Relations and Functions Quick Lesson: Class 12 Maths Chapter 1 (NCERT)

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The chapter in one minute

What relations and functions are, and what the exam asks.

A relation says which pairs of elements are linked, like "has the same remainder as". You test a relation for three properties: reflexive, symmetric and transitive.

A function sends each input to exactly one output. You test a function for two properties: one-one and onto.

To show that a property holds, it must work for every element: check them all in a small list, or argue with letters for a big set. To show that it fails, one example with numbers is enough.

Most exam questions on this chapter are one of eight kinds, and each kind has a fixed recipe. You will learn all eight below.

Composition and inverse functions (Section 1.4 of the book) are not in the CBSE syllabus for 2025-26 or 2026-27, so they are not examined and this page leaves them out.

8question types

9cards to learn

3tests for an equivalence relation

1example with numbers is enough to say no

Definitions

The definitions you must know

Learn these cards. Every question in the chapter is a test of one of them.

Reflexive

\begin{aligned} &(a, a) \in R \\ &\text{for every } a \in A \end{aligned}

Every element is related to itself.

Symmetric

\begin{aligned} &(a, b) \in R \\ &\Rightarrow (b, a) \in R \end{aligned}

Every pair comes with its reverse.

Transitive

\begin{aligned} &(a, b), (b, c) \in R \\ &\Rightarrow (a, c) \in R \end{aligned}

Every chain from $a$ to $b$ to $c$ needs the short cut from $a$ to $c$ ⁠.

Equivalence relation

\begin{aligned} &\text{reflexive} \\ &+ \text{symmetric} \\ &+ \text{transitive} \end{aligned}

All three at once. Missing one means it is not an equivalence relation.

Equivalence class

\begin{aligned} [a] = \{&b \in A : \\ &(b, a) \in R\} \end{aligned}

All the elements related to $a$ ⁠. For an equivalence relation, the classes split $A$ into groups that do not overlap.

One-one (injective)

\begin{aligned} &f(x_1) = f(x_2) \\ &\Rightarrow x_1 = x_2 \end{aligned}

Different inputs always give different outputs.

Onto (surjective)

\begin{aligned} &\text{for every } y \in Y, \\ &f(x) = y \text{ for some } x \in X \end{aligned}

Every element of the codomain $Y$ is an output: range = codomain.

Bijective

\text{one-one} + \text{onto}

Prove the two tests one after the other, then write "so $f$ is bijective".

Counting one-one maps

n! = n(n - 1)\cdots 2\cdot 1

The number of one-one maps from a set of $n$ elements to itself. They are also the onto maps. For example, $3! = 6$ ⁠.

Question types

The question types and how to crack each one

For each type: how to spot it, what to do, and one real NCERT question worked the exam way.

1. A relation given as a list of pairs: check each property

Spot it: $R$ is written out pair by pair, on a small set like $\{1, 2, 3\}$ ⁠.

Reflexive: is $(a, a)$ in $R$ for every element $a$ of the set? One missing pair means no.
Symmetric: for each pair in $R$ ⁠, is its reverse in $R$ too?
Transitive: for each chain $(a, b), (b, c)$ in $R$ ⁠, is $(a, c)$ in $R$ ⁠? Here $c$ may be the same as $a$ ⁠.

Exercise 1.1 Q6. Show that $R = \{(1, 2), (2, 1)\}$ in the set $\{1, 2, 3\}$ is symmetric but neither reflexive nor transitive.

Symmetric: $(1, 2), (2, 1) \in R$ ⁠, and each is the reverse of the other. Thus $R$ is symmetric.

Reflexive: $1$ is in the set, but $(1, 1) \notin R$ ⁠. Thus $R$ is not reflexive.

Transitive: $(1, 2), (2, 1) \in R$ ⁠, but $(1, 1) \notin R$ ⁠. Thus $R$ is not transitive.

∴ $R$ is symmetric but neither reflexive nor transitive.

See a full exam answer · Go deeper: see it visually · Basics: reflexive, symmetric, transitive

In past papers (4): click to see

Board 2025 Q37 · Sample 2025-26 Q36 · Sample 2024-25 Q37(iii)(a) · Sample 2020-21 Q2

2. A relation given by a rule: letters to prove, numbers to disprove

Spot it: $R$ is given by a condition, such as $a \le b$ on $\R$ ⁠, and the set is too big to list every pair.

To show a property holds, start from any element (or any pair in $R$ ⁠) and use the rule.
To show a property fails, give one pair of numbers where it breaks, and show the arithmetic.
End each part with "Thus $R$ is …" or "Thus $R$ is not …".

Exercise 1.1 Q4. Show that $R = \{(a, b) : a \le b\}$ in $\R$ is reflexive and transitive but not symmetric.

Reflexive: $a \le a$ for every $a$ ⁠, so $(a, a) \in R$ ⁠. Thus $R$ is reflexive.

Transitive: $a \le b$ and $b \le c$ give $a \le c$ ⁠, so $(a, c) \in R$ ⁠. Thus $R$ is transitive.

Symmetric: $(1, 2) \in R$ ⁠, as $1 \le 2$ ⁠, but $(2, 1) \notin R$ ⁠, as $2 > 1$ ⁠. Thus $R$ is not symmetric.

∴ $R$ is reflexive and transitive but not symmetric.

See a full exam answer · Go deeper: see it visually · Basics: how to prove and disprove

In past papers (3): click to see

Board 2024 Q26(a) · Board 2023 Q32 · Sample 2022-23 Q33(b)

3. Show that a relation is an equivalence relation

Spot it: the words "is an equivalence relation", often with "2 divides $a - b$ " or " $|a - b|$ is even".

Reflexive: show $(a, a) \in R$ for any $a$ ⁠.
Symmetric: suppose $(a, b) \in R$ ⁠, and show $(b, a) \in R$ ⁠.
Transitive: suppose $(a, b), (b, c) \in R$ ⁠, and show $(a, c) \in R$ ⁠. With differences, use $a - c = (a - b) + (b - c)$ ⁠.
Finish with "Hence $R$ is an equivalence relation".

Example 5. In $\Z$ ⁠, $(a, b) \in R$ when 2 divides $a - b$ ⁠. Show that $R$ is an equivalence relation.

Reflexive: $a - a = 0 = 2 \times 0$ ⁠, so $(a, a) \in R$ ⁠. Thus $R$ is reflexive.

Symmetric: if $a - b = 2k$ ⁠, with $k \in \Z$ ⁠, then $b - a = 2(-k)$ ⁠, so $(b, a) \in R$ ⁠. Thus $R$ is symmetric.

Transitive: if $a - b = 2k$ and $b - c = 2m$ ⁠, with $k, m \in \Z$ ⁠, then

\begin{aligned} a - c &= (a - b) + (b - c) \\ &= 2(k + m) \end{aligned}

so $(a, c) \in R$ ⁠. Thus $R$ is transitive.

∴ $R$ is an equivalence relation.

See a full exam answer · Go deeper: see it visually · Basics: even, odd and remainders

In past papers (5): click to see

Board 2026 Q32(a) · Sample 2026-27 Q32(a) · Sample 2023-24 Q33(a) · Sample 2022-23 Q33(a) · Sample 2020-21 Q29

4. Equivalence classes: everything related to one element

Spot it: "find the set of all elements related to 1", or "show that the elements of $\{1, 3, 5\}$ are related to each other".

Write the rule with the given element in it, for example " $|1 - b|$ is a multiple of 4".
Find every $b$ in the set that fits. Throw out values that are not in the set.
Write the answer as a set.

Exercise 1.1 Q9. In $A = \{0, 1, \dots, 12\}$ ⁠, $(a, b) \in R$ when $|a - b|$ is a multiple of 4. Find the set of all elements related to 1. (Part (i).)

For $b \in A$ ⁠, $|1 - b| \le 11$ ⁠. So $|1 - b| = 0, 4 \text{ or } 8$ ⁠.

This gives $b = 1, 5, 9$ ⁠. The other values, $-3$ and $-7$ ⁠, are not in $A$ ⁠.

∴ The set of elements related to 1 is $\{1, 5, 9\}$ ⁠.

See a full exam answer · Go deeper: see it visually · Basics: equivalence classes

In past papers (2): click to see

Sample 2023-24 Q33(a) · Sample 2020-21 Q29

5. Prove one-one and onto: the bijective function

Spot it: "show that $f$ is one-one and onto", or "is $f$ bijective?", with a formula like $f(x) = 3 - 4x$ ⁠.

One-one: start from $f(x_1) = f(x_2)$ and simplify until you reach $x_1 = x_2$ ⁠.
Onto: take any $y$ in the codomain and solve $f(x) = y$ for $x$ ⁠. Check this $x$ is in the domain.
Put your $x$ back in and show that $f(x) = y$ ⁠.

Exercise 1.2 Q7. Is $f : \R \to \R$ ⁠, $f(x) = 3 - 4x$ ⁠, one-one, onto or bijective? (Part (i).)

\begin{aligned} f(x_1) &= f(x_2) \\ \Rightarrow 3 - 4x_1 &= 3 - 4x_2 \\ \Rightarrow x_1 &= x_2 \end{aligned}

So $f$ is one-one. For $y \in \R$ ⁠, take $x = \frac{3 - y}{4} \in \R$ ⁠. Then

\begin{aligned} f(x) &= 3 - (3 - y) \\ &= y \end{aligned}

∴ $f$ is one-one and onto, so it is bijective.

See a full exam answer · Go deeper: see it visually · Basics: one-one and onto

In past papers (7): click to see

Board 2025 Q19 · Board 2024 Q1 · Board 2024 Q26(b) · Sample 2024-25 Q37(iii)(b) · Sample 2023-24 Q33(b) · Sample 2022-23 Q21(b) · Sample 2020-21 Q1(a)

6. Not one-one or not onto: one example is enough

Spot it: functions such as $x^2$ ⁠, $|x|$ or $[x]$ on $\R$ ⁠, where two inputs share an output, or some numbers are never reached.

Not one-one: find two different inputs with the same output, such as $f(-1) = f(1)$ ⁠.
Not onto: find one number in the codomain that is never an output, and say why.

Exercise 1.2 Q4. Show that $f : \R \to \R$ ⁠, $f(x) = |x|$ ⁠, is neither one-one nor onto.

$f(-1) = 1 = f(1)$ ⁠, but $-1 \ne 1$ ⁠. So $f$ is not one-one.

$|x| \ge 0$ for every $x$ ⁠, so $-1$ is not an output. So $f$ is not onto.

∴ $f$ is neither one-one nor onto.

See a full exam answer · Go deeper: see it visually · Basics: one-one and onto

In past papers (3): click to see

Sample 2026-27 Q32(b) · Sample 2024-25 Q20 · Sample 2022-23 Q21(b)

7. A fraction function with a point left out

Spot it: a fraction such as $f(x) = \frac{x - 2}{x - 3}$ ⁠, with one number taken out of the domain and one out of the codomain.

One-one: write $f(x_1) = f(x_2)$ ⁠, cross-multiply, expand and cancel. You reach $x_1 = x_2$ ⁠.
Onto: put $f(x) = y$ and make $x$ the subject.
Check that this $x$ is allowed, that is, it is not the number left out.

Exercise 1.2 Q10. Is $f : \R - \{3\} \to \R - \{1\}$ ⁠, $f(x) = \frac{x - 2}{x - 3}$ ⁠, one-one and onto?

One-one: cross-multiply $f(x_1) = f(x_2)$ ⁠.

\begin{aligned} &(x_1 - 2)(x_2 - 3) \\ &= (x_2 - 2)(x_1 - 3) \\ \Rightarrow\ &-3x_1 - 2x_2 \\ &= -2x_1 - 3x_2 \\ \Rightarrow\ &x_1 = x_2 \end{aligned}

So $f$ is one-one. Onto: for $y \ne 1$ ⁠, solve $f(x) = y$ ⁠.

\begin{aligned} x - 2 &= y(x - 3) \\ \Rightarrow x &= \frac{3y - 2}{y - 1} \end{aligned}

This $x$ is never 3, because $x = 3$ would need $3y - 2 = 3y - 3$ ⁠. So $x$ is in the domain, $f(x) = y$ ⁠, and $f$ is onto.

∴ Yes, $f$ is one-one and onto.

See a full exam answer · Go deeper: see it visually · Basics: what a function is

In past papers (1): click to see

Board 2026 Q32(b)

8. Counting questions: list the possibilities

Spot it: "the number of relations (or functions) …" on a small set like $\{1, 2, 3\}$ ⁠, usually an MCQ.

Start with the smallest relation: the pairs you are given, plus every pair the properties force.
Try adding each remaining pair, and see what else it forces.
Count the relations that work. For one-one maps of a set of $n$ elements to itself, the count is $n!$ ⁠.

Miscellaneous Q7 (MCQ). Let $A = \{1, 2, 3\}$ ⁠. The number of equivalence relations containing $(1, 2)$ is (A) 1 (B) 2 (C) 3 (D) 4.

The smallest one is

\begin{aligned} S_1 = \{&(1, 1), (2, 2), \\ &(3, 3), (1, 2), \\ &(2, 1)\} \end{aligned}

Adding any other pair, such as $(2, 3)$ ⁠, forces all the rest (by symmetry, then transitivity), which gives $A \times A$ ⁠. So there are 2: $S_1$ and $A \times A$ ⁠.

∴ Answer: (B) 2

See a full exam answer · Go deeper: see it visually · Basics: counting choices

In past papers (2): click to see

Sample 2024-25 Q37(i) · Sample 2020-21 Q1(b)

Shortcuts

Shortcuts and quick tips

Small habits that save time and marks.

Test reflexive first. One missing pair $(a, a)$ settles it in one line.
For an equivalence relation on a small set, write out the classes. Every element must be in exactly one class: a quick check on your answer.
If $R$ has no chain $(a, b), (b, c)$ at all, then $R$ is transitive, because nothing can fail.
A rule of the form "same something" (same remainder, same number of pages, same distance from the origin) always gives an equivalence relation.
With "2 divides $a - b$ ", write $a - b = 2k$ and work with $k$ ⁠.
From a finite set to itself, one-one and onto go together. Prove one and you have both.
Onto depends on the codomain: $f(x) = 2x$ is onto from $\R$ to $\R$ ⁠, but not from $\N$ to $\N$ ⁠.
Squares and modulus on $\R$ ⁠: $f(-1) = f(1)$ breaks one-one, and a negative number breaks onto.
In an MCQ on a list of pairs, test reflexive, symmetric and transitive in turn, and cross out options as you go.

Watch out

Mistakes that cost marks

Each one is common, and each has an easy fix.

Showing that a property holds with a few numbers → use letters $a, b, c$ ⁠. Numbers are only for showing that it fails.
Saying "not symmetric" with no pair → name the pair: $(1, 2) \in R$ but $(2, 1) \notin R$ ⁠.
Missing chains that come back to the start → $(1, 2), (2, 1)$ needs $(1, 1)$ ⁠.
Checking reflexive only for elements that appear in $R$ → check every element of the set, even one that is in no pair.
Testing onto against $\R$ when the codomain is different → every element of the given codomain must be an output.
Solving $f(x) = y$ and stopping → check that the $x$ you found is in the domain.

Marks

How the examiner gives marks

From CBSE's own marking schemes. Write these lines and the marks follow.

"Show that $R$ is an equivalence relation" (5 marks in the 2026-27 sample paper): reflexive, symmetric and transitive earn 1½ marks each, and the line "Hence $R$ is an equivalence relation" earns ½.
Each property needs a proof with letters, ending in words: "Thus $R$ is symmetric".
Not one-one: give a clear pair, like $f(-1) = f(1)$ ⁠. Not onto: name a number in the codomain that is not in the range. In that paper each of these earned 1½ marks.
Write the conclusion in words, such as "so $f$ is one-one" or "so $f$ is not onto". Conclusions carry marks.
An MCQ or assertion-reason question earns 1 mark for the right option only. Working earns nothing there, so keep it short.
A slip is penalised only once, so keep going. Any correct method gets full marks.

Practice

Practise these first

Easiest first. Together they cover every type above.

Exercise 1.1 Q15 (MCQ) a list of pairs: test each property
Exercise 1.2 Q12 (MCQ) a straight-line function that is bijective
Exercise 1.2 Q11 (MCQ) a fourth power: neither one-one nor onto
Exercise 1.1 Q3 write the pairs out, then test them
Exercise 1.1 Q2 a rule that fails all three tests: three counterexamples
Exercise 1.1 Q8 an equivalence relation with even differences
Exercise 1.1 Q14 parallel lines, and the lines related to one line
Exercise 1.2 Q2 squares and cubes on different sets
Exercise 1.2 Q1 one over x, with a point left out
Miscellaneous Q6 (MCQ) counting relations
Miscellaneous Q1 a harder fraction function that is bijective

Eight recipes, one at a time. Work through the practice list, prove with letters and disprove with numbers, and you are ready for the exam.

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Relations and functions, the quick way