CBSE Important Questions Class 11 Maths Chapter 1 – Sets

Class 11 Maths Chapter 1 Sets is the first chapter of the CBSE Class 11 Maths syllabus. It explains what a set is and how sets are written in roster form and set-builder form. Students learn about empty sets, finite and infinite sets, equal sets, subsets, the power set and the universal set, along with the operations of union, intersection, difference and complement, and how Venn diagrams make these easy to see.

These CBSE important questions for Class 11 Maths Chapter 1 cover finding the difference of two sets, assertion-reason questions on roster form, disjoint sets and complements, word problems on teachers teaching Mathematics and Physics, students choosing activities by roll number, writing complements of sets, identifying null sets, and using the symbols of subset. Every question has a clear step-by-step solution in simple English with figures, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. Let V = {a, e, i, o, u} and B = {a, i, k, u}. Find V − B and B − V.

Venn diagram showing the difference of two sets A and B shaded

Answer: The difference A − B is the set of elements that belong to A but not to B.
V − B = {e, o}, because e and o belong to V but not to B.
B − V = {k}, because k belongs to B but not to V.
Note that V − B ≠ B − V, so the difference of sets is not commutative.
In set-builder form: A − B = {x : x ∈ A and x ∉ B}

Q2. Read the statements and select the correct option.
Assertion (A): The set A = {x : x is an odd prime number less than 20} in roster form is {3, 5, 7, 11, 13, 17, 19}.
Reason (R): The set B = {x : y² = x, x is odd} is not an empty set.
(a) Both A and R are true, and R is the correct explanation of A
(b) Both A and R are true, but R is not the correct explanation of A
(c) A is true, but R is false
(d) A is false, but R is true

Answer: (b) Both A and R are true, but R is not the correct explanation of A
The odd primes below 20 are 3, 5, 7, 11, 13, 17 and 19, so the Assertion is correct. The set B also has elements (for example x = 1 with y = 1), so it is not empty and the Reason is also true. But the two statements are about different sets, so R does not explain A.

Q3. Read the statements and select the correct option.
Assertion (A): If n(A) = 7, n(B) = 4 and A ∩ B = φ, then the number of elements in A ∪ B is 11.
Reason (R): If A and B are disjoint, then n(A ∪ B) = n(A) + n(B).
(a) Both A and R are true, and R is the correct explanation of A
(b) Both A and R are true, but R is not the correct explanation of A
(c) A is true, but R is false
(d) A is false, but R is true

Answer: (a) Both A and R are true, and R is the correct explanation of A
Since A ∩ B = φ, the two sets are disjoint and share no element. So n(A ∪ B) = 7 + 4 = 11. The Reason states exactly the rule used, so it explains the Assertion.

Q4. Read the statements and select the correct option.
Assertion (A): Let A = {p, q} and B = {q, r, s}. Then A ∪ B = {p, q, r, s}.
Reason (R): If A ⊂ B, then A ∪ B = B.
(a) Both A and R are true, and R is the correct explanation of A
(b) Both A and R are true, but R is not the correct explanation of A
(c) A is true, but R is false
(d) A is false, but R is true

Answer: (b) Both A and R are true, but R is not the correct explanation of A
The union collects all the elements of both sets without repeating, so A ∪ B = {p, q, r, s} and the Assertion is correct. The Reason is also a true property. But here A is not a subset of B (because p ∉ B), so that property was not used, and R does not explain A.

Q5. Read the statements and select the correct option.
Assertion (A): If A = {1, 2, 3, 4, 5} and B = {3, 4, 5, 6, 7}, then A − B = {1, 2}.
Reason (R): A − B represents the set of all elements which belong to A but do not belong to B.
(a) Both A and R are true, and R is the correct explanation of A
(b) Both A and R are true, but R is not the correct explanation of A
(c) A is true, but R is false
(d) A is false, but R is true

Answer: (a) Both A and R are true, and R is the correct explanation of A
Removing from A every element that also lies in B (3, 4 and 5) leaves {1, 2}. The Reason gives the very definition of the difference of sets, so it correctly explains the Assertion.

Q6. In a school there are 20 teachers who teach Mathematics or Physics. Of these, 12 teach Mathematics and 4 teach both Physics and Mathematics. How many teach Physics only?

Answer: Let M be the set of Mathematics teachers and P the set of Physics teachers.
Given: n(M ∪ P) = 20, n(M) = 12 and n(M ∩ P) = 4.
Using the formula n(M ∪ P) = n(M) + n(P) − n(M ∩ P):
20 = 12 + n(P) − 4
20 = 8 + n(P)
n(P) = 12
Teachers who teach Physics only = n(P) − n(M ∩ P) = 12 − 4 = 8

Q7. In a class of 40 students, students with even roll numbers choose dance, those with roll numbers divisible by 3 choose martial arts, and those divisible by 5 choose painting. How many students choose none of these activities?

Answer: A student chooses none of the activities only if the roll number is not divisible by 2, 3 or 5.
Numbers from 1 to 40 divisible by 2 = 20, by 3 = 13, by 5 = 8
By 6 = 6, by 10 = 4, by 15 = 2, by 30 = 1
Using the inclusion-exclusion principle:
n(2 or 3 or 5) = 20 + 13 + 8 − 6 − 4 − 2 + 1 = 30
Students choosing none = 40 − 30 = 10

Q8. Taking U = {x : x is a natural number}, write the complements of the following sets:
1. A = {x : x is an even natural number}
2. B = {x : 2x + 5 = 9}
3. C = {x : x ≥ 7}
4. D = {x : x is a prime number}

Answer: The complement A′ contains all the elements of U that are not in A.
1. A′ = {x : x is an odd natural number}
2. Solving 2x + 5 = 9 gives x = 2, so B = {2} and B′ = {x : x ∈ N, x ≠ 2}
3. C′ = {x : x ∈ N and x < 7}
4. D′ = {x : x is a composite number or x = 1}

Q9. Which of the following sets are null sets?
(i) {x : x < 6 and x > 7}
(ii) {x : x² = 36, x ∈ N}
(iii) The set of all prime numbers lying between 16 and 19
(iv) {x : |x| < −3, x ∈ N}
(a) (i) and (iii)
(b) (ii) and (iii)
(c) (i) and (iv)
(d) (iii) and (iv)

Answer: (c) (i) and (iv)
(i) No number can be both less than 6 and greater than 7, so this is a null set.
(ii) x = 6 satisfies x² = 36 and 6 is a natural number, so the set is not empty.
(iii) 17 is a prime number between 16 and 19, so the set is not empty.
(iv) The modulus of a number can never be negative, so this is a null set.

Q10. Which of the following sets are null sets?
(i) The set of all prime numbers lying between 15 and 19
(ii) {x : x < 5 and x > 6}
(iii) {x : x² = 16, x ∈ N}
(iv) {x : |x| < −2, x ∈ N}
(a) (i) and (iii)
(b) (ii) and (iii)
(c) (i) and (iv)
(d) (ii) and (iv)

Answer: (d) (ii) and (iv)
(i) 17 lies between 15 and 19 and is prime, so the set is not empty.
(ii) No number is both less than 5 and greater than 6, so this is a null set.
(iii) x = 4 satisfies x² = 16, so the set is not empty.
(iv) The modulus is never negative, so this is a null set.

Q11. Consider the sets Φ, A = {1, 3, 9} and B = {1, 5, 9, 11}. Fill in the blanks with ⊂ or ⊄.
1) Φ _____ B
2) A _____ B

Answer: 1) Φ ⊂ B, because the empty set is a subset of every set.
2) A ⊄ B, because 3 ∈ A but 3 ∉ B. For A to be a subset of B, every element of A must also be in B.

Q12. Define the power set. If A = {1, 2, 3}, write P(A) and state the number of elements in it.

Answer: The power set of a set A, written P(A), is the set of all possible subsets of A, including the empty set and A itself.
For A = {1, 2, 3}:
P(A) = { φ, {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3} }
The number of elements is 2n, where n is the number of elements in A.
Here n = 3, so n[P(A)] = 2³ = 8

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