Chapter 1 of 16

JEE Main Mathematics★ ~1 question (4 marks) in JEE Main, usually a counting question on relations or a domain question, and the function vocabulary defined here is assumed by every calculus chapter that follows

Sets, Relations and Functions

Every object in this chapter is a set of ordered pairs, and every property is a condition on which pairs are present. Read that way, the dozen counting formulas stop being a list to memorise and become one argument repeated.

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Timed practiceMCQs from this chapter with instant scoring

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JEE Main PYQsSolved previous-year questions

By the end of this chapter you'll be able to…

1Read every object in the chapter as a subset of a Cartesian product, and every property as a condition on which pairs are present
2Rebuild the counting formulas from independent binary choices rather than recalling them, including the power set, relations, reflexive and symmetric counts
3Test a relation for reflexivity, symmetry and transitivity, and produce the standard counterexample showing the last two do not force the first
4Use the correspondence between equivalence relations and partitions to count equivalence relations without a formula
5Classify a function as one-one, onto, into or bijective, and count injections and surjections with the feasibility check applied first
6Evaluate composites in the correct order, state precisely which properties transfer, and find domains by intersecting conditions and ranges by demanding a real solution

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Why this chapter matters in JEE Main

A three-element set carries 512 relations. Sixty-four are reflexive, sixty-four are symmetric, eight are both, and exactly five are equivalence relations. No formula produces that five, and the attempt to find one is what makes the chapter feel like an unmanageable list. The five comes from counting partitions instead: three singletons, a pair with a singleton left over in three ways, or one block of three. That is the whole chapter in one line, because every object here is a set of ordered pairs and every property is a condition on which pairs are present. Reflexive forces the diagonal in. Symmetric locks the rest into mirror couples. A function restricts how many pairs may leave each element. Once you see the cells, each counting formula is a fifteen-second argument you can rebuild under pressure rather than a fact you have to recall correctly.

Before you start — revise these

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Fluency with interval notation and inequalities on the real line

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The multiplication principle and combinations, from Permutations and Combinations

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Solving a quadratic and reading the sign of its discriminant

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Sketching simple graphs well enough to apply a horizontal-line test

Sets, Relations and Functions

A set with three elements. Its Cartesian product with itself has $9$ pairs, so it carries $2^{9} = 512$ relations.

Of those $512$ , how many are equivalence relations?

The instinct is to count each property and combine. Reflexive relations number $2^{6} = 64$ . Symmetric relations also number $64$ . Relations that are both come to $2^{3} = 8$ . So the equivalence relations, being reflexive and symmetric and transitive, must be some part of that $8$ .

The answer is $5$ .

Property demanded	Count on a $3$ -element set
nothing	$512$
reflexive	$64$
symmetric	$64$
reflexive and symmetric	$8$
all three, i.e. equivalence	$5$

Three of the eight die, and transitivity kills them. Take the relation containing the diagonal, the couple ${(1, 2), (2, 1)}$ and the couple ${(1, 3), (3, 1)}$ , but not ${(2, 3), (3, 2)}$ . Then $2 R 1$ and $1 R 3$ force $2 R 3$ , which is absent. The same accident happens for each of the three choices of which single element is left out, so exactly $3$ of the $8$ fail.

The right way to get $5$ never mentions relations. Cut ${1, 2, 3}$ into non-overlapping pieces: three singletons, or a pair with a singleton left over ( $3$ ways), or one block of three. That is $1 + 3 + 1 = 5$ .

Equivalence relations are partitions. That is the chapter in one line, and it is why no formula exists for counting them.

The organising idea	What it buys
Every object here is a set of ordered pairs	relations, functions and their counts are one subject
Every property is a condition on which pairs are present	each formula becomes a fifteen-second argument, not a memory

1. Sets, Representation and the Power Set

Set. A well-defined collection of distinct objects.

Well-defined is doing real work. "The set of tall students" is not a set, because membership is not decidable.

Form	Example
Roster	$A = {2, 3, 5, 7}$
Set-builder	$A = {x : x prime, x < 10}$

Two rules catch people out.

Order is irrelevant. ${1, 2}$ and ${2, 1}$ are the same set.

Repetition does not count. ${1, 1, 2}$ has two elements, not three.

Power set $P (A)$ : the set of all subsets of $A$ , including $\emptyset$ and $A$ itself.

Each element is in or out of a given subset, independently.

$n elements \Rightarrow n binary choices \Rightarrow ∣ P (A) ∣ = 2^{n}$

Notice the shape of that argument. Every counting formula below reuses this exact skeleton: find the independent choices, then raise $2$ to their number.

Illustration 1

How many subsets of a $6$ -element set have an even number of elements?

The tempting route is to add binomial coefficients: $(0 6) + (2 6) + (4 6) + (6 6) = 1 + 15 + 15 + 1 = 32$ .

The independent-choice route is faster and needs no arithmetic. Fix one element, say $x$ . Choose freely whether each of the other $5$ elements is in or out, giving $2^{5} = 32$ half-formed subsets. Now $x$ has no choice at all: it goes in exactly when it is needed to make the size even.

So even-sized subsets number $2^{5} = 32$ , and by the same argument so do odd-sized ones. The two counts are equal for every $n \geq 1$ , which the binomial sum hides.

2. Operations and Their Algebra

Law	Statement
Commutative	$A \cup B = B \cup A$ , same for $\cap$
Associative	$(A \cup B) \cup C = A \cup (B \cup C)$
Distributive	$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ , and the reverse
Identity	$A \cup \emptyset = A$ , $A \cap U = A$
Idempotent	$A \cup A = A$ , $A \cap A = A$

Trap. In arithmetic, multiplication distributes over addition but addition does not distribute over multiplication. In set algebra each operation distributes over the other. JEE has exploited that asymmetry.

De Morgan's laws. Complementing swaps the two operations.

$(A \cup B)^{'} = A^{'} \cap B^{'}, (A \cap B)^{'} = A^{'} \cup B^{'}$

In words, and this fixes them permanently: not in either one means outside both; not in both means outside at least one.

Illustration 2

Check the second distributive law, $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$ , on $A = {1, 2}$ , $B = {2, 3}$ , $C = {3, 4}$ , and see why the arithmetic analogue fails.

Left side: $B \cap C = {3}$ , so $A \cup {3} = {1, 2, 3}$ .

Right side: $A \cup B = {1, 2, 3}$ and $A \cup C = {1, 2, 3, 4}$ , whose intersection is ${1, 2, 3}$ .

They agree. Now run the arithmetic analogue with $\cup$ as $+$ and $\cap$ as $\times$ : $2 + (3 \times 4) = 14$ , while $(2 + 3) \times (2 + 4) = 30$ .

Set operations are idempotent and arithmetic ones are not, and that is exactly the difference. Expanding the right side produces $A \cup A$ terms, which collapse back to $A$ ; expanding $(2 + 3) (2 + 4)$ produces $2 \times 2$ , which does not collapse to $2$ .

Counting with overlaps

Shared elements would otherwise be counted twice.

$n (A \cup B) = n (A) + n (B) - n (A \cap B)$

$n (A \cup B \cup C) = \sum n (A) - \sum n (A \cap B) + n (A \cap B \cap C)$

Inclusion-exclusion: add singles, subtract pairs, add the triple back.

Illustration 3

Of $200$ readers, $100$ take paper $A$ , $90$ take $B$ , $80$ take $C$ ; $40$ take $A$ and $B$ , $35$ take $B$ and $C$ , $30$ take $C$ and $A$ , and $15$ take all three. How many take exactly two?

The pairwise figures already include the triple-takers, and that is the whole difficulty.

At least one: $100 + 90 + 80 - 40 - 35 - 30 + 15 = 180$ .

Each of the three pairwise counts contains the $15$ triple-takers, so subtract them once from each pair.

$exactly two = (40 + 35 + 30) - 3 \times 15 = 105 - 45 = 60$

Exactly one is then $180 - 60 - 15 = 105$ , and the check holds: $105 + 60 + 15 = 180$ .

Trap. "Both $A$ and $B$ " means $n (A \cap B) = 40$ . "Only $A$ and $B$ " means $40 - 15 = 25$ . Read which one the question asked before computing anything.

3. The Cartesian Product: Where Pairs Come From

Cartesian product $A \times B$ : the set of all ordered pairs with first entry from $A$ and second from $B$ .

$∣ A \times B ∣ = mn$

Ordered means $(1, 2) \neq = (2, 1)$ , and therefore $A \times B \neq = B \times A$ unless $A = B$ or one of them is empty.

This is the pivot of the chapter. Everything from here is a subset of some Cartesian product; the topics differ only in which subsets are allowed.

Illustration 4

For $A = {1, 2, 3}$ and $B = {2, 3, 4}$ , find $(A \times B) \cap (B \times A)$ .

A pair $(x, y)$ lies in both products exactly when $x$ is in $A$ and in $B$ , and $y$ likewise.

So the intersection is $(A \cap B) \times (A \cap B)$ , here ${2, 3} \times {2, 3}$ , giving $(2, 2), (2, 3), (3, 2), (3, 3)$ .

$∣ (A \times B) \cap (B \times A) ∣ = ∣ A \cap B ∣^{2} = 2^{2} = 4$

The general identity is worth keeping: the product operation distributes over intersection on both sides.

4. Relations as Sets of Pairs

Relation from $A$ to $B$ : any subset of $A \times B$ .

That is the whole definition. No formula, no pattern and no rule is required.

$A \times B$ has $mn$ elements; a subset takes or leaves each one.

$N_{relations} = 2^{mn}$

On a single $n$ -element set the pairs live in $A \times A$ , so there are $2^{n^{2}}$ relations, which is the $512$ the chapter opened with.

Named extreme	Contents
Empty relation	no pairs at all
Universal relation	every pair of $A \times A$

Both are legitimate relations, and JEE tests whether you will accept them.

Domain of a relation: the first entries actually used. Range: the second entries actually used. Neither has to exhaust $A$ or $B$ .

5. Types of Relations

Each named type is a condition on which pairs must be present, so each definition converts directly into a count.

Type	Condition	Count on an $n$ -set
Reflexive	every $(a, a)$ present	$2^{n^{2} - n}$
Symmetric	$(a, b)$ present forces $(b, a)$	$2^{n (n + 1) /2}$
Transitive	$(a, b)$ and $(b, c)$ force $(a, c)$	no closed formula
Equivalence	all three	Bell number

$reflexive: n diagonal pairs forced \Rightarrow n^{2} - n free \Rightarrow 2^{n^{2} - n}$

$symmetric: n diagonal free + \frac{n ( n - 1 )}{2} couples free \Rightarrow 2^{n (n + 1) /2}$

$both: diagonal forced, couples free \Rightarrow 2^{n (n - 1) /2}$

Transitivity has no formula because it is not a partition of the cells into independent choices: whether one cell may be empty depends on which others are filled. If a question asks you to count transitive relations, the set will be small enough to enumerate.

Illustration 5

On $S = {1, 2, 3, 4, 6, 12}$ let $a R b$ mean " $a$ divides $b$ ". Test all three properties.

Reflexive: every number divides itself, so all six diagonal pairs are present. Yes.

Symmetric: $2 R 4$ holds because $2$ divides $4$ , but $4 R 2$ fails. No.

Transitive: if $a ∣ b$ and $b ∣ c$ then $b = k a$ and $c = l b = l k a$ , so $a ∣ c$ . Yes.

So divisibility is reflexive and transitive but not symmetric. It is in fact antisymmetric: $a ∣ b$ and $b ∣ a$ together force $a = b$ for positive integers.

Reflexive, antisymmetric and transitive is a partial order, not an equivalence relation. The word "partial" records that some pairs are simply incomparable: neither $4 ∣ 6$ nor $6 ∣ 4$ .

The trap that catches almost everyone

Does symmetric plus transitive imply reflexive?

The argument looks complete: $a R b$ gives $b R a$ by symmetry, and the two together give $a R a$ by transitivity.

It fails, because the argument needs some $b$ with $a R b$ to exist before it can start. An element related to nothing at all never enters the chain.

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

Symmetric, transitive, and not reflexive, because $(3, 3)$ is missing.

Trap. The identity relation is exactly the diagonal and nothing else. A reflexive relation must contain the diagonal but may contain anything else too. Every identity relation is reflexive; the converse is false.

6. Equivalence Relations and Partitions

Equivalence relation: reflexive, symmetric and transitive at once. This is the most examined idea in the chapter.

Its importance is structural. For each $a$ , collect everything related to $a$ into the equivalence class $[a]$ .

Those classes are either identical or completely disjoint. They never overlap partially, and the proof is one line: if $c$ lies in $[a]$ and in $[b]$ then $a R c$ and $c R b$ , so $a R b$ , so the two classes coincide.

$equivalence relation on A ⟷ partition of A$

Both directions hold. Every partition defines an equivalence relation, by relating two elements when they land in the same block.

That correspondence converts a hard question into an easy one: counting equivalence relations means counting partitions.

$n$	Equivalence relations (Bell number)
1	1
2	2
3	5
4	15
5	52

Illustration 6

Which three of the eight reflexive-and-symmetric relations on ${1, 2, 3}$ fail transitivity, and how do the surviving five match the five partitions?

Each of the eight is the diagonal plus a choice of which of the three couples ${1, 2}$ , ${1, 3}$ , ${2, 3}$ to include.

Couples included	Transitive?	Partition
none	yes	${1} {2} {3}$
any one of the three	yes	that pair, plus a singleton
any two of the three	no	none
all three	yes	${1, 2, 3}$

Take couples ${1, 2}$ and ${1, 3}$ but not ${2, 3}$ . Then $2 R 1$ and $1 R 3$ , so transitivity demands $2 R 3$ , which is absent. The three two-couple relations fail for exactly this reason, in each case because the missing couple is forced by the other two.

So $8 - 3 = 5$ , matching $1 + 3 + 1$ exactly, and the failure has a clean meaning: two couples sharing an element already merge all three elements into one block.

Illustration 7

On the integers define $a R b$ when $a - b$ is divisible by $5$ . Verify it is an equivalence relation and describe its classes.

Reflexive: $a - a = 0$ , divisible by $5$ .

Symmetric: $a - b = 5 k \Rightarrow b - a = - 5 k$ .

Transitive: $a - b = 5 k$ and $b - c = 5 l \Rightarrow a - c = 5 (k + l)$ .

The classes group integers by remainder on division by $5$ , giving exactly five classes: $[0], [1], [2], [3], [4]$ .

Every integer lies in precisely one, and together they cover $Z$ . That is the partition the theory promised, and it is why this relation is written $a \equiv b (mod 5)$ .

7. Functions: A Relation with a Rule

Function $f : A \to B$ : a relation in which every element of $A$ is the first entry of exactly one pair.

Two conditions hide in that sentence, and separating them makes the definition usable.

Condition	What it forbids	Example ruled out
every $a$ is used	an input with no image	$f (x) = 1/ x$ on all of $R$
no $a$ is used twice	an input with two images	$x^{2} + y^{2} = 1$ as $y$ of $x$

Nothing at all is required of $B$ . Codomain elements may be hit many times or not at all, and that freedom is the entire content of the next section.

Each of the $m$ elements of $A$ chooses independently among the $n$ elements of $B$ .

$N_{functions} = n^{m}$

Trap. The exponent is the size of the domain, because the domain elements do the choosing. Reversing it is the single most common slip in this chapter.

Range: the values actually taken, always a subset of the codomain. Keeping range and codomain apart is essential, because the gap between them is the whole meaning of onto.

8. One-One, Onto and Into

Term	Meaning	Condition
One-one (injective)	distinct inputs, distinct outputs	$f (a_{1}) = f (a_{2}) \Rightarrow a_{1} = a_{2}$
Onto (surjective)	every codomain element is hit	range $=$ codomain
Into	at least one is missed	range is a proper subset
Bijective	both at once	a perfect pairing

Vocabulary warning. In Indian textbooks and in the JEE syllabus, into means specifically not onto. Much of the international literature uses it merely to mean a function with a stated codomain, with no such implication. Follow the JEE usage in the exam.

To prove one-one: assume $f (a_{1}) = f (a_{2})$ , derive $a_{1} = a_{2}$ . Graphically, no horizontal line may cut the curve twice. Strict monotonicity gives it free.

To prove onto: take arbitrary $y$ in the codomain and produce an $x$ mapping to it. To disprove, exhibit one unreachable value.

$N_{injective} = \frac{n !}{( n - m )!}, m \leq n$

The first domain element has $n$ choices, the next $n - 1$ because it must avoid the first, and so on for $m$ factors.

$N_{surjective} = \sum_{r = 0}^{n} (- 1)^{r} (r n) (n - r)^{m}$

That is inclusion-exclusion again: subtract the functions missing at least one codomain element, then correct for double subtraction.

When $m = n$ the two notions coincide and the count of bijections is $n!$ . That coincidence is a finite-set fact and does not survive on infinite sets.

Illustration 8

From a $3$ -element set to a $5$ -element set, count the one-one functions and the onto functions.

One-one: the three inputs pick distinct images in order, giving $5 \times 4 \times 3 = 60$ , which is $5! /2!$ .

Onto: zero. Three inputs produce at most three distinct outputs, so at least two of the five codomain elements are never reached. No arithmetic is needed; the pigeonhole principle settles it.

Reverse the sets and both answers flip. From a $5$ -set to a $3$ -set there are no injections at all, and the surjections number $3^{5} - 3 \cdot 2^{5} + 3 \cdot 1^{5} = 243 - 96 + 3 = 150$ .

Trap. Check feasibility before reaching for a formula. The injective formula is meaningless when $m > n$ , and the surjective sum silently returns $0$ when $m < n$ .

Illustration 9

Show that $f : R \to R$ given by $f (x) = \frac{x}{1 + ∣ x ∣}$ is one-one but not onto, even though domain and codomain are the same set.

One-one. For $x \geq 0$ the function is $x / (1 + x) = 1 - 1/ (1 + x)$ , which strictly increases. For $x \leq 0$ it is $x / (1 - x)$ , also strictly increasing. The two pieces agree at $0$ , so $f$ increases strictly on all of $R$ and is therefore injective.

Not onto. Since $∣ f (x) ∣ = ∣ x ∣/ (1 + ∣ x ∣) < 1$ always, no value outside $(- 1, 1)$ is ever attained. The range is $(- 1, 1)$ , a proper subset of $R$ .

So an injective self-map of an infinite set need not be surjective. On a finite set that cannot happen, which is precisely what makes infinite sets different, and JEE has used this gap to build "one-one but into" options.

9. Composition of Functions

For $f : A \to B$ and $g : B \to C$ :

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

Read right to left. The function written first is applied last, and misreading the order is the commonest error in composition questions.

Property	Holds?
Associative, $h \circ (g \circ f) = (h \circ g) \circ f$	yes
Commutative, $f \circ g = g \circ f$	no, in general

How properties travel is one-directional, and the asymmetry is examinable.

If	Then
$f$ and $g$ both one-one	$g \circ f$ is one-one
$f$ and $g$ both onto	$g \circ f$ is onto
$g \circ f$ is one-one	$f$ is one-one; $g$ need not be
$g \circ f$ is onto	$g$ is onto; $f$ need not be

The inner function controls injectivity of the composite; the outer function controls surjectivity.

Illustration 10

For $f (x) = \frac{x - 1}{x + 1}$ , find $f \circ f$ and hence $f$ applied four times.

$f (f (x)) = \frac{\frac{x - 1}{x + 1} - 1}{\frac{x - 1}{x + 1} + 1} = \frac{( x - 1 ) - ( x + 1 )}{( x - 1 ) + ( x + 1 )} = \frac{- 2}{2 x} = - \frac{1}{x}$

The common denominator $x + 1$ cancels top and bottom, which is why the result is so much simpler than the starting expression.

Now apply the same map twice more.

$f^{4} (x) = f^{2} (f^{2} (x)) = - \frac{1}{- 1/ x} = x$

So $f$ composed with itself four times is the identity, and $f$ has order $4$ under composition. Asking for $f^{2026}$ is then trivial: $2026 = 4 \times 506 + 2$ , so the answer is $f^{2} (x) = - 1/ x$ .

10. Finding Domains and Ranges

Domain questions are reliable marks and reduce to a checklist of what a real expression forbids.

Expression	Requirement
$g (x)$	$g (x) \geq 0$
$1/ g (x)$	$g (x) \neq = 0$
$lo g g (x)$	$g (x) > 0$
$1/ g (x)$	$g (x) > 0$ , strictly

When several appear together, impose every condition and intersect the resulting sets.

Trap. Solving the conditions separately and reporting their union is a standard and costly mistake. Each condition is a requirement, not an alternative.

Ranges have no checklist. Two methods cover almost everything in JEE.

Invert. Write $y = f (x)$ , solve for $x$ , and ask which $y$ give a real answer.

Use monotonicity. On an interval where $f$ is monotonic, the range runs between the endpoint values.

Illustration 11

Find the range of $f (x) = \frac{x - 1}{x ^{2} - 3 x + 3}$ over all real $x$ .

First check the denominator never vanishes: its discriminant is $9 - 12 = - 3 < 0$ , so the domain is all of $R$ and no exclusion is needed.

Set $y = f (x)$ and clear the fraction into a quadratic in $x$ .

$y x^{2} - (3 y + 1) x + (3 y + 1) = 0$

For a real $x$ to exist the discriminant must be non-negative.

$(3 y + 1)^{2} - 4 y (3 y + 1) \geq 0 \Rightarrow (3 y + 1) [(3 y + 1) - 4 y] \geq 0 \Rightarrow (3 y + 1) (1 - y) \geq 0$

$- \frac{1}{3} \leq y \leq 1$

The case $y = 0$ has to be checked separately, because the quadratic degenerates to a linear equation there. It gives $x = 1$ , which is legitimate, so $0$ stays in the range.

$range = [- \frac{1}{3}, 1]$

Both endpoints are attained, at the $y$ values where the discriminant is exactly zero, which is the signature of a maximum or minimum reached rather than approached.

Summary

Every object here is a set of ordered pairs, and every property is a condition on which pairs are present.

Independent binary choices generate every count: $∣ P (A) ∣ = 2^{n}$ , and each formula below is the same argument with different cells forced or locked.

Each set operation distributes over the other, unlike arithmetic, because set operations are idempotent. De Morgan's laws say complementing swaps union with intersection.

Inclusion-exclusion adds singles, subtracts pairs, adds the triple back; "exactly two" needs the triple removed three times over.

A relation from $A$ to $B$ is any subset of $A \times B$ , so there are $2^{mn}$ of them, with no pattern required.

Reflexive relations number $2^{n^{2} - n}$ , symmetric ones $2^{n (n + 1) /2}$ , both together $2^{n (n - 1) /2}$ ; transitivity has no closed formula because its cells are not independent.

Symmetric plus transitive does not give reflexive: an element related to nothing never enters the argument.

Equivalence relations correspond exactly to partitions, which is why a $3$ -element set carries $5$ of them and not $8$ or $512$ .

A function is a relation using every domain element exactly once, so there are $n^{m}$ of them, with the domain size in the exponent.

Injections number $n! / (n - m)!$ and need $m \leq n$ ; surjections need inclusion-exclusion and need $m \geq n$ ; when the finite sets match in size the two coincide, and on infinite sets they do not.

In a composite the function written first is applied last. A one-one composite forces the inner function one-one, an onto composite forces the outer one onto, and neither converse holds.

Domain conditions are imposed and then intersected, never united. Ranges come from inverting and demanding a real solution, or from monotonicity.

Key formulas & results

Everything to memorise for the exam hall, in one card. Screenshot this for revision.

The organising principle

every object here is a set of ordered pairs, and every property is a condition on which pairs are present

Reflexive forces the diagonal in, symmetric locks the rest into mirror couples, and a function restricts how many pairs leave each element. Every counting formula is a consequence.

Size of the power set

|P(A)| = 2^n for a set A with n elements

Each element is independently in or out, giving n binary choices. Every other count in the chapter reuses this skeleton: find the free choices, raise 2 to their number.

De Morgan's laws

(A \cup B)^{'} = A^{'} \cap B^{'} an d (A \cap B)^{'} = A^{'} \cup B^{'}

Read in words: not in either one means outside both; not in both means outside at least one. Complementing swaps the two operations.

Inclusion-exclusion for three sets

n (A \cup B \cup C) = n (A) + n (B) + n (C) - n (A \cap B) - n (B \cap C) - n (C \cap A) + n (A \cap B \cap C)

Add singles, subtract pairs, add the triple back. For exactly two, take the sum of the pairwise counts and subtract three times the triple, because each pairwise figure already contains it.

Size of a Cartesian product

∣ A \times B ∣ = mn, an d t h e p ai r s a r eor d er e d

Ordered means A \times B is not B \times A unless the sets are equal. Their intersection is (A \cap B) \times (A \cap B), of size |A \cap B| squared.

Number of relations

N = 2^{mn} f r o m A t o B, an d 2^{n^{2}} o na s in g l e n - e l e m e n t se t

A relation is any subset of the product, with no requirement of a formula or pattern. The empty and universal relations both count.

Reflexive, symmetric, and both

2^{n^{2} - n} r e f l e x i v e, 2^{n (n + 1) /2} sy mm e t r i c, 2^{n (n - 1) /2} b o t h

Diagonal forced leaves n^2-n free; symmetry leaves the diagonal free plus n(n-1)/2 couples; both together leave only the couples. On n = 3 that is 64, 64 and 8.

Why transitivity has no formula

the cells are not independent: whether one may be empty depends on which others are filled

On a three-element set only 5 of the 8 reflexive-and-symmetric relations survive transitivity, because two couples sharing an element force the third.

Equivalence relations and partitions

equivalence relations on a set correspond exactly to partitions of it

Count partitions instead. The Bell numbers run 1, 2, 5, 15, 52 for n = 1 to 5, and each is obtained by listing block structures rather than by any formula.

Number of functions

N = n^m from a domain of size m to a codomain of size n

The domain size is the exponent, because the domain elements do the choosing. Reversing it is the commonest slip in the chapter.

Injections and surjections

n!/(n-m)! injections, valid only for m at most n; sum over r of (-1)^r C(n,r)(n-r)^m surjections, needing m at least n

Check feasibility before computing. From a 3-set to a 5-set there are 60 injections and zero surjections, by pigeonhole and with no arithmetic.

Composition and what it transfers

(g \circ f) (x) = g (f (x)); a o n e - o n eco m p os i t e f or ces f o n e - o n e, an o n t oco m p os i t e f or ces g o n t o

Read right to left: the function written first is applied last. Neither converse holds, because the composite never evaluates g outside the range of f.

Domain checklist, then intersect

even root needs argument at least 0; denominator non-zero; logarithm strictly positive; reciprocal of a root strictly positive

Every condition is a requirement, so intersect them. Reporting the union is a standard and costly error.

JEE Main test series

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Sets, Relations and Functions shows up in every JEE Main paper — sit a timed one and find out where you actually stand.

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Traps JEE Main sets — and how to dodge them

These are the exact option-traps and misreads that cost marks under negative marking.

WATCH OUT

✗ Putting the codomain size in the exponent when counting functions

✓ Ask who is doing the choosing. Each domain element must pick exactly one image, so the number of independent choices equals the domain size and the count is n^m. Rebuilding that sentence is faster than remembering which way round it went.

Why it happens: Both set sizes are in play and the formula is recalled rather than rebuilt.

WATCH OUT

✗ Concluding that a symmetric and transitive relation must be reflexive

✓ The chain never starts unless some b with aRb exists, and an element related to nothing is untouched by it. On {1,2,3}, R = {(1,1),(1,2),(2,1),(2,2)} is symmetric and transitive but misses (3,3). Keep that counterexample ready, because JEE has asked this directly.

Why it happens: The chain aRb, then bRa, then aRa looks like a complete proof.

WATCH OUT

✗ Reaching for a formula to count equivalence relations

✓ There is none. Count partitions instead. On a three-element set the answer is 1 + 3 + 1 = 5, obtained by listing block structures, and the Bell numbers 1, 2, 5, 15, 52 cover every case JEE can ask.

Why it happens: Reflexive and symmetric each had a clean power of two, so transitivity looks as though it should too.

WATCH OUT

✗ Confusing exactly two with at least two in inclusion-exclusion

✓ Subtract the triple count once from each of the three pairwise counts, so exactly two equals the sum of the pairs minus three times the triple. Check the arithmetic by confirming exactly one plus exactly two plus all three returns the at-least-one total.

Why it happens: The pairwise figures given in the question already include everyone who took all three.

WATCH OUT

✗ Uniting the conditions when finding a domain

✓ Each condition is something the expression requires, not an alternative it offers, so intersect. For a square root divided by a logarithm both requirements hold simultaneously, and the domain is where every one of them is satisfied at once.

Why it happens: Several conditions appear and the word or slips in where and belongs.

WATCH OUT

✗ Assuming a one-one function on the reals must be onto

✓ That coincidence is a finite-set fact. The function x/(1 + |x|) from R to R is strictly increasing, hence injective, but its range is only the open interval from minus one to one. Infinite domains break the equivalence, and JEE builds one-one-but-into options on exactly this gap.

Why it happens: On a finite set of matching size, injective and surjective really do coincide.

Exam-pattern practice

PYQ-style questions with full solutions. Work through them as a readiness check — mark yourself honestly and get your gap report at the end.

Readiness check

Are you exam-ready for Sets, Relations and Functions?

12 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

12 questions~8 min

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•512 relations on a three-element set, 64 reflexive, 64 symmetric, 8 both, and only 5 equivalence relations
•Every count is a tally of cells: which are forced in, which are locked into couples, which are free
•Set operations distribute over each other because they are idempotent; arithmetic ones do not
•De Morgan: complementing swaps union and intersection
•Exactly two equals the sum of the pairwise counts minus three times the triple
•A relation is any subset of the product; the empty and universal relations are legitimate
•Symmetric plus transitive does not force reflexive, because an element related to nothing never enters the chain
•Equivalence relations are partitions, which is the only practical way to count them
•Functions number n^m with the domain size in the exponent; injections need m at most n and surjections need m at least n
•In a composite the function written first is applied last; one-one travels inward, onto travels outward
•Domain conditions are intersected, never united; ranges come from inverting and demanding a real root

JEE Main question blueprint

How this topic is asked, tier by tier — so you can prep to the pattern.

Typical weightage: 4

Question style	Marks each	Typical count
Relations and their types	2	1
Functions, domain and range	1	1
Sets, counting and composition	1	1

Exam-hall strategy

Battle-tested tips from mentors and toppers for this topic under the sectional clock.

For any counting question, draw or imagine the grid of pairs and ask of each cell whether it is forced in, locked to another, or free. Almost every formula in this chapter is recoverable in fifteen seconds that way, which is faster and safer than recalling four similar-looking powers of two.
Check feasibility before applying an injection or surjection formula. If the domain is bigger than the codomain there are no injections; if it is smaller there are no surjections. Both answers are zero with no arithmetic, and JEE sets options that reward spotting it.
When a question asks for equivalence relations, stop counting relations and start listing partitions. For n at most 4 the enumeration takes under a minute and the Bell numbers 1, 2, 5, 15, 52 confirm it.
In survey questions, underline whether the wording says both or only, and at least or exactly, before writing a single number. The arithmetic is short; the reading is where the marks are lost.
For domains, list every condition the expression imposes, solve each on its own line, then take the intersection on the number line rather than in your head. For ranges, invert and demand a real solution, remembering to test the degenerate value separately.

Beyond the exam

Where this skill shows up in the job you're competing for — and in life.

Equivalence relations are how any classification scheme i…

Equivalence relations are how any classification scheme is made rigorous: blood groups, congruence classes in modular arithmetic, and the residue classes behind every checksum and cryptographic key all rely on the guarantee that classes never partially overlap

Relational databases take the definition literally

Relational databases take the definition literally, storing a relation as a table of tuples, so a join is an operation on subsets of a Cartesian product and inclusion-exclusion is what a query planner uses to estimate result sizes

Hash tables and file deduplication depend on injectivity:…

Hash tables and file deduplication depend on injectivity: a hash that is one-one on the data actually stored gives collision-free lookup, and the whole design problem is that a finite codomain makes true injectivity impossible on unbounded input

Where else this topic is tested

Prepare once, score in every exam that asks it.

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Questions aspirants ask

Pulled from the Q&A community and mentor sessions.

Because the reflexive and symmetric counts come from partitioning the cells of the grid into independent choices, and transitivity does not partition that way. Reflexivity forces the three diagonal cells and leaves the other six entirely free, giving 2 to the sixth. Symmetry locks the off-diagonal cells into mirror couples, each wholly in or wholly out. In both cases whether one cell is filled has no bearing on whether another may be. Transitivity is different: whether the cell (2,3) may be empty depends on whether (2,1) and (1,3) are filled, so the choices interlock and no product of twos describes them. On a three-element set this is why only 5 of the 8 reflexive-and-symmetric relations survive, and the three that die are exactly those containing two couples that share an element.

The codomain is the target set you declare when you write f from A to B, and it is a matter of choice. The range is the set of values the function actually produces, and it is a matter of fact. The range is always contained in the codomain, and onto means precisely that the two are equal. This matters because the same rule can be onto or into depending only on how you declared the codomain: x squared from R to R is into, since negatives are never produced, while x squared from R to the non-negative reals is onto with no change to the rule itself. When a JEE option says into, it is a claim about that declared codomain, so read the arrow before judging the formula.

Use monotonicity where you can. A function that is strictly increasing everywhere, or strictly decreasing everywhere, is automatically one-one, because distinct inputs must give distinct outputs. That usually means computing a derivative and checking it keeps one sign, which is often a two-line job where the algebraic route is long. The graphical version is the horizontal-line test: if some horizontal line meets the curve twice, two inputs share an output. Watch for functions built in pieces, such as x/(1 + |x|), where each piece increases and the pieces agree at the join, so the whole thing still increases. Piecewise definitions that jump back down are exactly where the shortcut fails.

Because the composite only reports what comes out at the far end, and the outer function can repair anything the inner one missed. Let f send 1 to 1 and 2 to 2 from {1,2} into {1,2,3}, so f misses 3 and is not onto. Let g send 1 to 1, 2 to 2 and 3 to 2 from {1,2,3} onto {1,2}. Then the composite hits both 1 and 2, so it is onto, while f is not. The general statement is that surjectivity travels outward and injectivity travels inward: an onto composite forces the outer function onto, and a one-one composite forces the inner function one-one. Neither converse holds, and the reason is the same in both directions, that the composite never evaluates g outside the range of f.

Invert. Write y equal to the expression, clear denominators into an equation in x, and ask for which y a real solution exists. For a rational function with a quadratic on top or bottom this produces a quadratic in x, and the condition is that its discriminant be non-negative, which is an inequality in y that you solve directly. For (x-1)/(x^2-3x+3) this gives (3y+1)(1-y) at least 0, so the range is the closed interval from minus one third to one. Two checks matter. Handle separately any y that makes the quadratic degenerate to a linear equation, since the discriminant test does not apply there. And note that endpoints where the discriminant is exactly zero are attained rather than approached, which is how you decide between a closed and an open bracket.