Chapter 1 of 13

CBSE Class 12 Mathematics★ Unit I · 8 marks, shared with Inverse Trigonometric Functions

Relations and Functions

This chapter picks up relations and functions where Class 11 left off — reflexive, symmetric, transitive and equivalence relations; one-one and onto functions; composition and invertibility. The book's own introduction and summary still mention "binary operations," but the actual teaching section on it has been removed from the current edition — a leftover reference, not a live topic. CBSE Class 12 Mathematics Chapter 1.

⏱ 11 min readMedium difficulty✓ Free · NCERT-aligned

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Key points, glossary and the questions most likely to show up.

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

1Classify a relation on a set as reflexive, symmetric, and/or transitive, and identify when all three combine into an equivalence relation
2Find the equivalence classes of a given equivalence relation and describe the partition they form
3Prove a function is one-one, onto, or bijective directly from the definitions
4Use the finite-set shortcut (one-one iff onto) where it applies, and recognise why it fails for infinite domains
5Compute the composition of two functions and check whether composition is commutative in a given case
6Prove a function is invertible by showing it is bijective, and construct its inverse by solving y=f(x) for x

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Why this chapter matters

This chapter extends the Class 11 idea of a function into the tools that the rest of Class 12 leans on: equivalence relations show up again in group-like structures, and 'invertible iff bijective' is exactly the property Inverse Trigonometric Functions needs two sections later to define principal value branches at all.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Relations and functions basics (Class 11, Chapter on Relations and Functions): domain, co-domain, range, and specific real-valued functions and their graphs, which this chapter assumes without re-teaching

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Set operations (Class 11, Sets): subsets, Cartesian product A x B, since a relation is defined as a subset of A x B

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Algebra of functions (Class 11): addition, subtraction, multiplication, and division of two functions, used without re-derivation here

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Sets and Cartesian products (Class 11): a relation on A is by definition a subset of A x A, so subset notation and the size of a Cartesian product are assumed

Relations and Functions

1. Check this before you revise anything

"Binary operations" is mentioned but not actually taught in this chapter anymore. The book's own opening paragraph says the chapter covers "different types of relations and functions, composition of functions, invertible functions and binary operations," and the Summary repeats the same list — but between the introduction and the summary there is no section on binary operations at all. The chapter goes straight from "Composition of Functions and Invertible Function" (section 1.4) to the Miscellaneous Exercise.

No definition, no example, no exercise question anywhere in the current book actually teaches binary operations. This is left-over wording from an older edition that the syllabus rationalisation dropped the content for but didn't fully scrub from the framing text — treat it as noise, not a topic to revise.

The old stub taught binary operations as a full section anyway, complete with a table of properties and its own line in the exam-focus list. That entire section has been removed from this rebuild, along with the invented five-question "Self-Test" that stood in for real exercise coverage.

The book has three real problem sets instead — Exercise 1.1 (16 questions, all about classifying relations), Exercise 1.2 (12 questions, all about one-one/onto functions), and a Miscellaneous Exercise (7 questions) — 35 questions in total, none of which the old stub touched.

2. What this chapter covers

Textbook section	Topic
1.2	Types of relations: empty, universal, reflexive, symmetric, transitive, equivalence; equivalence classes
1.3	Types of functions: one-one (injective), onto (surjective), bijective
1.4	Composition of functions; invertible functions

3. Types of relations

A relation $R$ in a set $A$ is any subset of $A \times A$ . Two extreme cases get names of their own: the empty relation $R = \emptyset$ (nothing is related to anything), and the universal relation $R = A \times A$ (everything is related to everything).

Property	Definition
Reflexive	$(a, a) \in R$ for every $a \in A$
Symmetric	$(a_{1}, a_{2}) \in R ⟹ (a_{2}, a_{1}) \in R$
Transitive	$(a_{1}, a_{2}) \in R$ and $(a_{2}, a_{3}) \in R ⟹ (a_{1}, a_{3}) \in R$
Equivalence	Reflexive and symmetric and transitive

Worked, mirroring the textbook's own Example 5. $R = {(a, b) : 2 divides a - b}$ on the set of integers $Z$ . Reflexive since $2$ divides $a - a = 0$ . Symmetric since $2 ∣ (a - b) ⟹ 2 ∣ (b - a)$ . Transitive since $(a - b) + (b - c) = a - c$ , so if $2$ divides both $a - b$ and $b - c$ it divides their sum. So $R$ is an equivalence relation — and it splits $Z$ into exactly two equivalence classes: the even integers $[0]$ and the odd integers $[1]$ .

That splitting isn't a coincidence of this one example. Any equivalence relation $R$ on a set $X$ partitions $X$ into disjoint equivalence classes $A_{i}$ whose union is all of $X$ — every element sits in exactly one class, and two elements are related exactly when they're in the same class.

This is a two-way street: an equivalence relation gives you a partition, and a partition (like splitting $Z$ into "multiples of 3," "one more than a multiple of 3," "two more than a multiple of 3") gives you back an equivalence relation.

The two extreme relations, classified. The universal relation $R = A \times A$ is an equivalence relation — it is trivially reflexive, symmetric and transitive, since every pair is present. The empty relation $R = \emptyset$ is symmetric and transitive vacuously (there are no pairs to violate the conditions), but it is not reflexive unless $A$ itself is empty, because $(a, a)$ is missing for every $a$ .

This asymmetry is a favourite exam question, and the reasoning is worth holding onto: symmetry and transitivity are conditional statements, so having no pairs satisfies them automatically, whereas reflexivity demands that certain pairs actually be present.

Why the three properties are independent. No two of them imply the third, and Exercise 1.1 is built almost entirely on constructing examples that have some but not others.

Example on $A = {1, 2, 3}$	R	S	T
${(1, 1), (2, 2), (3, 3), (1, 2)}$	yes	no	yes
${(1, 2), (2, 1)}$	no	yes	no
${(1, 2), (2, 3), (1, 3)}$	no	no	yes

Checking the second row is instructive: it is symmetric, but transitivity would require $(1, 1)$ to be present (from $(1, 2)$ and $(2, 1)$ ), and it is not.

Counting equivalence classes. For the relation " $a - b$ is divisible by $n$ " on $Z$ , there are exactly $n$ equivalence classes, one for each possible remainder $0, 1, \dots, n - 1$ . The example above with $n = 2$ giving the evens and the odds is the smallest case.

4. Types of functions

Type	Condition	Book's own test
One-one (injective)	$f (x_{1}) = f (x_{2}) ⟹ x_{1} = x_{2}$	Distinct inputs give distinct outputs
Onto (surjective)	Range $=$ Co-domain	Every $y$ in the co-domain has some $x$ with $f (x) = y$
Bijective	One-one and onto	Both hold together

A quirk worth knowing for finite sets specifically: for $f : X \to X$ on a finite set $X$ , one-one and onto are the same condition. If $f$ is onto but not one-one, two inputs would collapse onto one output, leaving too few images to cover all of $X$ — a contradiction. If $f$ is one-one but not onto, the images would fill fewer than $∣ X ∣$ slots while there are $∣ X ∣$ inputs demanding distinct images — also a contradiction.

This shortcut fails immediately for infinite sets: $f : N \to N$ , $f (x) = 2 x$ is one-one but not onto (no odd number is ever hit).

Worked, mirroring the textbook's own Example 12. $f : N \to N$ defined by $f (x) = x + 1$ if $x$ is odd, $f (x) = x - 1$ if $x$ is even. Every odd input and its paired even input map to each other's neighbours, so no two inputs share an image (one-one), and every natural number appears as someone's image (onto) — bijective, even though the formula looks asymmetric at first glance.

5. Composition and invertibility

For $f : A \to B$ and $g : B \to C$ , the composition $g \circ f : A \to C$ is defined by $(g \circ f) (x) = g (f (x))$ — apply $f$ first, then $g$ .

A function $f : X \to Y$ is invertible if there exists $g : Y \to X$ with $g \circ f = I_{X}$ and $f \circ g = I_{Y}$ (the identity functions). The single most useful fact in this section:

$f is invertible ⟺ f is bijective$

This is what makes invertibility checkable without ever constructing the inverse: prove one-one and onto, and invertibility follows for free.

Properties of composition. Composition is associative — $(f \circ g) \circ h = f \circ (g \circ h)$ — so a chain of three functions needs no brackets. It is not commutative: $g \circ f$ and $f \circ g$ are generally different functions, and may not even both be defined.

For the composition to make sense at all, the co-domain of the inner function must match the domain of the outer one. Checking this before computing is the first step in every composition question.

Composition preserves bijectivity. If $f : A \to B$ and $g : B \to C$ are both one-one, so is $g \circ f$ ; if both are onto, so is $g \circ f$ . Hence the composition of two bijections is a bijection, and therefore invertible.

The reversal law. When $f$ and $g$ are both invertible:

$(g \circ f)^{- 1} = f^{- 1} \circ g^{- 1}$

Note the order reverses. The intuition is the everyday one: to undo "put on socks, then shoes," you take off the shoes first. Writing $g^{- 1} \circ f^{- 1}$ instead is a standard error.

Inverting twice returns the original, $(f^{- 1})^{- 1} = f$ , and the inverse of a bijection is itself a bijection.

The inverse is unique. If $g_{1}$ and $g_{2}$ both invert $f$ , then $g_{1} = g_{1} \circ I = g_{1} \circ (f \circ g_{2}) = (g_{1} \circ f) \circ g_{2} = I \circ g_{2} = g_{2}$ . This is why the notation $f^{- 1}$ is legitimate — there is only one thing it can mean.

A warning about notation. The symbol $f^{- 1}$ means the inverse function here, not the reciprocal $1/ f$ . The two are unrelated, and the resemblance causes real confusion in later chapters.

Worked, mirroring the textbook's own Example 17. $f : N \to Y$ , $f (x) = 4 x + 3$ , where $Y = {y \in N : y = 4 x + 3 for some x \in N}$ . Solving $y = 4 x + 3$ for $x$ gives $x = \frac{y - 3}{4}$ , so define $g (y) = \frac{y - 3}{4}$ . Checking both compositions: $g (f (x)) = g (4 x + 3) = \frac{( 4 x + 3 ) - 3}{4} = x$ , and $f (g (y)) = f (\frac{y - 3}{4}) = 4 \cdot \frac{y - 3}{4} + 3 = y$ . Both identities hold, so $f$ is invertible with $f^{- 1} (y) = \frac{y - 3}{4}$ .

Summary

A relation $R$ on $A$ is a subset of $A \times A$ ; reflexive, symmetric, transitive are independent properties, and all three together make an equivalence relation.
An equivalence relation on $X$ partitions $X$ into disjoint equivalence classes covering all of $X$ .
One-one: distinct inputs, distinct outputs. Onto: every co-domain element is hit. Bijective: both.
For a function from a finite set to itself, one-one and onto are equivalent conditions — this fails for infinite sets.
$(g \circ f) (x) = g (f (x))$ : apply the inner function first.
$f$ is invertible if and only if $f$ is bijective; when it is, solve $y = f (x)$ for $x$ to get $f^{- 1} (y)$ .
The universal relation is an equivalence relation; the empty relation is symmetric and transitive vacuously but not reflexive (unless $A$ is empty).
Reflexive, symmetric and transitive are mutually independent — no two imply the third, which is what Exercise 1.1 is built on.
" $a - b$ divisible by $n$ " on $Z$ has exactly $n$ equivalence classes, one per remainder.
Composition is associative but not commutative, and needs the inner co-domain to match the outer domain.
The composition of two bijections is a bijection; $(g \circ f)^{- 1} = f^{- 1} \circ g^{- 1}$ , with the order reversed.
The inverse is unique, $(f^{- 1})^{- 1} = f$ , and $f^{- 1}$ means the inverse function, never the reciprocal $1/ f$ .
"Binary operations" is named in the book's introduction and summary paragraphs but has no actual teaching section, example, or exercise question in the current edition — a leftover reference, not something to revise.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Reflexive, symmetric, transitive

Reflexive: (a,a) in R for every a in A. Symmetric: (a1,a2) in R implies (a2,a1) in R. Transitive: (a1,a2) in R and (a2,a3) in R implies (a1,a3) in R

All three together define an equivalence relation; each is checked independently

One-one (injective)

f(x1) = f(x2) implies x1 = x2, for all x1, x2 in the domain

Equivalently: distinct inputs always give distinct outputs

Onto (surjective)

For every y in the co-domain Y, there exists x in the domain X such that f(x) = y

Equivalently: Range of f = Y

Finite-set shortcut

For f: X to X on a finite set X, f is one-one if and only if f is onto

Fails for infinite sets — f(x)=2x on N is one-one but not onto

Composition of functions

(g o f)(x) = g(f(x))

Apply f first, then g; composition is associative but not generally commutative

Invertibility criterion

f is invertible if and only if (iff) f is bijective

The standard route to proving invertibility without constructing the inverse first

Constructing the inverse

Solve y = f(x) for x to get x = f^{-1}(y)

Works once bijectivity is confirmed; check the result satisfies g o f = I_X and f o g = I_Y

The two extreme relations

Empty relation R = empty set; universal relation R = A x A

The universal relation is an equivalence relation; the empty relation is symmetric and transitive vacuously but NOT reflexive unless A is empty

Equivalence classes partition the set

An equivalence relation on X splits X into disjoint classes whose union is X; two elements are related exactly when they share a class

The correspondence runs both ways — a partition also defines an equivalence relation

Counting equivalence classes

The relation 'a - b is divisible by n' on the integers has exactly n classes, one per remainder 0, 1, ..., n-1

n = 2 gives the evens and the odds, which is the textbook's own worked example

Composition is associative but not commutative

(f o g) o h = f o (g o h), but g o f and f o g are generally different functions

The co-domain of the inner function must match the domain of the outer one for the composition to be defined at all

Composition preserves bijectivity

If f and g are both one-one so is g o f; if both are onto so is g o f; hence a composition of bijections is a bijection

This is what guarantees the composition of two invertible functions is itself invertible

Reversal law for inverses

(g o f)^-1 = f^-1 o g^-1

The order reverses — to undo 'socks then shoes' you remove the shoes first. Writing g^-1 o f^-1 is a standard error

Inverse of an inverse

(f^-1)^-1 = f, and the inverse of a bijection is itself a bijection

Inverting twice returns the original function exactly

Uniqueness of the inverse

If g1 and g2 both invert f then g1 = g1 o (f o g2) = (g1 o f) o g2 = g2

This is what makes the notation f^-1 legitimate — there is only one thing it can denote

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Assuming a relation is symmetric just because it 'sounds' two-directional, without checking a specific counterexample

✓ Test symmetry with an actual pair: find (a,b) in R and check whether (b,a) is also in R. One missing reverse pair disproves symmetry entirely.

WATCH OUT

✗ Concluding transitivity holds after checking only one or two pairs

✓ Transitivity must hold for every chain (a,b),(b,c) in R. Systematically list every pair whose second coordinate matches another pair's first coordinate before declaring transitivity confirmed.

WATCH OUT

✗ Confusing range and co-domain when checking 'onto'

✓ Onto means Range = Co-domain, not merely that Range is non-empty or large. Always compare the actual set of outputs to the stated co-domain.

WATCH OUT

✗ Assuming composition of functions is commutative

✓ In general (g o f)(x) is not equal to (f o g)(x) — verify with the book's own example f(x)=cos x, g(x)=3x^2, where gof(x)=3cos^2 x but fog(x)=cos(3x^2), clearly different.

WATCH OUT

✗ Trying to find f^{-1} directly before confirming f is bijective

✓ An inverse only exists for a bijective function. Prove one-one and onto first — if either fails, no inverse exists and solving y=f(x) for x is pointless.

WATCH OUT

✗ Revising 'binary operations' from an old source for this chapter

✓ The current book's introduction and summary paragraphs still mention binary operations, but there is no actual section, example, or exercise on it anywhere in this edition. It has been dropped from the syllabus rationalisation; skip it entirely.

WATCH OUT

✗ Calling the empty relation reflexive because 'there is nothing to violate'

✓ Symmetry and transitivity are conditional statements, so an empty relation satisfies them vacuously. Reflexivity instead demands that (a,a) be present for every a, so it fails unless A itself is empty.

WATCH OUT

✗ Writing (g o f)^-1 = g^-1 o f^-1 without reversing the order

✓ The correct law is (g o f)^-1 = f^-1 o g^-1. Undo the outermost operation first, exactly as you take off shoes before socks.

WATCH OUT

✗ Reading f^-1 as the reciprocal 1/f

✓ For functions, f^-1 denotes the inverse function and has nothing to do with the reciprocal. The resemblance to index notation is purely accidental.

WATCH OUT

✗ Composing two functions without checking that the domains and co-domains line up

✓ For g o f to exist, the co-domain of f must be contained in the domain of g. Check this before computing anything, since the composition may not be defined at all in one of the two orders.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Ex 1.1

Exercise 1.1

Classifying relations as reflexive, symmetric, transitive, and/or equivalence relations, including finding equivalence classes

Questions

Ex 1.2

Exercise 1.2

Checking one-one, onto, and bijective functions, including the greatest integer, modulus, and signum functions

Questions

Miscellaneous Ex

Miscellaneous Exercise

Mixed problems on injective functions, equality of functions, counting onto functions on a finite set, and equivalence relation counting

Questions

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Relations and Functions?

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

8 questions~6 min worth ~8 marks in CBSE exams

5-minute revision

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

•A relation R on A is a subset of A x A; reflexive, symmetric, transitive are checked independently, and all three together define an equivalence relation
•An equivalence relation partitions its set into disjoint equivalence classes whose union is the whole set
•One-one: distinct inputs give distinct outputs. Onto: every co-domain element has a preimage. Bijective: both at once
•For a function from a finite set to itself, one-one and onto are equivalent — this shortcut fails on infinite sets
•(g o f)(x) = g(f(x)): the inner function is applied first; composition is not generally commutative
•f is invertible if and only if f is bijective — prove bijectivity first, then solve y=f(x) for x to get the inverse
•Binary operations is named in the book's introduction and summary text but has no actual section, example, or exercise in the current edition

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: Unit I: 8 marks, shared with Inverse Trigonometric Functions

Question type	Marks each	Typical count	What it tests
Classifying Relations and Equivalence Classes	2-4	1	Reflexive/symmetric/transitive checks, finding equivalence classes, describing the resulting partition
One-One, Onto, Bijective Functions and the Finite-Set Shortcut	2-4	1	Proving injectivity/surjectivity directly from the definition, standard functions like greatest integer, modulus, signum
Composition and Invertibility	4-6	1	Computing gof and fog, proving invertibility via bijectivity, constructing f inverse

Prep strategy

For 'show R is an equivalence relation' questions, structure the answer as three clearly labelled parts: Reflexive, Symmetric, Transitive — each with its own one-line justification
For invertibility proofs, always prove bijectivity first and only then solve y=f(x) for x — never attempt to construct an inverse before confirming it exists
When a question restricts the domain or co-domain (e.g. R minus a point), check that any solved-for value actually lands back inside the stated set before declaring the function onto

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Database relations and keys

A database 'relation' linking, say, students to their enrolled courses is literally a subset of Students x Courses; a one-one, onto mapping (like a bijection between employee IDs and employees) is exactly what a database primary key guarantees.

Encryption and decryption

An encryption function must be invertible — bijective — so that the exact original message can always be recovered by applying the inverse (decryption) function; a many-one encryption scheme would make some messages impossible to decrypt uniquely.

Sorting and equivalence classes in data processing

Grouping records by a shared attribute (same postal code, same birth year) is applying an equivalence relation to partition a dataset into equivalence classes, the same structure used to group students by remainder classes in this chapter's examples.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Structure every 'show R is an equivalence relation' answer as three explicitly labelled checks — Reflexive, Symmetric, Transitive — examiners award marks per property, not for the conclusion alone

For 'one-one and onto' proofs, write the one-one argument and the onto argument as two clearly separate paragraphs rather than mixing them

When solving y=f(x) for x to find an inverse, always state which values of y are excluded (matching the function's actual range) before finalising f inverse

For finite-set questions, look for language like 'from {1,...,n} to itself' — that is the signal to invoke the one-one-iff-onto shortcut instead of proving both directions from scratch

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

STRETCH

Counting all equivalence relations on a set of size n is counting the Bell number B(n) — the number of ways to partition an n-element set — which grows extremely fast and has no simple closed form, unlike counting bijections (which is just n!)

STRETCH

The number of one-one functions from an n-element set to an m-element set (n <= m) is the falling factorial m(m-1)...(m-n+1), a natural generalisation of the book's own Miscellaneous Exercise Q22 count of one-one functions from a set to itself

STRETCH

Constructing the integers from pairs of natural numbers using an equivalence relation (a,b)~(c,d) iff a+d=b+c, as in one of the JEE problems above, is the standard formal construction used to build Z from N in foundational mathematics

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JEE Main & Advanced practice

Competitive-level problems on this chapter, above the board pattern. Try each one on paper before opening the solution.

JEE MainCounting equivalence relations containing a given pairDirect combinatorial counting on a small set

Let $A = {1, 2, 3, 4}$ . Find the number of equivalence relations on $A$ that contain the pair $(1, 2)$ .

Stuck? Show the approach

An equivalence relation containing $(1, 2)$ must also contain $(2, 1)$ by symmetry and $(1, 1), (2, 2)$ by reflexivity. Beyond that, ${1, 2}$ can either stay merged with the rest of the equivalence classes in various ways — enumerate the possible partitions of ${1, 2, 3, 4}$ in which $1$ and $2$ lie in the same block.

Show the full solution

Since $1$ and $2$ must be in the same equivalence class, treat ${1, 2}$ as one unit alongside the remaining elements $3, 4$ . Partitioning ${{1, 2}, 3, 4}$ (3 effective items) into blocks: all-separate ( ${1, 2}, {3}, {4}$ ), $3$ merged in ( ${1, 2, 3}, {4}$ ), $4$ merged in ( ${1, 2, 4}, {3}$ ), both merged separately-but-together with ${1, 2}$ ( ${1, 2, 3, 4}$ ), or $3, 4$ paired on their own ( ${1, 2}, {3, 4}$ ). That gives the Bell-number-style count for 3 effective items, which is 5.

Answer: 5

The trap

Forgetting that $3$ and $4$ can also form their own separate block from ${1, 2}$ while still being merged with each other — this is the partition ${1, 2}, {3, 4}$ , easy to miss when only checking whether $3$ and $4$ join the ${1, 2}$ block individually.

JEE MainComposition producing the identitySolving for an unknown in a composed function

If $f (x) = \frac{2 x + 3}{5}$ and $g (x) = \frac{5 x - 3}{2}$ , verify that $g$ is the inverse of $f$ , and find $(f \circ g \circ f) (1)$ .

Stuck? Show the approach

Check $g (f (x)) = x$ directly, then use that fact to simplify the triple composition without computing every intermediate value from scratch.

Show the full solution

$g (f (x)) = \frac{5 ( \frac{2 x + 3}{5} ) - 3}{2} = \frac{( 2 x + 3 ) - 3}{2} = \frac{2 x}{2} = x$ , confirming $g = f^{- 1}$ . So $(f \circ g) (y) = y$ for all $y$ , meaning $(f \circ g \circ f) (1) = f (1) = \frac{2 ( 1 ) + 3}{5} = 1$ .

Answer: 1

The trap

Computing all three functions in sequence numerically (finding $f (1)$ , then $g$ of that, then $f$ of that) works but wastes time — recognising $f \circ g = I$ collapses the middle two steps immediately.

JEE MainBijectivity on a restricted domainProving one-one and onto for a rational function

Let $f : R - {- 1} \to R - {1}$ be defined by $f (x) = \frac{x}{x + 1}$ . Show $f$ is bijective.

Stuck? Show the approach

Prove one-one directly from the definition, then solve $y = f (x)$ for $x$ to establish onto and confirm the result always lands back in the domain.

Show the full solution

One-one: $f (x_{1}) = f (x_{2}) \Rightarrow \frac{x _{1}}{x _{1} + 1} = \frac{x _{2}}{x _{2} + 1} \Rightarrow x_{1} (x_{2} + 1) = x_{2} (x_{1} + 1) \Rightarrow x_{1} x_{2} + x_{1} = x_{1} x_{2} + x_{2} \Rightarrow x_{1} = x_{2}$ . Onto: solving $y = \frac{x}{x + 1}$ gives $y (x + 1) = x \Rightarrow y x + y = x \Rightarrow x (y - 1) = - y \Rightarrow x = \frac{- y}{y - 1} = \frac{y}{1 - y}$ , well-defined since $y \neq = 1$ , and this $x \neq = - 1$ (since $x = - 1$ would need $- 1 = y / (1 - y) \Rightarrow - 1 + y = y$ , impossible).

Answer: f is bijective, with f^{-1}(y) = y/(1-y)

The trap

Forgetting to verify that the solved-for $x$ never equals the excluded value $- 1$ — without that check, onto is not fully established since $x$ must land back inside the stated domain.

JEE MainRelation properties on a set of ordered pairsProving equivalence on a product-style set

A relation $R$ is defined on $N \times N$ by $(a, b) R (c, d) ⟺ a + d = b + c$ . Show $R$ is an equivalence relation.

Stuck? Show the approach

Check reflexivity, symmetry, and transitivity directly from the defining equation, treating each ordered pair as a single object being compared.

Show the full solution

Reflexive: $(a, b) R (a, b)$ needs $a + b = b + a$ , true. Symmetric: $(a, b) R (c, d) \Rightarrow a + d = b + c \Rightarrow c + b = d + a \Rightarrow (c, d) R (a, b)$ . Transitive: $(a, b) R (c, d)$ and $(c, d) R (e, f)$ give $a + d = b + c$ and $c + f = d + e$ ; adding, $a + d + c + f = b + c + d + e \Rightarrow a + f = b + e \Rightarrow (a, b) R (e, f)$ .

Answer: R is an equivalence relation (this is the standard construction of the integers as equivalence classes of pairs of naturals)

The trap

Trying to prove transitivity by treating $a, b, c, d, e, f$ as arbitrary numbers and cross-multiplying instead of adding the two defining equations directly — addition is what cancels $c$ and $d$ cleanly.

JEE AdvancedCounting bijections satisfying an extra constraintCombining the finite-set shortcut with derangement-style counting

Let $f : {1, 2, 3, 4, 5} \to {1, 2, 3, 4, 5}$ be a one-one function such that $f (i) \neq = i$ for every $i$ . Find the number of such functions.

Stuck? Show the approach

Since the domain and co-domain are the same finite set, one-one automatically forces onto (the finite-set shortcut), so every such $f$ is a permutation of ${1, ..., 5}$ — specifically, one with no fixed point, i.e. a derangement of 5 objects.

Show the full solution

The number of derangements of $n$ objects is $D_{n} = n! \sum_{k = 0}^{n} \frac{( - 1 ) ^{k}}{k !}$ . For $n = 5$ : $D_{5} = 120 (1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120}) = 120 - 120 + 60 - 20 + 5 - 1 = 44$ .

Answer: 44

The trap

Missing that the finite-set shortcut (one-one implies onto here) is what justifies treating $f$ as a full permutation in the first place — on an infinite set, 'one-one with no fixed point' would not reduce to counting derangements this way.

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Questions students ask

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It is mentioned in the book's opening paragraph and its Summary, but there is no actual teaching section on it anywhere in the current edition — no definitions, no worked examples, no exercise questions. Treat any material on binary operations as leftover text, not something to revise.

Check whether it is bijective (one-one and onto). If it is, invertibility is guaranteed — you don't need to construct f inverse first to know it exists.

The shortcut only applies when the domain and co-domain are the same finite set. Natural numbers are infinite, so a one-one function can still miss values in the co-domain — f(x)=2x never outputs an odd number.

No — composition is not generally commutative. The book's own example, f(x)=cos x and g(x)=3x^2, gives gof(x)=3cos^2 x but fog(x)=cos(3x^2), which are different functions.

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