Sets — Class 11 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 11 Mathematics chapter "Sets" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 11 Mathematics chapter "Sets" — 8 important questions with detailed answers for CBSE board exam preparatio…
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NCERT Solutions for Class 11 Mathematics — Sets. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Sets — All 8 Questions Solved
Q1. If A = {x : x is a natural number and x < 6} and B = {x : x is an even natural number and x ≤ 8}, find A ∪ B and A ∩ B.
Step 1: Write set A in roster form.
A = {1, 2, 3, 4, 5}
Step 2: Write set B in roster form.
B = {2, 4, 6, 8}
Step 3: Find A ∪ B (all elements in either A or B).
A ∪ B = {1, 2, 3, 4, 5, 6, 8}
Step 4: Find A ∩ B (elements common to both A and B).
A ∩ B = {2, 4}
Final Answer: A ∪ B = {1, 2, 3, 4, 5, 6, 8}, A ∩ B = {2, 4}
Q2. For sets A, B, and C, verify De Morgan's Law: (A ∪ B)' = A' ∩ B' when U = {1, 2, 3, 4, 5, 6}, A = {1, 2, 3}, B = {3, 4, 5}.
Step 1: Find A' and B' with respect to U.
A' = U - A = {4, 5, 6}
B' = U - B = {1, 2, 6}
Step 2: Find A ∪ B.
A ∪ B = {1, 2, 3, 4, 5}
Step 3: Find (A ∪ B)' = U - (A ∪ B).
(A ∪ B)' = {6}
Step 4: Find A' ∩ B'.
A' ∩ B' = {4, 5, 6} ∩ {1, 2, 6} = {6}
Step 5: Verify (A ∪ B)' = A' ∩ B'.
{6} = {6} ✓
Final Answer: De Morgan's Law (A ∪ B)' = A' ∩ B' is verified.
Q3. Find the number of subsets and proper subsets of set S = {a, b, c, d}.
Step 1: Count the elements in set S.
Number of elements n = 4
Step 2: Find total number of subsets.
Total subsets = 2ⁿ = 2⁴ = 16
Step 3: List some subsets to verify.
∅, {a}, {b}, {c}, {d}, {a,b}, {a,c}, {a,d}, {b,c}, {b,d}, {c,d}, {a,b,c}, {a,b,d}, {a,c,d}, {b,c,d}, {a,b,c,d}
Step 4: Find number of proper subsets.
Proper subsets = Total subsets - 1 (excluding S itself) = 16 - 1 = 15
Final Answer: Total subsets = 16, Proper subsets = 15
Q4. Verify A ⊆ B and find A × B when A = {1, 2}, B = {1, 2, 3, 4}.
Step 1: Check if A ⊆ B.
Every element of A must be in B.
1 ∈ B? Yes
2 ∈ B? Yes
Therefore, A ⊆ B ✓
Step 2: Find A × B (Cartesian product).
A × B = {(a, b) : a ∈ A, b ∈ B}
Step 3: List all ordered pairs.
(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)
Step 4: Write A × B in set notation.
A × B = {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4)}
Step 5: Note that |A × B| = |A| × |B| = 2 × 4 = 8
Final Answer: A ⊆ B verified. A × B has 8 ordered pairs as listed above.
Q5. If A = {x : x² - 5x + 6 = 0} and B = {x : x² - 3x + 2 = 0}, find A ∪ B and verify if A ⊆ B.
Step 1: Solve x² - 5x + 6 = 0 to find set A.
(x - 2)(x - 3) = 0
x = 2 or x = 3
A = {2, 3}
Step 2: Solve x² - 3x + 2 = 0 to find set B.
(x - 1)(x - 2) = 0
x = 1 or x = 2
B = {1, 2}
Step 3: Find A ∪ B.
A ∪ B = {1, 2, 3}
Step 4: Verify if A ⊆ B.
Is 2 ∈ B? Yes
Is 3 ∈ B? No
Therefore, A ⊄ B (A is not a subset of B)
Final Answer: A ∪ B = {1, 2, 3}, and A ⊄ B
Q6. In a class of 50 students, 30 like Mathematics and 25 like Science. If 10 students like both, how many students like at least one subject?
Step 1: Define sets.
Let M = set of students who like Mathematics, |M| = 30
Let S = set of students who like Science, |S| = 25
|M ∩ S| = 10 (students liking both)
Step 2: Apply the principle of inclusion-exclusion.
|M ∪ S| = |M| + |S| - |M ∩ S|
|M ∪ S| = 30 + 25 - 10 = 45
Step 3: Interpret the result.
45 students like at least one subject.
Step 4: Find students liking neither.
Students liking neither = Total - |M ∪ S| = 50 - 45 = 5
Final Answer: 45 students like at least one subject; 5 students like neither.
Q7. For sets X = {1, 2, 3} and Y = {2, 3, 4}, find (X - Y) ∪ (Y - X) and verify it equals (X ∪ Y) - (X ∩ Y).
Step 1: Find X - Y (elements in X but not in Y).
X - Y = {1}
Step 2: Find Y - X (elements in Y but not in X).
Y - X = {4}
Step 3: Find (X - Y) ∪ (Y - X).
(X - Y) ∪ (Y - X) = {1} ∪ {4} = {1, 4}
Step 4: Find X ∪ Y.
X ∪ Y = {1, 2, 3, 4}
Step 5: Find X ∩ Y.
X ∩ Y = {2, 3}
Step 6: Find (X ∪ Y) - (X ∩ Y).
(X ∪ Y) - (X ∩ Y) = {1, 2, 3, 4} - {2, 3} = {1, 4}
Step 7: Verify equality.
{1, 4} = {1, 4} ✓
Final Answer: (X - Y) ∪ (Y - X) = {1, 4} = (X ∪ Y) - (X ∩ Y)
Q8. Show that the set {x : x ∈ ℝ, x² = -1} is empty by definition.
Step 1: State the definition of set membership.
An element x belongs to the set {x : x ∈ ℝ, x² = -1} if and only if x is a real number AND x² = -1.
Step 2: Analyze the condition x² = -1.
For any real number x, x² ≥ 0 (squares of real numbers are non-negative).
Step 3: Compare with the required value.
Since x² ≥ 0 for all x ∈ ℝ, there cannot exist any real x such that x² = -1.
The condition x² = -1 is impossible for real numbers.
Step 4: Conclusion.
No real number satisfies both conditions.
Therefore, the set is empty: {x : x ∈ ℝ, x² = -1} = ∅
Final Answer: The set is empty (∅) because no real number has a negative square.
Sets — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. What defines a well-defined collection in set theory?
- A. The elements are arranged in a specific order.
- B. The elements can be counted.
- C. It is clear whether an object belongs to the collection or not.
- D. The collection contains only numbers.
Answer: C. It is clear whether an object belongs to the collection or not. — A collection is well-defined if we can definitively determine whether any given object is an element of the collection or not. Order and countability are not defining characteristics of a set.
MCQ 2. Which of the following is an empty set?
- A. {x : x is a natural number, x < 5 and x > 7}
- B. {x : x is an even prime number}
- C. {x : x is a natural number, x < 5}
- D. {x : x is a real number, x² = 9}
Answer: A. {x : x is a natural number, x < 5 and x > 7} — There is no natural number that is simultaneously less than 5 and greater than 7. Therefore, the set contains no elements, making it an empty set.
MCQ 3. If A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7, 8}, find A ∪ B.
- A. {4, 5}
- B. {1, 2, 3, 6, 7, 8}
- C. {1, 2, 3, 4, 5, 6, 7, 8}
- D. { }
Answer: C. {1, 2, 3, 4, 5, 6, 7, 8} — The union of two sets A and B, denoted by A ∪ B, is the set of all elements which are in A or in B or in both. So, A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8}.
MCQ 4. If U = {1, 2, 3, 4, 5, 6, 7, 8, 9} is the universal set and A = {1, 3, 5, 7, 9}, find A'.
- A. {2, 4, 6, 8}
- B. {1, 3, 5, 7, 9}
- C. { }
- D. {1, 2, 3, 4, 5, 6, 7, 8, 9}
Answer: A. {2, 4, 6, 8} — The complement of set A, denoted by A', is the set of all elements in the universal set U that are not in A. Therefore, A' = U - A = {2, 4, 6, 8}.
MCQ 5. If n(A) = 20, n(B) = 28, and n(A ∪ B) = 36, find n(A ∩ B).
- A. 8
- B. 12
- C. 48
- D. 10
Answer: B. 12 — Using the formula n(A ∪ B) = n(A) + n(B) - n(A ∩ B), we have 36 = 20 + 28 - n(A ∩ B). This gives 36 = 48 - n(A ∩ B), so n(A ∩ B) = 48 - 36 = 12.
MCQ 6. Which of the following sets is infinite?
- A. The set of days in a week.
- B. The set of points on a line segment.
- C. The set of even prime numbers.
- D. The set of students in your class.
Answer: B. The set of points on a line segment. — A line segment consists of infinitely many points, so it is an infinite set. The other options represent finite collections.
MCQ 7. Let A = {x : x is an integer and x² < 10}. Write A in roster form.
- A. {0, 1, 2, 3}
- B. {-3, -2, -1, 0, 1, 2, 3}
- C. {1, 2, 3}
- D. {-2, -1, 0, 1, 2}
Answer: B. {-3, -2, -1, 0, 1, 2, 3} — Integers whose squares are less than 10 are -3 ((-3)²=9), -2 ((-2)²=4), -1 ((-1)²=1), 0 (0²=0), 1 (1²=1), 2 (2²=4), 3 (3²=9). Thus, A = {-3, -2, -1, 0, 1, 2, 3}.
MCQ 8. If A = {a, b, c}, how many subsets does A have?
- A. 3
- B. 6
- C. 8
- D. 2
Answer: C. 8 — The number of subsets of a set with n elements is 2ⁿ. Since A has 3 elements, it has 2³ = 8 subsets.
MCQ 9. If P = {x : x is a letter in the word 'MATHEMATICS'} and Q = {x : x is a letter in the word 'STATISTICS'}, find P ∩ Q.
- A. {M, H, E}
- B. {S, T, A, I, C}
- C. {M, A, T, H, E, I, C, S}
- D. {S, T, I, C}
Answer: B. {S, T, A, I, C} — First, list the distinct letters in each word: P = {M, A, T, H, E, I, C, S} and Q = {S, T, A, I, C}. The intersection P ∩ Q contains all elements that are common to both P and Q, which are {S, T, A, I, C}.
MCQ 10. Which of the following is a subset of {1, 2, 3}?
- A. {0, 1}
- B. {1, 2, 3, 4}
- C. {2}
- D. { } and {1, 2, 3} are the only subsets.
Answer: C. {2} — A set X is a subset of Y if all elements of X are also elements of Y. The set {2} has its element '2' present in {1, 2, 3}.
MCQ 11. If A and B are two sets such that A ⊆ B, what is A ∪ B?
- A. A
- B. B
- C. A ∩ B
- D. ∅
Answer: B. B — If A is a subset of B (A ⊆ B), it means every element of A is also an element of B. Therefore, the union of A and B will simply be the set B, as B already contains all elements of A.
MCQ 12. For any two sets A and B, A - (A ∩ B) is equal to:
- A. A
- B. A - B
- C. B - A
- D. A ∪ B
Answer: B. A - B — The expression A - (A ∩ B) represents all elements in A that are not in the intersection of A and B. This is precisely the definition of A - B, which includes elements in A but not in B.
MCQ 13. The set of prime numbers less than 10 in roster form is:
- A. {1, 2, 3, 5, 7}
- B. {2, 3, 5, 7}
- C. {2, 3, 5, 7, 9}
- D. {1, 3, 5, 7}
Answer: B. {2, 3, 5, 7} — Prime numbers are natural numbers greater than 1 that have no positive divisors other than 1 and themselves. The prime numbers less than 10 are 2, 3, 5, and 7.
MCQ 14. Let U = {1, 2, 3, 4, 5, 6}, A = {2, 3}, and B = {3, 4, 5}. Find (A ∪ B)'.
- A. {1, 6}
- B. {1, 2, 3, 4, 5, 6}
- C. {2, 3, 4, 5}
- D. {1, 2, 6}
Answer: A. {1, 6} — First, find A ∪ B = {2, 3, 4, 5}. Then, the complement (A ∪ B)' refers to elements in the universal set U that are not in A ∪ B. So, (A ∪ B)' = U - (A ∪ B) = {1, 6}.
MCQ 15. If P(A) denotes the power set of A, and A = {1, 2}, what is n(P(P(A)))?
- A. 4
- B. 8
- C. 16
- D. 256
Answer: C. 16 — If n(A) = m, then n(P(A)) = 2^m. Here, n(A) = 2, so n(P(A)) = 2² = 4. The set P(A) contains 4 elements. Now, we need n(P(P(A))), which is 2⁴ = 16.
MCQ 16. Which of the following statements is true for any set A?
- A. A ∈ A
- B. ∅ ∈ A
- C. A ⊆ A
- D. {A} ⊆ A
Answer: C. A ⊆ A — Every set is a subset of itself. A ∈ A is generally false unless A is a set of sets containing itself. ∅ ∈ A is false unless the empty set is explicitly an element of A. {A} ⊆ A is false because {A} is a set containing A as an element, not elements of A.
MCQ 17. If A = {x : x is a natural number and 1 < x ≤ 6}, B = {x : x is an even natural number and x < 10}, and C = {x : x is a prime number and x < 10}, then (A ∩ B) ∪ C is:
- A. {2, 3, 4, 5, 7}
- B. {2, 3, 5, 7}
- C. {2, 4}
- D. {2, 3, 4, 5, 6, 7}
Answer: D. {2, 3, 4, 5, 6, 7} — First, list the elements: A = {2, 3, 4, 5, 6}, B = {2, 4, 6, 8}, C = {2, 3, 5, 7}. Calculate A ∩ B = {2, 4, 6}. Finally, find the union of (A ∩ B) with C: {2, 4, 6} ∪ {2, 3, 5, 7} = {2, 3, 4, 5, 6, 7}.
MCQ 18. In a group of 800 people, 450 can speak Hindi and 300 can speak English. If 150 can speak both Hindi and English, how many can speak at least one of the two languages?
- A. 600
- B. 750
- C. 800
- D. 400
Answer: A. 600 — Let H be the set of people who speak Hindi and E be the set of people who speak English. Using the formula n(H ∪ E) = n(H) + n(E) - n(H ∩ E) = 450 + 300 - 150 = 600.
MCQ 19. If A = {x | x² = 4, x is odd}, then A is:
- A. {2, -2}
- B. {2}
- C. ∅
- D. { } and {2}
Answer: C. ∅ — The solutions to x² = 4 are x = 2 and x = -2. Neither of these is an odd number. Therefore, there are no elements that satisfy both conditions, making A an empty set.
MCQ 20. If A ⊆ B, then what is A ∩ B?
- A. A
- B. B
- C. ∅
- D. U (Universal Set)
Answer: A. A — If A is a subset of B, it means all elements of A are also elements of B. Therefore, the common elements (intersection) between A and B will simply be all the elements of A.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.