Written by : Anmol Gupta
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NCERT Exemplar Class 11 Maths Chapter 1 Sets Examples Solutions
Table of Contents
Solved Examples
Short Answer Type
Example 1 Write the following sets in the roster form.
(i) A = {x | x is a positive integer less than 10 and 2x – 1 is an odd number}
(ii) C = {x : x2 + 7x – 8 = 0, x ∈ R}
Answer 1 (i) Value of x can be 1,2,3,4,5,6,7,8,9
For above respective values of x, values of 2x – 1 are 1,3,7,15,31,63,127, 511 which are all odd numbers
Therefore, A = {1,2,3,4,5,6,7,8,9}
(ii) x2 + 7x – 8 = x2 + 8x – x – 8
(x + 8) (x – 1) = 0 giving x = – 8 or x = 1
Thus, C = {– 8, 1}
Example 2 State which of the following statements are true and which are false. Justify your answer.
(i) 37 ∉ {x | x has exactly two positive factors}
(ii) 28 ∈ {y | the sum of the all positive factors of y is 2y}
(iii) 7,747 ∈ {t | t is a multiple of 37}
Answer 2. (i) False
Since, 37 has exactly two positive factors, 1 and 37, 37 belongs to the set
(ii) True
Here, y is 28
The sum of positive factors of 28
= 1 + 2 + 4 + 7 + 14 + 28
= 56 = 2(28)
(iii) False
7,747 is not a multiple of 37.
Example 3. If X and Y are subsets of the universal set U, then show that
(i) Y ⊂ X ∪ Y (ii) X ∩ Y ⊂ X (iii) X ⊂ Y ⇒ X ∩ Y = X
Answer 3. (i) Let x ∈ Y
Then either x ∈ X or x ∉ X
In both cases x ∈ X ∪ Y
Therefore, Y ⊂ X ∪ Y
(ii) Let x ∈ X ∩ Y
⇒ x ∈ X and x ∈ Y
Means x ∈ X
Therefore, X ∩ Y ⊂ X
(iii) Given: X ⊂ Y
Let x ∈ X ∩ Y
⇒ x ∈ X and x ∈ Y
Means X ∩ Y ⊂ X
Now let x ∈ X and also by X ⊂ Y, x ∈ Y
⇒ X ⊂ X ∩ Y
Therefore, X ∩ Y = X
Example 4. Given that N = {1, 2, 3, …, 100}, then (i) Write the subset A of N, whose element are odd numbers.
(ii) Write the subset B of N, whose element are represented by x + 2, where x ∈ N
Answer 4. (i) A = {x | x ∈ N and x is odd}= {1, 3, 5, 7, …, 99}
(ii) B = {y | y = x + 2, x ∈ N}
So, for 1 ∈ N, y = 1 + 2 = 3
2 ∈ N, y = 2 + 2 = 4,
and so on. Therefore, B = {3, 4, 5, 6, … , 100}
Example 5. Given that E = {2, 4, 6, 8, 10}. If n represents any member of E, then, write the following sets containing all numbers represented by
(i) n + 1 (ii) n2
Answer 5. Given E = {2, 4, 6, 8, 10}
(i) Let A = {x | x = n + 1, n ∈ E}
Thus, for 2 ∈ E, x = 3
4 ∈ E, x = 5,
and so on. Therefore, A = {3, 5, 7, 9, 11}.
(ii) Let B = {x | x = n2, n ∈ E}
So, for 2 ∈ E, x = (2)2 = 4, 4 ∈ E, x = (4)2
= 16, 6 ∈ E, x = (6)2 = 36,
and so on. Hence, B = {4, 16, 36, 64, 100}
Example 6. Let X = {1, 2, 3, 4, 5, 6}. If n represent any member of X, express the following as sets:
(i) n ∈ X but 2n ∉ X
(ii) n + 5 = 8
(iii) n is greater than 4.
Answer 6. (i) For X = {1, 2, 3, 4, 5, 6}, it is the given that n ∈ X, but 2n ∉ X.
Let, A = {x | x ∈ X and 2x ∉ X}
Now, 1 ∉ A as 2.1 = 2 ∈ X
2 ∉ A as 2.2 = 4 ∈ X
3 ∉ A as 2.3 = 6 ∈ X
But 4 ∈ A as 2.4 = 8 ∉ X
5 ∈ A as 2.5 = 10 ∉ X
6 ∈ A as 2.6 = 12 ∉ X
So, A = {4, 5, 6}
(ii) Let B = {x | x ∈ X and x + 5 = 8}
Here, B = {3}
as x = 3 ∈ X and 3 + 5 = 8 and there is no other element belonging to X such that x + 5 = 8
(iii) Let C = {x | x ∈ X, x > 4}
Therefore, C = {5, 6}
Example 7. Draw the Venn diagrams to illustrate the followoing relationship among
sets E, M and U, where E is the set of students studying English in a school, M is the
set of students studying Mathematics in the same school, U is the set of all students in
that school.
(i) All the students who study Mathematics study English, but some students who
study English do not study Mathematics.
(ii) There is no student who studies both Mathematics and English.
(iii) Some of the students study Mathematics but do not study English, some study English
but do not study Mathematics, and some study both.
(iv) Not all students study Mathematics, but every students studying English studies Mathematics.
Answer 7. (i) M ⊂ E ⊂ U
(ii) E ∩ M = φ
(iii) Since there are some students who study both English and Mathematics, some
English only and some Mathematics only.
Thus, the Venn Diagram is
(iv) E ⊂ M ⊂ U
Example 8. For all sets A, B and C
Is (A ∩ B) ∪ C = A ∩ (B ∪ C)? Justify your statement
Answer 8. No. consider the following sets A, B and C :
A = {1, 2, 3}
B = {2, 3, 5}
C = {4, 5, 6}
Now (A ∩ B) ∪ C = ({1, 2, 3} ∩ {2, 3, 5}) ∪ {4, 5, 6}
= {2, 3} ∪ {4, 5, 6}
= {2, 3, 4, 5, 6}
And A ∩ (B ∪ C) = {1, 2, 3} ∩ [{2, 3, 5} ∪ {4, 5, 6}
= {1, 2, 3} ∩ {2, 3, 4, 5, 6}
= {2, 3}
Therefore, (A ∩ B) ∪ C ≠ A ∩ (B ∪ C)
Example 9. Use the properties of sets to prove that for all the sets A and B
A – (A ∩ B) = A – B
Answer 9. We have
A – (A ∩ B) = A ∩ (A ∩ B)′ (since A – B = A ∩ B′)
= A ∩ (A′ ∪ B′) [by De Morgan’s law)
= (A∩ A′) ∪ (A∩ B′) [by distributive law]
= φ ∪ (A ∩ B′)
= A ∩ B′ = A – B
Long Answer Type
Example 10. For all sets A, B and C Is (A – B) ∩ (C – B) = (A ∩ C) – B?
Justify your answer.
Answer 10. Yes
Let x ∈ (A – B) ∩ (C – B)
⇒ x ∈ A – B and x ∈ C – B
⇒ (x ∈ A and x ∉ B) and (x ∈ C and x ∉ B)
⇒ (x ∈ A and x ∈ C) and x ∉ B
⇒ (x ∈ A ∩ C) and x ∉ B
⇒ x ∈ (A ∩ C) – B
So (A – B) ∩ (C – B) ⊂ (A ∩ C) – B
Let y ∈ (A ∩ C) – B
⇒ y ∈ (A ∩ C) and y ∉ B
⇒ (y ∈ A and y ∈ C) and (y ∉ B)
⇒ (y ∈ A and y ∉ B) and (y ∈ C and y ∉ B)
⇒ y ∈ (A – B) and y ∈ (C – B)
⇒ y ∈ (A – B) ∩ (C – B)
So (A ∩ C) – B ⊂ (A – B) ∩ (C – B)
(A – B) ∩ (C – B) = (A ∩ C) – B
Example 11. Let A, B and C be sets. Then show that
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
Answer 11. We first show that A ∪ (B ∩ C) ⊂ (A ∪ B) ∩ (A ∪ C)
Let x ∈ A ∪ (B ∩ C). Then
x ∈ A or x ∈ B ∩ C
⇒ x ∈ A or (x ∈ B and x ∈ C)
⇒ (x ∈ A or x ∈ B) and (x ∈ A or x ∈ C)
⇒ (x ∈ A ∪ B) and (x ∈ A ∪ C)
⇒ x ∈ (A ∪ B) ∩ (A ∪ C)
Thus, A ∪ (B ∩ C) ⊂ (A ∪ B) ∩ (A ∪ C) … (1)
Now we will show that (A ∪ B) ∩ (A ∪ C) ⊂ (A ∪ C)
Let x ∈ (A ∪ B) ∩ (A ∪ C)
⇒ x ∈ A ∪ B and x ∈ A ∪ C
⇒ (x ∈ A or x ∈ B) and (x ∈ A or x ∈ C)
⇒ x ∈ A or (x ∈ B and x ∈ C)
⇒ x ∈ A or (x ∈ B ∩ C)
⇒ x ∈ A ∪ (B ∩ C)
Thus, (A ∪ B) ∩ (A ∪ C) ⊂ A ∪ (B ∩ C) … (2)
So, from (1) and (2), we have
A ∩ (B ∪ C) = (A ∪ B) ∩ (A ∪ C)
Example 12. Let P be the set of prime numbers and let S = {t | 2t – 1 is a prime}. Prove that S ⊂ P.
Answer 12. Now the equivalent contrapositive statement of x ∈ S ⇒ x ∈ P is x ∉ P ⇒ x ∉ S.
Now, we will prove the above contrapositive statement by contradiction method
Let x ∉ P
⇒ x is a composite number
Let us now assume that x ∈ S
⇒ 2x – 1 = m (where m is a prime number)
⇒ 2x = m + 1
Which is not true for all composite number, say for x = 4 because
24 = 16 which can not be equal to the sum of any prime number m and 1.
Thus, we arrive at a contradiction
x ∉ S
Thus, when x ∉ P, we arrive at x ∉ S
So S ⊂ P.
Example 13. From 50 students taking examinations in Mathematics, Physics and
Chemistry, each of the student has passed in at least one of the subject, 37 passed
Mathematics, 24 Physics and 43 Chemistry. At most 19 passed Mathematics and
Physics, at most 29 Mathematics and Chemistry and at most 20 Physics and Chemistry.
What is the largest possible number that could have passed all three examination?
Answer 13. Let M be the set of students passing in Mathematics
P be the set of students passing in Physics
C be the set of students passing in Chemistry
Now, n(M ∪ P ∪ C) = 50, n(M) = 37, n(P) = 24, n(C) = 43
n(M ∩ P) ≤ 19, n(M ∩ C) ≤ 29, n(P ∩ C) ≤ 20 (Given)
n(M ∪ P ∪ C) = n(M) + n(P) + n(C) – n(M ∩ P) – n(M ∩ C) – n(P ∩ C) + n(M ∩ P ∩ C) ≤ 50
⇒ 37 + 24 + 43 – 19 – 29 – 20 + n(M ∩ P ∩ C) ≤ 50
⇒ n(M ∩ P ∩ C) ≤ 50 – 36
⇒ n(M ∩ P ∩ C) ≤ 14
Thus, the largest possible number that could have passed all the three examinations is 14
Objective Type Questions
Choose the correct answer from the given four options in each of the Examples
14 to 16 : (M.C.Q.)
Example 14. Each set Xr contains 5 elements and each set Yr contains 2 elements
Answer 14. The correct answer is (B)
n(Xr) = 5 means n(X1) + n(X2) + n(X3) +…n(X20) = 5 × 20 = 100
This means: across all sets X1 to X20 , 100 elements are written (some repeated, some not)
Suppose the number of distinct elements in the union set S is s
Set S is union of Xr‘s means X1 U X2 U X3 …. U X20 = S
n(S) = s
10 × n(S) = 100
n(S) = s = 10
So, the total number of distinct elements in S is 10
Each of those 10 elements appears in 4 of the Yr sets
So total number of appearances = 10 × 4 = 40
But each Yr has 2 elements:
So, total element entries = n × 2
Equating: n × 2 = 40 ⇒ n = 20
Example 15. Two finite sets have m and n elements respectively. The total number of
subsets of first set is 56 more than the total number of subsets of the second set. The
values of m and n respectively are.
(A) 7, 6 (B) 5, 1 (C) 6, 3 (D) 8, 7
Answer 15. The correct answer is (C).
Since, let A and B be such sets, i.e., n (A) = m, n (B) = n
So n (P(A)) = 2m
n (P(B)) = 2n
Thus n (P(A)) – n (P(B)) = 56, i.e., 2m – 2n = 56
⇒ 2n(2m – n – 1) = 23 . 7
⇒ n = 3 , 2m – n – 1 = 7
⇒ 2m –3 = 8
⇒2m –3 = 23
⇒ m = 6
Example 16. The set (A ∪ B ∪ C) ∩ (A ∩ B′ ∩ C′)′ ∩ C′ is equal to
(A) B ∩ C′ (B) A ∩ C (C) B ∪ C′ (D) A ∩ C′
Answer 16. The correct choice is (A).
(A ∩ B′ ∩ C′)′ = (A′ ∪ (B ∪ C)) [ By De Morgan’s Law ]
(A ∪ B ∪ C) ∩ (A ∩ B′ ∩ C′)′ ∩ C′ = (A ∪ (B ∪ C)) ∩ (A′ ∪ (B ∪ C)) ∩ C′
Note that B ∪ C appears multiple times, so let’s let:B ∪ C = D
(A ∪ (B ∪ C)) ∩ (A′ ∪ (B ∪ C)) ∩ C′ = (A ∪ D) ∩ (A′ ∪ D) ∩ C′
= (A ∩ A′) ∪ (D) ∩ C′
= φ ∪ (B ∪ C) ∩ C′
= (B ∪ C) ∩ C′
Distributing intersection over union:
= (B ∩ C′) U (C ∩ C′)
= (B ∩ C′) U φ
= B ∩ C′
Fill in the blanks in Examples 17 and 18 :
Example 17. If A and B are two finite sets, then n(A) + n(B) is equal to _________
Answer 17. Since n(A ∪ B) = n (A) + n (B) – n (A ∩ B)
So n(A) + n (B) = n (A ∪ B) + n (A ∩ B)
Example 18. If A is a finite set containing n element, then number of subsets of A is _____
Answer 18. 2n
State true or false for the following statements given in Examples 19 and 20
Example 19. Let R and S be the sets defined as follows:
R = {x ∈ Z | x is divisible by 2}
S = {y ∈ Z | y is divisible by 3}
then R ∩ S = φ
Answer 19. False
Since 6 is divisible by both 3 and 2.
Thus R ∩ S ≠ φ
Example 20. Q ∩ R = Q, where Q is the set of rational numbers and R is the set of
real numbers.
Answer 20. True
Since Q ⊂ R
So Q ∩ R = Q