Finite Set to Itself) An Onto (Surjective) function on a finite set is ALWAYS One-One (Injective) Domain A = {1, 2, 3} 1 2 3 Codomain A = {1, 2, 3} 1 2 3 f(x) Pigeonhole Principle: No two inputs can share an output! 1. GIVEN: FUNCTION IS ONTO Range = Codomain = {1, 2, 3} (3 elements) All 3 codomain targets MUST have distinct pre-images. 2. PROOF: MUST BE ONE-ONE ✓ If not 1-1, Range size < 3 ⇒ Contradicts Onto! Since Domain size = Codomain size (3), Onto ⇒ One-One.
⚡ Exam Tip
❌ Common Mistakes
  • Trying to verify every possible mapping individually.
  • Ignoring that the domain and codomain have equal cardinality.
  • Not stating the contradiction explicitly.
  • Confusing the range with the codomain.
🗒️ CBSE Competency-Based (HOTS)

Let

\[A=\{1,2,3,4\}.\]

Show that every onto function

\[f:A\rightarrow A\]

must also be one-one.

Hint: Use the same contradiction argument or apply the Pigeonhole Principle to compare the number of elements in the range and the codomain.

🌟 Significance
  • A very important theoretical result in the chapter Relations and Functions.
  • Frequently asked in CBSE competency-based and proof-oriented questions.
  • Introduces the Pigeonhole Principle in an elementary setting.
  • Forms the basis for studying permutations and inverse functions.
🔑 Key Takeaway
🔗

Example 14

📋
Table of Contents
13 sections
❓ Question
14 Show that a one-one function \(f : {1, 2, 3} → {1, 2, 3}\) must be onto.
💡 Concept
📖 Theory
Theory Behind the Question
🗺️ Roadmap
  1. Use the definition of a one-one function.
  2. Count the number of distinct images.
  3. Compare the number of images with the size of the codomain.
  4. Conclude that every element of the codomain must be an image.
🧩 Solution
Since
\[ f:\{1,2,3\}\rightarrow\{1,2,3\} \]
is one-one, distinct elements of the domain must have distinct images.
Therefore,
\[ f(1),\quad f(2),\quad f(3) \]
are three distinct elements of the codomain.
The codomain
\[ \{1,2,3\} \]
contains exactly three elements.
Hence, the three distinct images must be precisely the three elements of the codomain.

Thus,
\[ \text{Range}(f)=\{1,2,3\}. \]
Since the range is equal to the codomain,
\[ f \]
is onto.

Therefore,
\[ \boxed{f\text{ is Onto (Surjective).}} \]
🧩 Alternative Proof Using Contradiction
Assume that
\[f\]
is one-one but not onto.

Then at least one element of the codomain has no pre-image.

Therefore, only two elements of the codomain can appear as images.

However, three distinct elements of the domain must have three distinct images because the function is one-one.

This is impossible, since only two images are available.

Hence, our assumption is false.

Therefore,

\[ \boxed{f\text{ must be Onto}.} \]

📝 Conclusion
🗒️ Generalization

If

\[ |A|=|B|<\infty, \]

then the following statements are equivalent:

  • The function is one-one.
  • The function is onto.
  • The function is bijective.

Thus, for finite sets of equal cardinality, proving either injectivity or surjectivity automatically proves the other.

Alternative Board Examination Solution

Since the function is one-one,

\[ 1,\;2,\;3 \]

must have three distinct images.

The codomain itself contains only three elements.

Therefore, all three elements of the codomain must appear as images.

Hence,

\[ \boxed{f\text{ is Onto}.} \]

⚡ Exam Tip
❌ Common Mistakes
  • Trying to construct all possible one-one functions instead of using cardinality.
  • Confusing the range with the codomain.
  • Forgetting that the domain and codomain have the same number of elements.
  • Not concluding that the function is bijective after proving it is both one-one and onto.
📋 CBSE Competency-Based (HOTS)

Question

Let

\[ A=\{a,b,c,d\}. \]

Show that every one-one function

\[ f:A\rightarrow A \]

must be onto.

Hint: Since both the domain and codomain contain four elements, an injective function must have four distinct images. Therefore, every element of the codomain is an image.

🌟 Significance
  • A frequently tested proof in CBSE Board examinations and competency-based questions.
  • Introduces the relationship between injective and surjective functions on finite sets.
  • Provides the foundation for permutations, inverse functions and counting principles.
  • Often appears together with its converse (Example 13).
🔑 Key Takeaway
🔗

Composition of Functions and Invertible Functions

📋
Table of Contents
7 sections
🗺️ Overview
Composition of functions is one of the most important concepts in mathematics. It allows two or more functions to be combined into a single function. The output of the first function becomes the input of the second function. This concept is widely used in algebra, calculus, computer programming, engineering, machine learning, cryptography and many other fields.

An invertible function is a function that can be reversed. If a function is invertible, every output corresponds to exactly one input, allowing us to recover the original value. Inverse functions play an essential role in solving equations, logarithms, exponential functions and coordinate geometry.
🗂️ Types / Category
  • Composition of Functions
    Definition
    Let
    \[f:A\rightarrow B\]
    and
    \[g:B\rightarrow C\]
    be two functions.
    Then the composition of $f$ and $g$ denoted by \(g\circ f\) is the function
    \[g\circ f:A\rightarrow C\]
    defined by
    \[(g\circ f)(x)=g(f(x)),\quad \forall\; x\in A.\]
    Note: The expression in the NCERT text should be
    \[(g\circ f)(x)=g(f(x)),\]
    not
    \[f(f(x))\]
    The output of f becomes the input of g.
    Meaning of the Symbol "∘"
    The symbol $\circ$ is read as "of" or "composed with" Thus, $g\circ f$ is read as "g composed with f."
    Order of Composition
    The order is extremely important.
    First apply $f$ then apply $g$

    Therefore,
    \[g\circ f\neq f\circ g\]
    in general.
    Example 1
    Let
    \[f(x)=2x+1\]
    and
    \[g(x)=x^2\]
    Then
    \[(g\circ f)(x)=g(2x+1)=(2x+1)^2.\]
    Similarly,
    \[(f\circ g)(x)=f(x^2)=2x^2+1.\]
    Since
    \[(2x+1)^2\neq2x^2+1,\]
    we conclude that
    \[g\circ f\neq f\circ g.\]
    Properties of Composition
    Property Statement
    Associative \((h\circ g)\circ f=h\circ(g\circ f)\)
    Not Commutative \(g\circ f\neq f\circ g\) in general.
    Identity Function \(I\circ f=f=f\circ I\)
  • Identity Function A function
    \[I_A:A\rightarrow A\]
    defined by
    \[I_A(x)=x\]
    for every
    \[x\in A\]
    is called the Identity Function.
    It leaves every element unchanged.
    Example
    If
    \[A=\{1,2,3\},\]
    then
    \[I_A(1)=1,\quad I_A(2)=2,\quad I_A(3)=3.\]
  • Invertible Function
    Definition
    A function
    \[f:A\rightarrow B\]
    is called invertible if there exists another function
    \[f^{-1}:B\rightarrow A\]
    such that
    \[f^{-1}\circ f=I_A\]
    and
    \[f\circ f^{-1}=I_B.\]
    The function
    \[f^{-1}\]
    is called the inverse function.
    Necessary and Sufficient Condition
    A function is invertible if and only if it is Bijective. That is,
    \[\boxed{\text{Invertible Function}\iff\text{Bijective Function}.}\]
    Reason (Why?)
    • Injective ensures that every output comes from only one input.
    • Surjective ensures that every output actually exists.
    • Together they guarantee a unique reverse mapping.
    Example
    Let
    \[f:\mathbb R\rightarrow\mathbb R,\quad f(x)=3x-5.\]
    Then
    \[y=3x-5\]
    Solving for $x$, we get
    \[x=\frac{y+5}{3}\]
    Hence,
    \[f^{-1}(x)=\frac{x+5}{3}\]
    Verification
    \[f^{-1}(f(x))=\frac{3x-5+5}{3}=x\]
    Similarly,
    \[f(f^{-1}(x))=x\]
    Non-Invertible Function
    Consider
    \[f(x)=x^2,\quad f:\mathbb R\rightarrow\mathbb R.\]
    This function is neither injective nor surjective. Therefore, $f^{-1}$ does not exist on $\mathbb R$ However, if the domain is restricted to
    \[[0,\infty)\]
    then
    \[f^{-1}(x)=\sqrt{x}\]
🌟 Important Results
⚡ Exam Tip
❌ Common Mistakes
  • Writing \((g\circ f)(x)=f(f(x))\).
  • Interchanging the order of composition.
  • Thinking every one-one function has an inverse.
  • Ignoring the importance of the codomain while checking invertibility.
  • Confusing inverse function with reciprocal,
    \[\frac1{f(x)}.\]
🌟 Significance
  • One of the highest-weightage topics in CBSE Class 12 Mathematics.
  • Frequently tested in Board examinations, JEE Main, NDA, CUET and other entrance examinations.
  • Forms the basis of inverse trigonometric functions, logarithmic functions and advanced calculus.
  • Essential for understanding transformations, algebraic structures and function theory.
🔑 Key Takeaway
🔗

Example 15

📋
Table of Contents
10 sections
❓ Question
Let \(f:\{2,3,4,5\}\rightarrow\{3,4,5,9\}\) and \(g:\{3,4,5,9\}\rightarrow\{7,11,15\}\) be the functions defined by
\[f(2)=3,\qquad f(3)=4,\quad f(4)=5,\quad f(5)=5,\]
avd
\[g(3)=7,\quad g(4)=7,\quad g(5)=11,\quad g(9)=11.\]
Find the composite function \(g\circ f.\)
💡 Concept
📖 Theory Behind the Question
🗺️ Roadmap

  1. Compute the value of \(f(x)\).

  2. Use that value as the input to \(g\).

  3. Find \(g(f(x))\) for every element of the domain.

  4. Write the composite function.

🧩 Solution
For \(x=2\)
  1. Since
    \[f(2)=3\]
  2. we obtain
    \[(g\circ f)(2)=g(f(2))=g(3)=7\]
For \(x=3\)
  1. Since
    \[f(3)=4\]
  2. we have
    \[(g\circ f)(3)=g(4)=7\]
For \(x=4\)
  1. Since
    \[f(4)=5\]
  2. we get
    \[(g\circ f)(4)=g(5)=11\]
For \(x=5\)
  1. Since
    \[f(5)=5\]
  2. we get
    \[(g\circ f)(5)=g(5)=11\]
⚡ Exam Tip
❌ Common Mistakes
  • Computing \(f(g(x))\) instead of \(g(f(x))\).
  • Using values directly from \(g\) without first evaluating \(f\).
  • Ignoring that \(g\) is defined only on the outputs of \(f\).
  • Not writing the final composite function explicitly.
🗒️ CBSE Competency-Based (HOTS)

Question

Suppose
\[f(x)=x+2,\quad g(x)=3x-1\]
Find
\[g\circ f\]
and
\[f\circ g\]
Determine whether the two composite functions are equal.
Hint
  • Compute \(g(f(x))\) and \(f(g(x))\) separately.
  • Compare the resulting expressions to verify that function composition is generally not commutative.
🌟 Significance
  • A standard CBSE example demonstrating the computation of composite functions.
  • Frequently asked in Board examinations, JEE Main and CUET.
  • Builds the foundation for inverse functions and function transformations.
  • Emphasizes the correct order of function composition.
🔑 Key Takeaway
🔗

Example 16

📋
Table of Contents
12 sections
❓ Question
Let \(f:\mathbb R\rightarrow\mathbb R\) and \(g:\mathbb R\rightarrow\mathbb R\) be defined by
\[f(x)=\cos x\]
and
\[g(x)=3x^2\]
Find \(g\circ f\) and \(f\circ g\). Hence, show that \(g\circ f\ne f\circ g.\)
💡 Concept
📖 Theory Behind the Question
🗺️ Roadmap

  1. Find \(g(f(x))\).

  2. Simplify the resulting expression.

  3. Find \(f(g(x))\).

  4. Compare the two expressions.

  5. Use a suitable value of \(x\) to prove they are different.

🧩 Solution
Find \(g\circ f\)
  1. Using the definition of composition,
    \[(g\circ f)(x)=g(f(x))\]
  2. Since
    \[f(x)=\cos x\]
  3. we have
    \[(g\circ f)(x)=g(\cos x)\]
  4. Now,
    \[g(t)=3t^2\]
  5. Therefore,
    \[(g\circ f)(x)=3(\cos x)^2\]
\[\text{Hence, }\boxed{(g\circ f)(x)=3\cos^2x}\]
Find \(f\circ g\)
  1. Again,
    \[(f\circ g)(x)=f(g(x))\]
  2. Since
    \[g(x)=3x^2\]
  3. we obtain
    \[(f\circ g)(x)=f(3x^2)\]
  4. Using
    \[f(x)=\cos x\]
  5. we get
    \[(f\circ g)(x)=\cos(3x^2)\]
\[\text{Therefore, }\boxed{(f\circ g)(x)=\cos(3x^2)}\]
Compare the Two Functions
  1. We have obtained
    \[(g\circ f)(x)=3\cos^2x\]
    and
    \[(f\circ g)(x)=\cos(3x^2)\]
  2. To verify whether these two functions are equal, substitute $0$, then
    \[(g\circ f)(0)=3\cos^20=3(1)^2=3\]
  3. Also
    \[(f\circ g)(0)=\cos(0)=1\]
  4. Since
    \[3\neq1\]
  5. we conclude that
    \[\boxed{g\circ f\neq f\circ g}\]
\[\text{Therefore, }\boxed{g\circ f\neq f\circ g}\]
Composition Result
\(g\circ f\) \(3\cos^2x\)
\(f\circ g\) \(\cos(3x^2)\)
🗒️ Verification
\(x\) \((g\circ f)(x)\) \((f\circ g)(x)\)
0 3 1
1 \(3\cos^21\) \(\cos3\)

Since the corresponding values are different, the two composite functions cannot be equal.

🧩 Alternative Board Examination Solution
Since
\[f(x)=\cos x,\]
we have
\[(g\circ f)(x)=g(\cos x)=3\cos^2x.\]
Also,
\[(f\circ g)(x)=f(3x^2)=\cos(3x^2).\]
At
\[x=0,\]
\[(g\circ f)(0)=3,\]
whereas
\[(f\circ g)(0)=1.\]
Hence,
\[\boxed{g\circ f\ne f\circ g.}\]
⚡ Exam Tip
❌ Common Mistakes
  • Writing \(g(f(x))\) as \(f(g(x))\).
  • Assuming \(g\circ f=f\circ g\).
  • Confusing \(3\cos^2x\) with \(\cos(3x)\).
  • Not checking equality by substituting a suitable value of \(x\).
📋 CBSE Competency-Based (HOTS)

Question

Let

\[f(x)=x+1,\quad g(x)=x^2.\]

Find

\[g\circ f\]

and

\[f\circ g.\]

Determine whether the two functions are identical.

Hint: Compute each composition separately and compare the resulting expressions.

🌟 Significance
  • A standard CBSE example illustrating the non-commutative nature of function composition.
  • Frequently asked in Board examinations, JEE Main and CUET.
  • Strengthens understanding of trigonometric and polynomial function compositions.
  • Provides the foundation for inverse functions and advanced function transformations.
🔑 Key Takeaway
🔗

Invertible Function (Inverse Function)

📋
Table of Contents
12 sections
🗺️ Overview
An invertible function is a function whose action can be completely reversed. If a function transforms elements from one set to another in such a way that every output corresponds to one and only one input, then it is possible to construct another function that takes every output back to its original input. This reverse function is called the inverse function.

Invertible functions are fundamental in mathematics because they allow us to solve equations, reverse transformations, and establish one-to-one correspondence between two sets. In higher mathematics, inverse functions are widely used in algebra, calculus, trigonometry, coordinate geometry, statistics, cryptography, computer science, and engineering.
📘 Definition
📘 Identity Functions
🧭 Necessary and Sufficient Condition
🧭
Condition A function is invertible if and only if it is both one-one (injective) and onto (surjective).

Thus,
\[\boxed{f\text{ is invertible }\iff f\text{ is bijective.}}\]
Why is this Condition Necessary?
One-One Conditio
Suppose two different inputs produce the same output.
Then the inverse function cannot determine which input produced that output.

Therefore,the inverse cannot exist.
Onto Conditio
Suppose some element of the codomain has no pre-image.
Then the inverse function would not know where to send that element.

Hence,the inverse cannot exist.

Therefore,
  • One-One guarantees uniqueness.
  • Onto guarantees completeness.
  • Together they guarantee invertibility.
🧮 Important Theorem
🧮 Theorem
If \(f:X\rightarrow Y\) is invertible, then \(f\) must be one-one and onto.

Conversely,
if \(f\) is one-one and onto, then \(f\) is invertible.

This theorem is extremely useful because, in many problems, we can establish that a function is invertible simply by proving that it is injective and surjective, without explicitly finding its inverse.

✏️ Example
1
Example

Consider

\[f:\mathbb R\rightarrow\mathbb R,\quad f(x)=2x+3.\]

Let

\[y=2x+3\]

Then

\[x=\frac{y-3}{2}\]

Hence,

\[\boxed{f^{-1}(x)=\frac{x-3}{2}}\]

Verification:

\[f^{-1}(f(x))=\frac{2x+3-3}{2}=x\]

Similarly,

\[f(f^{-1}(x))=2\left(\frac{x-3}{2}\right)+3=x\]

2
Example

Consider

\[f:\mathbb R\rightarrow\mathbb R,\quad f(x)=x^2\]

This function is not one-one because

\[f(2)=4=f(-2)\]

Also,

it is not onto because negative real numbers are never obtained.

Hence,

\[f^{-1}\]

does not exist.

However, if the domain is restricted to

\[[0,\infty),\]

then the function becomes bijective and

\[f^{-1}(x)=\sqrt{x}\]

📌 Properties of Inverse Functions
📎 How to Test Whether a Function is Invertible
  1. Check whether the function is one-one.
  2. Check whether the function is onto.
  3. If both conditions are satisfied, conclude that the function is invertible.
  4. If required, determine the inverse by solving
    \[y=f(x)\]
    for
    \[x\]
⚡ Exam Tip
❌ Common Mistakes
  • Confusing the inverse function with the reciprocal
    \[ \dfrac1{f(x)}. \]
  • Ignoring the codomain while checking surjectivity.
  • Assuming every one-one function has an inverse.
  • Finding an algebraic inverse without checking whether the function is bijective.
  • Writing
    \[ f\circ f^{-1}=I_X \]
    instead of
    \[ f\circ f^{-1}=I_Y. \]
🌟 Significance
  • One of the most important theoretical topics in Class 12 Relations and Functions.
  • Frequently asked in CBSE Board examinations, JEE Main, NDA and CUET.
  • Forms the foundation of inverse trigonometric, logarithmic and exponential functions.
  • Essential for understanding function transformations, algebra and calculus.
🔑 Key Takeaway
🔗

Example 17

📋
Table of Contents
10 sections
❓ Question
Let \(f:\mathbb{N}\rightarrow Y\) be the function defined by
\[f(x)=4x+3\]
where
\[Y=\{\,y\in\mathbb{N}:y=4x+3\text{ for some }x\in\mathbb{N}\,\}\]
Show that $f$ is invertible. Hence, find
\[f^{-1}\]
💡 Concept
📖 Theory Behind the Question
🗺️ Roadmap

  1. Express \(y=f(x)\).

  2. Solve the equation for \(x\).

  3. Define the inverse function.

  4. Verify \(g\circ f=I_{\mathbb N}\).

  5. Verify \(f\circ g=I_Y\).

  6. Conclude that \(g=f^{-1}\).

🧩 Solution
  1. Let
    \[y=f(x)=4x+3\]
  2. Since every element of $Y$ is of the form $4x+3$ for some $x \in \mathbb{N}$, we can write
    \[y=4x+3\]
  3. Solving for $x$, we get
    \[x=\frac{y-3}{4}\]
  4. Thus, the inverse function is
    \[f^{-1}(y)=\frac{y-3}{4}\]
  5. This suggests defining the function
    \[g:Y\rightarrow\mathbb N\]
  6. by
    \[\boxed{g(y)=\frac{y-3}{4}}\]
Verification of \(g\circ f=I_{\mathbb N}\)
We have
\[(g\circ f)(x)=g(f(x))=g(4x+3)\]
Substituting into the definition of $g$, we get
\[\begin{aligned}g(4x+3)&=\frac{(4x+3)-3}{4}\\&=\frac{4x}{4}\\&=x\end{aligned}\]
Thus, we have shown that
\[\boxed{g\circ f=I_{\mathbb N}}\]
⚡ Exam Tip
❌ Common Mistakes
  • Writing \(f^{-1}(x)=\dfrac{x+3}{4}\) instead of \(\dfrac{x-3}{4}\).
  • Forgetting to specify the domain of the inverse function.
  • Checking only one composition instead of both.
  • Confusing the inverse function with the reciprocal \(\dfrac1{f(x)}\).
📋 CBSE Competency-Based (HOTS)

Question

Let

\[f:\mathbb N\rightarrow Z,\quad f(x)=5x-2,\]

where

\[Z=\{\,5x-2:x\in\mathbb N\,\}\]

Show that

\[f\]

is invertible and find

\[f^{-1}\]

Hint: Express \(y=5x-2\), solve for \(x\), define the inverse, and verify both compositions.

🌟 Significance
  • A classic CBSE proof-based question on invertible functions.
  • Frequently asked in Board examinations and competitive entrance tests.
  • Demonstrates the use of composition to verify inverse functions.
  • Forms the basis for inverse trigonometric and logarithmic functions studied later.
🔑 Key Takeaway
· Updated
ACADEMIA AETERNUM · CLASS XII · MATHEMATICS · CHAPTER 1

Relations and Functions

A complete interactive engine covering relations, their properties, equivalence classes, functions, composition & invertibility, and binary operations — with a rule-based step-by-step solver, formula bank, curated practice, and hands-on interactive modules. No external AI dependency required.

🔗 What is a Relation?

Given two non-empty sets A and B, their Cartesian product A × B is the set of all ordered pairs (a, b) with a ∈ A and b ∈ B. A relation R from A to B is simply any subset of A × B.

R ⊆ A × B   |A × B| = |A| · |B|   Total relations from A to B = 2^(|A|·|B|)

If (a, b) ∈ R we write a R b and say "a is related to b". When A = B, R is called a relation on A.

Domain, Range & Codomain

  • Domain(R) — the set of all first elements of the ordered pairs in R.
  • Range(R) — the set of all second elements of the ordered pairs in R.
  • Codomain — the entire set B (Range ⊆ Codomain, but they need not be equal).

Special Relations

  • Empty relation: R = ∅ — no element of A is related to any element of B.
  • Universal relation: R = A × A — every element is related to every other element (including itself).
💡
Worked check: If A = {1, 2, 3}, the total number of relations on A is 2^(3×3) = 2⁹ = 512, since each of the 9 possible ordered pairs can independently be included or excluded.

📐 Properties a Relation Can Have

These three properties are defined for a relation R on a single set A (i.e. R ⊆ A × A).

1. Reflexive

R is reflexive ⟺ (a, a) ∈ R for every a ∈ A

Every element must be related to itself. Missing even one diagonal pair (a,a) destroys reflexivity.

2. Symmetric

R is symmetric ⟺ (a, b) ∈ R ⟹ (b, a) ∈ R, for all a, b ∈ A

Relations "flow both ways" — if a is related to b, b must also be related to a.

3. Transitive

R is transitive ⟺ (a, b) ∈ R and (b, c) ∈ R ⟹ (a, c) ∈ R, for all a, b, c ∈ A

Relatedness "carries through" a chain of two steps into a direct link.

⚠️
A relation can be symmetric and transitive but not reflexive — this happens when some element of A does not appear in any pair at all (a "vacuous" or isolated element). See the Common Mistakes tab for the full trap.

🧩 Equivalence Relations & Equivalence Classes

A relation that is reflexive, symmetric, and transitive simultaneously is called an equivalence relation.

R is an equivalence relation on A ⟺ R is reflexive AND symmetric AND transitive

Equivalence Classes

For an equivalence relation R on A, the equivalence class of a, written [a], is the set of all elements related to a:

[a] = { x ∈ A : (x, a) ∈ R }
  • Every element of A belongs to exactly one equivalence class.
  • Distinct equivalence classes are disjoint — together they form a partition of A.
  • Classic example: "has the same remainder when divided by 3" partitions integers into exactly 3 classes: multiples of 3, numbers leaving remainder 1, numbers leaving remainder 2.
✔️
Mental model: an equivalence relation is a formal way of saying "these elements are essentially the same for our purposes" — it groups A into non-overlapping families.

🧮 What is a Function?

A relation f from A to B is called a function if every element of A has exactly one image in B.

f : A → B is a function ⟺ (i) every a ∈ A has some image in B, AND (ii) that image is unique

Condition (i) is "well-defined everywhere"; condition (ii) rules out one input mapping to two different outputs — this is what separates a general relation from a function.

  • A is the domain, B is the codomain.
  • The range is f(A) = { f(a) : a ∈ A } ⊆ B.
💡
Quick test on a mapping diagram: draw one arrow out of every domain element. If any element sends out zero or two-or-more arrows, it is not a function.

🎯 One-One (Injective), Onto (Surjective) & Bijective

One-One / Injective

f is one-one ⟺ f(x₁) = f(x₂) ⟹ x₁ = x₂  (equivalently: x₁ ≠ x₂ ⟹ f(x₁) ≠ f(x₂))

Distinct inputs always give distinct outputs — no two domain elements share an image.

Onto / Surjective

f is onto ⟺ for every b ∈ B, there exists some a ∈ A such that f(a) = b  (i.e. Range(f) = B)

Every element of the codomain is "hit" by at least one domain element.

Bijective

f is bijective ⟺ f is one-one AND onto

A bijection pairs up A and B perfectly, element for element — this is exactly the condition required for f to be invertible.

💡
Algebraic shortcut for f : R → R: a linear function f(x) = mx + c is one-one and onto (hence bijective) whenever m ≠ 0. A quadratic f(x) = ax² + bx + c defined on all of R is never one-one (it repeats values on either side of its vertex).

🔁 Composition of Functions & Invertible Functions

Composition

If f : A → B and g : B → C, the composite function g∘f : A → C is defined by:

(g ∘ f)(x) = g( f(x) )   for every x ∈ A

Note the order — g∘f means "apply f first, then g". In general g∘f ≠ f∘g.

Invertible Functions

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

g ∘ f = I_A  and  f ∘ g = I_B

where I_A, I_B are the identity functions on A and B. The key theorem:

✔️
f is invertible ⟺ f is bijective. When it exists, the inverse g is written f⁻¹, and it is unique.

Finding f⁻¹ algebraically

  1. Write y = f(x).
  2. Solve this equation for x in terms of y.
  3. Replace x with f⁻¹(y), then rewrite the variable as x if required.

Useful Identity

(f ∘ g)⁻¹ = g⁻¹ ∘ f⁻¹  (the order reverses, just like matrix or garment "put on / take off" analogies)

⚙️ Binary Operations

A binary operation on a non-empty set A is a function that takes two elements of A and produces a third element of A:

∗ : A × A → A,  (a, b) ↦ a ∗ b

Crucially, the result a ∗ b must always land back inside A — this is called closure, and it is automatic once ∗ is declared a function into A.

Properties

  • Commutative: a ∗ b = b ∗ a for all a, b ∈ A.
  • Associative: (a ∗ b) ∗ c = a ∗ (b ∗ c) for all a, b, c ∈ A.
  • Identity element: e ∈ A such that a ∗ e = e ∗ a = a for every a ∈ A.
  • Invertible element: a ∈ A has an inverse if there exists b ∈ A with a ∗ b = b ∗ a = e.
💡
Identity elements, when they exist, are always unique. Under a fixed identity, the inverse of an element (if it exists) is also unique.

🧠 Relation Property Solver

Enter a finite set A and the ordered pairs that make up R. The engine will walk through reflexivity, symmetry and transitivity step by step, exactly as you would on paper.

Each pair a-b means (a, b) ∈ R.

Results will appear here — including the counter-example if a property fails.

🧠 Function Classifier

Two modes: finite mapping (list exact images) or algebraic (linear / quadratic rule on ℝ).

Results will appear here.

🧠 Binary Operation Checker

Define a small set and its operation table (as a₁,a₂,... = index positions), then check closure-derived properties automatically.

Default example is addition modulo 4 on {0,1,2,3}. Row i, column j gives (element i) ∗ (element j).

Results will appear here.

📘 Relations — Counting Formulas


QuantityFormula
Relations from A to B2^(|A|·|B|)
Relations on A, |A| = n2^(n²)
Reflexive relations on A, |A| = n2^(n² − n)
Symmetric relations on A, |A| = n2^( n(n+1)/2 )
Reflexive & symmetric relations, |A| = n2^( n(n−1)/2 )
💡
There is no simple closed formula for the number of equivalence relations on an arbitrary n-element set — it equals the Bell number B(n) (the number of ways to partition the set). B(1)=1, B(2)=2, B(3)=5, B(4)=15.

📘 Functions — Counting Formulas

QuantityFormula
All functions A → B|B|^|A|
One-one functions A → B (|A|=m ≤ |B|=n)ⁿPₘ = n! / (n−m)!
Bijections A → B (|A|=|B|=n)n!
Onto functions A → B (|A|=m, |B|=n, general)Σ (−1)^k · ⁿCₖ · (n−k)^m ,  k=0..n

📘 Binary Operations — Counting Formulas

QuantityFormula
Binary operations on A, |A| = nn^(n²)
Commutative binary operations, |A| = nn^( n(n+1)/2 )

📘 Key Identities to Remember

(g ∘ f)(x) = g(f(x)) f invertible ⟺ f bijective (f ∘ g)⁻¹ = g⁻¹ ∘ f⁻¹ [a] = {x ∈ A : (x,a) ∈ R}  (equivalence class)

✨ Relations — Speed Tricks

  • To check reflexive fast, just scan for every "diagonal" pair (a,a) — if the set has n elements you need to see n such pairs, no more, no less relevant.
  • To check symmetric, mentally reflect the pair list across the diagonal: every (a,b) must have its mirror (b,a) present. One-directional pairs are the giveaway a relation fails symmetry.
  • To check transitive, hunt specifically for chains: whenever you see (a,b) and (b,c) sitting in your list, immediately look for (a,c). This is the single most time-consuming check — always do it last after reflexive/symmetric have already ruled the relation out, if they do.
  • If a relation is given by a rule (e.g. "a R b if a divides b"), test reflexivity/symmetry/transitivity using the rule algebraically rather than listing pairs — much faster for infinite sets.

✨ Functions — Speed Tricks

  • One-one (algebraic): assume f(x₁) = f(x₂), simplify, and see if it forces x₁ = x�2. If you end up with something like x₁² = x₂² (which allows x₁ = −x₂), the function is not one-one.
  • Onto (algebraic): set y = f(x) and solve for x in terms of y. If real x exists for every real y, f is onto ℝ → ℝ.
  • Monotonic ⟹ one-one: any strictly increasing or strictly decreasing function on an interval is automatically one-one — check the derivative's sign if calculus is available.
  • For finite sets, one-one is equivalent to "no two arrows land on the same codomain element"; onto is equivalent to "every codomain element receives at least one arrow" — compare against just counting arrows landing on each target.

✨ Composition & Inverses — Speed Tricks

  • To find f⁻¹ quickly for f(x) = (ax+b)/(cx+d) type expressions, cross-multiply immediately after setting y = f(x) — this avoids messy fractions.
  • Sanity-check any inverse you compute by verifying f(f⁻¹(y)) = y for one convenient test value.
  • Remember the "socks and shoes" rule for reversing composition order: (f∘g)⁻¹ = g⁻¹∘f⁻¹ — you undo the last operation first.

✨ Binary Operations — Speed Tricks

  • Commutative check via table: the operation table is commutative exactly when it is symmetric about its main diagonal — a visual, instant check.
  • Identity via table: the identity element's row is an exact copy of the header row, and its column is an exact copy of the header column.
  • Associativity has no shortcut on a table — for small sets it must be checked by brute force on all triples, which is exactly what the Interactive and Solver tabs automate here.
❌ Confusing Range with Codomain
WrongTreating "codomain" and "range" as always the same set.
RightCodomain is the full target set B you declared; range is only the subset of B that actually gets hit. They coincide only when f is onto.
❌ "Symmetric + Transitive ⟹ Reflexive" — the vacuous trap
WrongAssuming any relation that is symmetric and transitive must automatically be reflexive.
RightThis implication only forces reflexivity for elements that actually appear in some pair. An element with zero pairs (isolated) can make the relation fail reflexivity while still technically satisfying symmetry/transitivity vacuously.
❌ One counter-example is enough to kill a property
WrongChecking a handful of pairs, seeing they behave, and concluding the relation is symmetric/transitive.
RightEvery single pair combination must satisfy the property. Just one violating pair — even hidden among many "good" ones — disproves the entire property.
❌ Mixing up One-One and Onto
WrongBelieving a function that "covers" the whole codomain (onto) must also be one-one, or vice versa.
RightThe two are independent properties. f(x) = x² on ℝ → [0,∞) is onto but not one-one; f(x) = 2x on ℤ → ℤ is one-one but not onto.
❌ Squaring both sides while proving one-one
WrongFrom f(x₁) = f(x₂) ⟹ x₁² = x₂², jumping straight to "x₁ = x₂, hence one-one" — forgetting the negative root.
Rightx₁² = x₂² gives x₁ = x₂ OR x₁ = −x₂. Unless the domain is restricted to exclude negatives, the function is not one-one.
❌ Using the n! bijection formula when |A| ≠ |B|
WrongApplying the "n!" bijection count even when the domain and codomain have different sizes.
RightA bijection can only exist between sets of the same cardinality. If |A| ≠ |B|, the number of bijections is 0 — check sizes match before reaching for n!.
❌ Forgetting to verify closure fails silently for binary operations
WrongDefining a*b by a formula (e.g. a − b on natural numbers) and assuming it's automatically a valid binary operation on that set.
RightA rule is a binary operation on A only if the output always lands back inside A. a − b on ℕ fails closure whenever b > a, since the result is negative.
❌ Wrong composition order
WrongComputing f∘g when the question asked for g∘f (or reading (g∘f)(x) as "g, then f" instead of "f, then g").
Right(g∘f)(x) = g(f(x)) — the function written on the right is applied first. Always apply innermost-first, exactly like nested brackets.
① Relations & Their Properties
Q1. Let A = {1, 2, 3, 4, 5} and let R = {(a, b) : |a − b| is a multiple of 2}. Show that R is an equivalence relation and list its equivalence classes.
View Solution
Reflexive

|a − a| = 0, and 0 is a multiple of 2 (0 = 2×0). So (a,a) ∈ R for every a. Reflexive ✔

Symmetric

|a − b| = |b − a| always, so if |a−b| is a multiple of 2, so is |b−a|. Symmetric ✔

Transitive

If |a−b| and |b−c| are both multiples of 2, then a and b have the same parity, and b and c have the same parity, so a and c have the same parity, making |a−c| a multiple of 2. Transitive ✔

Conclusion

R is reflexive, symmetric and transitive, hence an equivalence relation. It groups numbers by parity: [1] = [3] = [5] = {1,3,5} and [2] = [4] = {2,4}.

Q2. On the set of all lines in a plane, define R by: line ℓ₁ R ℓ₂ if ℓ₁ is perpendicular to ℓ₂. Check R for reflexivity, symmetry and transitivity.
View Solution
Reflexive

A line is never perpendicular to itself, so (ℓ, ℓ) ∉ R for any ℓ. Not reflexive ✘

Symmetric

If ℓ₁ ⊥ ℓ₂, then automatically ℓ₂ ⊥ ℓ₁ (perpendicularity is a mutual geometric property). Symmetric ✔

Transitive

Counter-example: let ℓ₁ ⊥ ℓ₂ and ℓ₂ ⊥ ℓ₃. Then ℓ₁ is actually parallel to ℓ₃, not perpendicular. So ℓ₁ R ℓ₃ fails. Not transitive ✘

Conclusion

R is symmetric only — neither reflexive nor transitive, so it is not an equivalence relation.

② Functions — One-One, Onto, Bijective
Q3. Show that f : ℝ → ℝ defined by f(x) = 3 − 4x is bijective, and find f⁻¹.
View Solution
One-one

Let f(x₁) = f(x₂): 3 − 4x₁ = 3 − 4x₂ ⟹ −4x₁ = −4x₂ ⟹ x₁ = x₂. So f is one-one.

Onto

Let y ∈ ℝ be arbitrary. Solve y = 3 − 4x for x: x = (3 − y)/4, which is a real number for every real y. So every y has a preimage — f is onto.

Bijective

One-one + onto ⟹ f is bijective, hence invertible.

Finding f⁻¹

From y = 3 − 4x, x = (3 − y)/4. So f⁻¹(y) = (3 − y)/4, i.e. f⁻¹(x) = (3 − x)/4.

Q4. Let A = {1, 2, 3, 4} and B = {p, q, r}. How many functions from A to B are onto? How many are one-one? Justify using the counting formulas.
View Solution
One-one check

|A| = 4 > |B| = 3. A one-one function needs distinct images for distinct inputs, which is impossible when the domain is larger than the codomain (pigeonhole principle). So the number of one-one functions is 0.

Onto count — setup

Use inclusion-exclusion: number of onto functions from an m-set to an n-set is Σ(−1)^k · ⁿCₖ · (n−k)^m, here m=4, n=3.

Onto count — computation

Σ = ³C₀·3⁴ − ³C₁·2⁴ + ³C₂·1⁴ − ³C₃·0⁴ = 81 − 3(16) + 3(1) − 0 = 81 − 48 + 3 = 36.

Conclusion

Onto functions: 36. One-one functions: 0 (domain larger than codomain rules this out entirely).

Q5. Is f : ℤ → ℤ given by f(x) = x² one-one? Is it onto? What if the domain is restricted to ℕ ∪ {0}?
View Solution
One-one on ℤ

f(2) = 4 and f(−2) = 4, but 2 ≠ −2. So f is not one-one on ℤ.

Onto on ℤ

Range of f is {0, 1, 4, 9, 16, ...} (perfect squares only), which is a strict subset of ℤ — negative integers and non-squares are never hit. Not onto.

Restricted domain ℕ ∪ {0}

Here f(x₁) = f(x₂) ⟹ x₁² = x₂² ⟹ x₁ = x₂ (since both are non-negative, the negative root is excluded). So f becomes one-one on this restricted domain, though it is still not onto ℤ (codomain would need to shrink to the perfect squares to make it onto).

③ Composition & Invertible Functions
Q6. If f(x) = 2x + 1 and g(x) = x² − 1, find (g∘f)(x) and (f∘g)(x), and verify they are not equal.
View Solution
Computing g∘f

(g∘f)(x) = g(f(x)) = g(2x+1) = (2x+1)² − 1 = 4x² + 4x + 1 − 1 = 4x² + 4x.

Computing f∘g

(f∘g)(x) = f(g(x)) = f(x² − 1) = 2(x² − 1) + 1 = 2x² − 2 + 1 = 2x² − 1.

Comparison

4x² + 4x ≠ 2x² − 1 in general (e.g. at x=1: g∘f gives 8, f∘g gives 1). This confirms composition is not commutative.

Q7. Show that f : ℝ − {2} → ℝ − {1} defined by f(x) = (x + 1)/(x − 2) is invertible, and find f⁻¹.
View Solution
One-one

Let f(x₁) = f(x₂): (x₁+1)/(x₁−2) = (x₂+1)/(x₂−2). Cross-multiplying: (x₁+1)(x₂−2) = (x₂+1)(x₁−2). Expanding both sides and simplifying gives −2x₁ + x₂ = −2x₂ + x₁, i.e. 3x₂ = 3x₁, so x₁ = x₂. One-one ✔

Onto

Let y = (x+1)/(x−2) with y ≠ 1. Cross-multiply: y(x−2) = x+1 ⟹ xy − 2y = x + 1 ⟹ x(y−1) = 1 + 2y ⟹ x = (1+2y)/(y−1). Since y ≠ 1, this is always a valid real x ≠ 2, so every y in the codomain has a preimage. Onto ✔

Inverse

From the onto step, x = (1+2y)/(y−1), so f⁻¹(x) = (1 + 2x)/(x − 1).

④ Binary Operations
Q8. On Q (rationals), define a ∗ b = a + b − ab. Show ∗ is commutative and associative, find the identity element, and find the inverse of an arbitrary element a ≠ 1.
View Solution
Commutative

a ∗ b = a + b − ab, and b ∗ a = b + a − ba = a + b − ab (since ab = ba). Equal, so ∗ is commutative.

Associative

(a∗b)∗c = (a+b−ab)∗c = (a+b−ab) + c − (a+b−ab)c = a+b+c−ab−ac−bc+abc. By symmetry of the expression in a,b,c, a∗(b∗c) simplifies to the same result. Associative ✔

Identity

Need a ∗ e = a: a + e − ae = a ⟹ e(1 − a) = 0 ⟹ e = 0 (for this to hold for all a, e = 0 is forced). Check: a ∗ 0 = a + 0 − 0 = a ✔. Identity element is e = 0.

Inverse

Need a ∗ b = 0: a + b − ab = 0 ⟹ b(1 − a) = −a ⟹ b = a/(a − 1), defined whenever a ≠ 1. So the inverse of a is a/(a−1).

Q9. Let A = {0, 1, 2, 3, 4, 5} and define ∗ as addition modulo 6. Verify closure, find the identity, and state which elements are invertible.
View Solution
Closure

For any a, b ∈ A, (a+b) mod 6 always lies in {0,1,2,3,4,5} = A by definition of the mod operation. Closed ✔

Identity

a ∗ 0 = (a+0) mod 6 = a for every a. Identity element is e = 0.

Invertibility

For each a, we need b with (a+b) mod 6 = 0, i.e. b = (6−a) mod 6. This always exists in A: 0↔0, 1↔5, 2↔4, 3↔3. Every element of A is invertible.

🕸️ Relation Graph Visualizer

Enter a set and its ordered pairs to draw a directed graph. Self-loops indicate reflexive pairs; the checklist below updates live once you draw.

node directed edge self-loop (reflexive pair)

🧩 Function Mapping Builder

Set your domain and codomain, then choose an image for every domain element using the dropdowns. The mapping diagram and classification update instantly.

🧪 Binary Operation Table Lab

Set a small base set (max 5 elements), then fill in every cell of the operation table using the dropdowns. The lab checks commutativity, associativity, identity and inverses live.

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ACADEMIA AETERNUM तमसो मा ज्योतिर्गमय · Est. 2025
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NCERT Class 12 Relations and Functions Notes
NCERT Class 12 Relations and Functions Notes — Complete Notes & Solutions · academia-aeternum.com
The chapter Relations and Functions lays the foundation for many advanced topics in mathematics, including calculus, inverse trigonometric functions, matrices, probability, and mathematical modelling. It introduces the concepts of relations, different types of functions, composition of functions, and invertible functions, enabling students to understand how mathematical objects are connected and transformed. In these notes, every topic is explained with detailed definitions, intuitive concepts,…
🎓 Class 12 📐 Mathematics 📖 NCERT ✅ Free Access 🏆 CBSE · JEE
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    Relation and Functions — Learning Resources

    Frequently Asked Questions

    A relation is a subset of the Cartesian product of two sets. It defines how elements of one set are associated with elements of another set.

    The main types are empty relation, universal relation, reflexive relation, symmetric relation, transitive relation, and equivalence relation.

    A relation can associate one element with multiple elements, whereas a function assigns exactly one image in the codomain to every element of the domain.

    An equivalence relation is a relation that is simultaneously reflexive, symmetric, and transitive.

    An injective function maps distinct elements to distinct images, a surjective function covers every element of the codomain, and a bijective function is both injective and surjective.

    The composition of functions combines two functions into one and is defined as \((g\circ f)(x)=g(f(x))\), where the output of the first function becomes the input of the second.

    An invertible function is a bijective function that has an inverse function \(f^{-1}\), satisfying \(f^{-1}\circ f=I_X\) and \(f\circ f^{-1}=I_Y\).

    It is the foundation for inverse trigonometric functions, calculus, matrices, probability, and advanced algebra, making it highly important for CBSE Boards, JEE Main, CUET, NDA, and other entrance examinations.

    A function is invertible if and only if it is both one-one (injective) and onto (surjective). This means every output has a unique pre-image.

    Key formulas include \(R\subseteq A\times B\), \((g\circ f)(x)=g(f(x))\), \(f^{-1}\circ f=I_X\), \(f\circ f^{-1}=I_Y\), and the conditions for injective, surjective, bijective, and equivalence relations.

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