NCERT Solutions for Class 11 Maths Chapter 1: Sets

These Class 11 Maths Chapter 1 solutions cover Sets from the NCERT textbook (Reprint 2026–27). Every question of Exercise 1.1, 1.2, 1.3, 1.4 and 1.5 and the Miscellaneous Exercise is reproduced verbatim and solved step by step — roster and set-builder form, empty/finite/infinite sets, subsets and intervals, union, intersection, difference and complement, with answers verified against the book’s answer key.

Class: 11 Subject: Mathematics Chapter: 1 – Sets Exercises: 1.1, 1.2, 1.3, 1.4, 1.5 & Miscellaneous Session: 2026–27
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Chapter 1 Overview

Chapter 1 of Class 11 Mathematics, Sets, lays the foundation for almost every later topic — relations, functions, probability, sequences and more. Developed by Georg Cantor (1845–1918), set theory begins with the idea of a well-defined collection of objects. The chapter teaches you to write a set in roster form and set-builder form; classify sets as empty, finite or infinite; test for equal sets and subsets (including intervals as subsets of R); and perform the four key operations — union (∪), intersection (∩), difference (−) and complement (′) — using Venn diagrams and De Morgan’s laws. The Class 11 Maths Chapter 1 solutions below work through every exercise and the Miscellaneous Exercise.

Key Concepts & Definitions

Set: a well-defined collection of distinct objects. If a is in set A we write a ∈ A; if not, a ∉ A.

Roster form: all elements listed inside braces, separated by commas, e.g. {2, 4, 6}. Order and repetition do not matter.

Set-builder form: elements described by a common property, e.g. V = {x : x is a vowel in the English alphabet}.

Empty (null/void) set: a set with no element, denoted φ or { }.

Finite / infinite set: a set is finite if it is empty or has a definite number n(S) of elements; otherwise it is infinite.

Equal sets: A = B if they have exactly the same elements.

Subset: A ⊂ B if every element of A is also in B. φ is a subset of every set; every set is a subset of itself. A proper subset is A ⊂ B with A ≠ B.

Universal set (U): the basic set containing all elements relevant to a context; its subsets are A, B, C, …

Union, intersection, difference, complement: A ∪ B = {x : x ∈ A or x ∈ B}; A ∩ B = {x : x ∈ A and x ∈ B}; A − B = {x : x ∈ A and x ∉ B}; A′ = U − A.

Disjoint sets: A and B with A ∩ B = φ.

Important Formulas & Laws (Chapter 1)

Number of subsets: a set with n elements has 2n subsets and 2n − 1 proper subsets.

Intervals: (a, b) = {x : a < x < b}; [a, b] = {x : a ≤ x ≤ b}; [a, b) and (a, b] are half-open; length = b − a.

Commutative: A ∪ B = B ∪ A, A ∩ B = B ∩ A.

Associative: (A ∪ B) ∪ C = A ∪ (B ∪ C); (A ∩ B) ∩ C = A ∩ (B ∩ C).

Distributive: A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C); A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).

Identity / U / idempotent: A ∪ φ = A, A ∩ U = A, A ∪ A = A, A ∩ A = A, U ∪ A = U, φ ∩ A = φ.

Complement laws: A ∪ A′ = U, A ∩ A′ = φ, (A′)′ = A, φ′ = U, U′ = φ.

De Morgan’s laws: (A ∪ B)′ = A′ ∩ B′ and (A ∩ B)′ = A′ ∪ B′.

Exercise 1.1 Solutions

Questions are reproduced verbatim from the NCERT textbook; the worked solutions are original and verified against the answers at the back of the book.

1. Which of the following are sets ? Justify your answer. (i) The collection of all the months of a year beginning with the letter J. (ii) The collection of ten most talented writers of India. (iii) A team of eleven best-cricket batsmen of the world. (iv) The collection of all boys in your class. (v) The collection of all natural numbers less than 100. (vi) A collection of novels written by the writer Munshi Prem Chand. (vii) The collection of all even integers. (viii) The collection of questions in this Chapter. (ix) A collection of most dangerous animals of the world.

SOLUTION A collection is a set only if it is well-defined — we can decide for any object whether it belongs. (i) Set — the months January, June, July are clearly determined. (ii) Not a set — “most talented” varies from person to person. (iii) Not a set — “best” is not well-defined. (iv) Set — the boys in your class are clearly determined. (v) Set — natural numbers less than 100 are definite. (vi) Set — the novels of Premchand are determinable. (vii) Set — even integers are well-defined. (viii) Set — the questions in this chapter are definite. (ix) Not a set — “most dangerous” is a matter of opinion. ∴ (i), (iv), (v), (vi), (vii) and (viii) are sets.

2. Let A = {1, 2, 3, 4, 5, 6}. Insert the appropriate symbol ∈ or ∉ in the blank spaces: (i) 5 . . . A   (ii) 8 . . . A   (iii) 0 . . . A (iv) 4 . . . A   (v) 2 . . . A   (vi) 10 . . . A

SOLUTION (i) 5 ∈ A   (ii) 8 ∉ A   (iii) 0 ∉ A (iv) 4 ∈ A   (v) 2 ∈ A   (vi) 10 ∉ A

3. Write the following sets in roster form: (i) A = {x : x is an integer and −3 ≤ x < 7} (ii) B = {x : x is a natural number less than 6} (iii) C = {x : x is a two-digit natural number such that the sum of its digits is 8} (iv) D = {x : x is a prime number which is divisor of 60} (v) E = The set of all letters in the word TRIGONOMETRY (vi) F = The set of all letters in the word BETTER

SOLUTION (i) Integers from −3 up to (but not including) 7: A = {−3, −2, −1, 0, 1, 2, 3, 4, 5, 6}. (ii) B = {1, 2, 3, 4, 5}. (iii) Two-digit numbers whose digits add to 8: C = {17, 26, 35, 44, 53, 62, 71, 80}. (iv) 60 = 22 × 3 × 5, so the prime divisors are D = {2, 3, 5}. (v) Distinct letters of TRIGONOMETRY: E = {T, R, I, G, O, N, M, E, Y}. (vi) Distinct letters of BETTER: F = {B, E, T, R}.

4. Write the following sets in the set-builder form : (i) {3, 6, 9, 12}   (ii) {2, 4, 8, 16, 32}   (iii) {5, 25, 125, 625} (iv) {2, 4, 6, . . .}   (v) {1, 4, 9, . . ., 100}

SOLUTION (i) Multiples of 3 from 3 to 12: {x : x = 3n, n ∈ N and 1 ≤ n ≤ 4}. (ii) Powers of 2 from 21 to 25: {x : x = 2n, n ∈ N and 1 ≤ n ≤ 5}. (iii) Powers of 5 from 51 to 54: {x : x = 5n, n ∈ N and 1 ≤ n ≤ 4}. (iv) {x : x is an even natural number}. (v) Squares from 12 to 102: {x : x = n2, n ∈ N and 1 ≤ n ≤ 10}.

5. List all the elements of the following sets : (i) A = {x : x is an odd natural number} (ii) B = {x : x is an integer, −½ < x < 9/2} (iii) C = {x : x is an integer, x2 ≤ 4} (iv) D = {x : x is a letter in the word “LOYAL”} (v) E = {x : x is a month of a year not having 31 days} (vi) F = {x : x is a consonant in the English alphabet which precedes k}.

SOLUTION (i) A = {1, 3, 5, . . .} (infinite). (ii) Integers strictly between −0.5 and 4.5: B = {0, 1, 2, 3, 4}. (iii) x2 ≤ 4 means −2 ≤ x ≤ 2, so C = {−2, −1, 0, 1, 2}. (iv) Distinct letters of LOYAL: D = {L, O, Y, A}. (v) Months with fewer than 31 days: E = {February, April, June, September, November}. (vi) Consonants before k (a–j): F = {b, c, d, f, g, h, j}.

6. Match each of the set on the left in the roster form with the same set on the right described in set-builder form: (i) {1, 2, 3, 6}    (a) {x : x is a prime number and a divisor of 6} (ii) {2, 3}    (b) {x : x is an odd natural number less than 10} (iii) {M, A, T, H, E, I, C, S}    (c) {x : x is natural number and divisor of 6} (iv) {1, 3, 5, 7, 9}    (d) {x : x is a letter of the word MATHEMATICS}.

SOLUTION (i) {1, 2, 3, 6} are the divisors of 6 → (c). (ii) {2, 3} are the prime divisors of 6 → (a). (iii) {M, A, T, H, E, I, C, S} are the distinct letters of MATHEMATICS → (d). (iv) {1, 3, 5, 7, 9} are odd naturals less than 10 → (b). ∴ (i)↔(c), (ii)↔(a), (iii)↔(d), (iv)↔(b).

Exercise 1.2 Solutions

1. Which of the following are examples of the null set (i) Set of odd natural numbers divisible by 2 (ii) Set of even prime numbers (iii) {x : x is a natural numbers, x < 5 and x > 7} (iv) {y : y is a point common to any two parallel lines}

SOLUTION (i) Null set — no odd number is divisible by 2. (ii) Not null — 2 is an even prime, so the set is {2}. (iii) Null set — no number is both < 5 and > 7. (iv) Null set — parallel lines never meet, so there is no common point. ∴ (i), (iii) and (iv) are null sets.

2. Which of the following sets are finite or infinite (i) The set of months of a year (ii) {1, 2, 3, . . .} (iii) {1, 2, 3, . . . 99, 100} (iv) The set of positive integers greater than 100 (v) The set of prime numbers less than 99

SOLUTION (i) Finite (12 months).   (ii) Infinite (all naturals).   (iii) Finite (100 elements). (iv) Infinite (101, 102, … never end).   (v) Finite (a definite list of primes below 99).

3. State whether each of the following set is finite or infinite: (i) The set of lines which are parallel to the x-axis (ii) The set of letters in the English alphabet (iii) The set of numbers which are multiple of 5 (iv) The set of animals living on the earth (v) The set of circles passing through the origin (0, 0)

SOLUTION (i) Infinite — infinitely many parallel lines.   (ii) Finite (26 letters). (iii) Infinite — multiples of 5 never end.   (iv) Finite — a definite (though huge) number. (v) Infinite — infinitely many circles can pass through one point.

4. In the following, state whether A = B or not: (i) A = {a, b, c, d}   B = {d, c, b, a} (ii) A = {4, 8, 12, 16}   B = {8, 4, 16, 18} (iii) A = {2, 4, 6, 8, 10}   B = {x : x is positive even integer and x ≤ 10} (iv) A = {x : x is a multiple of 10},   B = {10, 15, 20, 25, 30, . . .}

SOLUTION (i) A = B — same elements, order does not matter. (ii) A ≠ B — 12 ∈ A but 12 ∉ B (and 18 ∈ B but 18 ∉ A). (iii) B = {2, 4, 6, 8, 10}, so A = B. (iv) A ≠ B — B contains 15, 25, … which are not multiples of 10.

5. Are the following pair of sets equal ? Give reasons. (i) A = {2, 3}, B = {x : x is solution of x2 + 5x + 6 = 0} (ii) A = {x : x is a letter in the word FOLLOW}   B = {y : y is a letter in the word WOLF}

SOLUTION (i) Solve x2 + 5x + 6 = 0 ⇒ (x + 2)(x + 3) = 0 ⇒ x = −2, −3, so B = {−2, −3}. Since A = {2, 3}, A ≠ B. (ii) Letters of FOLLOW = {F, O, L, W}; letters of WOLF = {W, O, L, F}. Same elements, so A = B (equal).

6. From the sets given below, select equal sets : A = {2, 4, 8, 12}, B = {1, 2, 3, 4}, C = {4, 8, 12, 14}, D = {3, 1, 4, 2} E = {−1, 1}, F = {0, a}, G = {1, −1}, H = {0, 1}

SOLUTION B = {1, 2, 3, 4} and D = {3, 1, 4, 2} have the same elements → B = D. E = {−1, 1} and G = {1, −1} have the same elements → E = G. A, C, F, H share no full match with any other set. ∴ the equal sets are B = D and E = G.

Exercise 1.3 Solutions

1. Make correct statements by filling in the symbols ⊂ or ⊄ in the blank spaces : (i) {2, 3, 4} . . . {1, 2, 3, 4, 5} (ii) {a, b, c} . . . {b, c, d} (iii) {x : x is a student of Class XI of your school} . . . {x : x student of your school} (iv) {x : x is a circle in the plane} . . . {x : x is a circle in the same plane with radius 1 unit} (v) {x : x is a triangle in a plane} . . . {x : x is a rectangle in the plane} (vi) {x : x is an equilateral triangle in a plane} . . . {x : x is a triangle in the same plane} (vii) {x : x is an even natural number} . . . {x : x is an integer}

SOLUTION (i) ⊂ — every element of {2,3,4} is in {1,2,3,4,5}. (ii) ⊄ — a ∈ {a,b,c} but a ∉ {b,c,d}. (iii) ⊂ — Class XI students are students of the school. (iv) ⊄ — not every circle has radius 1. (v) ⊄ — a triangle is not a rectangle. (vi) ⊂ — every equilateral triangle is a triangle. (vii) ⊂ — every even natural number is an integer.

2. Examine whether the following statements are true or false: (i) {a, b} ⊄ {b, c, a} (ii) {a, e} ⊂ {x : x is a vowel in the English alphabet} (iii) {1, 2, 3} ⊂ {1, 3, 5} (iv) {a} ⊂ {a, b, c} (v) {a} ∈ {a, b, c} (vi) {x : x is an even natural number less than 6} ⊂ {x : x is a natural number which divides 36}

SOLUTION (i) False — both a and b lie in {b, c, a}, so {a, b} is a subset. (ii) True — a and e are vowels. (iii) False — 2 ∈ {1,2,3} but 2 ∉ {1,3,5}. (iv) True — a is in {a, b, c}. (v) False — {a} is a subset, not an element, of {a, b, c}; the elements are a, b, c (not {a}). (vi) Left set = {2, 4}; divisors of 36 = {1, 2, 3, 4, 6, 9, 12, 18, 36} contain both 2 and 4 → True.

3. Let A = {1, 2, {3, 4}, 5}. Which of the following statements are incorrect and why? (i) {3, 4} ⊂ A   (ii) {3, 4} ∈ A   (iii) {{3, 4}} ⊂ A (iv) 1 ∈ A   (v) 1 ⊂ A   (vi) {1, 2, 5} ⊂ A (vii) {1, 2, 5} ∈ A   (viii) {1, 2, 3} ⊂ A   (ix) φ ∈ A (x) φ ⊂ A   (xi) {φ} ⊂ A

SOLUTION The elements of A are 1, 2, the set {3, 4}, and 5. (i) Incorrect — 3 and 4 are not separate elements of A; {3, 4} is one element, so {3,4} ∈ A, not ⊂ A. (ii) Correct — {3, 4} is an element of A. (iii) Correct — the single-element set {{3, 4}} is a subset, since {3, 4} ∈ A. (iv) Correct — 1 is an element of A. (v) Incorrect — 1 is an element, not a set, so “1 ⊂ A” is meaningless. (vi) Correct — 1, 2, 5 are all elements of A. (vii) Incorrect — {1, 2, 5} is not a single element of A; it is a subset (⊂), not an element (∈). (viii) Incorrect — 3 is not an element of A (only {3, 4} is), so {1, 2, 3} ⊄ A. (ix) Incorrect — φ is not listed as an element of A. (x) Correct — φ is a subset of every set. (xi) Incorrect — {φ} ⊂ A would need φ ∈ A, but φ is not an element of A. ∴ The incorrect statements are (i), (v), (vii), (viii), (ix) and (xi).

4. Write down all the subsets of the following sets (i) {a}   (ii) {a, b}   (iii) {1, 2, 3}   (iv) φ

SOLUTION A set with n elements has 2n subsets. (i) φ, {a} (2 subsets). (ii) φ, {a}, {b}, {a, b} (4 subsets). (iii) φ, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3} (8 subsets). (iv) The only subset of φ is φ itself (1 subset).

5. Write the following as intervals : (i) {x : x ∈ R, −4 < x ≤ 6}   (ii) {x : x ∈ R, −12 < x < −10} (iii) {x : x ∈ R, 0 ≤ x < 7}   (iv) {x : x ∈ R, 3 ≤ x ≤ 4}

SOLUTION A square bracket includes the end-point; a round bracket excludes it. (i) (−4, 6]   (ii) (−12, −10)   (iii) [0, 7)   (iv) [3, 4].

6. Write the following intervals in set-builder form : (i) (−3, 0)   (ii) [6, 12]   (iii) (6, 12]   (iv) [−23, 5)

SOLUTION (i) {x : x ∈ R, −3 < x < 0}. (ii) {x : x ∈ R, 6 ≤ x ≤ 12}. (iii) {x : x ∈ R, 6 < x ≤ 12}. (iv) {x : x ∈ R, −23 ≤ x < 5}.

7. What universal set(s) would you propose for each of the following : (i) The set of right triangles.   (ii) The set of isosceles triangles.

SOLUTION (i) A suitable universal set is the set of all triangles in a plane (or the set of all polygons). (ii) Again, the set of all triangles in a plane serves as a universal set.

8. Given the sets A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 6, 8}, which of the following may be considered as universal set(s) for all the three sets A, B and C (i) {0, 1, 2, 3, 4, 5, 6} (ii) φ (iii) {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} (iv) {1, 2, 3, 4, 5, 6, 7, 8}

SOLUTION A universal set must contain every element of A, B and C — in particular 8 (from C). (i) Missing 8 → no.   (ii) Empty → no.   (iv) Missing 0 → no. (iii) Contains 0, 1, 2, 3, 4, 5, 6, 8 (all needed elements) → (iii) is the universal set.

Exercise 1.4 Solutions

1. Find the union of each of the following pairs of sets : (i) X = {1, 3, 5}   Y = {1, 2, 3} (ii) A = {a, e, i, o, u}   B = {a, b, c} (iii) A = {x : x is a natural number and multiple of 3}   B = {x : x is a natural number less than 6} (iv) A = {x : x is a natural number and 1 < x ≤ 6}   B = {x : x is a natural number and 6 < x < 10} (v) A = {1, 2, 3}, B = φ

SOLUTION (i) X ∪ Y = {1, 2, 3, 5}. (ii) A ∪ B = {a, b, c, e, i, o, u}. (iii) A = {3, 6, 9, …}, B = {1, 2, 3, 4, 5}, so A ∪ B = {x : x = 1, 2, 4, 5 or a multiple of 3}. (iv) A = {2, 3, 4, 5, 6}, B = {7, 8, 9}, so A ∪ B = {2, 3, 4, 5, 6, 7, 8, 9} = {x : 1 < x < 10, x ∈ N}. (v) A ∪ φ = {1, 2, 3}.

2. Let A = {a, b}, B = {a, b, c}. Is A ⊂ B ? What is A ∪ B ?

SOLUTION Both a and b lie in B, so yes, A ⊂ B. A ∪ B = {a, b, c} = B (since A ⊂ B).

3. If A and B are two sets such that A ⊂ B, then what is A ∪ B ?

SOLUTION If A ⊂ B, every element of A is already in B, so adding A contributes nothing new. ∴ A ∪ B = B.

4. If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8} and D = {7, 8, 9, 10}; find (i) A ∪ B   (ii) A ∪ C   (iii) B ∪ C   (iv) B ∪ D (v) A ∪ B ∪ C   (vi) A ∪ B ∪ D   (vii) B ∪ C ∪ D

SOLUTION (i) A ∪ B = {1, 2, 3, 4, 5, 6}. (ii) A ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}. (iii) B ∪ C = {3, 4, 5, 6, 7, 8}. (iv) B ∪ D = {3, 4, 5, 6, 7, 8, 9, 10}. (v) A ∪ B ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}. (vi) A ∪ B ∪ D = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. (vii) B ∪ C ∪ D = {3, 4, 5, 6, 7, 8, 9, 10}.

5. Find the intersection of each pair of sets of question 1 above.

SOLUTION (i) X ∩ Y = {1, 3}. (ii) A ∩ B = {a}. (iii) Multiples of 3 less than 6 common to both: {3}. (iv) A = {2,3,4,5,6}, B = {7,8,9}; no common element, so φ. (v) A ∩ φ = φ.

6. If A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13}, C = {11, 13, 15} and D = {15, 17}; find (i) A ∩ B   (ii) B ∩ C   (iii) A ∩ C ∩ D (iv) A ∩ C   (v) B ∩ D   (vi) A ∩ (B ∪ C) (vii) A ∩ D   (viii) A ∩ (B ∪ D)   (ix) (A ∩ B) ∩ (B ∪ C) (x) (A ∪ D) ∩ (B ∪ C)

SOLUTION (i) A ∩ B = {7, 9, 11}. (ii) B ∩ C = {11, 13}. (iii) A ∩ C = {11}; (A ∩ C) ∩ D = {11} ∩ {15, 17} = φ. (iv) A ∩ C = {11}. (v) B ∩ D = φ (no common element). (vi) B ∪ C = {7, 9, 11, 13, 15}; A ∩ (B ∪ C) = {7, 9, 11}. (vii) A ∩ D = φ. (viii) B ∪ D = {7, 9, 11, 13, 15, 17}; A ∩ (B ∪ D) = {7, 9, 11}. (ix) A ∩ B = {7, 9, 11}; B ∪ C = {7, 9, 11, 13, 15}; intersection = {7, 9, 11}. (x) A ∪ D = {3, 5, 7, 9, 11, 15, 17}; B ∪ C = {7, 9, 11, 13, 15}; intersection = {7, 9, 11, 15}.

7. If A = {x : x is a natural number}, B = {x : x is an even natural number}, C = {x : x is an odd natural number} and D = {x : x is a prime number}, find (i) A ∩ B   (ii) A ∩ C   (iii) A ∩ D (iv) B ∩ C   (v) B ∩ D   (vi) C ∩ D

SOLUTION (i) Even naturals are a subset of naturals, so A ∩ B = B (set of even natural numbers). (ii) A ∩ C = C (set of odd natural numbers). (iii) A ∩ D = D (set of prime numbers). (iv) No number is both even and odd, so B ∩ C = φ. (v) The only even prime is 2, so B ∩ D = {2}. (vi) Odd primes (3, 5, 7, …): C ∩ D = {x : x is an odd prime number}.

8. Which of the following pairs of sets are disjoint (i) {1, 2, 3, 4} and {x : x is a natural number and 4 ≤ x ≤ 6} (ii) {a, e, i, o, u} and {c, d, e, f} (iii) {x : x is an even integer} and {x : x is an odd integer}

SOLUTION (i) Second set = {4, 5, 6}; common element 4, so not disjoint. (ii) Common element e, so not disjoint. (iii) No integer is both even and odd, so the sets are disjoint.

9. If A = {3, 6, 9, 12, 15, 18, 21}, B = {4, 8, 12, 16, 20}, C = {2, 4, 6, 8, 10, 12, 14, 16}, D = {5, 10, 15, 20}; find (i) A − B   (ii) A − C   (iii) A − D   (iv) B − A (v) C − A   (vi) D − A   (vii) B − C   (viii) B − D (ix) C − B   (x) D − B   (xi) C − D   (xii) D − C

SOLUTION (i) A − B (remove 12) = {3, 6, 9, 15, 18, 21}. (ii) A − C (remove 6, 12) = {3, 9, 15, 18, 21}. (iii) A − D (remove 15) = {3, 6, 9, 12, 18, 21}. (iv) B − A (remove 12) = {4, 8, 16, 20}. (v) C − A (remove 6, 12) = {2, 4, 8, 10, 14, 16}. (vi) D − A (remove 15) = {5, 10, 20}. (vii) B − C (remove 4, 8, 12, 16) = {20}. (viii) B − D (remove 20) = {4, 8, 12, 16}. (ix) C − B (remove 4, 8, 12, 16) = {2, 6, 10, 14}. (x) D − B (remove 20) = {5, 10, 15}. (xi) C − D (remove 10) = {2, 4, 6, 8, 12, 14, 16}. (xii) D − C (remove 4, 10) = {5, 15, 20}.

10. If X = {a, b, c, d} and Y = {f, b, d, g}, find (i) X − Y   (ii) Y − X   (iii) X ∩ Y

SOLUTION (i) X − Y (remove b, d) = {a, c}. (ii) Y − X (remove b, d) = {f, g}. (iii) X ∩ Y = {b, d}.

11. If R is the set of real numbers and Q is the set of rational numbers, then what is R − Q?

SOLUTION R − Q removes every rational number from R, leaving the numbers that are real but not rational. ∴ R − Q = the set of irrational numbers.

12. State whether each of the following statement is true or false. Justify your answer. (i) {2, 3, 4, 5} and {3, 6} are disjoint sets. (ii) {a, e, i, o, u} and {a, b, c, d} are disjoint sets. (iii) {2, 6, 10, 14} and {3, 7, 11, 15} are disjoint sets. (iv) {2, 6, 10} and {3, 7, 11} are disjoint sets.

SOLUTION (i) False — 3 is common to both. (ii) False — a is common to both. (iii) True — no common element. (iv) True — no common element.

Exercise 1.5 Solutions

1. Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4}, B = {2, 4, 6, 8} and C = {3, 4, 5, 6}. Find (i) A′ (ii) B′ (iii) (A ∪ C)′ (iv) (A ∪ B)′ (v) (A′)′ (vi) (B − C)′

SOLUTION (i) A′ = U − A = {5, 6, 7, 8, 9}. (ii) B′ = U − B = {1, 3, 5, 7, 9}. (iii) A ∪ C = {1, 2, 3, 4, 5, 6}, so (A ∪ C)′ = {7, 8, 9}. (iv) A ∪ B = {1, 2, 3, 4, 6, 8}, so (A ∪ B)′ = {5, 7, 9}. (v) (A′)′ = A = {1, 2, 3, 4}. (vi) B − C = {2, 8}, so (B − C)′ = {1, 3, 4, 5, 6, 7, 9}.

2. If U = {a, b, c, d, e, f, g, h}, find the complements of the following sets : (i) A = {a, b, c}   (ii) B = {d, e, f, g} (iii) C = {a, c, e, g}   (iv) D = {f, g, h, a}

SOLUTION (i) A′ = {d, e, f, g, h}. (ii) B′ = {a, b, c, h}. (iii) C′ = {b, d, f, h}. (iv) D′ = {b, c, d, e}.

3. Taking the set of natural numbers as the universal set, write down the complements of the following sets: (i) {x : x is an even natural number}   (ii) {x : x is an odd natural number} (iii) {x : x is a positive multiple of 3}   (iv) {x : x is a prime number} (v) {x : x is a natural number divisible by 3 and 5} (vi) {x : x is a perfect square}   (vii) {x : x is a perfect cube} (viii) {x : x + 5 = 8}   (ix) {x : 2x + 5 = 9} (x) {x : x ≥ 7}   (xi) {x : x ∈ N and 2x + 1 > 10}

SOLUTION Here U = N. The complement is “all natural numbers not in the set”. (i) {x : x is an odd natural number}. (ii) {x : x is an even natural number}. (iii) {x : x ∈ N and x is not a multiple of 3}. (iv) Non-primes: {x : x is a positive composite number or x = 1}. (v) Divisible by 3 and 5 means divisible by 15; complement = {x : x is a positive integer not divisible by 15} (i.e. not divisible by 3 or not divisible by 5). (vi) {x : x ∈ N and x is not a perfect square}. (vii) {x : x ∈ N and x is not a perfect cube}. (viii) The set is {3}; complement = {x : x ∈ N and x ≠ 3}. (ix) 2x + 5 = 9 ⇒ x = 2; complement = {x : x ∈ N and x ≠ 2}. (x) The set is {7, 8, 9, …}; complement = {x : x ∈ N and x < 7}. (xi) 2x + 1 > 10 ⇒ x > 4.5, i.e. x ≥ 5; complement = {x : x ∈ N and x ≤ 9/2}, i.e. {1, 2, 3, 4}.

4. If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 4, 6, 8} and B = {2, 3, 5, 7}. Verify that (i) (A ∪ B)′ = A′ ∩ B′   (ii) (A ∩ B)′ = A′ ∪ B′

SOLUTION A′ = {1, 3, 5, 7, 9}, B′ = {1, 4, 6, 8, 9}. (i) A ∪ B = {2, 3, 4, 5, 6, 7, 8}, so (A ∪ B)′ = {1, 9}. Also A′ ∩ B′ = {1, 9}.  ⇒ verified. (ii) A ∩ B = {2}, so (A ∩ B)′ = {1, 3, 4, 5, 6, 7, 8, 9}. Also A′ ∪ B′ = {1, 3, 4, 5, 6, 7, 8, 9}.  ⇒ verified.

5. Draw appropriate Venn diagram for each of the following : (i) (A ∪ B)′,   (ii) A′ ∩ B′,   (iii) (A ∩ B)′,   (iv) A′ ∪ B′

SOLUTION Draw a rectangle for U with two overlapping circles A and B. (i) (A ∪ B)′: shade only the region outside both circles. (ii) A′ ∩ B′: shade the region outside A AND outside B — this is the same shaded region as (i), confirming De Morgan’s law (A ∪ B)′ = A′ ∩ B′. (iii) (A ∩ B)′: shade everything except the overlap (the central lens). (iv) A′ ∪ B′: shade (outside A) together with (outside B) — this equals the region in (iii), confirming (A ∩ B)′ = A′ ∪ B′.

6. Let U be the set of all triangles in a plane. If A is the set of all triangles with at least one angle different from 60°, what is A′?

SOLUTION A′ consists of triangles with no angle different from 60° — i.e. all three angles equal to 60°. ∴ A′ = the set of all equilateral triangles.

7. Fill in the blanks to make each of the following a true statement : (i) A ∪ A′ = . . .   (ii) φ′ ∩ A = . . . (iii) A ∩ A′ = . . .   (iv) U′ ∩ A = . . .

SOLUTION (i) A ∪ A′ = U. (ii) φ′ = U, and U ∩ A = A. (iii) A ∩ A′ = φ. (iv) U′ = φ, and φ ∩ A = φ.

Miscellaneous Exercise on Chapter 1 Solutions

1. Decide, among the following sets, which sets are subsets of one and another: A = {x : x ∈ R and x satisfy x2 − 8x + 12 = 0}, B = {2, 4, 6}, C = {2, 4, 6, 8, . . .}, D = {6}.

SOLUTION Solve x2 − 8x + 12 = 0 ⇒ (x − 2)(x − 6) = 0 ⇒ x = 2, 6, so A = {2, 6}. A = {2, 6} ⊂ B = {2, 4, 6}; B ⊂ C (even naturals); D = {6} ⊂ A, B and C. ∴ A ⊂ B, A ⊂ C, B ⊂ C, D ⊂ A, D ⊂ B, D ⊂ C.

2. In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example. (i) If x ∈ A and A ∈ B, then x ∈ B (ii) If A ⊂ B and B ∈ C, then A ∈ C (iii) If A ⊂ B and B ⊂ C, then A ⊂ C (iv) If A ⊄ B and B ⊄ C, then A ⊄ C (v) If x ∈ A and A ⊄ B, then x ∈ B (vi) If A ⊂ B and x ∉ B, then x ∉ A

SOLUTION (i) False. Let A = {1}, B = {{1}, 2}. Then 1 ∈ A and A ∈ B, but 1 ∉ B. (ii) False. Let A = {1}, B = {1, 2}, C = {{1, 2}, 3}. Then A ⊂ B and B ∈ C, but A ∉ C. (iii) True. If a ∈ A, then a ∈ B (since A ⊂ B), and then a ∈ C (since B ⊂ C). So every element of A is in C, i.e. A ⊂ C. (iv) False. Let A = {1, 2}, B = {3}, C = {1, 2, 4}. Then A ⊄ B and B ⊄ C, but A ⊂ C. (v) False. Let A = {1, 2}, B = {2, 3}. Here 1 ∈ A and A ⊄ B, but 1 ∉ B. (vi) True. Suppose x ∈ A. Since A ⊂ B, x ∈ B — contradicting x ∉ B. Hence x ∉ A.

3. Let A, B, and C be the sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C. Show that B = C.

SOLUTION Let b ∈ B. Then b ∈ A ∪ B = A ∪ C, so b ∈ A or b ∈ C. Case 1: b ∈ A. Then b ∈ A ∩ B = A ∩ C, so b ∈ C. Case 2: b ∉ A. Then from b ∈ A ∪ C we must have b ∈ C. Either way b ∈ C, so B ⊂ C. By symmetry (swap B and C) C ⊂ B. Hence B = C.

4. Show that the following four conditions are equivalent : (i) A ⊂ B   (ii) A − B = φ   (iii) A ∪ B = B   (iv) A ∩ B = A

SOLUTION We prove (i) ⇒ (ii) ⇒ (iii) ⇒ (iv) ⇒ (i). (i) ⇒ (ii): If A ⊂ B, no element of A is outside B, so A − B = φ. (ii) ⇒ (iii): A − B = φ means every element of A is in B, so A ∪ B = B. (iii) ⇒ (iv): If A ∪ B = B, then A ⊂ B, so the common part A ∩ B = A. (iv) ⇒ (i): If A ∩ B = A, then every element of A is also in B, i.e. A ⊂ B. The chain closes, so all four conditions are equivalent.

5. Show that if A ⊂ B, then C − B ⊂ C − A.

SOLUTION Let x ∈ C − B. Then x ∈ C and x ∉ B. Since A ⊂ B, if x were in A it would be in B; but x ∉ B, so x ∉ A. Thus x ∈ C and x ∉ A, i.e. x ∈ C − A. Hence C − B ⊂ C − A.

6. Show that for any sets A and B, A = (A ∩ B) ∪ (A − B) and A ∪ (B − A) = (A ∪ B)

SOLUTION First result. (A ∩ B) ∪ (A − B) = (A ∩ B) ∪ (A ∩ B′) = A ∩ (B ∪ B′) = A ∩ U = A. Second result. A ∪ (B − A) = A ∪ (B ∩ A′) = (A ∪ B) ∩ (A ∪ A′) = (A ∪ B) ∩ U = A ∪ B.

7. Using properties of sets, show that (i) A ∪ (A ∩ B) = A   (ii) A ∩ (A ∪ B) = A.

SOLUTION (i) A ∪ (A ∩ B) = (A ∩ U) ∪ (A ∩ B) = A ∩ (U ∪ B) = A ∩ U = A (absorption law). (ii) A ∩ (A ∪ B) = (A ∪ φ) ∩ (A ∪ B) = A ∪ (φ ∩ B) = A ∪ φ = A (absorption law).

8. Show that A ∩ B = A ∩ C need not imply B = C.

SOLUTION Take A = {1, 2}, B = {2, 3}, C = {2, 4}. Then A ∩ B = {2} and A ∩ C = {2}, so A ∩ B = A ∩ C. But B = {2, 3} ≠ {2, 4} = C. Hence A ∩ B = A ∩ C does not imply B = C.

9. Let A and B be sets. If A ∩ X = B ∩ X = φ and A ∪ X = B ∪ X for some set X, show that A = B. (Hint: A = A ∩ (A ∪ X), B = B ∩ (B ∪ X) and use Distributive law)

SOLUTION A = A ∩ (A ∪ X) = A ∩ (B ∪ X) = (A ∩ B) ∪ (A ∩ X) = (A ∩ B) ∪ φ = A ∩ B. Similarly, B = B ∩ (B ∪ X) = B ∩ (A ∪ X) = (B ∩ A) ∪ (B ∩ X) = (A ∩ B) ∪ φ = A ∩ B. Hence A = A ∩ B = B, i.e. A = B.

10. Find sets A, B and C such that A ∩ B, B ∩ C and A ∩ C are non-empty sets and A ∩ B ∩ C = φ.

SOLUTION Take A = {1, 2}, B = {1, 3}, C = {2, 3}. A ∩ B = {1}, B ∩ C = {3}, A ∩ C = {2} — all non-empty. A ∩ B ∩ C = φ, since no single element lies in all three sets.

Common Mistakes to Avoid

Watch out for these

  • Calling a vague collection (“best players”, “most talented”) a set — a set must be well-defined.
  • Confusing ∈ (element of) with ⊂ (subset of). In A = {1, 2, {3, 4}, 5}, {3, 4} ∈ A but {3, 4} ⊄ A.
  • Repeating elements or worrying about order in roster form — both are irrelevant.
  • Mixing up interval brackets: round = excluded end-point, square = included.
  • Thinking “divisible by 3 and 5” means divisible by 3 or 5 — it means divisible by 15.
  • Forgetting that φ is a subset of every set, and every set is a subset of itself.
  • Assuming A ∩ B = A ∩ C forces B = C — it does not.

Practice MCQs & Assertion–Reason

1. Which of the following is a well-defined set?

(a) The collection of clever students    (b) The collection of beautiful flowers    (c) The collection of even integers    (d) The collection of difficult sums

2. The set {x : x ∈ N and x2 ≤ 4} in roster form is:

(a) {1, 2}    (b) {−2, −1, 1, 2}    (c) {0, 1, 2}    (d) {1, 2, 3, 4}

3. The number of subsets of a set having 4 elements is:

(a) 8    (b) 12    (c) 16    (d) 4

4. The interval {x : x ∈ R, −4 < x ≤ 6} is written as:

(a) [−4, 6]    (b) (−4, 6]    (c) [−4, 6)    (d) (−4, 6)

5. If A and B are disjoint sets, then A ∩ B is:

(a) A    (b) B    (c) φ    (d) A ∪ B

6. For any two sets A and B, (A ∪ B)′ equals:

(a) A′ ∪ B′    (b) A′ ∩ B′    (c) A ∩ B    (d) A′ − B′

7. If A ⊂ B, then A ∪ B is:

(a) A    (b) B    (c) φ    (d) A ∩ B

8. If R is the set of reals and Q the set of rationals, then R − Q is the set of:

(a) integers    (b) natural numbers    (c) irrational numbers    (d) whole numbers

9. The set of even prime numbers is:

(a) φ    (b) {2}    (c) {2, 3}    (d) infinite

10. For any set A, A ∩ A′ is equal to:

(a) U    (b) A    (c) A′    (d) φ

Answer key: 1-(c), 2-(a), 3-(c), 4-(b), 5-(c), 6-(b), 7-(b), 8-(c), 9-(b), 10-(d).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: The collection of the ten most talented writers of India is not a set.

Reason: A set must be a well-defined collection of objects.

A-R 2. Assertion: The empty set φ is a subset of every set.

Reason: Every set is a subset of itself.

A-R 3. Assertion: A set with 3 elements has 8 subsets.

Reason: A set with n elements has 2n subsets.

A-R 4. Assertion: For sets A and B, (A ∩ B)′ = A′ ∩ B′.

Reason: De Morgan’s law states that the complement of an intersection is the union of the complements.

A-R 5. Assertion: If A ∩ B = A ∩ C, then B = C.

Reason: For A = {1, 2}, B = {2, 3}, C = {2, 4}, A ∩ B = A ∩ C = {2} but B ≠ C.

Answer key: 1-(A), 2-(B), 3-(A), 4-(D), 5-(D).

Quick Revision Summary

  • A set is a well-defined collection of distinct objects; write it in roster or set-builder form.
  • Empty set φ has no elements; a set is finite if it has a definite count n(S), else infinite.
  • A = B iff they have exactly the same elements; A ⊂ B iff every element of A is in B.
  • A set with n elements has 2n subsets; φ is a subset of every set.
  • Intervals are subsets of R: ( ) excludes, [ ] includes; length = b − a.
  • A ∪ B = elements in A or B; A ∩ B = in both; A − B = in A not B; A′ = U − A.
  • De Morgan’s laws: (A ∪ B)′ = A′ ∩ B′ and (A ∩ B)′ = A′ ∪ B′.

How to score full marks in this chapter

Read each collection carefully to test if it is well-defined before calling it a set, and always distinguish ∈ from ⊂. When listing subsets, work systematically by size (0, 1, 2, … elements) so you reach the full 2n count. For operation problems, write each set out fully first, then apply ∪, ∩, − or ′ step by step. Quote De Morgan’s and the distributive laws by name in proofs, and verify identities with a small numerical example to catch slips.

Frequently Asked Questions

What is Class 11 Maths Chapter 1 Sets about?

Chapter 1 introduces sets as well-defined collections of objects. It covers roster and set-builder forms, empty/finite/infinite and equal sets, subsets and intervals of R, the universal set, Venn diagrams, and the operations of union, intersection, difference and complement, including De Morgan’s laws.

How many exercises are there in Class 11 Maths Chapter 1?

There are five exercises — Exercise 1.1, 1.2, 1.3, 1.4 and 1.5 — plus a Miscellaneous Exercise on Chapter 1. Every question of all six is solved step by step on this page.

What is the difference between ∈ and ⊂ in sets?

The symbol ∈ means “is an element of” and relates an object to a set (e.g. 2 ∈ {1, 2, 3}). The symbol ⊂ means “is a subset of” and relates one set to another (e.g. {1, 2} ⊂ {1, 2, 3}). A set can be an element of another set, which is why {3, 4} ∈ {1, {3, 4}} but {3, 4} ⊄ it.

Are these Class 11 Maths Chapter 1 solutions free?

Yes. All solutions are free and follow the official NCERT Mathematics textbook for the 2026–27 session, with every answer verified against the book’s answer key.

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