JEE Maths — Sets Relations and Functions Complete Chapter Guide

Chapter Overview & Weightage

Sets, Relations and Functions is one of those chapters that looks deceptively simple in Class 11 but shows up in JEE in ways that catch unprepared students off guard. The questions rarely test raw definitions — they test whether you actually understand what a function does.

Weightage: 3–4% in JEE Main (1–2 questions per paper). JEE Advanced tests this indirectly through other chapters like Calculus and Vectors, so your investment here compounds over the paper.

Year	JEE Main Questions	Marks	Topics Tested
2024	1–2	4–8	Equivalence relations, inverse functions
2023	1	4	Composition of functions, domain
2022	2	8	Types of functions, relations on sets
2021	1	4	Equivalence classes, onto functions
2020	1–2	4–8	Bijections, set operations

The pattern is clear: JEE Main favors equivalence relations and function type questions. Get these two areas right and you’ve covered roughly 70% of what this chapter contributes.

Key Concepts You Must Know

Ranked by how often they show up in PYQs:

Relations (High Priority)

Reflexive, symmetric, transitive properties — and how to check them systematically
Equivalence relations and equivalence classes (this is a JEE favorite)
Number of relations vs. number of equivalence relations on a set of $n$ elements

Functions (High Priority)

One-one (injective): every element in codomain has at most one preimage
Onto (surjective): every element in codomain has at least one preimage
Bijection: both injective and surjective simultaneously
Composition $f \circ g$ and its domain restrictions

Inverse Functions (Medium Priority)

Inverse exists only for bijections
$(f \circ g)^{-1} = g^{-1} \circ f^{-1}$ — the reversal rule
Graphical interpretation: reflection about $y = x$

Sets (Lower Priority in JEE Main, but foundational)

De Morgan’s laws: $(A \cup B)' = A' \cap B'$ and $(A \cap B)' = A' \cup B'$
Number of subsets of an $n$ -element set: $2^n$
Power set, Cartesian product

Important Formulas

For a set $A$ with $n$ elements:

Total number of relations on $A$ = $2^{n^2}$
Total number of reflexive relations = $2^{n^2 - n}$
Total number of symmetric relations = $2^{n(n+1)/2}$
Total number of reflexive + symmetric relations = $2^{n(n-1)/2}$

When to use: Questions of the form “How many relations on $\{1,2,3\}$ are reflexive?” — plug in $n=3$ .

If $|A| = m$ and $|B| = n$ :

Total functions from $A$ to $B$ : $n^m$
One-one functions (requires $n \geq m$ ): $\frac{n!}{(n-m)!} = {}^nP_m$
Onto functions from $A$ to $B$ (requires $m \geq n$ ): use inclusion-exclusion $\sum_{k=0}^{n} (-1)^k \binom{n}{k}(n-k)^m$

When to use: Any question asking “how many functions satisfy…” — identify the type first, then pick the right formula.

$(f \circ g)(x) = f(g(x))$ — apply $g$ first, then $f$
$f \circ g \neq g \circ f$ in general (not commutative)
$f \circ (g \circ h) = (f \circ g) \circ h$ (associative)
If $f$ and $g$ are both one-one, so is $f \circ g$
If $f$ and $g$ are both onto, so is $f \circ g$

When to use: Whenever the question gives you two functions and asks about their composition’s type.

$f^{-1}$ exists $\iff$ $f$ is a bijection.

If $f: A \to B$ and $g: B \to A$ satisfy $f \circ g = I_B$ and $g \circ f = I_A$ , then $g = f^{-1}$ .

When to use: Questions that give you a function and ask whether its inverse exists, or ask you to find $f^{-1}(x)$ .

Solved Previous Year Questions

PYQ 1 — JEE Main 2023 (Equivalence Relation)

Question: Let $R$ be a relation on $\mathbb{Z}$ defined by $aRb \iff a - b$ is divisible by 5. Show that $R$ is an equivalence relation and find the equivalence class of 2.

Solution:

We check all three properties.

Reflexive: $a - a = 0 = 5 \times 0$ . So $aRa$ holds for all $a \in \mathbb{Z}$ . ✓

Symmetric: If $aRb$ , then $5 \mid (a-b)$ . This means $5 \mid (-(a-b)) = (b-a)$ , so $bRa$ . ✓

Transitive: If $aRb$ and $bRc$ , then $5 \mid (a-b)$ and $5 \mid (b-c)$ . Adding: $5 \mid (a-b+b-c) = (a-c)$ , so $aRc$ . ✓

Since all three hold, $R$ is an equivalence relation.

Equivalence class of 2: We need all integers $a$ such that $a - 2$ is divisible by 5.

[2] = \{..., -8, -3, 2, 7, 12, 17, ...\} = \{a \in \mathbb{Z} : a \equiv 2 \pmod{5}\}

The five equivalence classes here are $[0], [1], [2], [3], [4]$ — they partition $\mathbb{Z}$ completely. This is the key idea: equivalence classes always form a partition of the set.

PYQ 2 — JEE Main 2024 Shift 1 (Type of Function)

Question: Let $f: \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = \frac{x}{1+x^2}$ . Then $f$ is:

(A) one-one and onto
(B) neither one-one nor onto
(C) one-one but not onto
(D) onto but not one-one

Solution:

Check one-one: Suppose $f(x_1) = f(x_2)$ .

\frac{x_1}{1+x_1^2} = \frac{x_2}{1+x_2^2}

x_1(1+x_2^2) = x_2(1+x_1^2)

x_1 - x_2 = x_1 x_2^2 - x_2 x_1^2 = x_1 x_2(x_2 - x_1)

x_1 - x_2 + x_1 x_2(x_1 - x_2) = 0

(x_1 - x_2)(1 + x_1 x_2) = 0

So either $x_1 = x_2$ or $x_1 x_2 = -1$ . Since $x_1 x_2 = -1$ gives $x_1 \neq x_2$ (take $x_1 = 1, x_2 = -1$ : $f(1) = \frac{1}{2}$ and $f(-1) = \frac{-1}{2}$ , which are different anyway — wait, let’s check directly: $f(1) = \frac{1}{2}$ , $f(2) = \frac{2}{5}$ , $f(-1) = -\frac{1}{2}$ ).

Actually $f(1) = \frac{1}{2}$ and $f(-1) = -\frac{1}{2}$ , so those give different outputs. The factor $(1 + x_1 x_2) = 0$ case: let $x_1 = t, x_2 = -1/t$ . Then $f(t) = \frac{t}{1+t^2}$ and $f(-1/t) = \frac{-1/t}{1 + 1/t^2} = \frac{-1/t}{(t^2+1)/t^2} = \frac{-t}{t^2+1}$ . These are negatives of each other — not equal unless $t = 0$ . So actually $f$ is one-one? Let me re-examine.

The equation $(x_1 - x_2)(1 + x_1 x_2) = 0$ means $f(x_1) = f(x_2) \implies x_1 = x_2$ or $x_1 x_2 = -1$ . But when $x_1 x_2 = -1$ , we showed the outputs are negatives — not equal. So the only case forcing equality is $x_1 = x_2$ . Therefore $f$ is one-one.

Check onto: The range of $f(x) = \frac{x}{1+x^2}$ . By AM-GM, for $x > 0$ : $1 + x^2 \geq 2x$ , so $\frac{x}{1+x^2} \leq \frac{1}{2}$ . Similarly for $x < 0$ , $f(x) \geq -\frac{1}{2}$ . The range is $\left[-\frac{1}{2}, \frac{1}{2}\right]$ , not all of $\mathbb{R}$ .

So $f$ is one-one but not onto. Answer: (C)

PYQ 3 — JEE Main 2022 (Composition & Inverse)

Question: If $f(x) = \frac{2x+3}{x-2}$ , find $f^{-1}(x)$ and verify that $f(f^{-1}(x)) = x$ .

Solution:

Let $y = \frac{2x+3}{x-2}$ . Solve for $x$ :

y(x-2) = 2x + 3

xy - 2y = 2x + 3

xy - 2x = 2y + 3

x(y-2) = 2y + 3

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

So $f^{-1}(x) = \frac{2x+3}{x-2}$ .

Notice $f^{-1} = f$ here — $f$ is its own inverse (an involution). This is a famous JEE trick pattern.

Verify: $f(f^{-1}(x)) = f\!\left(\frac{2x+3}{x-2}\right) = \frac{2 \cdot \frac{2x+3}{x-2} + 3}{\frac{2x+3}{x-2} - 2} = \frac{\frac{4x+6+3x-6}{x-2}}{\frac{2x+3-2x+4}{x-2}} = \frac{7x}{7} = x$ ✓

When you find $f^{-1}(x) = f(x)$ , the function is called an involution. Functions of the form $f(x) = \frac{ax+b}{cx-a}$ (notice the $a$ and $-a$ pattern in numerator and denominator) are always involutions. Spotting this saves you 2 minutes in the exam.

Difficulty Distribution

For JEE Main questions from this chapter over the last 5 years:

Difficulty	Percentage	What It Tests
Easy	40%	Basic relation properties (reflexive/symmetric/transitive check), counting subsets
Medium	45%	Equivalence classes, type of function (one-one/onto), simple composition
Hard	15%	Number of onto functions (inclusion-exclusion), complex composition domains

The “Hard” 15% usually appears as the tricky MCQ in Set B papers. If you’re targeting 90+ in Maths, you need the inclusion-exclusion formula for onto functions cold. If you’re targeting 70+, skip it and secure the Easy/Medium questions perfectly.

Expert Strategy

Week 1 — Crack the definitions properly. Don’t memorize the properties of relations as rules — understand them geometrically. Draw the relation as a directed graph on a small set like $\{1,2,3\}$ . Reflexive means every node has a self-loop. Symmetric means every arrow has a reverse arrow. Transitive means if $a \to b \to c$ , then $a \to c$ exists.

Week 2 — Equivalence relations and classes. Practice 10 PYQs specifically on this. The pattern is always: prove it’s an equivalence relation (3 checkboxes), then find the equivalence class of a specific element. Both parts appear in JEE regularly.

Week 3 — Functions. Focus on the logic: to disprove one-one, you need to exhibit two different inputs giving the same output. To prove one-one, you assume $f(x_1) = f(x_2)$ and derive $x_1 = x_2$ . This direction confusion is where most marks are lost.

For function type questions: check the graph using the horizontal line test (one-one ↔ every horizontal line intersects the graph at most once) and check whether the range equals the codomain (onto). For polynomial and rational functions, algebra is faster than graphing.

Revision Pattern: This chapter rewards 2 hours of focused PYQ practice more than 6 hours of re-reading theory. After your first pass, shift entirely to solving. Target: 20 PYQs from the last 10 years, timed.

Common Traps

Trap 1 — Confusing Range and Codomain. ” $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x^2$ ” — the codomain is $\mathbb{R}$ , but the range is $[0, \infty)$ . Since range $\neq$ codomain, this $f$ is not onto. Students who memorize ” $x^2$ is onto” without checking the codomain lose this mark every time.

Trap 2 — Transitive Vacuously. If a relation has no pair $(a,b)$ and $(b,c)$ with $b$ in common, transitive is vacuously true. A relation $R = \{(1,2)\}$ on $\{1,2,3\}$ is transitive — there’s no chain to violate. Students mark it “not transitive” without checking whether a violation actually exists.

Trap 3 — Composition Order. $(f \circ g)(x) = f(g(x))$ . The function written on the right acts first. JEE questions often ask for $(g \circ f)(x)$ when your instinct is to compute $f(g(x))$ . Read the notation once, slowly.

Trap 4 — Assuming Bijection Without Checking. A common question type: “Find $f^{-1}$ for $f(x) = 3x - 5$ .” Students find it correctly but don’t state why it exists (because $f$ is a bijection on $\mathbb{R}$ ). In JEE Advanced, showing conditions matters. In JEE Main, the trap is applying inverse to a function that isn’t bijective — always verify first.

Trap 5 — The Symmetric vs Antisymmetric Confusion. Antisymmetric means: if $aRb$ and $bRa$ , then $a = b$ . It does NOT mean “not symmetric.” The relation of equality is both symmetric and antisymmetric. This distinction appears in questions about the “less than or equal to” relation ( $\leq$ ), which is antisymmetric but not symmetric.