NCERT Solutions for Class 11 Maths Chapter 1 - Sets
Exercise 1.1
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 question in this Chapter.
(ix) A collection of most dangerous animals of the world.
Sol. (i) We know the months of a year beginning with the letter J are January, June and July. So, the collection of all months of a year beginning with the letter J are well-defined.
Hence, it is a set.
(ii) Since, the criteria for determining the talent of a writer is vary from person to person.
So, the collection of ten most talented writer in India is not well defined.
Hence, it is not a set.
(iii) Since the criteria for determining talent of a batsman vary from person to person.
So, the collection of a team of eleven bestcricket batsmen of the world is not well defined.
Hence, it is not a set.
(iv) Since, we know the number of boys and their names in our class. So, the collection of all boys in our class is well defined.
Hence, it is a set.
(v) Since, we know the collection of natural numbers less than 100.
i.e., {1, 2, …, 99}
so, this collection is well defined.
Hence, it is a set.
(vi) We can easily collect the names of novels written by the writer Munshi Prem Chand. So, the collection is well defined.
Hence, it is a set.
(vii) We can easily collect the numbers of even integers i.e., 2, 4, 6, 8, ….. so, this collection is well-defined.
Hence, it is a set.
(viii) We can easily collect the questions in this chapter. So, this collection is well-defined.
Hence it is a set.
(ix) Since, the criteria for determining the dangerousness of an animal vary from person to person.
So the collection of most dangerous animals of the world is not well defined. Hence it is not a set.
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
Sol. (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.
Sol. (i) A = {– 2, – 1, 0, 1, 2, 3, 4, 5, 6}
(ii) B = {1, 2, 3, 4, 5}
(iii) C = {17, 26, 35, 44, 53, 62, 71, 80}
(iv) Since 60 = 2 × 2 × 3 × 5
∴ 2, 3 and 5 are the elements of the given set. D = {2, 3, 5}
(v) E = {T, R, I, G, O, N, M, E, Y}
(vi) 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}
Sol. Let A be the set-builder form. Thus,
(i) A = {x : x is a natural number multiple of 3 and x < 15}
or A = {x : x = 3n, n ∈ N and n ≤ 4}
(ii) A = {x : x = 2n, n ∈ N and n ≤ 5}
(iii) A = {x : x = 5n, n ∈ N and n ≤ 4}
(iv) A = {x : x is an even natural number}
(v) A = {x : x = n2, n ∈ N and n ≤ 10}
5. List all the elements of the following sets:
(i) A = {x : x is an odd natural number}
$$\textbf{(ii) B =} \lbrace\textbf{x : x is an integer,}-\frac{\textbf1}{\textbf2}\lt \textbf{x}\lt\frac{\textbf9}{\textbf2}\rbrace$$
(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}
Sol. (i) A = {1, 3, 5, 7, …}
(ii) B = {0, 1, 2, 3, 4}
(iii) C = {– 2, – 1, 0, 1, 2}
(iv) D = { L, O, Y, A}
(v) E = {February, April, June, September, November}
(vi) 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}
(ii) {2, 3}
(iii) {M, A, T, H, E, I, C, S}
(iv) {1, 3, 5, 7, 9}
(a) {x : x is a prime number and a divisor of 6}
(b) {x : x is an odd natural number less than 10}
(c) {x : x is natural number and divisor of 6}
(d) {x : x is a letter of the world MATHEMATICS}
Sol. (i) (c)
(ii) (a)
(iii) (d)
(iv) (b)
Exercise 1.2
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}
Sol. (i) Since, odd numbers are not divisible by 2.
Hence, the set is a null set.
(ii) We know that 2 is even prime number.
Hence, the set is not a null set.
(iii) There will be no such natural number which is less than 5 or greater than 7.
Hence, the set in a null set.
(iv) We know that, parallel lines never intersect each other.
Hence, the given set is a null set
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
Sol. (i) The number of months in a year is 12. Thus, the set of months of a year is a finite set.
(ii) The given set is a collection of infinite natural numbers. Thus, the set is an infinite set.
(iii) The given set is a collection of first hundred natural number which is countable. Thus, the given set is a finite set.
(iv) Since, the set of integers greater than 100 is an infinite set.
(v) We can esily find the set of all prime numbers less than 99.
i.e., {2, 3, 4, 7, …, 97}
Hence, the set is a finite set
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).
Sol. (i) Infinite set, because we can draw infinite number of lines parallel to the x-axis.
(ii) Finite set, because number of letter in english alphabet = 26.
(iii) Infinite set, because there are infinite numbers which are multiple of 5.
(iv) Finite set, because we can find the number of animals living on the earth and number of animals living on earth is finite.
(v) Infinite set, because we can draw infinite number of circles passing through the origin (0, 0).
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, …}
Sol. (i) Since both A and B have exactly the same elements.
∴ A = B.
(ii) Since, 12 ∈ A but 12 ∉ B or 18 ∈ B and 18 ∉ A
∴ A ≠ B
(iii) If we convert set B in Roaster form then element of B is equal to elements of A.
Thus, A = B.
(iv) A = {x : x is a multiple of 10} = {10, 20, 30, …}
Since, 15 ∈ B but 15 ∉ A.
∴ A ≠ B.
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 world WOLF}
Sol. (i) Solution of x2 + 5x + 6 = 0
⇒ x2 + 3x + 2x + 6 = 0
⇒ x(x + 3) + 2(x + 3) = 0
⇒ (x + 2) (x + 3) = 0
∴ x = – 2, – 3
∴ B = {– 2, – 3} and A = {2, 3}
Since elements of A is not equal to elements of B.
Hence, A ≠ B.
(ii) First we write the set A and B in roaster form.
∴ A = {F, O, L, W}
and B = {W, O, L, F}
Thus, element of A is equal to its element of B.
Hence, A = B.
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}.
Sol. Since 2 ∈ A but 2 ∉ C
∴ A ≠ C
Also,, 8 ∈ A but 8 ∉ B, 8 ∉ D, 8 ∉ E, 8 ∉ F, 8 ∉ G and 8 ∉ H
∴ A ≠ B, A ≠ D, A ≠ E, A ≠ F, A ≠ G, and A ≠ H Since, 2 ∈ B but 2 ∉ C, 2 ∉ E, 2 ∉ F, 2 ∉ G and 2 ∉ H
∴ B ≠ C, B ≠ E, B ≠ F, B ≠ F, B ≠ G and B ≠ H Since, both B and D have exactly the same elements.
Thus B = D
Since, 14 ∈ C but 14 ∉ D, 14 ∉ E, 14 ∉ F, 14 ∉ G and 14 ∉ H
∴ C ≠ D, C ≠ E, C ≠ F, C ≠ G and C ≠ H Since, 4 ∈ D but 4 ∉ E, 4 ∉ F, 4 ∉ G and 4 ∉ H
∴ D ≠ E, D ≠ F, D ≠ G and D ≠ H
Since, – 1 ∈ E but – 1 ∉ F and – 1 ∉ H
∴ E ≠ F and E ≠ H
Since both E and G have exactly the same elements.
Thus, E = G
Since, a ∈ F but a ∉ G and a ∉ H
∴ F ≠ G and F ≠ H
Since, – 1 ∈ G but – 1 ∉ H
∴ G ≠ H
Exercise 1.3
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 is a 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 is a triangle in the plane}
(vii) {x : x is an even natural number} … {x : x is an integer}.
Sol. A ⊂ B means that all the elements of set A, are present in set B, A ⊄ B means that all the elements of set A are not present in set B.
(i) {2, 3, 4} ⊂ {1, 2, 3, 4, 5}
{∵ all elements of Ist set are there in IInd set}
(ii) {a, b, c} ⊄ {b, c, d}
{∵ a ∉ II set}
(iii) {x : x is a student of class XI of your school} ⊂ {x : x is a student of your school}
{∵ If a student is of class XI of your school, then surely he will be a 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}
{∵ elements in the Ist can set be a circle of any radius, so it all element will not be there in the IInd set}
(v) {x : x is a triangle in a plane} ⊄ {x : x is a rectangle in the plane}
{∵ All elements of I set ∉ II set}
(vi) {x : x is an equilateral triangle in a plane} ⊂ {x : x is a traingle in the plane}
(∵ All elements of I set ∈ II set}.
(vii) {x : x is an even natural number} ⊂ {x : x is an integer}
(∵ all elements of Ist set will be there in the IInd set because integers have both even and odd natural number)
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}
Sol. (i) False; each element of {a, b} is also an element of {b, c, a}.
(ii) True; a, e are two vowels of the english
alphabet.
(iii) {1, 2, 3} ⊂ {1, 3, 5} is false because 2 ∉ {1, 3, 5}.
(iv) {a} ⊂ {a, b, c} is true as {a} is contained in the set {a, b, c}.
(v) False; The elements of {a, b, c} are a, b, c. Therefore , {a} ⊂ {a, b, c}.
(vi) {x : x is an even natural number less than 6} = {2, 4}
{x : x is a Natural Number which divided 36}
= {1, 2, 3, 6, 9, 12, 8, 36}
{2, 4} ⊂ {1, 2, 3, 6, 9, 12, 18, 36}
hence, the given statement is true.
3. Let A = {1, 2, {3, 4}, 5}. Which of the following statements are in correct 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) f ∈ A
(x) f ⊂ A
(xi) {f} ⊂ A
Sol. (i) It is incorrect because {3, 4} ⊄ A but {(3, 4)} ⊂ A is correct. First of all make it clear that the given set A has four elements , 2, {3, 4} and 5. Remember that {3, 4} is an element not a set.
(ii) {3, 4} ∈ A, it is a correct statement.
(iii) As we have discussed above, {{3, 4}} ⊂ A, it is a correct statement.
(iv) 1 ∈ A, it is a correct statement.
(v) It is incorrect, as 1 ∈ A.
(vi) {1, 2, 5} ⊂ A, it is a correct statement.
(vii) It is incorrect as {1, 2, 5} ∉ A.
(viii)It is incorrect as it is not possible.
(ix) It is incorrect because f ⊂ A.
(x) f ⊂ A, it is a correct statement because f is a subset of every set.
(xi) It is incorrect because f ⊂ A.
Hence, (i), v), (vii), (viii), (ix) and (xi) are the incorrect statements.
4. Write down all the subsets of the following sets:
(i) {a}
(ii) {a, b}
(iii) {1, 2, 3}
(iv) Φ
Sol. (i) {a}
Number of Subsets = Φ, {a}
(ii) {a, b}
Number of Subsets = Φ, {a}, {b}, {a, b}
(iii) {1, 2, 3}
Number of Subsets = Φ, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}
(iv) Φ Subet = Φ
Note: Φ is a subset of all sets
5. How many elements has P(A), if A = Φ?
Sol. If A = Φ, then by definition of power set, it has 20 = 1 element which is given by P(A) = P(Φ) = {Φ}.
6. 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}
Sol. (i) {x : x ∈ R, – 4 < x ≤ 6} = (– 4, 6]
(ii) {x : x ∈ R, – 12 < x < – 10} = (– 12, – 10)
(iii) {x : x ∈ R, 0 ≤ x < 7} = [0, 7)
(iv) {x : x ∈ R, 3 ≤ x ≤ 4} = [3, 4]
7. Write the following intervals in set builder form:
(i) (– 3, 0)
(ii) [6, 12]
(iii) (6, 12]
(iv) [– 23, 5)
Sol. (i) (– 3, 0) = {x : x ∈ R, – 3 < x < 0)
(ii) [6, 12] = {x : x ∈ R, 6 ≤ x ≤ 12}
(iii) (6, 12] = {x : x ∈ R, 6 < x ≤ 12}
(iv) [– 23, 5) = [x : x ∈ R, – 23 ≤ x < 5}
8. What universal set(s) would you propose for each of the following:
(i) The set of right triangles.
(ii) The set of isosceles triangles.
Sol. (i) For the set of right triangles, the universal set can be the set of triangles.
(ii) For the set of isosceles triangles, the universal set can be the set of triangles or the set of polygons.
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 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}
Sol. (i) A ⊂ {0, 1, 2, 3, 4, 5, 6},
B ⊂ {0, 1, 2, 3, 4, 5},
C ⊄ {0, 1, 2, 3, 4, 5, 6}
∴ The set {0, 1, 2, 3, 4, 5, 6} can not be the universal set for the sets A, B and C.
(ii) A ⊄ f
∴ The set f cannot be an universal set for A, B and C.
(iii) A ⊂ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10},
B ⊂ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10},
C ⊂ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
∴ The set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} is the universal set of A, B and C.
(iv) C ⊄ {1, 2, 3, 4, 5, 6, 7, 8}
∴ The set {1, 2, 3, 4, 5, 7, 8} is not the universal set of A, B and C.
Exercise 1.4
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 < x ≤ 6}
B = {x : x is a natural number 6 < x < 10}
(v) A = {1, 2, 3}, B = Φ.
Sol. (i) X = {1, 3, 5} and Y = {1, 2, 3}
Therefore,
X ∪ Y = {1, 2, 3, 5}
(ii) A = {a, e, i, o, u} and B = {a, b, c}
Therefore
⇒ A ∪ B = {a, e, i, o, u, b, c}
(iii) A = {x : x is a natural number and multiple of 3}
⇒ A = {3, 6, 9, 12, …}
and B = {x : x is a natural number less than 6}
= {1, 2, 3, 4, 5, 6}
⇒ A ∪ B = {1, 2, 3, 4, 5, 6, 9, 12, 15…}
∴ A ∪ B = {x : x = 1, 2, 4, 5 or a multiple of 3}
(iv) A = {x : x is a natural numbers and 1 < x ≤ 6}
⇒ A = {2, 3, 4, 5, 6}
B = {x : x is a natural number and 6 < x < 10}
⇒ B = {7, 8, 9}
⇒ A ∪ B = {2, 3, 4, 5, 6, 7, 8, 9}
(v) A = {1, 2, 3}, B = Φ
⇒ A ∪ B = {1, 2, 3}
2. Let A = {a, b}, B = {a, b, c} is A ⊂ B? What is A ∪ B?
Sol. Since, every element of A is also an element of B, therefore A ⊂ B.
And A ∪ B = {a, b, c} = B.
3. If A and B are two sets such that A ⊂ B, then what is A ∪ B?
Sol. Since, A is a subset of B. Hence, every element of set A is contained in the set B and hence, 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
Sol. (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 = (A ∪ B) ∪ C
= {1, 2, 3, 4, 5, 6} ∪ {5, 6, 7, 8}
= {1, 2, 3, 4, 5, 6, 7, 8}
(vi) A ∪ B ∪ D = (A ∪ B) ∪ D
= {1, 2, 3, 4, 5, 6} ∪ {7, 8, 9, 10}
= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(vii) B ∪ C ∪ D = (B ∪ C) ∪ D
= (3, 4, 5, 6, 7, 8) ∪ (7, 8, 9, 10)
= {3, 4, 5, 6, 7, 8, 9, 10}
5. Find the intersection of each pair of sets of question-1 above.
Sol. We know that the intersection of two or more sets is the set of common elements.
(i) X = {1, 3, 5} and Y = {1, 2, 3}
Therefore X ∩ Y = {1, 3, 5} ∩ {1, 2, 3}
= {1, 3}
(ii) A = {a, e, i, o, u} and B = {a, b, c}
Therefore A ∩ B = {a}
(iii) A = {x : x is a Natural Number and multiple of 3}
A = {3, 6, 9, …}
B = {x : x is a Natural Number less than 6}
B = {1, 2, 3, 4, 5}
Therefore A ∩ B = {3}
(iv) A = {x : x is a natural number and 1 < x ≤ 6} = {2, 3, 4, 5, 6}
B = {x : x is a natural number and 6 < x < 10} = {7, 8, 9}
∴ A ∩ B = Φ
(v) A = {1, 2, 3} and B = Φ
Therefore A ∩ B = {1, 2, 3} ∩ 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 ∪ C)
(x) (A ∪ D) ∩ (B ∪ C)
Sol. (i) A ∩ B = {3, 5, 7, 9, 11} ∩ {7, 9, 11, 13} = {7, 9, 11}
(ii) B ∩ C = {7, 9, 11, 13} ∩ {11, 13, 15} = {11, 13}
(iii) A ∩ C ∩ D = {3, 5, 7, 9, 11} ∩ {11, 13, 15} ∩ {15, 17} = {11} ∩ {15, 17} = Φ
(iv) A ∩ C = {3, 5, 7, 9, 11} ∩ {11, 13, 15} = {11}
(v) B ∩ D = {7, 9, 11, 13} ∩ {15, 17} = Φ
(vi) A ∩ (B ∪ C) = {3, 5, 7, 9, 11} ∩ [{7, 9, 11, 13} ∪ {11, 13, 15}]
= {3, 5, 7, 9, 11} ∩ {7, 9, 11, 13, 15} = {7, 9, 11}
(vii) A ∩ D = {3, 5, 7, 9, 11} ∩ {15, 17} = Φ
(viii) A ∩ (B ∪ D) = {3, 5, 7, 9, 11} ∩ [{7, 9, 11, 13} ∪ {15, 17}]
= {3, 5, 7, 9, 11} ∩ {7, 9, 11, 13, 15, 17} = {7, 9, 11}
(ix) (A ∩ B) ∩ (B ∪ C) = [{3, 5, 7, 9, 11} ∩ {7, 9, 11, 13}] ∩ [{7, 9, 11, 13} ∪ {11, 13, 15}]
= {7, 9, 11} ∩ {7, 9, 11, 13, 15} = {7, 9, 11}
(x) (A ∪ D) ∩ (B ∪ C) = [{3, 5, 7, 9, 11} ∪ {15, 17} ] ∩ [{7, 9, 11, 13} ∪ {11, 13, 15}]
= [{3, 5, 7, 9, 11, 17} ∩ {7, 9, 11, 13, 15}]
= {7, 9, 11, 15}
7. If A = {x : x is a natural number},
B = {x : x is an even natural number},
C = {x : x an odd natural number} and
D = {x : x is an prime number}, find
(i) A ∩ B
(ii) A ∩ C
(iii) A ∩ D
(iv) B ∩ C
(v) B ∩ D
(vi) C ∩ D
Sol. According to question
A = {1, 2, 3, 4, 5, …}, B = {2, 4, 6, 8, 10, …}, D = {2, 3, 5, 7, 11, …}, C = {1, 3, 5, 7, 9, …}
(i) A ∩ B = {1, 2, 3, 4, 5…} ∩ {2, 4, 6, 8, 10…} = {2, 4, 6, 8, 10…} = B
(ii) A ∩ C = {1, 2, 3, 4, 5…} ∩ {1, 3, 5, 7, 9…} = {1, 3, 5, 7, 9…} = C
(iii) A ∩ D = {1, 2, 3, 4, 5, 6, 7…} ∩ {2, 3, 5, 7, 11, …} = {2, 3, 5, 7…} = D
(iv) B ∩ C = {2, 4, 6, 8, 10…} ∩ {1, 3, 5, 7, 9…} = Φ
(v) B ∩ D = {2, 4, 6, 8, 10…} ∩ {2, 3, 5, 7, 11…} = {2}
(vi) C ∩ D = {1, 3, 5, 7, 9, 11} ∩ {2, 3, 5, 7, 11…}
= {3, 5, 7, 11…} = {3, 5, 7, 11…} = {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}
Sol. (i) Let A = {1, 2, 3, 4}
and B = {x : x is an natural number and 4 ≤x ≤ 6} = {4, 5, 6}
According to question
⇒ A ∩ B = {1, 2, 3, 4} ∩ {4, 5, 6} = {4}
Therefore, pair of sets are not disjoint.
(ii) {a, e, i, o, u} ∩ {c, d, e, f} = {e}
Therefore, pair of sets are not disjoint.
(iii) {x : x is an even integer} ∩ {x : x is an odd integer}
{2, 4, 6, 8, 10…} ∩ {1, 3, 5, 7…} = Φ
Hence, pair of 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} and 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
Sol. (i) A – B
= {3, 6, 9, 12, 15, 18, 21} – {4, 8, 12, 16, 20}
= {3, 6, 9, 15, 18, 21}
(ii) A – C
= {3, 6, 9, 12, 15, 18, 21} – {2, 4, 6, 8, 10, 12, 14, 16}
= {3, 9, 15, 18, 21}
(iii) A – D
= {3, 6, 9, 12, 15, 18, 21} – {5, 10, 15, 20}
= {3, 6, 9, 12, 18, 21}
(iv) B – A
= {4, 8, 12, 16, 20} – {3, 6, 9, 12, 15, 18, 21}
= {4, 8, 16, 20}
(v) C – A
= {2, 4, 6, 8, 10, 12, 14, 16} – {3, 6, 9, 12, 15, 18, 21}
= {2, 4, 8, 10, 14, 16}
(vi) D – A
= {5, 10, 15, 20} – {3, 6, 9, 12, 15, 18, 21}
= {5, 10, 20}
(vii) B – C
= {4, 8, 12, 16, 20} – {2, 4, 6, 8, 10, 12, 14, 16} = {20}
(viii)B – D
= {4, 8, 12, 16, 20} – {5, 10, 15, 20} = {4, 8, 12, 16}
(ix) C – B
= {2, 4, 6, 8, 10, 12, 14, 16} – {4, 8, 12, 16, 20}
= {2, 6, 10, 14}
(x) D – B
= {5, 10, 15, 20} – {4, 8, 12, 16, 20}
= {5, 10, 15}
(xi) C – D
= {2, 4, 6, 8, 10, 12, 14, 16} – {5, 10, 15, 20}
= {2, 4, 6, 8, 12, 14, 16}
(xii) D – C
= {5, 10, 15, 20} – {2, 4, 6, 8, 10, 12, 14, 16}
= {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
Sol. (i) X – Y = {a, b, c, d} – {f, b, d, g} = {a, c}
(ii) Y – X = {f, b, d, g} – {a, b, c, d} = {f, g}
(iii) X ∩ Y = {a, b, c, d} ∩ {f, b, d, g} = {b, d}
11. If R is the set of real numbers and Q is st of rational numbers, then what is R – Q?
Sol. R : set of real numbers, Q : set of rational numbers
Therefore, R – Q as a set of irratinoal numbers.
12. State whether each of the following statements 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.
Sol. (i) {2, 3, 4, 5} ∩ (3, 6) = {3] ≠ Φ
⇒ Given sets are not disjoint and hence, given statement is false.
(ii) {a, e, i, o, u} ∩ {a, b, c, d} = {a} ≠ Φ
⇒ Given sets are not disjoint and hence, given statement is false.
(iii) {2, 6, 10, 14) ∩ {3, 7, 11, 15} = Φ
⇒ Given sets are disjoint and hence, given statement is true.
(iv) {2, 6, 10} ∩ {3, 7, 11} = Φ
⇒ Given sets are disjoint and hence, given statement is true.
Exercise 1.5
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)′
Sol. (i) A′ = U – A
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4}
= {5, 6, 7, 8, 9}
(ii) B′ = U – B
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 4, 6, 8}
= {1, 3, 5, 7, 9}
(iii) (A ∪ C)′ = U – (A ∪ C)
= {1, 2, 3, 4, 5, 6, 7, 8, 9}– {1, 2, 3, 4, 5, 6}
= {7, 8, 9}
(iv) (A ∪ B)′ = U – (A ∪ B)
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4, 6, 8}
= {5, 7, 9}
(v) (A′)′ = A = {1, 2, 3, 4}
(vi) (B – C)′ = U – (B – C)
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 8}
= {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}
Sol. (i) A′ = U – A
= {a, b, c, d, e, f, g, h} – {a, b, c} = {d, e, f, g, h}
(ii) B′ = U – B
= {a, b, c, d, e, f, g, h) – {d, e, f, g} = {a, b, c, h}
(iii) C′ = U – C
= {a, b, c, d, e, f, g, h} – {a, c, e, g} = {b, d, f, h}
(iv) D′ = U – D
= {a, b, c, d, e, f, g, h} – {f, g, h, a} = {b, c, d, e}
3. Taking the set of a 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}
Sol. Given, universal set U = {Natural Numbers}
U = {1, 2, 3, 4, 5, 6, …}
We know that
(i) Let A = {x : x is an even natural numbers}
A′ = U – A = {x : x is an odd natural numbers}
= {1, 3, 5, 7, 9, …}
(ii) Let A = {x : x is an odd natural numbers}
A′ = U – A = {x : is a even natural numbers}
= {2, 4, 6, 8, …}
(iii) Let A = {x : x is a positive multiple of 3}
A = {3, 6, 9, 12, 15, …}
A′ = U – A = {1, 2, 4, 5, 7, …}
A′ = {x : x ∈ N and x is not a multiple of 3}
(iv) Let A = {x : x is a prime number} = {2, 3, 5, 7, 11, …}
Therefore, complement of A = A′ = U – A
= {1, 2, 3, 4, 5, 6, 7, 8, …} – {2, 3, 5, 7, 11, …}
= {1, 4, 6, 8, …}
= {x : x is a positive composite number and x = 1}
(v) Let A = {x : x is a natural number divisible by 3 and 5}
= {15, 30, 45, 60, …}
Therefore, complement of A
= A′ = U – A
= {1, 2, 3, 4, …, 5, 16, …, 30, 31, …}
– {5, 30, 45, 60, …}
= {x : x is a positive integer which is not divisible by 3 or not divisible by 5}
(vi) Let A = {x : x is a perfect square}
Therefore, complement of A = A′
= {x : x ∈ N and x is not a perfect square}
(vii) Let A = {x : x is a perfect cube}
Therefore, complement of A = A′
= {x : x ∈ N and x is not a perfect cube}
(viii)Let A = {x : x + 5 = 8} = {x : x = 3}
Therefore, complement of A = A′ = U – A = {x : x ∈ N and x ≠ 3}
(ix) Let A = {x : 2x + 5 – 9}
⇒ A = {x : x = 2}
Therefore, complement of A = A′ = U – A = {x : x ∈ N and x ≠ 2}
(x) Let A = {x : x ≥ 7}
Therefore, complement of A = A′ = U – A = {x : x ∈ N and x < 7}
(xi) Let A = {x : x ∈ N and 2x + 1 > 10}
$$\Rarr\text{A =}\lbrace x : x \epsilon \text{N} \space\text{and}\space x \gt\frac{9}{2}\rbrace$$
Therefore, complement of A = A′ = U – A =
$$\lbrace \text{x}:\text{x} \epsilon \text{N}\space\text{and} \space\text{x}\le \frac{9}{2}\rbrace$$
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′
Sol. (i) A ∪ B = {2, 4, 6, 8} ∪ {2, 3, 5, 7}
A ∪ B = {2, 3, 4, 5, 6, 7, 8}
⇒ (A ∪ B)′ = U – (A ∪ B)
⇒ (A ∪ B)′ = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 3, 4, 5, 6, 7, 8}
= {1, 9} …(i)
Now, complement of A = A′ = U – A
A′ = {, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 4, 6, 8}
= {1, 3, 5, 7, 9}
And, complement of B = B′ = U – B
B′ = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 3, 5, 7}
B′ = {1, 4, 6, 8, 9}
⇒ A′ ∩ B′ = {1, 3, 5, 7, 9} ∩ {, 4, 6, 8, 9}
= {1, 9} …(ii)
From Eqs. (i) and (ii),
⇒ (A ∪ B)′ = A′ ∩ B′ Hence Verified.
(ii) A ∩ B = {2, 4, 6, 8} ∩ {2, 3, 5, 7} = {2}
(A ∩ B)′ = U – (A ∩ B)
= {, 2, 3, 4, 5, 6, 7, 8, 9} – {2}
= {1, 3, 4, 5, 6, 7, 8, 9} …(i)
Now, complement of A = A′ = {1, 3, 5, 7, 9}
and complement of B = B′ = {1, 4, 6, 8, 9} as in
part (i)
⇒ A′ ∪ B′ = {1, 3, 4, 5, 6, 7, 8, 9} …(ii)
From Eqs. (i) and (ii),
⇒ (A ∩ B)′ = A′ ∪ B′. Hence Verified.
5. Draw appropriate Venn diagram for each of the following:
(i) (A ∪ B)′
(ii) A′ ∩ B′
(iii) (A ∩ B)′
(iv) A′ ∪ B′
Sol. (i) (A ∪ B)′ = I – (A ∪ B) = shaded area
(ii) A′ ∩ B′ = (A ∪ B)′ (By De-Morgan’s law)
= 1 – (A ∪ B) = shaded area fig. (i)
(iii) (A ∩ B)′ = U – (A ∩ B) = Shaded area
(iv) A′ ∪ B′ = (A ∩ B)′ (By De-Morgan’s law)
= U – (A ∩ B) = shaded area fig. (ii)
6. Let U be the set of all triangles in a plane. If A is the set of all triangles with atleast one angle different from 60°, what is A′?
Sol. A is the set of triangles in which no triangle is equilateral because one angle is different trom 60° and we know that equilateral traingle has angle 60°.
∴ A′ = set of elements of U which are not in A.
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 = …
Sol. (i) A ∪ A′ = U
Since, union of a set and its complement is equal to the universal set.
(ii) Φ′ ∩ A
= U ∩ A (·.· complement of null set is universal set)
= A (intersection of any set A and universal set is equal to the set A)
(iii) A ∩ A′ = Φ
Since, intersection of a set and its complement is equal to the null set.
(iv) U′ ∩ A
= Φ ∩ A (·.· complement of set is null set)
= Φ (intersection of any set A and null set is equal to the null set)
Exercise 1.6
1. If X and Y are two sets such that n(X) = 17, n(Y) = 23 and n(X ∪ Y) = 38. Find n(X ∩ Y).
Sol. Given that,
n(X) = 17, n(Y) = 23, n(X ∪ Y) = 38
To find n(X ∩ Y) = ?
We know that
n(X ∪ Y) = n(X) + n(Y) – N(X ∩ Y)
⇒ n(X ∩ Y) = n(X) + n(Y) – n(X ∪ Y)
= 7 + 23 – 38 = 40 – 38 = 2
2. If X and Y are two sets such that X ∪ Y has 8 elements. X has 8 elements and Y has 15 elements, how many elements does X ∩ Y have?
Sol. Given that
n(X) = 8, n(Y) = 15, n(X ∪ Y) = 18
To find n(X ∩ Y) = ?
We know that
n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y), we have
18 = 8 + 15 – n(X ∩ Y)
⇒ n(X ∩ Y) = 23 – 18 = 5
3. In a group of 400 people, 250 can speak Hindi and 200 can speak English. How many people can speak both Hindi and English?
Sol. Let the set of people who are speaking Hindi are denoted by H and the set of people who are speaking English de denoted by E.
Given that, n(H) = 250, n(E) = 200, n(H ∩ E) = 400
To find:
n(H ∩ E) = ?
We know that
n(H ∪ E) = n(H) + n(E) – n(H ∩ E)
n(H ∩ E) = 250 + 200 – 400
= 50
∴ 50 people can speak both English and Hindi
4. If S and T are two sets such that S has 21 elements, T has 32 elements and S ∩ T has 11 elements, how many elements does S ∪ T have?
Sol. We have, n(S) = 21, n(T) = 32, n(S ∩ T) = 1
We know that
n(S ∪ T) = n(S) + n(T) – n(S ∩ T), we get
∴ n(S ∪ T) = 21 + 32 – 11 = 53 – 11 = 42.
5. If X and Y are two sets such taht X has 40 elements, X ∪ Y has 60 elements and X ∩ Y has 10 elements, how many elements does Y have?
Sol. We have, n(X) = 40, n(X ∪ Y) = 60, n(X ∩ Y) = 10
n(Y) = ?
We know that
n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)
⇒ 60 = 40 + n(Y) – 10
⇒ 60 = 30 + n(Y)
⇒ n(Y) = 30.
6. In a group of 70 people, 37 like office, 52 like tea and each person likes atleast one of the two drinks. How many people like both coffee and tea?
Sol. Let C and T denote the people who like coffee and tea, respectively.
Then, n(C ∪ T) = 70, n(C) = 37, n(T) = 52
n(C ∩ T) = ?
We know that
nC ∪ T) = n(C) + n(T) – n(C ∩ T), we have
∴ 70 = 37 + 52 – n(C ∩ T)
⇒ 70 = 89 – n(C ∩ T)
⇒ n(C ∩ T) = 89 – 70 = 19
∴ The number of people who like both coffee and tea are 19 people
7. In a group of 65 people, 40 like circket, 10 like both cricket and tennis. How many like tennis only and not cricket? How many like tennis?
Sol. Let C and T denote the people who like the games cricket and tennis, respectively.
Then, n(C ∪ T) = 65, n(C) = 40, n(C ∩ T) = 10
We know that
n(C ∪ T) = n(C) + n(T) – n(C ∩ T)
∴ 65 = 40 + n(T) – 10
⇒ 65 = 30 + n(T)
⇒ 65 – 30 = n(T)
⇒ n(T) = 35
Now, number of people who like tennis only but not cricket
= n(T ∩ C′)
= n(T) – n(C ∩ T) = 35 – 10 = 25.
8. In a communittee, 50 people speak French, 20 speak Spanish and 10 speak both Spanish and French. How many speak atleast one of these two languages?
Sol. Given that,
n(F) = 50, n(S) = 20, n(S ∩ F) = 10
To find: n(S ∪ F) = ?
We know that
n(S ∩ F) = n(S) + n(F) – n(S ∩ F)
n(S ∪ F) = 20 + 50 – 10
= 70 – 10 = 60
∴ The number of people who can speak at least one of the languages is 60 people.
Miscellaneous Exercise
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}.
Sol. Given, A = {x : x ∈ R and x satisfy x2 – 8x + 12 = 0}
∴ x2 – 8x + 12 = 0
⇒ x2 – 6x – 2x + 12 = 0
⇒ x(x – 6) – 2(x – 6) = 0
⇒ (x – 6) (x – 2) = 0
∴ x = 2, 6
So, A = {2, 6}, B = {2, 4, 6}
C = {2, 4, 6, 8, …} and D = {6}
Now, we can see that,
every element of A is in B and C.
⇒ A ⊂ B and A ⊂ C
D ⊂ A ⊂ B ⊂ C.
every element of B is in C.
⇒ B ⊂ C
and every element of D is in A, B and C.
⇒ D ⊂ A, D ⊂ B and D ⊂ C
∴ D ⊂ A ⊂ B ⊂ C
2. In each of the following, determine whether the statement is true or false. If it is ture, prove it. If it is false, given 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.
Sol. (i) False
Let A = {1, 4} and B = {{2}, {1, 4}, {5}}
⇒ 1 ∈ A and A ∈ B but 1 ∉ B.
(ii) False
Let A = {1}, B = {1, 2} and C = {{1, 2}, 3}
⇒ A ⊂ B and B ∈ C but A ∉ C.
(iii) True
Let a ∈ A, then
a ∈ B (∵ A ⊂ B)
and a ∈ C (∵ B ⊂ C)
so, a ∈ A implies a ∈ C
Hence, A ⊂ C.
(iv) False
Let A = {1, 2}, B = {2, 3} and C = {0, L, 2}
Clearly we can see that
A ⊄ B and B ⊄ C but A ⊂ C.
(v) False
Let A = {4, 5} and B = {0, 1, 2, 3, 4}
So, 5 ∈ A and A ⊄ B but 5 ∉ B.
(vi) True
Let x ∈ A
⇒ x ∈ B (∵ A ⊂ B)
Thus, if x ∉ B then 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.
Sol. Given, A ∪ B = A ∪ C …(i)
A ∩ B = A ∩ C …(ii)
From (i), we have
(A ∪ B) ∩ C = (A ∪ C) ∩ C
⇒ (A ∩ C) ∪ (B ∩ C) = C [∵ (A ∪ C) ∩ C = C]
⇒ (A ∩ B) ∪ (B ∩ C) = C …(iii)
[from (ii)]
Again, A ∪ B = A ∪ C
⇒ (A ∪ B) ∩ B = (A ∪ C) ∩ B
⇒ B = (A ∩ B) ∪ (C ∩ B)
[∵ (A ∪ B) ∩ B = B]
⇒ B = (A ∩ B) ∪ (B ∩ C) …(iv)
From (iii) and (iv), we have
B = C. Hence Proved
4. Show that the following four conditions are equivalent:
(i) A ⊂ B
(ii) A – B = Φ
(iii) A ∪ B = B
(iv) A ∩ B = A
Sol. Let, A ⊂ B
We have to prove that
A – B = Φ
or (i) ⇒ (ii)
Suppose, A – B ≠ Φ
Then, ∃ x ∈ A s.t. x ∉ B
which is not possible as A ⊂ B
∴ A – B = Φ
Now, let A – B = Φ
We have to prove that
A ⊂ B
or (ii) ⇒ (i)
Let x ∈ A
Clearly, x ∈ B because if x ∉ B, then
A – B ≠ f
∴ A – B = f
⇒ A ⊂ B
Hence, (i) ⇒ (ii)
Again, let A ⊂ B
We have to prove that
A ∪ B = B
or (i) ⇒ (iii)
Clearly, B ⊂ A ∪ B
Let a ∈ A ∪ B
⇒ a ∈ A or a ∈ B
(1) If a ∈ A, then
a ∈ B (∵ A ⊂ B)
⇒ (A ∪ B) ⊂ B
(2) If a ∈ B then
(A ∪ B) ⊂ B
Thus, (A ∪ B) = B
Now, we have to prove that (iii) ⇒ (i)
Let A ∪ B = B
Also let a ∈ A
⇒ a ∈ A ∪ B
⇒ a ∈ B (∵ A ∪ B = B)
Thus, A ⊂ B
Hence, (i) ⇒ (iii)
Again, let A ⊂ B
We have to prove that
A ∩ B = A
or (i) ⇒ (iv)
Clearly, A ∩ B ⊂ A
Let a ∈ A
a ∈ (A ∩ B)
Thus, A ⊂ A ∩ B
Thus, A ∩ B = A
Now, we have to prove that (iv) ⇒ (i)
Let, A ∩ B = A
Also, let a ∈ A
∴ a ∈ A ∩ B (∵ A = A ∩ B)
∴ a ∈ A and x ∈ B
∴ a ∈ B
Thus, A ⊂ B
Hence, (i) ⇒ (iv).
5. Show that if A ⊂ B then C – B ⊂ C – A.
Sol. To prove: C – B ⊂ C – A
x ∈ C – B
∴ x ∈ C and x ∉ B
⇒ x ∈ C and x ∉ A (∵ A ⊂ B)
⇒ x ∈ C – A
Thus, C – B ⊂ C – A.
6. Assume that P(A) = P(B). Show that A = B.
Sol. Let x be any arbitrary element of A.
Let X be a subset of A then $ x of set A s.t.
x ∈ X
i.e., X ⊂ A
⇒ X ∈ P(A)
⇒ X ∈ P(B) [∵ P(A) = P(B)]
⇒ X ⊂ B
⇒ x ∈ B
Thus, x ∈ A
⇒ x ∈ B
Hence, A ⊂ B …(i)
Again, let y be an arbitrary element of B.
Let y be an subset of B.
then $ y of set B s.t.
y ∈ y
i.e., y ⊂ B
⇒ y ⊂ P(B)
⇒ y ⊂ P(A) [∵ P(A) = P(B)]
⇒ y ⊂ A
⇒ y ∈ A
Thus y ∈ B
⇒ y ∈ A
Hence, B ⊂ A …(ii)
Since, we know that if two sets are subsets of each other than they are equal.
From (i) and (ii), we get
A = B. Hence Proved.
7. It is true that for any sets A and B.
P(A) ∪ P(B) = P(A ∪ B) ? Justify your answer.
Sol. For any sets A and B,
P(A) ∪ P(B) = P(A ∪ B) is not true.
Let A = {1} and B = {2}
⇒ A ∪ B = {1, 2}
∴ P(A) {Φ, {1}},
and P(B) = {Φ, {2}}
∴ P(A ∪ B) = {f, {1}, {1, 2}} …(i)
and P(A) ∪ P(B) = {f, {1}, {2}} …(ii)
From (i) and (ii), we get
P(A ∪ B) ≠ P(A) ∪ P(B).
8. Show that for any sets A and B, A = (A ∩ B) ∪ (A – B) and A ∪ (B – A) = (A ∪ B).
Sol. Now, (A ∩ B) ∪ (A – B)
= (A ∩ B) ∪ (A ∩ B′)
(∵ A – B = A ∩ B′)
0 = A ∩ (B ∪ B′)
[∵ (A ∩ B) ∪ (A ∩ B′) = A ∩ (B ∪ B′)
= A ∩ U (∵ B ∪ B′ = U)
= A (∵ A ∩ X = A)
Again, A ∪ (B – A)
= A ∪ (B ∩ A′) (∵ B – A = B ∩ A′)
= (A ∪ B) ∩ (A ∪ A′)
= (A ∪ B) ∩ U (∵ A ∪ A′ = U)
= A ∪ B (∵ A ∩ X = A)
Hence Proved.
9. Using properties of sets, show that:
(i) A ∪ (A ∩ B) = A
(ii) A ∩ (A ∪ B) = A
Sol. (i) We have to show that
A ∪ (A ∩ B) = A
As we know that
A ⊂ A
∴ A ∩ B ⊂ A
∴ A ∪ (A ∩ B) ⊂ A …(1)
Also, A ⊂ A ∪ (A ∩ B) …(2)
Thus, from (1) and (2),
A ∪ (A ∩ B) = A …(3)
(ii) We have to show that,
A ∩ (A ∪ B) = A
Since, A ∩ (A ∪ B)
= (A ∩ A) ∪ (A ∪ B)
(Distributive law)
= A ∪ (A ∪ B) (∵ A ∩ A = A)
= A [from (3)]
10. Show that A ∩ B = A ∩ C need not imply B = C.
Sol. Let A = {a, b}
B = {a, c}
and C = {a, d}
Then, A ∩ B = {a}
and A ∩ C = {a}
So, it is clear that if A ∩ B = A ∩ C, then
B ≠ C {∵ d ∈ C but d ∉ B}
11. 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.
Sol. We have
A ∪ X = B ∪ X for some set X …(i)
Since, A = A ∩ (A ∪ X) = A ∩ (B ∪ X)
[from (i)]
⇒ A = (A ∩ B) ∪ (A ∩ X)
(Distributive law)
⇒ A = (A ∩ B) ∪ Φ (∵ A ∩ X = Φ)
⇒ A = A ∩ B …(ii)
B = B ∩ (B ∪ X) = B ∩ (A ∪ X)
[from (i)]
⇒ B = (B ∩ A) ∪ (B ∩ X)
(Distributive law)
⇒ B = (B ∩ A) ∪ f (·.· B ∩ X = Φ)
⇒ B = B ∩ A
or B = A ∩ B
It is only possible when A = B
So, from (ii) and (iii), we get
A = B
12. Find sets A, B and C such that A ∩ B, B ∩ C and A ∩ C are non-empty sets and A ∩ B ∩ C = Φ.
Sol. Let A = {x, y}, B = {y, z} and C = {x, z}
Now A ∩ B = {y},
B ∩ C = {z},
A ∩ C = {x}
Clearly, A ∩ B, B ∩ C and A ∩ C are non-empty
sets.
Thus, A ∩ B ∩ C = (A ∩ B) ∩ C
= {y} ∩ C
= {y} ∩ {x, z}
= Φ.
Note: x, y, z are any numbers or variables.
13. In a survey of 600 students in a school, 150 students were found to be taking tea and 225 taking coffee, 100 were taking both tea and coffee. Find how many students were taking neither tea nor coffee?
Sol. Given total number of students n(U) = 600
Let C and T represent the students taking coffee
and tea respectively.
Number of students taking coffee = nC) = 225,
Number of students taking tea
= n(T) = 150
and Number of students taking both
= n(C ∩ T) = 100
We know that,
n(C ∪ T) = n(C) + n(T) – n(C ∩ T)
⇒ n(C ∪ T) = 225 + 150 – 100
= 375 – 100
= 275
Thus, n(C ∪ T) is the no. of student taking neither coffee nor tea.
⇒ n(C ∪ T)′ = n(U) – n(C ∪ T)
= 600 – 275
= 325.
14. In a group of students, 100 students know Hindi, 50 know English and 25 know both. Each of the students knows either Hindi or English, how many students are there in the group?
Sol. Let H and E represents the students who knows
Hindi and English respectively.
Number of students who know Hindi
= n(H) = 100
Number of students who know English
= n(E) = 50
and Number of students who know both
= n(H ∩ E) = 25
Thus, we know that
Total number of students
n(H ∪ E) = n(H) + n(E) – n(H ∩ E)
= 100 + 50 – 25
= 125.
15. In a survey of 60 people, it was found that 25 people read newspaper H, 26 read newspaper T, 26 read newspaper I, 9 read both H and I. 11 read both H and T, 8 read both T and I, 3 red all three newspapers. Find:
(i) The number of people who read atleast one of the newspapers.
(ii) The number of people who read exactly one newspaper.
Sol. Let H , I and T represent the people who read H, I and T newspapers respectively.
Given, number of people who read newspaper H
= n(H) = 25
Number of people who read newspaper I
= n(I) = 26
Number of people who read newspaper T
= n(T) = 26
Number of people who read both newspaper H and I
= n(H ∩ I) = 9
Number of people who read both newspapers H and I
= n(H ∩ T) = 11
Number of people who read both newspaper H and T
= n(T ∩ I) = 8
Number of people who read all three newspaper
= n(H ∩ T ∩ I) = 3
(i) Number of people who read atleast one newspaper
= n(H ∪ T ∪ I)
= n(H) + n(T) + n(I) – n(H ∩ T)– n(H ∩ I) – n(T ∩ I) + n(H ∩ T ∩ I)
= 25 + 26 + 26 – 11 – 9 – 9 + 3
= 52.
(ii) Number of people who read exactly one newspaper
= n(H) – n(H ∩ T) – n(H ∩ I) + n(H ∩ T ∩ I) + n(T) – n(T ∩ H) – n(T ∩ I) + n(H ∩ T ∩ I) + n(I) – n(I ∩ H) – n(I ∩ T) + n(H ∩ T ∩ I)
= n(H) + n(T) + n(I) – 2[n(H ∩ T) + n(H ∩ I) + n(T ∩ I)] + 3[n(H ∩ T ∩ I)]
= 25 + 26 + 26 – 2[11 + 9 + 8] + 3 × 3
= 77 – 2 × 28 + 9 = 30.
16. In a survey it was found that 21 people liked product A, 26 liked product B and 29 liked product C. If 14 people liked product A and B, 12 people liked products C and A, 14 people liked products B and C and 8 liked all the three products, find how many liked product C only?
Sol. Let A, B and C represents the people who liked the product A, B and C respectively.
According to question,
Number of people who liked product A
= n(A) = 21
Number of people who liked product B
= n(B) = 26
Number of people who liked product C
= n(C) = 29
Number of people who liked both A and B
= n(A ∩ B) = 14
Number of people who liked both C and A
= n(C ∩ A) = 12
Number of people who liked both B and C
= n(B ∩ C) = 14
Number of people who liked all three products
= n(A ∩ B ∩ C) = 8
Number of people who liked only product C
= n(C only) = n(C) – n(C ∩ A) – n(C ∩ B) + n(C ∩ B ∩ A)
= 29 – 12 – 14 + 8
= 11.
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