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State Board · Class 11 · Mathematics Part 2

Lesson 5: Sets and Relations

5 concepts · 40 practice questions with answers · Free to read

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Sets, subsets and intervals

From daily life

The list of children in a class, the days of the week, a group of vowels: a clearly defined collection is a set. "Clever children" is not a set because who is clever cannot be decided. Building on classes 9–10 : Since classes 9–10 we know sets and their operations; here we add subsets, the power set and intervals.

Simple definition

A set is a collection of clearly defined objects. It is written by the roster method {1, 2, 3} or the set-builder method {x : x a natural number, x < 4}. If every element of A lies in B, A is a subset of B (A ⊆ B). The set of all subsets is the power set P(A).

Rule 1: Number of subsets — A set of n elements has 2ⁿ subsets; proper subsets 2ⁿ − 1; the empty set φ is a subset of every set; subsets containing a particular element 2ⁿ⁻¹.

Rule 2: Intervals — Open (a, b) = {x : a < x < b}; closed [a, b] = {x : a ≤ x ≤ b}; half-open (a, b] or [a, b). An empty circle means the end is excluded, a filled one means it is included.

ABUA ⊆ B15(1, 5]

In the figure : A ⊆ B and the interval (1, 5] on the number line.

How is the formula obtained? : Each element has two choices: put it in the subset or leave it out. For n elements 2 × 2 × … (n times) = 2ⁿ.

  • Easy : The subsets of {1, 2, 3} are 2³ = 8: φ, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}.
  • Medium : A = {x : x a natural number, x² < 30} = {1, 2, 3, 4, 5}; 5 elements, so the power set has 2⁵ = 32 elements.
  • Exam-type : Non-empty proper subsets of a 4-element set: 2⁴ − 2 = 14. (2, 7] = {x : 2 < x ≤ 7}.
Remember

φ and the set itself are subsets of every set; φ ≠ {0}.

Common mistake

Mixing the signs for element and subset: 2 ∈ {1, 2}, but {2} ⊆ {1, 2}.

Shortcut

To count subsets raise 2 to the number of elements: 2ⁿ; proper subsets one fewer. Memory clue : "Two choices for each".

CET trap

Proper subsets and non-empty proper subsets differ: 2ⁿ − 1 and 2ⁿ − 2.

CET-style example : The subsets of a 4-element set that contain a particular element? (A) 8 (B) 16 (C) 4 (D) 12

Answer : (A). 2⁴⁻¹ = 8. Time-saving method : That element is fixed; 2³ for the other 3.

Spot the difference

SymbolMeaningExample
∈is an element of2 ∈ {1, 2}
⊆is a subset of{2} ⊆ {1, 2}
φempty setn(φ) = 0

Practice questions

Easy
  1. 1Write the set of letters of the word BOOK by the roster method.
  2. 2How many elements does the power set of a 5-element set have?
  3. 3Write (−1, 3] by the set-builder method.
Medium
  1. 4How many subsets of {1, 2, 3, 4, 5, 6} contain 1?
  2. 5How many elements are in {x : x an integer, −2 ≤ x < 3}?
  3. 6Is {1, 2} a proper subset of {1, 2, 3}?
Challenging
  1. 7How many subsets of {1, 2, 3, 4, 5} have exactly 2 elements?
  2. 8How many proper subsets does a 6-element set have?
Show answers (try them yourself first)
  1. 1{B, O, K}.
  2. 22⁵ = 32.
  3. 3{x : x वास्तव संख्या, −1 < x ≤ 3}.
  4. 42⁵ = 32.
  5. 5−2, −1, 0, 1, 2 : 5 घटक.
  6. 6हो ; सर्व घटक आहेत आणि संचापेक्षा वेगळा.
  7. 7⁵C₂ = 10.
  8. 82⁶ − 1 = 63.

Operations on sets and laws

From daily life

The combined list of players of two teams, the players in both, those in one team but not the other: these three actions are union, intersection and difference. Building on classes 9–10 : Since classes 9–10 we know A ∪ B and A ∩ B; here we add the complement, the difference and De Morgan’s laws.

Simple definition

A ∪ B: elements in either set. A ∩ B: elements in both. A − B: elements in A but not in B. A′: elements of the universal set U not in A.

Rule 1: De Morgan and other laws — De Morgan: (A ∪ B)′ = A′ ∩ B′ and (A ∩ B)′ = A′ ∪ B′. Distributive: A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C). A − B = A ∩ B′. (A′)′ = A.

A ∪ BA ∩ BA − BA′

In the figure : The shaded part in four diagrams: A ∪ B, A ∩ B, A − B, A′.

How is the formula obtained? : x ∈ (A ∪ B)′ means x is not in A and not in B, that is x ∈ A′ and x ∈ B′; hence (A ∪ B)′ = A′ ∩ B′.

  • Easy : U = {1, …, 10}, A = {1, …, 6}, B = {4, …, 8}: A ∪ B = {1, …, 8}; A ∩ B = {4, 5, 6}.
  • Medium : For the same sets A − B = {1, 2, 3}; A′ = {7, 8, 9, 10}.
  • Exam-type : De Morgan check: (A ∪ B)′ = {9, 10}; A′ ∩ B′ = {7, 8, 9, 10} ∩ {1, 2, 3, 9, 10} = {9, 10}.
Remember

A − B ≠ B − A; but A ∪ B and A ∩ B stay the same when the order is swapped.

Common mistake

Taking the complement within some other set instead of U. A′ is always taken in U.

Shortcut

De Morgan: taking the complement turns ∪ into ∩ and ∩ into ∪. Memory clue : "Complement comes, sign flips".

CET trap

"Or" means union ∪; "and" means intersection ∩; "only A" means A − B.

CET-style example : A = even numbers up to 20, B = multiples of 3 up to 20; how many elements are in A ∩ B? (A) 3 (B) 4 (C) 6 (D) 2

Answer : (A). Multiples of 6: 6, 12, 18. Time-saving method : Even and a multiple of 3 means a multiple of 6; the whole part of 20/6 is 3.

Spot the difference

OperationMeaningExample above
A ∪ Bin either{1, …, 8}
A ∩ Bin both{4, 5, 6}
A − Bonly A{1, 2, 3}
A′in U but not in A{7, 8, 9, 10}

Practice questions

Easy
  1. 1P = {1, …, 5}, Q = {4, 5, 6, 7}: write P ∪ Q.
  2. 2Write P ∩ Q.
  3. 3Write P − Q.
Medium
  1. 4If U = {1, …, 8} write P′.
  2. 5Write Q − P.
  3. 6Write (P ∪ Q)′ with U = {1, …, 8}.
Challenging
  1. 7Write (P ∩ Q)′ and check it against P′ ∪ Q′.
  2. 8Compare n(P ∪ Q) + n(P ∩ Q) and n(P) + n(Q).
Show answers (try them yourself first)
  1. 1{1, 2, 3, 4, 5, 6, 7}.
  2. 2{4, 5}.
  3. 3{1, 2, 3}.
  4. 4{6, 7, 8}.
  5. 5{6, 7}.
  6. 6{8}.
  7. 7{1, 2, 3, 6, 7, 8} ; P′ ∪ Q′ = {6, 7, 8} ∪ {1, 2, 3, 8} = तेच.
  8. 87 + 2 = 9 आणि 5 + 4 = 9 ; समान.

Counting elements of sets

From daily life

In a group of 40, 25 drink tea and 20 drink coffee, and some drink both. Adding 25 + 20 = 45 counts the ones who drink both twice. Building on classes 9–10 : Since classes 9–10 we have counted groups with Venn diagrams; here we use a formula.

Simple definition

n(A ∪ B) = n(A) + n(B) − n(A ∩ B). For three sets n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A ∩ B) − n(B ∩ C) − n(A ∩ C) + n(A ∩ B ∩ C).

Rule 1: Useful relations — only A = n(A) − n(A ∩ B); neither = n(U) − n(A ∪ B); exactly one = only A + only B.

1258ABn(A ∪ B) = n(A) + n(B) − n(A ∩ B)17 + 13 − 5 = 25

In the figure : Only A 12, both 5, only B 8; n(A) = 17, n(B) = 13, n(A ∪ B) = 25.

How is the formula obtained? : In n(A) + n(B) the elements of A ∩ B appear twice; subtracting them once makes every element counted once.

  • Easy : In 40 children 25 drink tea, 20 coffee, 8 both: tea or coffee = 25 + 20 − 8 = 37; neither = 40 − 37 = 3.
  • Medium : n(A) = 17, n(B) = 13, n(A ∩ B) = 5: n(A ∪ B) = 25; only A = 12.
  • Exam-type : n(A) = 30, n(B) = 25, n(C) = 20, n(A ∩ B) = 10, n(B ∩ C) = 8, n(A ∩ C) = 6, n(A ∩ B ∩ C) = 4: 30 + 25 + 20 − 24 + 4 = 55.
Remember

In the three-set formula the signs are + − +: singles, pairs, all three.

Common mistake

Taking n(A) + n(B) as the union. Do not forget to subtract the intersection.

Shortcut

Draw the Venn diagram first and fill it from the innermost region (all three) outward. Memory clue : "Inside out".

CET trap

If n(A ∩ B) comes out negative the data is inconsistent; check before answering.

CET-style example : If n(A) = 20, n(B) = 15 and n(A ∪ B) = 28, then n(A ∩ B) = ? (A) 7 (B) 8 (C) 5 (D) 13

Answer : (A). 20 + 15 − 28 = 7. Time-saving method : Reverse the formula: n(A ∩ B) = n(A) + n(B) − n(A ∪ B).

Spot the difference

GroupFormulaExample above
In eithern(A) + n(B) − n(A ∩ B)37
Only in onen(A) − n(A ∩ B)17
In neithern(U) − n(A ∪ B)3

Practice questions

Easy
  1. 1If n(A) = 12, n(B) = 9, n(A ∩ B) = 4 find n(A ∪ B).
  2. 2If n(A ∪ B) = 20, n(A) = 11, n(B) = 14 find n(A ∩ B).
  3. 3If n(U) = 60 and n(A) = 35 find n(A′).
Medium
  1. 4Each of 45 children likes maths or science; 28 like maths and 22 like science. How many like both?
  2. 5Among 100 people 60 speak Marathi, 45 English and 20 both. How many speak only Marathi?
  3. 6In the previous question how many speak exactly one language?
Challenging
  1. 7If n(A) = 10, n(B) = 12, n(C) = 8, n(A ∩ B) = 4, n(B ∩ C) = 3, n(A ∩ C) = 2, n(A ∩ B ∩ C) = 1 find n(A ∪ B ∪ C).
  2. 8Among 80 people 50 like A, 40 like B and 10 like neither. How many like both?
Show answers (try them yourself first)
  1. 112 + 9 − 4 = 17.
  2. 211 + 14 − 20 = 5.
  3. 360 − 35 = 25.
  4. 428 + 22 − 45 = 5.
  5. 560 − 20 = 40.
  6. 640 + (45 − 20) = 65.
  7. 730 − 9 + 1 = 22.
  8. 8n(A ∪ B) = 70 ; 50 + 40 − 70 = 20.

Cartesian product and relations

From daily life

If all pairs of 3 shirts and 2 trousers are listed, each pair can be written in order as (shirt, trouser). The set of such pairs is the Cartesian product. Building on classes 9–10 : Since classes 9–10 we have plotted ordered pairs (x, y) on a graph; here we study them as relations.

Simple definition

A × B = {(a, b) : a ∈ A, b ∈ B}; it has n(A) × n(B) elements. A relation R from A to B is any subset of A × B. The domain is the set of first components, the co-domain is B and the range is the set of second components.

Rule 1: Counting — If n(A) = m and n(B) = n then A × B has mn pairs; there are 2ᵐⁿ relations from A to B. A × B ≠ B × A (except when A = B or one is empty).

123126 pointsa < b : 1(1, 2)R ⊆ A × B

In the figure : A = {1, 2, 3}, B = {1, 2}: 6 points; the relation a < b is only (1, 2).

How is the formula obtained? : Every first element goes with all second elements: m × n. Each pair is either in the relation or not: 2 × 2 × … (mn times) = 2ᵐⁿ.

  • Easy : A = {1, 2}, B = {3, 4, 5}: A × B has 2 × 3 = 6 pairs.
  • Medium : A = {1, 2, 3}, B = {1, 2, 3, 4, 5}, R: b = a + 2 gives R = {(1, 3), (2, 4), (3, 5)}; domain {1, 2, 3}, range {3, 4, 5}.
  • Exam-type : On A = {1, 2, 3, 4} the relation a < b has 6 pairs, domain {1, 2, 3}, range {2, 3, 4}.
Remember

The order in a pair matters: (1, 2) ≠ (2, 1).

Common mistake

Taking range and co-domain as the same. The range is a subset of the co-domain.

Shortcut

List the pairs of the relation: first components = domain, second components = range. Memory clue : "First is domain, second is range".

CET trap

The number of elements of A × B is m × n, not the sum.

CET-style example : If n(A) = 3 and n(B) = 2, how many relations are there from A to B? (A) 64 (B) 6 (C) 8 (D) 32

Answer : (A). 2³ˣ² = 2⁶ = 64. Time-saving method : 6 pairs; each is taken or not: 2⁶.

Spot the difference

TermMeaningExample above (b = a + 2)
Domainfirst components{1, 2, 3}
Co-domainB{1, 2, 3, 4, 5}
Rangesecond components{3, 4, 5}

Practice questions

Easy
  1. 1A = {1, 2, 3}, B = {1, 2}: find n(A × B).
  2. 2A = {1, 2}, B = {1, 2, 3, 4}: find the number of elements of A × B.
  3. 3Are (1, 2) and (2, 1) equal?
Medium
  1. 4Write the relation a + b = 5 on {1, 2, 3, 4} as pairs.
  2. 5Write the domain of the relation above.
  3. 6How many relations are there on a 2-element set?
Challenging
  1. 7Write the relation b = 2a on A = {1, …, 6} as pairs.
  2. 8How many subsets of {1, 2} × {1, 2} have exactly 2 pairs?
Show answers (try them yourself first)
  1. 13 × 2 = 6.
  2. 22 × 4 = 8.
  3. 3नाही.
  4. 4{(1, 4), (2, 3), (3, 2), (4, 1)}.
  5. 5{1, 2, 3, 4}.
  6. 62⁴ = 16.
  7. 7{(1, 2), (2, 4), (3, 6)}.
  8. 8⁴C₂ = 6.

Types of relations

From daily life

"Is of the same age as" is true of everyone with themselves, true in both directions and can be carried along a chain. Such relations split people into groups. Building on classes 9–10 : Since classes 9–10 we have grouped objects by a common property (e.g. similar triangles); here are the rules.

Simple definition

A relation R on A is reflexive if (a, a) ∈ R for every a; symmetric if (a, b) ∈ R gives (b, a) ∈ R; transitive if (a, b), (b, c) ∈ R give (a, c) ∈ R. If all three hold, R is an equivalence relation.

Rule 1: Equivalence classes — An equivalence relation splits the set into groups so that every element lies in exactly one group; the groups do not overlap.

rem 03, 6, 9rem 11, 4, 7rem 22, 5, 8a R b if a − b is divisible by 3reflexive, symmetric, transitiveclasses do not overlap

In the figure : a R b if a − b is divisible by 3: three groups with remainder 0, 1, 2.

How is the formula obtained? : a − a = 0 is divisible by 3 (reflexive); if a − b is, so is b − a (symmetric); if a − b and b − c are, so is their sum a − c (transitive).

  • Easy : On {1, 2, 3}, R = {(1, 1), (2, 2), (3, 3)} is reflexive, symmetric and transitive, so it is an equivalence relation.
  • Medium : R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)} has all three properties (e.g. (1, 2), (2, 1) give (1, 1)); it is an equivalence relation.
  • Exam-type : On natural numbers a < b is not reflexive, not symmetric, but transitive. "a divides b" is reflexive and transitive but not symmetric (2 divides 4 but 4 does not divide 2).
Remember

Equivalence relation = reflexive + symmetric + transitive; missing any one means it is not.

Common mistake

Treating symmetric as reflexive. {(1, 2), (2, 1)} is symmetric but not reflexive.

Shortcut

Test one property at a time: (a, a) for all? (a, b) gives (b, a)? chain a → c? Memory clue : "Self, mirror, chain".

CET trap

Transitivity needs (a, c) only when (a, b), (b, c) are both present; with no such pairs the condition holds automatically.

CET-style example : The relation "a − b is even" on natural numbers has how many equivalence classes? (A) 2 (B) 1 (C) 3 (D) 4

Answer : (A). Even and odd: 2 classes. Time-saving method : Division by 2 leaves remainder 0 or 1, hence 2.

Spot the difference

RelationReflexiveSymmetricTransitive
a − b divisible by 3yesyesyes
a < bnonoyes
a divides byesnoyes

Practice questions

Easy
  1. 1Is R = {(1, 1), (2, 2), (1, 2)} on {1, 2} reflexive?
  2. 2Is the same R symmetric?
  3. 3Is the same R transitive?
Medium
  1. 4On {1, 2, 3} is R = {(1, 2), (2, 1)} symmetric, and is it reflexive?
  2. 5Which pair is missing for transitivity of this R?
  3. 6Is the relation "same units digit" an equivalence relation?
Challenging
  1. 7On 1 to 20 with a − b divisible by 5: how many equivalence classes and how many elements in each?
  2. 8The smallest reflexive and symmetric relation on {1, 2, 3} containing (1, 2): how many pairs?
Show answers (try them yourself first)
  1. 1हो.
  2. 2नाही ; (1, 2) आहे पण (2, 1) नाही.
  3. 3हो.
  4. 4सममित आहे ; स्वतःशी नाही.
  5. 5(1, 1) आणि (2, 2).
  6. 6हो ; तिन्ही गुण आहेत.
  7. 75 वर्ग, प्रत्येकात 4 घटक.
  8. 8(1, 1), (2, 2), (3, 3), (1, 2), (2, 1) : 5.

Lesson 5: Sets and relations — quick revision

TopicFormula / point
Subsets2ⁿ for n elements; proper 2ⁿ − 1; with one fixed element 2ⁿ⁻¹
OperationsA ∪ B, A ∩ B, A − B = A ∩ B′, A′
De Morgan(A ∪ B)′ = A′ ∩ B′; (A ∩ B)′ = A′ ∪ B′
Countingn(A ∪ B) = n(A) + n(B) − n(A ∩ B)
Three setssum − pairs + all three
Productn(A × B) = mn; relations 2ᵐⁿ
Equivalencereflexive + symmetric + transitive; classes do not overlap

🎯 Most important points for the exam

  • Do not mix ∈ (element) and ⊆ (subset).
  • Or = ∪, and = ∩.
  • Fill a Venn diagram from the inside out.
  • The order of a pair matters.
  • For equivalence test all three properties.
Study mantra

"Inside out, mind the sign".