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Mathematics Grade XII

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Chapter 1: Relations and Functions

Class 12 • Mathematics • Chapter 1

Relations and Functions

How elements of one set connect to another, and what makes a connection a true function.

CBSE / NCERT Aligned

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Chapter Roadmap

Types of Relations • Equivalence Relations • One-One Functions • Onto and Bijective • Composition • Invertible Functions • Key Results • Applications

Why Relations and Functions Matter

In Class 11 you learned what a relation and a function are. In Class 12 we look more closely at the kinds of relations and functions, because the differences between them decide what we are allowed to do with them. A relation can be reflexive, symmetric or transitive; a function can be one-one, onto, or both. These properties are the deciding factor for whether a function can be reversed, which matters throughout calculus and algebra.

This chapter sharpens that understanding. You will classify relations, recognise the very important equivalence relations that split a set into neat groups, tell apart one-one and onto functions, combine functions by composition, and find the inverse of a function when one exists. Each idea is small on its own, but together they form the language used in every later chapter.

Key idea

A relation just links elements; a function links each input to exactly one output. A function can be reversed only when it is both one-one (distinct inputs give distinct outputs) and onto (every output is reached). Such a function is called bijective.

Key Terms You Must Know

Term

Meaning

Example

Relation

A set of ordered pairs linking elements of one set to another.

R = {(1, 2), (2, 3)}

Reflexive

Every element is related to itself: (a, a) is in R.

(1, 1), (2, 2), … all present

Symmetric

If (a, b) is in R then (b, a) is too.

(1, 2) in R needs (2, 1) in R

Transitive

If (a, b) and (b, c) are in R then (a, c) is too.

links chain together

Equivalence relation

A relation that is reflexive, symmetric and transitive.

‘is parallel to’ on lines

One-one (injective)

Distinct inputs give distinct outputs.

f(x) = 2x + 3

Onto (surjective)

Every element of the codomain is an output.

f(x) = 2x + 3 on the reals

Core Concepts, Step by Step

1. Relations and Their Types

A relation R on a set A is just a collection of ordered pairs of elements of A. Three properties matter most. R is reflexive if every element is related to itself, that is (a, a) is in R for all a. R is symmetric if whenever (a, b) is in R, so is (b, a). R is transitive if whenever (a, b) and (b, c) are in R, then (a, c) is in R as well. The table below lays out the three properties with examples.

The three key properties of a relation

Property

Meaning

Quick example on {1, 2, 3}

Reflexive

(a, a) in R for every a

must contain (1,1), (2,2), (3,3)

Symmetric

(a, b) in R forces (b, a) in R

(1,2) present needs (2,1) present

Transitive

(a, b) and (b, c) force (a, c)

(1,2) and (2,3) need (1,3)

2. Equivalence Relations

A relation that is reflexive, symmetric and transitive all at once is called an equivalence relation. These are special because they sort the whole set into neat, non-overlapping groups called equivalence classes, where everything in a group is related to everything else in it. Everyday examples include ‘is parallel to’ for lines, ‘has the same remainder when divided by 5’ for integers, and ‘is the same age as’ for people.

3. One-One (Injective) Functions

A function f is one-one, or injective, if different inputs always give different outputs. The neat way to test this is to assume f(x₁) = f(x₂) and show that this forces x₁ = x₂. For example f(x) = 2x + 3 is one-one, because 2x₁ + 3 = 2x₂ + 3 leads straight to x₁ = x₂. By contrast f(x) = x² is not one-one, since 2 and −2 both map to 4.

A one-one function: distinct inputs, distinct outputs

One-one function: distinct inputs map to distinct outputs

4. Onto (Surjective) and Bijective Functions

A function f : A → B is onto, or surjective, if every element of the codomain B is the image of at least one element of A, so nothing in B is left out. To test it, take a general y in B and try to find an x with f(x) = y. A function that is both one-one and onto is called bijective. Bijective functions are the well-behaved ones that can be reversed, as the next concepts show.

Comparing the three kinds of function

Type

Condition

Example (on the real numbers)

One-one

distinct inputs give distinct outputs

f(x) = 2x + 3

Onto

every codomain value is reached

f(x) = 2x + 3

Bijective

both one-one and onto

f(x) = 3x − 4

Neither

fails one or both

f(x) = x² on the reals

A bijection: f(x) = x³ is one-one and onto on the real numbers

Graph of the cubic y = x cubed, a bijection

5. Composition of Functions

Two functions can be applied one after another. If f : A → B and g : B → C, the composition g ∘ f means ‘do f first, then g’, and is defined by (g ∘ f)(x) = g(f(x)). The order matters: in general g ∘ f and f ∘ g are not the same. For example, with f(x) = 2x + 3 and g(x) = x², we get (g ∘ f)(x) = (2x + 3)² but (f ∘ g)(x) = 2x² + 3, which are clearly different.

6. Invertible Functions

A function f is invertible if it can be undone by another function f⁻¹, so that applying f and then f⁻¹ returns the original input. The central fact is that a function is invertible if and only if it is bijective, that is one-one and onto. To find the inverse, write y = f(x), make x the subject, and then swap the roles of x and y. For instance, f(x) = 2x + 3 has inverse f⁻¹(x) = (x − 3)/2.

Key Results & Proofs

Three results capture the heart of this chapter, and each is proved by careful, plain reasoning.

Result 1: A Remainder Relation is an Equivalence Relation

Statement. On the integers, the relation a R b defined by ‘(a − b) is divisible by n’ is an equivalence relation.

Proof

Check the three properties one by one for the given relation.

Reflexive: a − a = 0, which is divisible by n, so a R a. – reflexive holds

Symmetric: if (a − b) is divisible by n, then so is (b − a), so b R a. – symmetric holds

Transitive: if (a − b) and (b − c) are divisible by n, their sum (a − c) is too, so a R c. – transitive holds

All three properties hold, so R is an equivalence relation. – conclusion

This relation sorts the integers into n classes, one for each possible remainder on division by n.

Result 2: Composition of One-One Functions is One-One

Statement. If f and g are both one-one, then g ∘ f is one-one.

Proof

Assume two inputs give the same output, then peel off g and then f.

(g ∘ f)(x₁)

=

(g ∘ f)(x₂)

assume the outputs are equal

g(f(x₁))

=

g(f(x₂))

meaning of composition

f(x₁)

=

f(x₂)

because g is one-one

x₁

=

x₂

because f is one-one

Equal outputs force equal inputs, so g ∘ f is one-one.

Result 3: Invertible Means Bijective

Statement. A function f : A → B is invertible if and only if it is one-one and onto.

Proof

If f is bijective, we can build its inverse directly.

Since f is onto, every y in B equals f(x) for some x in A. – onto gives a pre-image

Since f is one-one, that x is the only one, so it is unique. – one-one makes it unique

Define g(y) to be that unique x. Then g(f(x)) = x and f(g(y)) = y. – build the inverse g

So g undoes f, which means f is invertible. – conclusion

Conversely, any function with an inverse must be one-one and onto, so invertible and bijective mean the same thing.

Worked Examples

Example 1

Question: Is the relation R = {(1, 1), (2, 2), (3, 3)} on A = {1, 2, 3} reflexive?

▶ Show full working

A relation is reflexive when every element is related to itself.

For A = {1, 2, 3} we need (1,1), (2,2) and (3,3) all in R. – what reflexive needs

All three pairs are present. – check R

Answer: Yes, R is reflexive.

Example 2

Question: Is the relation R = {(1, 2), (2, 1)} on {1, 2} symmetric?

▶ Show full working

Symmetric means each pair has its reverse in R.

(1, 2) is in R, and its reverse (2, 1) is also in R. – check both pairs

Answer: Yes, R is symmetric.

Example 3

Question: On the integers, a R b means (a − b) is divisible by 3. Show that R is reflexive.

▶ Show full working

Check whether a R a holds for every integer a.

a − a

=

0

difference of a with itself

0 is divisible by 3, so a R a for every integer a. – reflexive holds

Answer: R is reflexive.

Example 4

Question: Show that ‘is parallel to’ on the set of lines in a plane is an equivalence relation.

▶ Show full working

Check the three properties for parallel lines.

Reflexive: every line is parallel to itself. – reflexive

Symmetric: if l is parallel to m, then m is parallel to l. – symmetric

Transitive: if l is parallel to m and m to n, then l is parallel to n. – transitive

Answer: It is reflexive, symmetric and transitive, so an equivalence relation.

Example 5

Question: Is f : R → R, f(x) = 2x + 3, one-one?

▶ Show full working

Assume f(x₁) = f(x₂) and see whether x₁ must equal x₂.

2x₁ + 3

=

2x₂ + 3

set the outputs equal

2x₁

=

2x₂

subtract 3

x₁

=

x₂

divide by 2

Answer: Yes, f is one-one.

Example 6

Question: Is f : R → R, f(x) = x², one-one?

▶ Show full working

Look for two different inputs with the same output.

f(2)

=

4

compute one value

f(−2)

=

4

compute another

Two different inputs give the same output. – one-one fails

Answer: No, f is not one-one.

Example 7

Question: Is f : R → R, f(x) = 2x + 3, onto?

▶ Show full working

Take any y in the codomain and try to find an x with f(x) = y.

2x + 3

=

y

set f(x) = y

x

=

(y − 3)/2

solve for x

This x is real for every real y, so every y is reached. – onto check

Answer: Yes, f is onto.

Example 8

Question: Is f : N → N, f(x) = 2x, onto?

▶ Show full working

List the outputs and see whether any codomain value is missed.

The outputs are 2, 4, 6, …, all even. – what f produces

Odd numbers such as 3 are never outputs. – a value is missed

Answer: No, f is not onto.

Example 9

Question: If f(x) = 2x + 3 and g(x) = x², find (g ∘ f)(x) and (f ∘ g)(x).

▶ Show full working

Composition means substitute one function into the other.

(g ∘ f)(x)

=

g(f(x)) = (2x + 3)²

put f inside g

(f ∘ g)(x)

=

f(g(x)) = 2x² + 3

put g inside f

Answer: (g ∘ f)(x) = (2x + 3)² and (f ∘ g)(x) = 2x² + 3.

Example 10

Question: Find the inverse of f(x) = 2x + 3.

▶ Show full working

Write y = f(x), make x the subject, then swap x and y.

y

=

2x + 3

start

x

=

(y − 3)/2

solve for x

f⁻¹(x)

=

(x − 3)/2

replace y by x

Answer: f⁻¹(x) = (x − 3)/2.

Example 11

Question: Show that f : R → R, f(x) = 3x − 4, is bijective and find its inverse.

▶ Show full working

Check one-one and onto, then invert.

One-one: 3x₁ − 4 = 3x₂ − 4 gives x₁ = x₂. – one-one

Onto: for any y, x = (y + 4)/3 gives f(x) = y. – onto

f⁻¹(x)

=

(x + 4)/3

invert as before

Answer: f is bijective and f⁻¹(x) = (x + 4)/3.

Example 12

Question: On {1, 2, 3, 4}, a R b means a = b. Identify the type of relation.

▶ Show full working

Write out R and check the three properties.

R = {(1,1), (2,2), (3,3), (4,4)}, the identity relation. – list the pairs

It is reflexive, symmetric and transitive. – check all three

Answer: It is an equivalence relation (the identity relation).

Where You Meet This in Real Life

Sorting and grouping

Equivalence relations are exactly how we sort things into groups, such as students by house or items by category, where everything in a group shares a property.

Encoding and decoding

An invertible function is one that can be undone, which is the basis of every code that must be read back, from passwords to file compression.

Databases

Relations in the mathematical sense are the foundation of relational databases, which link records in one table to records in another.

Computer functions

A program function takes inputs to outputs; whether it is one-one or onto decides whether its result can be reversed or repeated safely.

Conversions

Changing units, currencies or temperatures uses one-one functions, so the conversion can always be reversed exactly.

Practice Sets A to D

Practice Set A – Basics

A1. Is R = {(1, 1), (2, 2)} on {1, 2} reflexive?

▶ Reveal full working

Check for (1,1) and (2,2).

Both (1,1) and (2,2) are present.

So every element relates to itself.

Answer: Yes, it is reflexive.

A2. Is f : R → R, f(x) = x + 1, one-one?

▶ Reveal full working

Assume equal outputs.

x₁ + 1

=

x₂ + 1

set outputs equal

x₁

=

x₂

subtract 1

Answer: Yes, it is one-one.

A3. Is f : R → R, f(x) = x², onto?

▶ Reveal full working

Can a negative number be an output?

Squares are never negative, so −4 is never an output.

Answer: No, it is not onto.

A4. Find the inverse of f(x) = x − 5.

▶ Reveal full working

Solve y = x − 5 for x.

f⁻¹(x)

=

x + 5

add 5 to undo the subtraction

Answer: f⁻¹(x) = x + 5.

Practice Set B – Conceptual

B1. Define a reflexive relation.

▶ Reveal full working

Think about an element and itself.

A relation R on A is reflexive if (a, a) is in R for every a in A.

Answer: Every element is related to itself.

B2. Which three properties make a relation an equivalence relation?

▶ Reveal full working

Recall the definition.

It must be reflexive, symmetric and transitive.

Answer: Reflexive, symmetric and transitive.

B3. What is the difference between a one-one and an onto function?

▶ Reveal full working

One looks at inputs, the other at the codomain.

One-one: distinct inputs give distinct outputs.

Onto: every codomain element is some output.

Answer: One-one is about distinct outputs; onto is about covering the codomain.

B4. When is a function invertible?

▶ Reveal full working

Recall the key condition.

Exactly when it is both one-one and onto, that is bijective.

Answer: When it is bijective.

Practice Set C – Application / Numerical

C1. On the integers, a R b means (a − b) is divisible by 5. Show that R is symmetric.

▶ Reveal full working

Start from a R b.

If (a − b) is divisible by 5, then (b − a) = −(a − b) is too.

So b R a, which means R is symmetric.

Answer: R is symmetric.

C2. If f(x) = 3x + 2 and g(x) = x − 1, find (g ∘ f)(x).

▶ Reveal full working

Substitute f into g.

(g ∘ f)(x)

=

g(3x + 2) = (3x + 2) − 1

put f inside g

(g ∘ f)(x)

=

3x + 1

simplify

Answer: (g ∘ f)(x) = 3x + 1.

C3. Is f : R → R, f(x) = x³, bijective?

▶ Reveal full working

Check one-one and onto.

One-one: x₁³ = x₂³ gives x₁ = x₂.

Onto: every real has a real cube root, so every y is reached.

Answer: Yes, it is bijective.

C4. Find the inverse of f(x) = (x + 1)/2.

▶ Reveal full working

Solve y = (x + 1)/2 for x.

2y

=

x + 1

multiply by 2

x

=

2y − 1

make x the subject

f⁻¹(x)

=

2x − 1

swap x and y

Answer: f⁻¹(x) = 2x − 1.

Practice Set D – HOTS / Word Problems

D1. Show that ‘has the same remainder on division by 4’ on the integers is an equivalence relation.

▶ Reveal full working

Check the three properties.

Reflexive: a has the same remainder as itself.

Symmetric: if a and b share a remainder, so do b and a.

Transitive: if a, b share a remainder and b, c share one, then a, c do.

So it is an equivalence relation.

Answer: It is reflexive, symmetric and transitive.

D2. Prove that f : R → R, f(x) = 5x + 7, is one-one.

▶ Reveal full working

Assume equal outputs.

5x₁ + 7

=

5x₂ + 7

set outputs equal

5x₁

=

5x₂

subtract 7

x₁

=

x₂

divide by 5

Answer: f is one-one.

D3. With f(x) = 2x and g(x) = x + 3, find (f ∘ g)(x) and (g ∘ f)(x), and show they differ.

▶ Reveal full working

Compose in each order.

(f ∘ g)(x)

=

2(x + 3) = 2x + 6

f after g

(g ∘ f)(x)

=

2x + 3

g after f

2x + 6 is not equal to 2x + 3. – so order matters

Answer: They differ: 2x + 6 versus 2x + 3.

D4. Find the inverse of f(x) = 4x − 3 and verify that f(f⁻¹(x)) = x.

▶ Reveal full working

Invert, then substitute back.

f⁻¹(x)

=

(x + 3)/4

solve y = 4x − 3 for x

f(f⁻¹(x))

=

4 · (x + 3)/4 − 3

substitute the inverse

f(f⁻¹(x))

=

x + 3 − 3 = x

simplify

Answer: f⁻¹(x) = (x + 3)/4, and the check gives x.

Chapter Summary

Everything in One Glance

Relations

Reflexive: (a, a) always in R. Symmetric: (a, b) forces (b, a). Transitive: (a, b) and (b, c) force (a, c).

Equivalence Relations

Reflexive, symmetric and transitive together; they sort a set into equivalence classes.

One-One

Distinct inputs give distinct outputs; test by assuming f(x₁) = f(x₂).

Onto and Bijective

Onto: every codomain value is reached. Bijective: both one-one and onto.

Composition

(g ∘ f)(x) = g(f(x)); order matters, so g ∘ f need not equal f ∘ g.

Invertible

A function is invertible exactly when it is bijective; find the inverse by solving y = f(x) for x.

Are You Exam-Ready?

8-Point Exam Quick-Check

Is R = {(1,1), (2,2), (3,3), (1,2)} on {1,2,3} reflexive?

Give an example of a symmetric relation on {1, 2}.

Is f(x) = 4x − 1 one-one? Justify in one line.

Is f : N → N, f(x) = x + 2, onto?

If f(x) = x + 1 and g(x) = 2x, find (g ∘ f)(x).

State the condition for a function to be invertible.

Find the inverse of f(x) = 3x.

Name the three properties of an equivalence relation.

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Class 12 Mathematics Chapter 1: Relations and Functions, Complete Notes and Practice

These free Class 12 Maths Chapter 1 Relations and Functions notes follow the latest NCERT 2026-27 syllabus and give complete, exam-ready coverage for board exams and competitive foundations. Includes worked examples, step-by-step proofs and graded practice, free on SchoolRevise.com.