NCERT Solutions Class 11 Maths

Chapter 1: Sets (Exercise 1.1)

Detailed Solutions with Expert Tips

This guide provides step-by-step solutions for Exercise 1.1 of Class 11 Maths Chapter 1 (Sets). Unlike standard answer keys, these solutions include "Expert Tips" and "Common Mistakes" to help you understand the logic and score higher in exams.

Question 1: Which of the following are sets? Justify your answer.

1. The collection of all the months of a year beginning with the letter J.

• Answer: It is a set.

• Reason: The collection is well-defined. We can list the specific months: {January, June, July}. Everyone will get the same answer.

2. The collection of ten most talented writers of India.

• Answer: It is not a set.

• Reason: The term "most talented" is vague and subjective. A writer I consider talented might not be on your list. Since the criteria vary from person to person, it is not well-defined.

• 💡 Expert Tip: Whenever you see subjective adjectives like "best," "most talented," "dangerous," or "beautiful," the collection is usually NOT a set.

3. A team of eleven best-cricket batsmen of the world.

• Answer: It is not a set.

• Reason: Similar to the previous question, "best" is a matter of opinion. There is no mathematical formula to define the "best" batsman.

4. The collection of all boys in your class.

• Answer: It is a set.

• Reason: It is well-defined. You can clearly identify exactly who belongs to this collection (the boys currently enrolled in your specific class).

5. The collection of all natural numbers less than 100.

• Answer: It is a set.

• Reason: The numbers are fixed: {1, 2, 3, ..., 99}. This is well-defined.

6. A collection of novels written by the writer Munshi Prem Chand.

• Answer: It is a set.

• Reason: This is fact-based. A book is either written by him or it isn't. There is no ambiguity.

7. The collection of all even integers.

• Answer: It is a set.

• Reason: Even integers follow a specific mathematical rule (multiples of 2).

8. The collection of questions in this Chapter.

• Answer: It is a set.

• Reason: The questions are fixed and countable within the textbook.

9. A collection of most dangerous animals of the world.

• Answer: It is not a set.

• Reason: "Dangerous" is subjective. Some might consider a mosquito dangerous (disease), while others think of a lion.

Question 2: Let A = {1, 2, 3, 4, 5, 6}. Insert the appropriate symbol ∈ or ∉ in the blank spaces.

(Note: ∈ means "belongs to" and ∉ means "does not belong to")

1. 5 ... A → 5 ∈ A (Since 5 is in the list)

2. 8 ... A → 8 ∉ A (8 is not in the list)

3. 0 ... A → 0 ∉ A (0 is not in the list)

4. 4 ... A → 4 ∈ A

5. 2 ... A → 2 ∈ A

6. 10 ... A → 10 ∉ A

Question 3: Write the following sets in roster form.

1. A = {x : x is an integer and –3 ≤ x < 7}

• Solution: A = {–3, –2, –1, 0, 1, 2, 3, 4, 5, 6}

• ⚠️ Common Mistake: Pay attention to the signs! "≤" means you include –3. "<" means you exclude 7.

2. B = {x : x is a natural number less than 6}

• Solution: B = {1, 2, 3, 4, 5}

• Note: Natural numbers start from 1, not 0.

3. C = {x : x is a two-digit natural number such that the sum of its digits is 8}

• Solution: C = {17, 26, 35, 44, 53, 62, 71, 80}

• ⚠️ Common Mistake: Students often forget 80 (8+0=8) or try to include 08 (which is not a two-digit number).

4. D = {x : x is a prime number which is divisor of 60}

• Solution: D = {2, 3, 5}

• Explanation: The divisors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60. Out of these, only 2, 3, and 5 are prime numbers. (1 is not prime).

5. E = The set of all letters in the word TRIGONOMETRY

• Solution: E = {T, R, I, G, O, N, M, E, Y}

• 💡 Expert Tip: In Roster form, elements must be unique. Do not repeat letters like 'O' or 'T' even if they appear twice in the word.

6. F = The set of all letters in the word BETTER

• Solution: F = {B, E, T, R}

Question 4: Write the following sets in the set-builder form.

1. {3, 6, 9, 12}

• Solution: {x : x = 3n, n ∈ N and 1 ≤ n ≤ 4}

• Logic: These are multiples of 3 (3×1, 3×2, 3×3, 3×4).

2. {2, 4, 8, 16, 32}

• Solution: {x : x = 2ⁿ, n ∈ N and 1 ≤ n ≤ 5}

• Logic: These are powers of 2 (2¹, 2², 2³, 2⁴, 2⁵).

3. {5, 25, 125, 625}

• Solution: {x : x = 5ⁿ, n ∈ N and 1 ≤ n ≤ 4}

• Logic: These are powers of 5.

4. {2, 4, 6, ...}

• Solution: {x : x is an even natural number}

5. {1, 4, 9, ..., 100}

• Solution: {x : x = n², n ∈ N and 1 ≤ n ≤ 10}

• Logic: These are perfect squares (1², 2², ... 10²).

Question 5: List all the elements of the following sets.

1. A = {x : x is an odd natural number}

• Solution: A = {1, 3, 5, 7, ...}

2. B = {x : x is an integer, –1/2 < x < 9/2}

• Solution: B = {0, 1, 2, 3, 4}

• 💡 Expert Tip: Convert fractions to decimals to make it easier. –0.5 < x < 4.5. The integers inside this range are 0, 1, 2, 3, 4.

3. C = {x : x is an integer, x² ≤ 4}

• Solution: C = {–2, –1, 0, 1, 2}

• Logic: The squares of these numbers are 4, 1, 0, 1, 4, all of which are less than or equal to 4.

4. D = {x : x is a letter in the word "LOYAL"}

• Solution: D = {L, O, Y, A}

• Note: We drop the repeated 'L'.

5. E = {x : x is a month of a year not having 31 days}

• Solution: E = {February, April, June, September, November}

• 💡 Expert Tip: Use the "Knuckle Trick" to remember which months have 31 days!

6. F = {x : x is a consonant in the English alphabet which precedes k}

• Solution: F = {b, c, d, f, g, h, j}

• Caution: You must exclude vowels (a, e, i) that come before 'k'.

Question 6: Match each of the set on the left with the same set on the right.

• (i) {1, 2, 3, 6} matches with (c) {x : x is a natural number and divisor of 6}

• (ii) {2, 3} matches with (a) {x : x is a prime number and a divisor of 6}

• (iii) {M, A, T, H, E, I, C, S} matches with (d) {x : x is a letter of the word MATHEMATICS}

• (iv) {1, 3, 5, 7, 9} matches with (b) {x : x is an odd natural number less than 10}

NCERT Solutions Class 11 Maths

Chapter 1: Sets (Exercise 1.2)

Detailed Solutions with Expert Tips

This guide provides step-by-step solutions for Exercise 1.2 of Class 11 Maths Chapter 1 (Sets). This exercise focuses on classifying sets as Empty (Null) Sets, Finite vs. Infinite Sets, and Equal Sets.

Question 1: Which of the following are examples of the null set?

Concept Refresher: A Null Set (or Empty Set, denoted by φ or { }) is a set that contains no elements at all.

1. Set of odd natural numbers divisible by 2.

• Answer: It is a Null Set.

• Reason: Odd numbers (1, 3, 5...) are by definition not divisible by 2. Therefore, no such number exists.

2. Set of even prime numbers.

• Answer: It is not a Null Set.

• Reason: The number 2 is both even and prime. Since the set contains an element {2}, it is not empty.

• ⚠️ Common Mistake: Students often think all prime numbers are odd. Remember, 2 is the only even prime number.

3. {x : x is a natural number, x < 5 and x > 7}

• Answer: It is a Null Set.

• Reason: No natural number can be simultaneously less than 5 AND greater than 7. It’s impossible.

4. {y : y is a point common to any two parallel lines}

• Answer: It is a Null Set.

• Reason: By definition, parallel lines never meet. Therefore, they have no common point.

Question 2: Which of the following sets are finite or infinite?

Concept Refresher: A Finite Set has a countable number of elements (it ends). An Infinite Set goes on forever.

1. The set of months of a year.

• Answer: Finite.

• Reason: There are exactly 12 months in a year. The set is countable.

2. {1, 2, 3, ...}

• Answer: Infinite.

• Reason: The three dots (...) indicate that the sequence of natural numbers goes on forever.

3. {1, 2, 3, ..., 99, 100}

• Answer: Finite.

• Reason: The set has a definite starting point (1) and a definite ending point (100). It contains exactly 100 elements.

4. The set of positive integers greater than 100.

• Answer: Infinite.

• Reason: The integers continue endlessly (101, 102, 103, ...). There is no last number.

5. The set of prime numbers less than 99.

• Answer: Finite.

• Reason: Although there are many prime numbers, we have a specific upper limit (99). We can count exactly how many primes exist below 99.

Question 3: State whether each of the following set is finite or infinite.

1. The set of lines which are parallel to the x-axis.

• Answer: Infinite.

• Reason: You can draw an endless number of lines parallel to the x-axis at different heights (y = 1, y = 1.5, y = 2, etc.).

• 💡 Expert Tip: Think of the geometric plane as having infinite space. Any set of lines or points describing a general geometric property is usually infinite.

2. The set of letters in the English alphabet.

• Answer: Finite.

• Reason: There are exactly 26 letters (A to Z).

3. The set of numbers which are multiple of 5.

• Answer: Infinite.

• Reason: Multiples of 5 are 5, 10, 15, 20... This sequence never stops.

4. The set of animals living on the earth.

• Answer: Finite.

• Reason: While the number is very large, it is still a specific, countable number at any given moment. It is not infinite.

5. The set of circles passing through the origin (0,0).

• Answer: Infinite.

• Reason: You can draw circles of varying sizes (radii) and centers that all pass through the origin point.

Question 4: In the following, state whether A = B or not.

Concept Refresher: Equal Sets must have exactly the same elements. The order of elements does not matter, and repetition does not matter.

1. A = {a, b, c, d}; B = {d, c, b, a}

• Answer: A = B

• Reason: Both sets contain the exact same elements. The order is different, but that doesn't change the set.

2. A = {4, 8, 12, 16}; B = {8, 4, 16, 18}

• Answer: A ≠ B

• Reason: Set A contains 12, which is not in Set B. Set B contains 18, which is not in Set A.

3. A = {2, 4, 6, 8, 10}; B = {x : x is positive even integer and x ≤ 10}

• Answer: A = B

• Reason: If we list the elements of set B, we get the positive even integers 2, 4, 6, 8, 10. This matches Set A perfectly.

4. A = {x : x is a multiple of 10}; B = {10, 15, 20, 25, 30, ...}

• Answer: A ≠ B

• Reason: Set A contains only multiples of 10 (10, 20, 30...). Set B contains numbers like 15 and 25, which are not multiples of 10.

Question 5: Are the following pair of sets equal? Give reasons.

1. A = {2, 3}; B = {x : x is solution of x² + 5x + 6 = 0}

• Answer: No, A ≠ B

• Reason: Let's solve the quadratic equation for set B: x² + 5x + 6 = 0 (x + 2)(x + 3) = 0 x = -2, x = -3 So, Set B = {-2, -3}. Since {2, 3} ≠ {-2, -3}, the sets are not equal.

• ⚠️ Common Mistake: Always solve the equation! Don't just look at the coefficients 2, 3, 5, 6 and assume they match. Signs matter!

2. A = {x : x is a letter in the word FOLLOW}; B = {y : y is a letter in the word WOLF}

• Answer: Yes, A = B

• Reason:

  • Elements of A = {F, O, L, W} (Remember, we don't repeat letters like L or O).

  • Elements of B = {W, O, L, F}.

  • Both sets have the exact same 4 distinct letters.

Question 6: From the sets given below, select equal sets.

Given 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:

1. B = D

• Set B has elements {1, 2, 3, 4}.

• Set D has elements {3, 1, 4, 2}.

• They contain the exact same numbers, just in a different order.

2. E = G

• Set E has elements {-1, 1}.

• Set G has elements {1, -1}.

• They contain the exact same numbers.

Note: No other sets among the list are equal.

NCERT Solutions Class 11 Maths

Chapter 1: Sets (Exercise 1.3)

Detailed Solutions with Expert Tips

This guide provides step-by-step solutions for Exercise 1.3 of Class 11 Maths Chapter 1 (Sets). This exercise introduces the critical concept of Subsets (⊂), Intervals, and Universal Sets.

Question 1: Make correct statements by filling in the symbols ⊂ or ⊄ in the blank spaces.

Concept Refresher: ⊂ (Subset) means the first set is fully contained inside the second set. ⊄ (Not a Subset) means at least one element in the first set is missing from the second.

1. {2, 3, 4} ... {1, 2, 3, 4, 5}

• Answer: ⊂

• Reason: Every element (2, 3, and 4) of the first set is present in the second set.

2. {a, b, c} ... {b, c, d}

• Answer: ⊄

• Reason: The element 'a' is in the first set but missing from the second set.

3. {x : x is a student of Class XI of your school} ... {x : x student of your school}

• Answer: ⊂

• Reason: All students in Class XI are definitely students of the school. The class is a part of the school.

4. {x : x is a circle in the plane} ... {x : x is a circle in the same plane with radius 1 unit}

• Answer: ⊄

• Reason: The first set contains all circles (big and small). The second set only contains tiny circles with radius 1. A circle with radius 5 is in the first set but not the second.

• ⚠️ Common Mistake: Don't confuse the order! If the sets were swapped, it would be a subset.

5. {x : x is a triangle in a plane} ... {x : x is a rectangle in the plane}

• Answer: ⊄

• Reason: A triangle is never a rectangle. They are completely different shapes.

6. {x : x is an equilateral triangle in a plane} ... {x : x is a triangle in the same plane}

• Answer: ⊂

• Reason: Every equilateral triangle is, by definition, a triangle.

7. {x : x is an even natural number} ... {x : x is an integer}

• Answer: ⊂

• Reason: Even natural numbers (2, 4, 6...) are part of the larger set of integers (..., -1, 0, 1, 2, ...).

Question 2: Examine whether the following statements are true or false.

1. {a, b} ⊄ {b, c, a}

• Answer: False

• Reason: Both 'a' and 'b' are present in the second set. So, {a, b} is a subset (⊂) of {b, c, a}. The statement claiming it is not a subset is false.

2. {a, e} ⊂ {x : x is a vowel in the English alphabet}

• Answer: True

• Reason: The vowels are {a, e, i, o, u}. Since 'a' and 'e' are in that list, it is a subset.

3. {1, 2, 3} ⊂ {1, 3, 5}

• Answer: False

• Reason: The element 2 is in the first set but missing from the second.

4. {a} ⊂ {a, b, c}

• Answer: True

• Reason: The set containing 'a' is a subset of the set containing 'a, b, c'.

5. {a} ∈ {a, b, c}

• Answer: False

• Reason: The symbol ∈ (belongs to) connects an element to a set. Here, '{a}' is a set, not an element. The element 'a' is inside, but the set '{a}' is not listed as an item inside the brackets.

• 💡 Expert Tip:

  • a ∈ {a, b, c} (Correct: 'a' is an element)

  • {a} ⊂ {a, b, c} (Correct: {a} is a subset)

  • {a} ∈ {{a}, b, c} (Correct only if the set specifically contains {a} as an item).

6. {x : x is an even natural number less than 6} ⊂ {x : x is a natural number which divides 36}

• Answer: True

• Reason:

  • Set A (Even natural < 6) = {2, 4}

  • Set B (Divisors of 36) = {1, 2, 3, 4, 6, 9, 12, 18, 36}

  • Since both 2 and 4 are divisors of 36, Set A is a subset of Set B.

Question 3: Let A = {1, 2, {3, 4}, 5}. Which of the following statements are incorrect and why?

Crucial Concept: This question tests your ability to distinguish between elements and subsets. In set A, the items are 1, 2, {3, 4}, and 5. The item {3, 4} is treated as a single "packet" or element.

1. {3, 4} ⊂ A

• Status: Incorrect

• Why: {3, 4} is an element inside A, not a subset. For it to be a subset, it would need its own outer brackets, like {{3, 4}}.

2. {3, 4} ∈ A

• Status: Correct

• Why: The packet {3, 4} is literally sitting inside set A.

3. {{3, 4}} ⊂ A

• Status: Correct

• Why: Here, we took the element {3, 4} and put it inside set brackets. This makes it a subset.

4. 1 ∈ A

• Status: Correct

• Why: 1 is an element listed in A.

5. 1 ⊂ A

• Status: Incorrect

• Why: 1 is a number (element), not a set. Only sets can be subsets.

6. {1, 2, 5} ⊂ A

• Status: Correct

• Why: 1, 2, and 5 are all individual elements in A. Grouping them creates a valid subset.

7. {1, 2, 5} ∈ A

• Status: Incorrect

• Why: The group {1, 2, 5} is not listed as a single "packet" inside A.

8. {1, 2, 3} ⊂ A

• Status: Incorrect

• Why: The number 3 is not an individual element in A. It is trapped inside the packet {3, 4}. You cannot pull it out separately.

9. φ ∈ A

• Status: Incorrect

• Why: The symbol φ (empty set) is not listed as an element inside A.

10. φ ⊂ A

• Status: Correct

• Why: The Empty Set is a subset of every set.

11. {φ} ⊂ A

• Status: Incorrect

• Why: This would mean that φ is an element inside A, which (as seen in #9) is false.

Question 4: Write down all the subsets of the following sets.

• Formula: If a set has n elements, it has 2ⁿ subsets.

1. {a}

• Subsets: φ, {a}

2. {a, b}

• Subsets: φ, {a}, {b}, {a, b}

3. {1, 2, 3}

• Subsets: φ, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}

4. φ

• Subsets: φ

• (The empty set has only one subset: itself.)

Question 5: How many elements has P(A), if A = φ?

• Solution:

  • A = φ means the set is empty.

  • Number of elements n(A) = 0.

  • Number of elements in Power Set P(A) = 2ⁿ = 2⁰ = 1.

• Answer: 1 element (That element is the empty set itself: {φ}).

Question 6: Write the following as intervals.

Concept Refresher:

• [ ] Square Brackets: Include the end number (≤ or ≥).

• ( ) Round Brackets: Exclude the end number (< or >).

1. {x : x ∈ R, –4 < x ≤ 6}

• Answer: (–4, 6]

• (Open at -4, Closed at 6)

2. {x : x ∈ R, –12 < x < –10}

• Answer: (–12, –10)

• (Open at both ends)

3. {x : x ∈ R, 0 ≤ x < 7}

• Answer: [0, 7)

• (Closed at 0, Open at 7)

4. {x : x ∈ R, 3 ≤ x ≤ 4}

• Answer: [3, 4]

• (Closed at both ends)

Question 7: Write the following intervals in set-builder form.

1. (–3, 0)

• Answer: {x : x ∈ R, –3 < x < 0}

2. [6, 12]

• Answer: {x : x ∈ R, 6 ≤ x ≤ 12}

3. (6, 12]

• Answer: {x : x ∈ R, 6 < x ≤ 12}

4. [–23, 5)

• Answer: {x : x ∈ R, –23 ≤ x < 5}

Question 8: What universal set(s) would you propose for each of the following?

1. The set of right triangles.

2. The set of isosceles triangles.

• Answer: For both cases, a good Universal Set (U) would be:

  • U = {x : x is a triangle in the plane} or

  • U = {x : x is a polygon in the plane}

• Reason: Both right triangles and isosceles triangles are specific types of triangles (and polygons).

Question 9: 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?

• Requirement: The Universal Set must contain every single element from A, B, and C combined.

  • Elements needed: 0, 1, 2, 3, 4, 5, 6, 8.

1. {0, 1, 2, 3, 4, 5, 6}

• Verdict: No (Missing 8).

2. φ

• Verdict: No (Empty).

3. {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

• Verdict: Yes. It contains all required numbers (0, 1, 2, 3, 4, 5, 6, 8) and extra ones, which is fine.

4. {1, 2, 3, 4, 5, 6, 7, 8}

• Verdict: No (Missing 0).

• Correct Answer: Option 3

NCERT Solutions Class 11 Maths

Chapter 1: Sets (Exercise 1.4)

Detailed Solutions with Expert Tips

This guide provides step-by-step solutions for Exercise 1.4 of Class 11 Maths Chapter 1 (Sets). This exercise teaches you how to combine sets using Union, find common elements using Intersection, and subtract sets using Difference.

Question 1: Find the union of each of the following pairs of sets.

Concept Refresher: The Union (∪) of two sets contains all the elements from both sets. If an element is repeated, write it only once.

1. X = {1, 3, 5} and Y = {1, 2, 3}

• Solution: X ∪ Y = {1, 2, 3, 5} 1

• Note: 1 and 3 are in both, but we list them just once.

2. A = {a, e, i, o, u} and B = {a, b, c}

• Solution: A ∪ B = {a, b, c, e, i, o, u} 2

3. A = {x : x is a natural number and multiple of 3} and B = {x : x is a natural number less than 6}

• First, decode the sets:

  • A = {3, 6, 9, 12, ...}

  • B = {1, 2, 3, 4, 5}

• Solution: A ∪ B = {1, 2, 3, 4, 5, 6, 9, 12, ...} 3

4. A = {x : x is a natural number and 1 < x ≤ 6} and B = {x : x is a natural number and 6 < x < 10}

• First, decode the sets:

  • A = {2, 3, 4, 5, 6} (Include 6 because of ≤)

  • B = {7, 8, 9} (Strictly between 6 and 10)

• Solution: A ∪ B = {2, 3, 4, 5, 6, 7, 8, 9} 4

5. A = {1, 2, 3} and B = φ

• Solution: A ∪ B = {1, 2, 3} 5

• Reason: Combining something with "nothing" leaves you with the original set.

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

• Is A ⊂ B?

  • Answer: Yes.

  • Reason: Every element of A (a, b) is present in B.

• What is A ∪ B?

  • Answer: A ∪ B = {a, b, c} = B 6

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

• Answer: A ∪ B = B 7

• Logic: If A is entirely inside B, then combining them doesn't add anything new to B. B already has everything.

Question 4: If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8} and D = {7, 8, 9, 10}; find:

1. A ∪ B

• Solution: {1, 2, 3, 4, 5, 6}

2. A ∪ C

• Solution: {1, 2, 3, 4, 5, 6, 7, 8}

3. B ∪ C

• Solution: {3, 4, 5, 6, 7, 8}

4. B ∪ D

• Solution: {3, 4, 5, 6, 7, 8, 9, 10}

5. A ∪ B ∪ C

• Solution: {1, 2, 3, 4, 5, 6, 7, 8}

• Tip: Just throw elements from all three sets into one big basket.

6. A ∪ B ∪ D

• Solution: {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

7. B ∪ C ∪ D

• Solution: {3, 4, 5, 6, 7, 8, 9, 10}

Question 5: Find the intersection of each pair of sets of Question 1 above.

Concept Refresher: The Intersection (∩) contains only the elements that are common to both sets (the overlap).

1. X = {1, 3, 5}, Y = {1, 2, 3}

• Solution: X ∩ Y = {1, 3} 9

2. A = {a, e, i, o, u}, B = {a, b, c}

• Solution: A ∩ B = {a}

3. A = {3, 6, 9...}, B = {1, 2, 3, 4, 5}

• Solution: A ∩ B = {3}

4. A = {2, 3, 4, 5, 6}, B = {7, 8, 9}

• Solution: A ∩ B = φ

• Reason: They have no numbers in common (Disjoint sets).

5. A = {1, 2, 3}, B = φ

• Solution: A ∩ B = φ

• Reason: You can't have overlap with an empty set.

Question 6: If A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13}, C = {11, 13, 15} and D = {15, 17}; find:

1. A ∩ B

• Solution: {7, 9, 11}

2. B ∩ C

• Solution: {11, 13}

3. A ∩ C ∩ D

• Solution: φ

• Reason: Let's look for a number common to ALL three.

  • A and C share {11}.

  • Does D have 11? No.

  • So, nothing is common to all three.

4. A ∩ C

• Solution: {11}

5. B ∩ D

• Solution: φ (No common elements)

6. A ∩ (B ∪ C)

• First: B ∪ C = {7, 9, 11, 13, 15}

• Then: Intersect with A {3, 5, 7, 9, 11}

• Solution: {7, 9, 11}

7. A ∩ D

• Solution: φ

8. A ∩ (B ∪ D)

• First: B ∪ D = {7, 9, 11, 13, 15, 17}

• Then: Intersect with A

• Solution: {7, 9, 11}

9. (A ∩ B) ∩ (B ∪ C)

• Step 1: (A ∩ B) = {7, 9, 11}

• Step 2: (B ∪ C) = {7, 9, 11, 13, 15}

• Step 3: Intersection of Step 1 & 2

• Solution: {7, 9, 11}

10. (A ∪ D) ∩ (B ∪ C)

• Step 1: (A ∪ D) = {3, 5, 7, 9, 11, 15, 17}

• Step 2: (B ∪ C) = {7, 9, 11, 13, 15}

• Step 3: Find common elements

• Solution: {7, 9, 11, 15}

Question 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:

1. A ∩ B

• Solution: B (Even natural numbers)

• Reason: All even numbers are natural numbers. So their overlap is just the even numbers.

2. A ∩ C

• Solution: C (Odd natural numbers)

3. A ∩ D

• Solution: D (Prime numbers)

4. B ∩ C

• Solution: φ

• Reason: A number cannot be both Even and Odd.

5. B ∩ D

• Solution: {2}

• 💡 Expert Tip: 2 is the only even prime number.

6. C ∩ D

• Solution: {x : x is an odd prime number}

• Reason: D contains all primes {2, 3, 5, 7...}. C contains all odds {1, 3, 5, 7...}. The overlap is all primes except 2.

Question 8: Which of the following pairs of sets are disjoint?

Concept Refresher: Disjoint Sets have an intersection of φ (No common elements).

1. {1, 2, 3, 4} and {x : x is a natural number and 4 ≤ x ≤ 6}

• Set 2: {4, 5, 6}

• Overlap: {4}

  Answer: Not Disjoint (They share 4) 12

2. {a, e, i, o, u} and {c, d, e, f}

• Overlap: {e}

• Answer: Not Disjoint 13

3. {x : x is an even integer} and {x : x is an odd integer}

• Overlap: None.

• Answer: Disjoint 14

Question 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:

Concept Refresher: Difference (A – B) means "Elements in A minus any that are also in B". Think of it as taking a bite out of A.

1. A – B

• Elements in A: 3, 6, 9, 12, 15, 18, 21

• Remove items found in B: Remove 12.

• Solution: {3, 6, 9, 15, 18, 21}

2. A – C

• Remove from A: 6, 12.

• Solution: {3, 9, 15, 18, 21}

3. A – D

• Remove from A: 15.

• Solution: {3, 6, 9, 12, 18, 21}

4. B – A

• Elements in B: 4, 8, 12, 16, 20

• Remove items found in A: Remove 12.

• Solution: {4, 8, 16, 20}

5. C – A

• Remove from C: 6, 12.

• Solution: {2, 4, 8, 10, 14, 16}

6. D – A

• Remove from D: 15.

• Solution: {5, 10, 20}

7. B – C

• Remove from B: 4, 8, 12, 16.

• Solution: {20}

8. B – D

• Remove from B: 20.

• Solution: {4, 8, 12, 16}

9. C – B

• Remove from C: 4, 8, 12, 16.

• Solution: {2, 6, 10, 14}

10. D – B

• Remove from D: 20.

• Solution: {5, 10, 15}

11. C – D

• Remove from C: 10.

• Solution: {2, 4, 6, 8, 12, 14, 16}

12. D – C

• Remove from D: 10.

• Solution: {5, 15, 20}

Question 10: If X = {a, b, c, d} and Y = {f, b, d, g}, find

1. X – Y

• Remove from X: b, d.

• Solution: {a, c}

2. Y – X

• Remove from Y: b, d.

• Solution: {f, g}

3. X ∩ Y

• Common elements:

• Solution: {b, d}

Question 11: If R is the set of real numbers and Q is the set of rational numbers, then what is R – Q?

• Answer: The set of Irrational Numbers (T)

• Reason: Real Numbers = Rational + Irrational. If you take away the Rational ones, only the Irrational ones remain.

Question 12: State whether each of the following statement is true or false. Justify your answer

1. {2, 3, 4, 5} and {3, 6} are disjoint sets.

• Answer: False

• Reason: They share the element 3.

2. {a, e, i, o, u} and {a, b, c, d} are disjoint sets.

• Answer: False

• Reason: They share the element a.

3. {2, 6, 10, 14} and {3, 7, 11, 15} are disjoint sets.

• Answer: True

• Reason: No common elements.

4. {2, 6, 10} and {3, 7, 11} are disjoint sets.

• Answer: True

• Reason: No common elements.

NCERT Solutions Class 11 Maths Chapter 1: Sets (Exercise 1.5)

Detailed Solutions with Expert Tips

This guide provides step-by-step solutions for Exercise 1.5 of Class 11 Maths Chapter 1 (Sets). This exercise covers the concept of the Complement of a Set (A'), which includes everything in the Universal Set that is not in Set A.

Question 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

Concept Refresher: A' (Complement of A) = U – A. It means "Everything in U except A".

1. A'

• Method: Remove {1, 2, 3, 4} from U.

• Solution: {5, 6, 7, 8, 9}

2. B'

• Method: Remove {2, 4, 6, 8} from U.

• Solution: {1, 3, 5, 7, 9}

3. (A ∪ C)'

• Step 1: Find A ∪ C = {1, 2, 3, 4, 5, 6}

• Step 2: Find the complement (Remove Step 1 from U).

• Solution: {7, 8, 9}

4. (A ∪ B)'

• Step 1: Find A ∪ B = {1, 2, 3, 4, 6, 8}

• Step 2: Find the complement.

• Solution: {5, 7, 9}

5. (A')'

• Solution: {1, 2, 3, 4}

• 💡 Expert Tip: The complement of a complement is the original set. (A')' = A.

6. (B – C)'

• Step 1: Find B – C. (Elements in B but not C).

  • B = {2, 4, 6, 8}

  • Remove 4, 6.

  • B – C = {2, 8}

• Step 2: Find the complement of {2, 8} in U.

• Solution: {1, 3, 4, 5, 6, 7, 9}

Question 2: If U = {a, b, c, d, e, f, g, h}, find the complements of the following sets:

1. A = {a, b, c}

• Solution: A' = {d, e, f, g, h}

2. B = {d, e, f, g}

• Solution: B' = {a, b, c, h}

3. C = {a, c, e, g}

• Solution: C' = {b, d, f, h}

4. D = {f, g, h, a}

• Solution: D' = {b, c, d, e}

Question 3: Taking the set of natural numbers as the universal set, write down the complements of the following sets:

Note: U = N = {1, 2, 3, 4, ...}

1. {x : x is an even natural number}

• Answer: {x : x is an odd natural number}

• Reason: Natural numbers are either even or odd. If you remove the evens, you are left with the odds.

2. {x : x is an odd natural number}

• Answer: {x : x is an even natural number}

3. {x : x is a positive multiple of 3}

• Answer: {x : x ∈ N and x is not a multiple of 3}

4. {x : x is a prime number}

• Answer: {x : x is a positive composite number and x = 1}

• ⚠️ Common Mistake: Don't just say "Composite numbers". 1 is neither prime nor composite, so it must be included in the complement.

5. {x : x is a natural number divisible by 3 and 5}

• Answer: {x : x ∈ N and x is not divisible by 15}

• Reason: "Divisible by 3 and 5" means divisible by 15. The complement is numbers NOT divisible by 15.

6. {x : x is a perfect square}

• Answer: {x : x ∈ N and x is not a perfect square}

7. {x : x is a perfect cube}

• Answer: {x : x ∈ N and x is not a perfect cube}

8. {x : x + 5 = 8}

• Step 1: Solve x + 5 = 8 → x = 3.

• Step 2: The set is {3}.

• Answer: {x : x ∈ N and x ≠ 3}

9. {x : 2x + 5 = 9}

• Step 1: Solve 2x + 5 = 9 → 2x = 4 → x = 2.

• Step 2: The set is {2}.

• Answer: {x : x ∈ N and x ≠ 2}

10. {x : x ≥ 7}

• Reason: The set is {7, 8, 9, ...}.

• Answer: {x : x ∈ N and x < 7} or {1, 2, 3, 4, 5, 6}

11. {x : x ∈ N and 2x + 1 > 10}

• Step 1: Solve 2x > 9 → x > 4.5.

• Step 2: Since x is natural, the set is {5, 6, 7, ...}.

• Answer: {x : x ∈ N and x ≤ 4} or {1, 2, 3, 4}

Question 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:

1. (A ∪ B)' = A' ∩ B'

• LHS (Left Hand Side):

  • A ∪ B = {2, 3, 4, 5, 6, 7, 8}

  • (A ∪ B)' = {1, 9}

• RHS (Right Hand Side):

  • A' = {1, 3, 5, 7, 9}

  • B' = {1, 4, 6, 8, 9}

  • A' ∩ B' = {1, 9}

• Result: LHS = RHS. Verified.

2. (A ∩ B)' = A' ∪ B'

• LHS:

  • A ∩ B = {2}

  • (A ∩ B)' = {1, 3, 4, 5, 6, 7, 8, 9}

• RHS:

  • A' ∪ B' = {1, 3, 4, 5, 6, 7, 8, 9}

• Result: LHS = RHS. Verified.

Question 5: Draw appropriate Venn diagram for each of the following:

1. (A ∪ B)'

• Visual: Draw a rectangle (U) and two overlapping circles (A and B). Shade the entire region outside both circles.

2. A' ∩ B'

• Visual: Same as above.

• Reason: By De Morgan's Law, A' ∩ B' is exactly the same as (A ∪ B)'. Shade the region outside both circles.

3. (A ∩ B)'

• Visual: Draw the circles. The intersection is the small "football" shape in the middle. Shade everything except that intersection.

4. A' ∪ B'

• Visual: Same as Question 3.

• Reason: By De Morgan's Law, A' ∪ B' is the same as (A ∩ B)'.

Question 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'?

• Reasoning:

  • Set A = Triangles where NOT all angles are 60°. (Because if even one is different, it's in A).

  • Complement A' = Triangles where it is FALSE that "at least one angle is different".

  • This means A' = Triangles where all angles are 60°.

• Answer: The set of all equilateral triangles.

Question 7: Fill in the blanks to make each of the following a true statement:

1. A ∪ A' = ...

• Answer: U

• Reason: Combining a set with everything NOT in that set gives you everything (Universal Set).

2. φ' ∩ A = ...

• Reason: φ' = U. So, U ∩ A = A.

• Answer: A

3. A ∩ A' = ...

• Answer: φ

• Reason: A set and its complement have nothing in common.

4. U' ∩ A = ...

• Reason: U' = φ (Empty Set). So, φ ∩ A = φ.

• Answer: φ

NCERT Solutions Class 11 Maths Chapter 1: Sets (Miscellaneous Exercise)

This exercise is designed to test your comprehensive understanding of sets, including complex proofs, properties, and logical reasoning.

Detailed Solutions with Expert Tips

The Miscellaneous Exercise in Chapter 1 combines all the concepts you've learned: Subsets, Power Sets, Venn Diagrams, and Set Operations. It is excellent practice for competitive exams like JEE.

Question 1: Decide, among the following sets, which sets are subsets of one and another:

A = {x : x ∈ R and x satisfies x² – 8x + 12 = 0}

B = {2, 4, 6}

C = {2, 4, 6, 8, ...}

D = {6}

• Step 1: Decode Set A.

  • Solve x² – 8x + 12 = 0

  • (x – 6)(x – 2) = 0

  • x = 2, 6.

  • So, A = {2, 6}

• Step 2: Decode other sets.

  • B = {2, 4, 6}

  • C = {2, 4, 6, 8, ...} (Set of even natural numbers)

  • D = {6}

• Step 3: Analyze Subsets (Who is inside whom?).

  • D ({6}) is inside A, B, and C. -> D ⊂ A, D ⊂ B, D ⊂ C.

  • A ({2, 6}) is inside B and C. -> A ⊂ B, A ⊂ C.

  • B ({2, 4, 6}) is inside C. -> B ⊂ C.

Question 2: Determine whether the statement is true or false. If true, prove it. If false, give an example.

1. If x ∈ A and A ∈ B, then x ∈ B.

• Answer: False

• Example: Let A = {1} and B = {{1}, 2}.

  • Here, 1 ∈ A.

  • A ({1}) is an element of B.

  • But 1 (the number) is not in B. Only {1} (the set) is in B.

2. If A ⊂ B and B ∈ C, then A ∈ C.

• Answer: False

• Example: Let A = {1}, B = {1, 2}, C = {{1, 2}, 3}.

  • A ⊂ B is true.

  • B ∈ C is true.

  • But A ({1}) is not an element of C.

3. If A ⊂ B and B ⊂ C, then A ⊂ C.

• Answer: True

• Proof: Let x be any element of A. Since A ⊂ B, x must be in B. Since B ⊂ C, x must be in C. Therefore, every element of A is in C.

4. If A ⊄ B and B ⊄ C, then A ⊄ C.

• Answer: False

• Example: Let A = {1}, B = {2, 3}, C = {1, 2, 4}.

  • A is not in B (1 missing).

  • B is not in C (3 missing).

  • But A is a subset of C (1 is present).

5. If x ∈ A and A ⊄ B, then x ∈ B.

• Answer: False

• Example: Let A = {1, 2} and B = {2, 3}.

  • A is not a subset of B.

  • 1 ∈ A.

  • But 1 ∉ B.

6. If A ⊂ B and x ∉ B, then x ∉ A.

• Answer: True

• Proof: This is the logical equivalent (contrapositive) of the definition of a subset. If A is inside B, then anything outside B must also be outside A.

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

• Proof:

  • Step 1: We want to show B = C. Let's start with set B.

  • Step 2: We know B = B ∩ (B ∪ A). (Intersection with union returns the set itself).

  • Step 3: Substitute A ∪ B with A ∪ C (Given).

    • B = B ∩ (A ∪ C)

  • Step 4: Use Distributive Law:

    • B = (B ∩ A) ∪ (B ∩ C)

  • Step 5: Substitute B ∩ A with C ∩ A (Given A ∩ B = A ∩ C).

    • B = (C ∩ A) ∪ (B ∩ C)

  • Step 6: Factor out C using Distributive Law in reverse:

    • B = C ∩ (A ∪ B)

  • Step 7: Substitute A ∪ B with A ∪ C again.

    • B = C ∩ (A ∪ C)

  • Step 8: C ∩ (A ∪ C) simplifies to C.

  • Conclusion: Therefore, B = C.

Question 4: Show that the following four conditions are equivalent:

(i) A ⊂ B (ii) A – B = φ (iii) A ∪ B = B (iv) A ∩ B = A

• Explanation: All these statements mean the same thing: "A is inside B".

  • (i) → (ii): If A is inside B, removing B from A leaves nothing (φ).

  • (ii) → (iii): If A has nothing unique (A - B = φ), then combining A and B just gives B.

  • (iii) → (iv): If A ∪ B = B, then A must be inside B. So, the overlap (intersection) is just A.

Question 5: Show that if A ⊂ B, then C – B ⊂ C – A

• Logic: If A is smaller than B, then subtracting B removes more than subtracting A.

• Proof:

  • Let x be an element of C – B.

  • This means x ∈ C and x ∉ B.

  • Since A ⊂ B, if x is not in B, it definitely cannot be in A (x ∉ A).

  • So, x ∈ C and x ∉ A.

  • This implies x ∈ C – A.

  • Therefore, C – B ⊂ C – A.

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

• Part 1: A = (A ∩ B) ∪ (A – B)

  • Logic: Set A consists of two parts: the part shared with B (Intersection) and the part NOT shared with B (Difference). Combining them gives back A.

  • Proof:

    • RHS = (A ∩ B) ∪ (A ∩ B')

    • = A ∩ (B ∪ B') [Distributive Law]

    • = A ∩ U

    • = A = LHS.

• Part 2: A ∪ (B – A) = (A ∪ B)

  • Logic: Combining A with "the part of B that isn't A" is the same as combining A and B entirely.

  • Proof:

    • LHS = A ∪ (B ∩ A')

    • = (A ∪ B) ∩ (A ∪ A') [Distributive Law]

    • = (A ∪ B) ∩ U

    • = A ∪ B = RHS.

Question 7: Using properties of sets, show that:

(i) A ∪ (A ∩ B) = A

(ii) A ∩ (A ∪ B) = A

• (i) A ∪ (A ∩ B) = A

  • Since (A ∩ B) is a subset of A, adding it to A adds nothing new. The result is A.

• (ii) A ∩ (A ∪ B) = A

  • Since A is a subset of (A ∪ B), the common part between them is just A itself.

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

• Answer:

• Counter Example:

  • Let A = {1, 2}

  • Let B = {1, 3}

  • Let C = {1, 4}

  • Check:

    • A ∩ B = {1}

    • A ∩ C = {1}

    • So, A ∩ B = A ∩ C.

  • Conclusion: However, B ({1, 3}) is NOT equal to C ({1, 4}). Thus proved.

Question 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.

• Proof:

  • Start with A = A ∩ (A ∪ X) (Absorption Property).

  • Substitute A ∪ X with B ∪ X (Given).

  • A = A ∩ (B ∪ X)

  • A = (A ∩ B) ∪ (A ∩ X) (Distributive Law).

  • Since A ∩ X = φ, we get A = A ∩ B. (This means A ⊂ B).

  • Now do the same for B:

  • B = B ∩ (B ∪ X) = B ∩ (A ∪ X).

  • B = (B ∩ A) ∪ (B ∩ X).

  • Since B ∩ X = φ, we get B = B ∩ A. (This means B ⊂ A).

  • Since A ⊂ B and B ⊂ A, therefore A = B.

Question 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: We need three sets that overlap in pairs but have no common element shared by all three.

• Example:

  • A = {1, 2}

  • B = {2, 3}

  • C = {1, 3}

• Check:

  • A ∩ B = {2} (Non-empty)

  • B ∩ C = {3} (Non-empty)

  • A ∩ C = {1} (Non-empty)

  • A ∩ B ∩ C = φ (No number is common to all three).