Chapter 1 of 14

Karnataka (KSEEB)Class 11 Mathematics★ Unit I · part of the 23-mark block covering Sets, Relations & Functions, and Trigonometric Functions in CBSE Class 11 Mathematics (80 marks total; CBSE states there is no official per-chapter split)

Sets

Every later chapter in this book — functions, trigonometry, probability, even calculus — is written in a language, and that language is set notation. This chapter builds that language from a single idea (a well-defined collection of objects) up through subsets, Venn diagrams, and the operations that combine sets, ending with De Morgan's laws. One formula that shows up in almost every coaching handout for this chapter is not actually in the 2026-27 book at all. CBSE Class 11 Mathematics Chapter 1.

⏱ 15 min readEasy difficulty✓ Free · NCERT-aligned

Worksheet

Podcast

Listen as a conversation

A teacher & a student talk through the chapter — in your language.

Mind Map

The chapter at a glance

Every concept and exam-favourite branch — revisable in 5 minutes.

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Study Guide

Night-before notes

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

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VocabularyBuild your vocabularyLexiGenome mines the key words and schedules them with spaced repetition.Mine words →

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

1Represent sets in roster form and set-builder form and convert between the two
2Classify sets as empty, finite, infinite, or equal
3Distinguish subset (⊆) from proper subset (⊂), and explain why ∅ is a subset of every set
4Write and read intervals as subsets of R, in both notations
5Perform union, intersection, and difference operations on sets
6Find the complement of a set relative to a chosen universal set
7State and apply De Morgan's laws, and verify them for a given pair of sets
8Distinguish an element of a set from a subset of a set — a frequent source of error when a set itself contains sets

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Why this chapter matters

Sets form the foundational language of all mathematics — every subsequent chapter in Class 11 and 12 uses set notation, from relations and functions through probability. One formula many coaching handouts still drill for this chapter, the cardinal number of a union, is not actually part of the 2026-27 book or syllabus for it.

Before you start — revise these

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

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Class 10 Mathematics — real numbers

Familiarity with natural numbers, integers, rationals and irrationals, used throughout this chapter's examples

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Class 9/10 Mathematics — basic logic and justification

Distinguishing a true statement from a false one with a reason, needed for the true/false exercises in this chapter

Sets

1. Check this before you revise anything

One formula that appears in almost every coaching note for this chapter is not in the 2026-27 NCERT text, and not in the CBSE syllabus for it.

Topic	NCERT 2026-27 chapter	CBSE 2026-27
Cardinal number of a union, $n (A \cup B) = n (A) + n (B) - n (A \cap B)$	Absent — no such formula appears	Not listed for this chapter
"Practical problems" / survey word problems (tea-and-coffee style)	Absent	Not listed
Everything else — representation, subsets, intervals, Venn diagrams, union, intersection, difference, complement, De Morgan's laws	Present in full	Listed

This is an older-edition holdover. An earlier edition of this book had a section on counting the size of a union, with exactly the survey-style word problems ("in a class of 60, 25 play cricket, 20 play tennis...") that many coaching sheets still drill. None of the current chapter's five exercises, and none of the ten-question miscellaneous exercise, ever ask for it. Spend your revision time on the eight topics that are actually still here.

2. What this chapter covers

Textbook section	Topic
1.1-1.2	Introduction; sets and their representations
1.3-1.5	The empty set; finite and infinite sets; equal sets
1.6-1.7	Subsets, intervals as subsets of $R$ ; the universal set
1.8-1.9	Venn diagrams; operations on sets (union, intersection, difference)
1.10	Complement of a set

3. What a set actually is

A set is a well-defined collection of distinct objects. "Well-defined" is doing real work in that sentence: given any object, you must be able to say, without ambiguity, whether it belongs to the set or not.

This is why "the collection of the ten best writers in India" is not a set. "Best" is a matter of opinion — different people would include different writers. "The collection of all months beginning with J" is a set: January, June, July, and nothing else, and nobody can argue about it.

Sets are named with capital letters ( $A$ , $B$ , $X$ ); their members, called elements, with lowercase letters. $a \in A$ reads " $a$ is an element of $A$ "; $b \in / A$ reads " $b$ is not an element of $A$ ."

Two ways to write the same set

Form	Method	Example
Roster (tabular)	List every element inside braces, each once, order irrelevant	$A = {2, 4, 6, 8}$
Set-builder	State the property that decides membership	$A = {x : x is an even natural number less than 10}$

Worked, mirroring the textbook's own Example 3. Write $A = {1, 4, 9, 16, 25, \dots}$ in set-builder form. Each term is a perfect square, so $A = {x : x = n^{2}, n \in N}$ .

Roster form never repeats an element and never depends on order — ${1, 2, 3}$ and ${3, 1, 2}$ are the exact same set, written two different ways.

4. Empty, finite, infinite, equal — the basic vocabulary

Type	Definition	Example
Empty (null) set	Has no elements at all; written $\emptyset$ or ${}$	${x : x \in N, 1 < x < 2}$
Finite set	Empty, or its elements can be counted off and the counting stops	${1, 2, \dots, 100}$
Infinite set	The counting never stops	$N = {1, 2, 3, \dots}$
Equal sets	Exactly the same elements, any order	${1, 2, 3} = {3, 1, 2}$

A set with one element, like ${0}$ , is not the same as the empty set. ${0}$ has one element — the number zero — while $\emptyset$ has none at all. This is a very common early confusion, since "zero" and "nothing" feel like the same idea outside mathematics.

The standard number sets, since they appear constantly from here on

$N$ (natural numbers), $Z$ (integers, from the German Zahlen), $Q$ (rationals, from quotient), $R$ (reals). Every one of these is infinite, and $N \subset Z \subset Q \subset R$ .

5. Subsets, intervals, and the universal set

$A$ is a subset of $B$ (written $A \subseteq B$ ) when every element of $A$ is also an element of $B$ . Two consequences follow immediately and are worth stating separately: every set is a subset of itself ( $A \subseteq A$ ), and the empty set is a subset of every set ( $\emptyset \subseteq A$ ) — vacuously true, since there is no element of $\emptyset$ that could ever fail to be in $A$ .

$A$ is a proper subset ( $A \subset B$ ) when $A \subseteq B$ but $A \neq = B$ — $B$ has at least one element $A$ lacks.

A subtlety the textbook's own Example 11 tests directly: if $A \in B$ and $B \subset C$ , is $A \in C$ ? Not necessarily — $A$ being an element of $B$ says nothing about $A$ 's relationship to $C$ 's elements. Membership ( $\in$ ) and subset ( $\subset$ ) are different relations and do not chain together automatically.

Intervals — subsets of $R$ you'll use constantly from here on

Notation	Set-builder form	Name
$(a, b)$	${x : a < x < b}$	Open interval
$[a, b]$	${x : a \leq x \leq b}$	Closed interval
$[a, b)$ , $(a, b]$	endpoints included/excluded as written	Half-open

The universal set

In any given context, the universal set $U$ is the basic set containing every element under discussion. Studying divisibility, $U$ might be $N$ ; studying a population survey, $U$ is everyone in the population. There is no single fixed universal set — it is chosen to fit the problem.

6. Venn diagrams

A Venn diagram represents sets as circles inside a rectangle (the universal set), developed by John Venn. They are a visual shortcut, not a proof method — but they make the operations in the next section easy to picture before computing them.

7. Union, intersection, and difference

Union — everything in either set

$A \cup B = {x : x \in A or x \in B}$

Worked, mirroring the textbook's own Example 14. Let $X = {Ram, Geeta, Akbar}$ be the Class XI students on the school's basketball team, and $Y = {Geeta, David, Ankit}$ the students on the hockey team. $X \cup Y = {Ram, Geeta, Akbar, David, Ankit}$ — the students on at least one team. Geeta, who plays both, is listed exactly once: a set never repeats an element even when an object qualifies for membership two different ways.

$A \cup B = B \cup A$ (commutative), $(A \cup B) \cup C = A \cup (B \cup C)$ (associative), and $A \cup \emptyset = A$ .

$A \cap B = {x : x \in A and x \in B}$

Two sets are disjoint when $A \cap B = \emptyset$ — they share nothing at all. In the team example above, if a third student, Rahim, plays only cricket, ${Rahim}$ and $Y$ are disjoint.

Difference — in one, not the other

$A - B = {x : x \in A and x \in / B}$

$A - B$ and $B - A$ are almost never equal. For $V = {a, e, i, o, u}$ and $B = {a, i, k, u}$ : $V - B = {e, o}$ (vowels not in $B$ ), while $B - V = {k}$ (the one element of $B$ that is not a vowel). Order matters completely.

8. Complement, and De Morgan's laws

Relative to a chosen universal set $U$ , the complement of $A$ is everything in $U$ that is not in $A$ :

$A^{'} = U - A = {x \in U : x \in / A}$

The complement laws, stated together

Law	Statement
Complement laws	$A \cup A^{'} = U$ and $A \cap A^{'} = \emptyset$
Double complement	$(A^{'})^{'} = A$
Laws of $\emptyset$ and $U$	$\emptyset^{'} = U$ and $U^{'} = \emptyset$

De Morgan's laws — the most tested result in the chapter

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

Read them as a translation rule, not two formulas to memorise separately: complementing a union turns it into an intersection of complements, and vice versa — the operation flips every time you push the complement inward.

Worked, mirroring the textbook's own Example 22. $U = {1, 2, 3, 4, 5, 6}$ , $A = {2, 3}$ , $B = {3, 4, 5}$ . Directly: $A \cup B = {2, 3, 4, 5}$ , so $(A \cup B)^{'} = {1, 6}$ . Separately: $A^{'} = {1, 4, 5, 6}$ , $B^{'} = {1, 2, 6}$ , so $A^{'} \cap B^{'} = {1, 6}$ . Both routes agree, as De Morgan's law guarantees they must.

9. Worked: an equality that looks surprising until you chase elements

Textbook Example 25. Show that $A \cup B = A \cap B$ implies $A = B$ .

At first glance this looks like it should need extra information — a union and an intersection being equal for two different sets doesn't sound impossible. Chasing individual elements settles it directly.

Take any $a \in A$ . Then certainly $a \in A \cup B$ . Since we are given $A \cup B = A \cap B$ , this means $a \in A \cap B$ too — and in particular, $a \in B$ . So every element of $A$ is in $B$ , i.e. $A \subseteq B$ .

Run the identical argument starting from an arbitrary $b \in B$ : $b \in A \cup B = A \cap B$ , so $b \in A$ . Hence $B \subseteq A$ .

$A \subseteq B$ and $B \subseteq A$ together mean $A = B$ — this is exactly the "iff" test from section 5. The surprising-looking equality forces the two sets to collapse into being the same set entirely; there was no room for them to differ.

10. Where set theory itself came from

Set theory as a formal subject began with the German mathematician Georg Cantor (1845-1918), who developed it while studying trigonometric series in the 1870s. His 1874 result — that the real numbers cannot be matched one-to-one with the integers, so some infinities are strictly larger than others — was radical enough that a contemporary, Kronecker, publicly attacked him for treating infinite sets the way finite ones are treated.

The subject nearly broke under its own weight in 1902, when the philosopher Bertrand Russell showed that allowing a genuinely unrestricted "set of all sets" leads straight to a contradiction — now called Russell's Paradox. That single result forced mathematicians to axiomatise set theory carefully rather than trust naive intuition about what a "collection" can be, and the careful, restricted definitions this chapter uses are a direct descendant of that fix.

Summary

A set is a well-defined collection of distinct objects; write it in roster form (list elements) or set-builder form (state the defining property).
Empty, finite, infinite, and equal sets are defined purely by counting and comparing elements — not by any notion of size or importance.
$A \subseteq B$ means every element of $A$ is in $B$ ; every set is a subset of itself, and $\emptyset$ is a subset of every set.
Intervals $(a, b)$ , $[a, b]$ , $[a, b)$ , $(a, b]$ are subsets of $R$ that will be used constantly from here on.
$A \cup B$ is everything in either set; $A \cap B$ is only what both share; $A - B$ is in $A$ but not $B$ , and is generally not equal to $B - A$ .
The complement $A^{'} = U - A$ depends on the chosen universal set; De Morgan's laws, $(A \cup B)^{'} = A^{'} \cap B^{'}$ and $(A \cap B)^{'} = A^{'} \cup B^{'}$ , are the single most-tested result in the chapter.
The cardinal-number-of-a-union formula and survey-style word problems are not part of the 2026-27 syllabus for this chapter, despite appearing in older coaching material.

Key formulas & results

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

Roster and set-builder form

list elements in {}, OR state the defining property {x : ...}

Two equivalent ways to write the same set

Subset

A ⊆ B if a ∈ A implies a ∈ B

Every set is a subset of itself; ∅ is a subset of every set

Proper subset

A ⊂ B if A ⊆ B and A ≠ B

B has at least one element A does not

Intervals as subsets of R

(a,b)={x:a<x<b}; [a,b]={x:a≤x≤b}; [a,b) and (a,b] half-open

Round bracket excludes the endpoint, square bracket includes it

Union

A ∪ B = {x : x ∈ A or x ∈ B}

Commutative and associative; A ∪ ∅ = A

Intersection

A ∩ B = {x : x ∈ A and x ∈ B}

A and B are disjoint when A ∩ B = ∅

Difference

A − B = {x : x ∈ A and x ∉ B}

A − B and B − A are generally NOT equal

Complement

A' = U − A = {x ∈ U : x ∉ A}

Depends on the chosen universal set U

Complement laws

A ∪ A' = U; A ∩ A' = ∅; (A')' = A

Also: ∅' = U and U' = ∅

De Morgan's laws

(A∪B)' = A'∩B' ; (A∩B)' = A'∪B'

The single most-tested result in the chapter

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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

✗ Writing the empty set as {∅} instead of ∅ or {}

✓ {∅} is a set CONTAINING the empty set — it has one element. The empty set itself is ∅ or {}, with no elements at all.

WATCH OUT

✗ Confusing subset (⊆) with proper subset (⊂)

✓ A⊆A is always true (every set is a subset of itself). A⊂A is NEVER true — a proper subset requires at least one extra element in the other set.

WATCH OUT

✗ Forgetting that ∅ is a subset of every set

✓ ∅⊆A is true for any set A, vacuously — there is no element of ∅ that could ever fail to be in A.

WATCH OUT

✗ Revising the cardinal-number-of-a-union formula n(A∪B)=n(A)+n(B)−n(A∩B) for this chapter

✓ It is not in the 2026-27 NCERT chapter and not listed in the CBSE syllabus for it — an older-edition holdover. None of the current exercises need it.

WATCH OUT

✗ Treating 'a is an element of A' and '{a} is an element of A' as the same statement

✓ They are different. {a}∈A requires the SET {a} itself to be one of A's elements, which is unusual; a∈A only requires the object a itself to be listed. Exercise 1.3 Q2(v) tests exactly this distinction.

WATCH OUT

✗ Treating an element of a set as automatically a subset of it

✓ If A={1,2,{3,4},5}, then {3,4} is an ELEMENT of A (so {3,4}∈A), not a subset of A — 3 and 4 are not individually elements of A, so {3,4}⊂A is false. The set containing that element, {{3,4}}, IS a valid subset.

WATCH OUT

✗ Assuming A−B and B−A give the same set

✓ They are generally different. A−B keeps A's elements that are missing from B; B−A keeps B's elements missing from A — swapping the order changes which set is being filtered.

WATCH OUT

✗ Computing a complement without first fixing the universal set

✓ A' depends entirely on which U is chosen. The same set A can have different complements in different contexts — always state U before computing A'.

WATCH OUT

✗ Misremembering which way De Morgan's laws flip the operation

✓ (A∪B)'=A'∩B' and (A∩B)'=A'∪B' — complementing a union gives an intersection, and complementing an intersection gives a union. It always flips, never stays the same.

WATCH OUT

✗ Assuming A⊂B and B∈C together imply A∈C

✓ They do not — membership (∈) and subset (⊂) do not chain together. A={1}, B={1,2}, C={{1,2},3} is a standard counterexample: A⊂B and B∈C, but A∉C.

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

Deciding what counts as a set; roster and set-builder form

Questions

Ex 1.2

Exercise 1.2

The empty set, finite and infinite sets, equal sets

Questions

Ex 1.3

Exercise 1.3

Subsets, proper subsets, intervals as subsets of R, universal sets

Questions

Ex 1.4

Exercise 1.4

Union, intersection, difference, and disjoint sets

Questions

Ex 1.5

Exercise 1.5

Complement of a set and De Morgan's laws

Questions

Miscellaneous Ex

Miscellaneous Exercise

Proofs and general properties of subset, union, intersection and difference

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 Sets?

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

9 questions~6 min worth ~23 marks in Karnataka (KSEEB) exams

5-minute revision

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

•A set is a well-defined collection; roster form lists elements, set-builder form states the defining property.
•Empty, finite, infinite and equal sets are classified purely by counting and comparing elements.
•A⊆B means every element of A is in B; every set is a subset of itself, and ∅ is a subset of every set.
•A⊂B (proper subset) additionally requires A≠B.
•Intervals (a,b), [a,b], [a,b), (a,b] are subsets of R — round excludes, square includes.
•The universal set U is chosen to fit the context; it is not a single fixed set.
•A∪B is everything in either set; A∩B is only what's shared; disjoint sets have A∩B=∅.
•A−B and B−A are generally different sets.
•A' = U−A depends entirely on the chosen universal set.
•Complement laws: A∪A'=U, A∩A'=∅, (A')'=A.
•De Morgan's laws: (A∪B)'=A'∩B' and (A∩B)'=A'∪B' — the complement always flips the operation.
•The cardinal-number-of-union formula is NOT part of the 2026-27 syllabus for this chapter.
•An element of a set is not automatically a subset — {3,4} being an element of A does not make {3,4} a subset of A unless 3 and 4 are themselves elements of A.

Karnataka (KSEEB) marks blueprint

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

Typical chapter weightage: Unit I sits inside the 23-mark block covering Sets, Relations & Functions, and Trigonometric Functions (CBSE Class 11 Mathematics, 80 marks; no official per-chapter split)

Question type	Marks each	Typical count	What it tests
Set representation, types of sets and subsets	1-2	1-2	Roster/set-builder conversion, classifying empty/finite/infinite/equal sets, subset vs proper subset, counting subsets of a given set
Intervals and set operations	2-3	1-2	Interval notation both ways, union, intersection, difference, and identifying disjoint sets
Complement of a set and De Morgan's laws	3-4	1	Finding a complement relative to a stated universal set, verifying De Morgan's laws for given sets, applying the complement laws

Prep strategy

Learn the chapter as three blocks: what a set IS and how to write it (1.1-1.5), subsets and intervals (1.6-1.7), and operations plus complement (1.8-1.10)
Practise converting between roster and set-builder form with varied examples — number patterns, letters of a word, geometric conditions
For any 'verify' question (De Morgan's laws especially), compute both sides completely independently before comparing, exactly as the textbook itself does
Do NOT revise the cardinal-number-of-union formula or survey word problems for this chapter — they are not in the 2026-27 syllabus here
Drill the difference between an element of a set and a subset of a set using a set that itself contains a set as one of its elements, like {1, {2,3}, 4}

Where this shows up in the real world

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

Database queries (SQL)

SQL's UNION and INTERSECT operations are direct implementations of set union and intersection, used in essentially every database-driven application.

Search engine filters

Combining search filters ('show results matching category A and price range B') is intersection; broadening a search ('show A or B') is union.

Logic and digital circuit design

De Morgan's laws are fundamental in digital electronics, letting AND gates be rewritten as OR gates with inverters, which simplifies circuit design.

Venn-diagram reasoning in everyday arguments

Any 'who qualifies for both conditions' or 'who is excluded from either' question is a disguised intersection or complement, whether the context is eligibility criteria, tax brackets, or scheduling.

Exam strategy

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

State the universal set explicitly before computing any complement — a complement without a stated U is not fully defined

For a 'verify' question, compute both sides of the claimed identity completely independently, then compare — never shortcut one side using the other

When a set contains another set as one of its own elements, pause and check element-vs-subset carefully before answering

List subsets systematically (by size: 0 elements, then 1, then 2, ...) to avoid missing any when asked for all of them

Do not spend revision time on the cardinal-number-of-union formula for this specific chapter — it is not examinable here under the 2026-27 syllabus

Going beyond the textbook

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

STRETCH

Symmetric difference: A△B = (A−B)∪(B−A) — prove it is associative, and that A△B=∅ if and only if A=B

STRETCH

Prove the inclusion-exclusion principle for three sets, n(A∪B∪C), by extending the two-set argument used implicitly in this chapter

STRETCH

Cantor's diagonal argument: show that the power set of any set A always has strictly more elements than A itself, even when A is infinite

STRETCH

Show that the set of rational numbers is countably infinite (can be listed in a sequence) while the set of real numbers between 0 and 1 cannot

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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 elements after removing a subsetFormula application

If a set A has 8 elements and a subset B of A has 3 elements, how many subsets of A contain none of the elements of B?

Stuck? Show the approach

A subset of A that avoids every element of B can only be built from A's remaining elements, so this reduces to counting all subsets of a smaller set.

Show the full solution

$A - B has 8 - 3 = 5 elements, so the subsets of A avoiding all of B are exactly the subsets of A - B$ $Number of such subsets = 2^{5} = 32$

Answer: 32

The trap

Trying to subtract 2^3 (subsets of B) from 2^8 (subsets of A) counts the wrong thing entirely — it does not isolate subsets built purely from A-B; the correct approach restricts attention to A-B from the start.

JEE MainSolving for unknown set sizes from union and intersectionAlgebraic set problem

Sets A and B are such that $n (A) = 20$ , $n (A \cup B) = 35$ , and $n (A \cap B) = 5$ . Find $n (B)$ and $n (B - A)$ .

Stuck? Show the approach

This uses the same counting logic as the cardinal-number-of-a-union idea (not itself part of this NCERT chapter, but a standard early JEE application of union and intersection) — rearrange the relationship between the four quantities to isolate the unknown.

Show the full solution

$n (A \cup B) = n (A) + n (B) - n (A \cap B) ⟹ 35 = 20 + n (B) - 5 ⟹ n (B) = 20$ $n (B - A) = n (B) - n (A \cap B) = 20 - 5 = 15$

Answer: n(B) = 20; n(B - A) = 15

The trap

This formula and this style of problem sit outside the current NCERT chapter's own exercises (see the chapter's opening note) — useful and common in competitive exams, but not something this specific chapter's own exercises will ask for.

JEE MainComplement of a symmetric-difference-style conditionSet identity

If $A$ and $B$ are subsets of a universal set $U$ , simplify $(A \cup B) \cap (A \cup B^{'})$ using set laws.

Stuck? Show the approach

Treat this as the distributive law run in reverse: factor out the common set $A$ from the union appearing in both bracketed terms.

Show the full solution

$(A \cup B) \cap (A \cup B^{'}) = A \cup (B \cap B^{'}) [distributive law]$ $= A \cup \emptyset = A$

Answer: A

The trap

Attempting to expand this by picking arbitrary elements case-by-case is slower and error-prone; recognising the distributive law's reverse pattern, A ∪ (B ∩ C) from (A∪B) ∩ (A∪C), solves it in two lines.

JEE AdvancedCounting subsets satisfying a compound conditionCombinatorial set-counting

A set $S$ has 10 elements. Find the number of subsets of $S$ that contain at least one of two fixed, distinct elements $a, b \in S$ .

Stuck? Show the approach

It is far easier to count the complementary case — subsets containing NEITHER $a$ nor $b$ — and subtract from the total number of subsets, than to count 'at least one' directly.

Show the full solution

Total subsets of $S$ : $2^{10}$ .

Subsets avoiding both $a$ and $b$ are exactly the subsets of the remaining 8 elements: $2^{8}$ .

$Subsets with at least one of a, b = 2^{10} - 2^{8} = 1024 - 256 = 768$

Answer: 768

The trap

Directly counting 'contains a, or contains b, or contains both' by cases and adding them risks double-counting subsets that contain both a and b unless done very carefully — the complement approach sidesteps that risk entirely.

JEE AdvancedProving a set identity for arbitrary setsElement-wise proof

For any three sets $A$ , $B$ , $C$ , prove that $A - (B \cup C) = (A - B) \cap (A - C)$ .

Stuck? Show the approach

Prove the two sets are equal by showing each is a subset of the other — start with an arbitrary element of the left-hand side and show it must lie in the right-hand side, then reverse the argument.

Show the full solution

Let $x \in A - (B \cup C)$ . Then $x \in A$ and $x \in / (B \cup C)$ , so $x \in / B$ and $x \in / C$ . Since $x \in A$ and $x \in / B$ , $x \in A - B$ ; since $x \in A$ and $x \in / C$ , $x \in A - C$ . So $x \in (A - B) \cap (A - C)$ .

Conversely, let $x \in (A - B) \cap (A - C)$ . Then $x \in A - B$ (so $x \in A, x \in / B$ ) and $x \in A - C$ (so $x \in / C$ ). So $x \in A$ and $x \in / B \cup C$ , giving $x \in A - (B \cup C)$ .

Since each side contains the other, the sets are equal.

Answer: A - (B union C) = (A - B) intersection (A - C), proved by mutual containment

The trap

This is essentially De Morgan's law dressed differently — recognising that 'not in B or C' is the same structural pattern as (B∪C)' makes the proof fall out of the same reasoning already used for De Morgan's laws in this chapter, rather than needing a fresh approach.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 11 Mathematics examMedium

JEE Main and Advanced (Sets and Relations)Medium

NDA MathematicsMedium

CUET MathematicsLow

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No. {0} is a set containing the number 0 — it has one element. The empty set ∅ has zero elements. This is a very natural confusion, since 'zero' and 'nothing' feel similar outside mathematics, but a set holding the number zero is not an empty collection.

2⁴ = 16 subsets in total, including ∅ and the set itself. Among these, 14 are proper subsets (every subset except the set itself, since a proper subset must differ from the original set).

No, and this is worth stating clearly because a lot of coaching material still drills it heavily for Chapter 1. The 2026-27 NCERT chapter never states this formula, and it does not appear anywhere in the CBSE 2026-27 syllabus line for Sets. None of the five numbered exercises or the miscellaneous exercise use it. It remains a genuinely useful and common idea in competitive exams like JEE, so it is not wasted knowledge — it simply is not examinable from THIS particular chapter under the current syllabus.

∈ (element of) means A itself is listed as one single object inside B; ⊂ (subset of) means every individual element of A also appears inside B. They can both be true, both false, or one true and the other false, completely independently — for A={1}, B={{1},2}, A∈B is true (A is literally one of B's two elements) but A⊂B is false (1 itself, A's only element, is not one of B's elements).

The definition of A⊆B is that every element of A is also in B. When A=∅, there are no elements to check at all, so the condition is satisfied automatically — there is no possible counterexample, since there is nothing in ∅ that could fail to be in B. This kind of 'true because there's nothing to check' reasoning is called vacuous truth, and it shows up again later in mathematics, not just here.

Compute both sides completely independently and compare at the end — never assume one side just because you computed the other. For (A∪B)'=A'∩B': first find A∪B, then take its complement in U for the left side; separately find A' and B' individually, then intersect them for the right side. If your working ever uses one side's intermediate result to shortcut the other side, you haven't actually verified anything, you've just assumed the law is true.

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Related chapters

Students who read Sets usually read these next.

Relations and Functions

A relation is any association between two sets at all — as loose or as strange as you like. A function is the one, very specific kind of relation where every input has exactly one output, and that single restriction is what makes it possible to build the rest of mathematics on top of it. One classification almost every coaching handout drills for this chapter — one-one, onto, bijective — has actually moved to Class 12. CBSE Class 11 Mathematics Chapter 2.

Trigonometric Functions

The unit-circle definition extends sine and cosine, first met as ratios inside a right triangle, to every real number — including negative numbers and numbers far larger than 360 degrees. One topic many coaching sheets drill hard for this chapter, the general solution of trigonometric equations, is graded formatively only under the 2026-27 CBSE syllabus and never appears in this chapter's own exercises. CBSE Class 11 Mathematics Chapter 3.

Complex Numbers and Quadratic Equations

The equation x² + 1 = 0 has no real solution, so this chapter builds a new number, i = √(-1), and a new algebra around it. Despite the chapter's own name, actually solving ax² + bx + c = 0 for complex roots — and the polar form its own section heading promises — are both graded formatively only under the 2026-27 CBSE syllabus. CBSE Class 11 Mathematics Chapter 4.

Linear Inequalities

Not every real relationship is an equality — Ravi's ₹200 rice budget and Reshma's stationery shopping both translate into inequalities, not equations. This chapter builds the two rules for solving them, chief among them the sign-flip rule. The entire two-variable graphical half-plane method most coaching material drills for this chapter — plotting ax+by+c>0 as a shaded region — is graded formatively only and does not appear in the current book at all. CBSE Class 11 Mathematics Chapter 5.

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