1. Sets Mathematics class 11 exercise Miscellaneous
1. Sets Mathematics class 11 exercise Miscellaneous ncert book solution in english-medium
NCERT Books Subjects for class 11th Hindi Medium
Exercise 1.1
Exercise 1.1
Q1. 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 questions in this Chapter.
(ix) A collection of most dangerous animals of the world.
Answers:
(i) Members of this collection are:
January, June, July
∴ this collection is well-defined and hence, It is a set.
(ii) A writer may be most talented for one person but may not be so for other.
∴ this collection is unclear term and not well-define, so it is not a set.
(iii) A batsman may be best to one person and may not be so for other. Hence the term 'best' is unclear and not well-define. So it is not a set.
(iv) The members of this collection is definite as a class room.
∴ This collection is well-defined and so it is a set.
(v) The collection of all natural numbers less than 100 is well-define as it can be kept in a set like 1, 2, 3, 4, .......... 99.
∴ this collection is a set.
(vi) All novels writen by the writer Munshi Prem Chand can be kept in a group and this collection is well-define. So it is a set.
(vii) All even number like 2, 4, 6, 8, ............. can be kept in a group.
∴ this collection is well-define and so it is a set.
(viii) Questions in this chapter are clearly can be kept in well-define group. Therefore this collection is set.
(ix) Any animal may be the most dangerous for someone but may not be for other. Therefore this term is unclear and not well-define. So it is not a set.
Q2. 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
Answers:
(i) 5 ∈ A
(ii) 8 ∉ A
(iii) 0 ∉ A
(iv) 4 ∈ A
(v) 2 ∈ A
(vi) 10 ∉ A
Q3. Write the following sets in roster form:
(i) A = {x : x is an integer and –3 ≤ x < 7}
Solution:
A = {-3, -2, -1, 0, 1, 2, 3, 4, 5, 6 }
(ii) B = {x : x is a natural number less than 6}
Solution:
B = {1, 2, 3, 4, 5}
(iii) 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}
(iv) D = {x : x is a prime number which is divisor of 60}
Solution:
Divisor of 60 ⇒ 2, 3, 4, 5, 6, 12, 15, 20, 30, 60
but looking for prime number, which are 2, 3, 5
∴ D = {2, 3, 5}
(v) E = The set of all letters in the word TRIGONOMETRY
Solution:
E = {T, R, I, G, O, N, M, E, Y}
Note: repeated letter can only be kept once.
(vi) F = The set of all letters in the word BETTER.
F = {B, E, T, R}
Q4. Write the following sets in the set-builder form :
(i) (3, 6, 9, 12}
Solution:
A = {x : x is a natural number multiple of 3 and x < 15}
Or
A = { x : x = 3n, n ∈ N and 1≤ n ≤4}
(ii) {2,4,8,16,32}
Solution:
B = { x : x = 2n , n ∈ N and 1≤ n ≤5}
(iii) {5, 25, 125, 625}
Solution:
C = {x : x = 5n , n ∈ N and 1≤ n ≤4}
(iv) {2, 4, 6, . . .}
Solution:
D = {x : x is an even natural number }
(v) {1,4,9, . . .,100}
Solution:
E = {x : x = n2, n ∈ N and n < 11 }
Or
E = {x : x = n2, n ∈ N and 1≤ n ≤10 }
Or
E = {x : x = n2, where n is the first 10 natural number }
Q5. List all the elements of the following sets :
(i) A = {x : x is an odd natural number}
Solution:
A = {1, 3, 5, 7, 9, ....... }
|
(ii) B = {x : x is an integer, - |
1 |
< x < |
9 |
} |
|
2 |
2 |
B = {0, 1, 2, 3, 4}
(iii) C = {x : x is an integer, x2 ≤ 4}
Solution:
x ≤ √4
x ≤ ±2
the number between -2 to 2.
∴C = {-2, -1, 0, 1, 2}
(iv) D = {x : x is a letter in the word “LOYAL”}
Solution:
D = {L, O, Y, A}
(v) E = {x : x is a month of a year not having 31 days}
Solution:
E = {February, April, June, September, November}
(vi) F = {x : x is a consonant in the English alphabet which precedes k }.
Solution:
consonant befor Letter 'k'
F = {b, c, d, f, g, h, j}
Q6. 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} | (a) {x : x is a prime number and a divisor of 6} |
| (ii) {2, 3} | (b) {x : x is an odd natural number less than 10} |
| (iii) {M,A,T,H,E,I,C,S} | (c) {x : x is natural number and divisor of 6} |
| (iv) {1, 3, 5, 7, 9} | (d) {x : x is a letter of the word MATHEMATICS}. |
Solution:
(i) ------------------ (c)
(ii) ------------------ (a)
(iii) ------------------ (d)
(iv) ------------------ (b)
Exercise 1.2
Exercise 1.2
Q1. 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}
Solution:
(i) There is any set of add number which is divisible by 2.
(ii) Let A is set which having even prime number
Therefore, A = {2}
So, this is not a null set
(iii) Let A = { x : x is a natural numbers, x < 5 and x > 7 }
therefore there is no natural numbers which is both x < 5 and x > 7
So, this is a null set.
(iv) Two parallel lines never meet on a point, therefore this is a null set.
Q2. 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
Solutions:
(i) The set of months of a year is a finite set because it has 12 elements.
(ii) {1, 2, 3 ...} is an infinite set as it has infinite number of natural numbers.
(iii) {1, 2, 3 ...99, 100} has definite elements so it is a finite set as the numbers from 1 to 100 are finite in number.
(iv) The set of positive integers greater than 100 is an infinite set because positive integers greater than 100 are infinite in number.
(v)The set of prime numbers less than 99 is a finite set because prime numbers less than 99 are finite in number.
Q3. 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)
Solutions:
(i) Infinite: The set of lines which are parallel to the x-axis can be drawn infinitive in number.
Therefore, it is a infinite set.
(ii) Finite: The set of letters in the English alphabet is a finite set because English alphabet has only 26 number which is finite.
(iii) Infinite: The set of numbers which are multiple of 5 is an infinite set because multiples of 5 are infinite in number.
(iv) Finite: Animals living on the earth is countable, therefore the set is finite.
(v) Infinite: Through the origin(0, 0) can be drawn infinite numbers of circle. Because many circle can be drawn through a point.
Q4. 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, . . . }
Solutions:
(i) A = {a, b, c, d}; B = {d, c, b, a}
All elements of set A also belong to set B.
∴ A = B
(ii) A = {4, 8, 12, 16}; B = {8, 4, 16, 18}
It can be seen that 12 ∈ A but 12 ∉ B.
∴ A ≠ B
(iii) A = {2, 4, 6, 8, 10}
B = {x: x is a positive even integer and x ≤ 10}
= {2, 4, 6, 8, 10}
All elements of set A also belong to set B.
∴ A = B
(iv) A = {x: x is a multiple of 10}
B = {10, 15, 20, 25, 30 ...}
It can be seen that 15 ∈ B but 15 ∉ A.
∴ A ≠ B
Q5. 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 word WOLF}
Solutions:
(i) A = {2, 3}; B = {x: x is a solution of x2 + 5x + 6 = 0}
The equation x2 + 5x + 6 = 0
=> x(x + 3) + 2(x + 3) = 0
=> (x + 2)(x + 3) = 0
=> x = –2 or x = –3
∴ A = {2, 3}; B = {–2, –3}
∴ A ≠ B
(ii)
A = {x: x is a letter in the word FOLLOW}
= {F, O, L, W}
B = {y: y is a letter in the word WOLF}
= {W, O, L, F}
All elements of set A also belong to set B.
∴ A = B
Q6. 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}
Solution:
B and D are equal sets and also E and G are equal sets.
Exercise 1.3
Exercise 1.3
Q1. 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 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 : x is a triangle in the same plane}
(vii) {x : x is an even natural number} . . . {x : x is an integer}
Solutions:
(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 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 : x is a triangle in the same plane}
(vii) {x : x is an even natural number} ⊂ {x : x is an integer}
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} ⊂ { x : x is a natural number which divides 36}
Solutions:
(i) False, Because Each element of { a, b } is also an element of { b, c, a }.
(ii) True, Because {a, e} is also vowels of English alphabet.
(iii) False, Hence 2 ∈ {1, 2, 3}; while, 2 ∉ {1, 3, 5}
(iv) True, Because each elements of set { a } is also element of { a, b, c }
Q3. Let A = { 1, 2, { 3, 4 }, 5 }. Which of the following statements are incorrect 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) φ ∈ A
(x) φ ⊂ A
(xi) {φ} ⊂ A
Solutions:
Given that A = {1, 2, {3, 4}, 5}
(i) The statement {3, 4} ⊂ A is incorrect because 3 ∈ {3, 4}; while, 3∉A.
(ii) The statement {3, 4} ∈ A is correct because {3, 4} is an element of A.
(iii) The statement {{3, 4}} ⊂ A is correct because {3, 4} ∈ {{3, 4}} and {3, 4} ∈ A.
(iv) The statement 1∈A is correct because 1 is an element of A.
(v) The statement 1⊂ A is incorrect because an element of a set can never be a subset of itself.
(vi) The statement {1, 2, 5} ⊂ A is correct because each element of {1, 2, 5} is also an element of A.
(vii)The statement {1, 2, 5} ∈ A is incorrect because {1, 2, 5} is not an
element of A.
(viii) The statement {1, 2, 3} ⊂ A is incorrect because 3 ∈ {1, 2, 3}; however, 3 ∉ A.
(ix) The statement Φ ∈ A is incorrect because Φ is not an element of A.
(x) The statement Φ ⊂ A is correct because Φ is a subset of every set.
(xi) The statement {Φ} ⊂ A is incorrect because Φ∈ {Φ}; however, Φ ∈ A.
Q4. Write down all the subsets of the following sets
(i) {a} (ii) {a, b} (iii) {1, 2, 3} (iv) φ
Solutions:
(i) The subsets of {a} are φ and {a}.
(ii) The subsets of {a, b} are φ, {a}, {b} and {a, b}.
(iii) The subsets of {1, 2, 3} are φ, {1}, {2}, {3}, {1, 2}, {1,3}, {2, 3} and {1, 2, 3}.
(iv) The subset of φ is φ.
Q5. How many elements has P(A), if A = φ?
Solution:
Given that A = φ
Therefore, no. of elements n(A) = 0
n[P(A)] = 2n = 20 = 1
Hence, P(A) has only 1 element.
Q6. 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}
Solutions:
(i) {x: x ∈ R, –4 < x ≤ 6} is an open interval from -4 to 6, including 6 but excluding -4.
Hence interval = (–4, 6]
(ii) {x: x ∈ R, –12 < x < –10} is an open interval from -12 to -10, excluding both -12 and -10.
Hence interval = (–12, –10)
(iii) {x: x ∈ R, 0 ≤ x < 7} is an open interval from 0 to 7, including 0 but excluding 7.
Hence interval = [0, 7)
(iv) {x: x ∈ R, 3 ≤ x ≤ 4} is an close interval from 3 to 4, including both 3 and 4.
Hence interval = [3, 4]
Q7. Write the following intervals in set-builder form :
(i) (– 3, 0)
(ii) [6 , 12]
(iii) (6, 12]
(iv) [–23, 5)
Solutions:
(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}
Q8. What universal set(s) would you propose for each of the following :
(i) The set of right triangles.
(ii) The set of isosceles triangles.
Solutions:
(i) The sets of all possible triangles and polygons can be universal set for the right triangles.
(ii) The sets of all possible triangles and polygons can be universal set for the isosceles triangles.
Q9. 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
(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}
Solution:
(iii) {0,1,2,3,4,5,6,7,8,9,10} can be universal set (s) for all the three sets A, B and C.
Because,
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}
Therefore, the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} is the universal set for the
sets A, B, and C.
Exercise 1.4
Exercise 1.4
Q1. 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 and 1 < x ≤ 6 }
B = {x : x is a natural number and 6 < x < 10 }
(v) A = {1, 2, 3}, B = φ
Solution:
(i) X = {1, 3, 5} Y = {1, 2, 3}
X ∪ Y= {1, 2, 3, 5}
(ii) A = {a, e, i, o, u} B = {a, b, c}
A ∪ B = {a, b, c, e, i, o, u}
(iii) A = {x: x is a natural number and multiple of 3} = {3, 6, 9 ...}
B = {x: x is a natural number less than 6} = {1, 2, 3, 4, 5, 6}
A ∪ B = {1, 2, 4, 5, 3, 6, 9, 12 ...}
∴ A ∪ B = {x: x = 1, 2, 4, 5 or a multiple of 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 = {2, 3, 4, 5, 6, 7, 8, 9}
∴ A∪ B = {x: x ∈ N and 1 < x < 10}
(v) A = {1, 2, 3}, B = Φ
A ∪ B = {1, 2, 3}
Q2. Let A = { a, b }, B = {a, b, c}. Is A ⊂ B ? What is A ∪ B ?
Solution:
Here, A = {a, b} and B = {a, b, c}
Yes, A ⊂ B.
As a ∈ B and b ∈ B
A ∪ B = {a, b, c} = B
{Rule: if A ∪ B = B then A ⊂ B Or if A ∪ B = A then B ⊂ A }
Q3. If A and B are two sets such that A ⊂ B, then what is A ∪ B ?
Solution:
Given that: A and B are two sets such that A ⊂ B
Then A ∪ B = B
Illustration by example:
Let A = {1, 2, 3} and B = {1, 2, 3, 4, 5}
Here A ⊂ B Because All elements of A 1, 2, 3 ∈ B
[B also contains 1, 2, 3]
Now, A ∪ B = {1, 2, 3, 4, 5} = B
Therefore, A ∪ B = B
Q4. 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
Solution:
Given A = {1, 2, 3, 4], B = {3, 4, 5, 6}, C = {5, 6, 7, 8} and D = {7, 8, 9, 10}
(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 = {1, 2, 3, 4, 5, 6, 7, 8}
(vi) A ∪ B ∪ D = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(vii) B ∪ C ∪ D = {3, 4, 5, 6, 7, 8, 9, 10}
Q5. Find the intersection of each pair of sets of question 1 above.
(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 and 1 < x ≤ 6 }
B = {x : x is a natural number and 6 < x < 10 }
(v) A = {1, 2, 3}, B = φ
Solution:
(i) X = {1, 3, 5}, Y = {1, 2, 3}
X ∩ Y = {1, 3}
(ii) A = {a, e, i, o, u}, B = {a, b, c}
A ∩ B = {a}
(iii) A = {x: x is a natural number and multiple of 3} = (3, 6, 9 ...}
B = {x: x is a natural number less than 6} = {1, 2, 3, 4, 5}
∴ A ∩ B = {3}
(iv) A = {x: x is a natural number and 1 < x ≤ 6}
Or A = {2, 3, 4, 5, 6}
B = {x: x is a natural number and 6 < x < 10}
Or B = {7, 8, 9}
A ∩ B = Φ
(v) A = {1, 2, 3}, B = Φ.
So, A ∩ B = Φ
Q6. 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 ) ∩ ( B ∪ C )
(x) ( A ∪ D) ∩ ( B ∪ C)
Solution:
(i) A = { 3, 5, 7, 9, 11 }, B = {7, 9, 11, 13},
Therefore, A ∩ B = {7, 9, 11}
Solution:
(ii) B = {7, 9, 11, 13}, and C = {11, 13, 15}
Therefore, B ∩ C = {11, 13}
Solution:
(iii) A = { 3, 5, 7, 9, 11 }, C = {11, 13, 15} and D = {15, 17};
A ∩ C ∩ D = { A ∩ C} ∩ D = {11} ∩ {15, 17} = Φ
Solution:
(iv) A = { 3, 5, 7, 9, 11 }, C = {11, 13, 15}
Therefore, A ∩ C = {11}
Solution:
(v) B = {7, 9, 11, 13}, and D = {15, 17};
Therefore, B ∩ D = Φ
Solution:
(vi) A ∩ (B C) = (A ∩ B) (A ∩ C)
= {7, 9, 11} {11} = {7, 9, 11}
Solution:
(vii) A ∩ D = Φ
Solution:
(viii) A = { 3, 5, 7, 9, 11 }, B = {7, 9, 11, 13}, and D = {15, 17};
We know; A ∩ (B ∪ D) = (A ∩ B) ∪ (A ∩ D)
= {7, 9, 11} Φ = {7, 9, 11}
Solution:
(ix) A = { 3, 5, 7, 9, 11 }, B = {7, 9, 11, 13}, and C = {11, 13, 15}
(A ∩ B) = {7, 9, 11}
(B ∪ C) = {7, 9, 11, 13, 15}
(A ∩ B) ∩ (B ∪ C)
= {7, 9, 11} ∩ {7, 9, 11, 13, 15}
= {7, 9, 11}
Solution:
(x) Given A = { 3, 5, 7, 9, 11 }, B = {7, 9, 11, 13}, C = {11, 13, 15}and D = {15, 17};
(A ∪ D) ∩ (B ∪ C)
= {3, 5, 7, 9, 11, 15, 17) ∩ {7, 9, 11, 13, 15}
= {7, 9, 11, 15}
Q7. 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
(i) A ∩ B
(ii) A ∩ C
(iii) A ∩ D
(iv) B ∩ C
(v) B ∩ D
(vi) C ∩ D
Solution 7:
A = {x: x is a natural number} = {1, 2, 3, 4, 5 ...}
B = {x: x is an even natural number} = {2, 4, 6, 8 ...}
C = {x: x is an odd natural number} = {1, 3, 5, 7, 9 ...}
D = {x: x is a prime number} = {2, 3, 5, 7 ...}
Solution:
(i) A ∩ B = {2, 4, 6, 8 .....}
A ∩ B = {x : x is a even natural number}
A ∩ B = B
Solution:
(ii) A ∩ C = {1, 3, 5, 7, 9 ...}
A ∩ C = {x : x is an odd natural number}
A ∩ C = C
Solution:
(iii) A ∩ D = {2, 3, 5, 7 ...}
A ∩ D = {x : x is a prime number}
A ∩ D = D
Solution:
(iv) B ∩ C = Φ
Solution:
(v) B ∩ D = {2}
Solution:
(vi) C ∩ D = {3, 5, 7, 11 ....}
C ∩ D = {x : x is odd prime number}
Q8. 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}
{Note : Disjoint means no elements matched in each others, in other words intersection of two sets give null set then it said to be disjoint.}
Solution:
(i) Let A = {1, 2, 3, 4} and
B = {x: x is a natural number and 4 ≤ x ≤ 6}
Or B = {4, 5, 6}
Now, A ∩ B
= {1, 2, 3, 4} ∩ {4, 5, 6}
= {4}
Therefore, this pair of sets is not disjoint.
Solution:
(ii) Let X = {a, e, i, o, u} and
Y = (c, d, e, f}
Now, X ∩ Y
= {a, e, i, o, u} ∩ (c, d, e, f}
= {e}
Therefore, {a, e, i, o, u} and (c, d, e, f} are not disjoint.
Solution:
(iii) Let A = {x : x is an even integer}
Or A = {2, 4, 6, 8 ... } and
B = {x : x is an odd integer}
Or B = {1, 3, 5, 7 ... }
Now, A ∩ B
= {2, 4, 6, 8 ... } ∩ {1, 3, 5, 7 ... }
= Φ
Therefore,
{x : x is an even integer} ∩ {x : x is an odd integer}
= Φ
Therefore, A ∩ B gives null set so this pair of sets is disjoint.
Q9. 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
(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
Solution:
(i) A – B = {3, 6, 9, 15, 18, 21}
(ii) A – C = {3, 9, 15, 18, 21}
(iii) A – D = {3, 6, 9, 12, 18, 21}
(iv) B – A = {4, 8, 16, 20}
(v) C – A = {2, 4, 8, 10, 14, 16}
(vi) D – A = {5, 10, 20}
(vii) B – C = {20}
(viii) B – D = {4, 8, 12, 16}
(ix) C – B = {2, 6, 10, 14}
(x) D – B = {5, 10, 15}
(xi) C – D = {2, 4, 6, 8, 12, 14, 16}
(xii) D – C = {5, 15, 20}
Q10. If X = { a, b, c, d } and Y = { f, b, d, g}, find
(i) X – Y
(ii) Y – X
(iii) X ∩ Y
Solution:
(i) X – Y = { a, c }
(ii) Y – X = {f, g}
(iii) X ∩ Y = {b, d}
Q11. If R is the set of real numbers and Q is the set of rational numbers, then what is R – Q?
Solution:
R = {x : x is set of real numbers}
Q = {x : x is set of rational numbers}
We know that,
Real numbers = Rational Numbers + Irrational numbers
Real numbers - Rational Numbers = Irrational numbers
R - Q = I
Therefore, R – Q is a set of irrational numbers.
Q12. State whether each of the following statement is true or false. Justify your answer.
(i) { 2, 3, 4, 5 } and { 3, 6} are disjoint sets.
Solution:
(i) False, Because { 2, 3, 4, 5 } ∩ { 3, 6} = {3}
(ii) { a, e, i, o, u } and { a, b, c, d } are disjoint sets.
Solution:
(ii) False, Because { a, e, i, o, u } ∩ { a, b, c, d } = {a}
(iii) { 2, 6, 10, 14 } and { 3, 7, 11, 15} are disjoint sets.
Solution:
(iii) True, Because { 2, 6, 10, 14 } ∩ { 3, 7, 11, 15} = Φ
(iv) { 2, 6, 10 } and { 3, 7, 11} are disjoint sets.
Solution:
(iv) True, Because { 2, 6, 10 } ∩ { 3, 7, 11} = Φ
Exercise 1.5
Exercise 1.5
Q1. 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)′
Solution: Given that
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 }.
(i) A' = {5, 6, 7, 8, 9}
(ii) B' = {1, 3, 5, 7, 9}
(iii) A ∪ C = {1, 2, 3, 4, 5, 6}
Therefore, (A ∪ C)′ = {7, 8, 9}
(iv) A ∪ B = {1, 2, 3, 4, 6, 8}
Therefore, (A ∪ B)′ = {5, 7, 9}
(v) A' = {5, 6, 7, 8, 9}
(A')' = A = {1, 2, 3, 4}
(vi) B - C = {2, 8}
(B - C)' = 1, 3, 4, 5, 6, 7, 9}
Q2. 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}
Solution: Given that
U = { a, b, c, d, e, f, g, h}
(i) A = {a, b, c}
A' = {d, e, f, g, h}
(ii) B = {d, e, f, g}
B' = {a, b, c, h}
(iii) C = {a, c, e, g}
C' = {b, d, f, h}
(iv) D = { f, g, h, a}
D' = {b, c, d e}
Q3. Taking the set of 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 }
Solution: Given that U = { 1, 2, 3, 4, 5, 6, 7 ....}
(i) Let A = {x : x is an even natural number}
Or A = {2, 4, 6, 8 .....}
A' = { 1, 3, 5, 7 .....}
= {x : x is an odd natural number}
(ii) Let B = { x : x is an odd natural number }
Or B = { 1, 3, 5, 7 .....}
B' = {2, 4, 6, 8 .....}
= {x : x is an even natural number}
(iii) Let C = {x : x is a positive multiple of 3}
Or C = {3, 6, 9 ....}
C' = {1, 2, 4, 5, 7, 8, 10 .....}
= {x: x N and x is not a multiple of 3}
(iv) Let D = { x : x is a prime number }
Or D = {2, 3, 5, 7, 11 ... }
D' = {1, 4, 6, 8, 9, 10 ...... }
= {x: x is a positive composite number and x = 1}
(v) Let E = {x : x is a natural number divisible by 3 and 5}
Or E = {15, 30, 45 .....}
E' = {x: x is a natural number that is not divisible by 3 or 5}
(vi) Let F = { x : x is a perfect square }
F' = {x: x N and x is not a perfect square}
(vii) Let G = {x: x is a perfect cube}
G' = {x: x N and x is not a perfect cube}
(viii) Let H = {x: x + 5 = 8}
H' = {x: x N and x ≠ 3}
(ix) Let I = {x: 2x + 5 = 9}
I' = {x: x N and x ≠ 2}
(x) Let J = {x: x ≥ 7}
J' = {x: x N and x < 7}
(xi) Let K = {x: x N and 2x + 1 > 10}
K = {x: x N and x ≤ 9/2}
Q4. 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′
Solution:
(i) U = {1, 2, 3, 4, 5, 6, 7, 8, 9 }, A = {2, 4, 6, 8} and B = { 2, 3, 5, 7}.
(A ∪ B)′ = A′ ∩ B′
A ∪ B = {2, 3, 4, 5, 6, 7, 8}
LHS = (A ∪ B)′ = {1, 9} ...(i)
RHS = A′ ∩ B′
= {1, 3, 5, 7, 9} ∩ {1, 4, 6, 8, 9}
= {1, 9} .... (ii)
LHS = RHS
Hence Verified.
Solution:
(ii) U = {1, 2, 3, 4, 5, 6, 7, 8, 9 }, A = {2, 4, 6, 8} and B = { 2, 3, 5, 7}.
(A ∩ B)′ = A′ ∪ B′
A ∩ B = {2}
LHS = (A ∩ B)′ = {1, 3, 4, 5, 6, 7, 8, 9 }
RHS = A′ ∪ B′
= {1, 3, 5, 7, 9} ∪ {1, 4, 6, 8, 9}
= {1, 3, 4, 5, 6, 7, 8, 9 }
LHS = RHS
Hence Verified
Q5. Draw appropriate Venn diagram for each of the following :
(i) (A ∪ B)′,
(ii) A′ ∩ B′,
(iii) (A ∩ B)′,
(iv) A′ ∪ B′
Solution:
(i) (A ∪ B)′
Venn diagram of (A ∪ B)′
(ii) A′ ∩ B′,
Venn diagram of A′ ∩ B′
Note: Venn diagram of A′ ∩ B′ will be same as (A ∪ B)′
Because (A ∪ B)′ = A′ ∩ B′
(iii) (A ∩ B)′
Venn diagram of (A ∩ B)′
(iv) A′ ∪ B′
Venn diagram of A′ ∪ B′
Q6. 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′?
Solution:
A = {the set of all triangles with at least one angle different from 60°}
A' = {the set of all equilateral triangles}
Q7. 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 = . . .
Solution:
(i) A ∪ A′ = U
(ii) φ′ = U
Therefore φ′ ∩ A = U ∩ A = A
so, φ′ ∩ A = A
(iii) A ∩ A′ = φ
(iv) U′ ∩ A = φ
Exercise 1.6
Exercise 1.6
Q1. If X and Y are two sets such that n ( X ) = 17, n ( Y ) = 23 and n ( X ∪ Y ) = 38, find n ( X ∩ Y ).
Solution: Given that
n ( X ) = 17, n ( Y ) = 23 and n ( X ∪ Y ) = 38
n ( X ∩ Y ) = ?
n(X ∪ Y ) = n(X) + n(Y) - n(X ∩ Y)
=> 38 = 17 + 23 - n(X ∩ Y)
=> 38 = 40 - n(X ∩ Y)
=> n(X ∩ Y) = 40 - 38
=> n(X ∩ Y) = 2
Q2. If X and Y are two sets such that X ∪ Y has 18 elements, X has 8 elements and Y has 15 elements ; how many elements does X ∩ Y have?
Solution: Given that;
n ( X ) = 8, n ( Y ) = 15 and n ( X ∪ Y ) = 18
n ( X ∩ Y ) = ?
n(X ∪ Y ) = n(X) + n(Y) - n(X ∩ Y)
=> 18 = 8 + 15 - n(X ∩ Y)
=> 18 = 23 - n(X ∩ Y)
=> n(X ∩ Y) = 23 - 18
=> n(X ∩ Y) = 5
Q3. In a group of 400 people, 250 can speak Hindi and 200 can speak English. How many people can speak both Hindi and English?
Solution:
Let the people can speak hindi be n(H) = 250,
The people can speak English be n(E) = 200 and
and Total people be n(H ∪ E) = 400
and people can speak both Hindi and English be n(H ∩ E) = ?
n(H ∪ E) = n(H) + n(E) – n(H ∩ E)
∴ 400 = 250 + 200 – n(H ∩ E)
⇒ 400 = 450 – n(H ∩ E)
⇒ n(H ∩ E) = 450 – 400
∴ n(H ∩ E) = 50
Thus, 50 people can speak both Hindi and English.
Q4. 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?
Solution: Given that: n(S) = 21, n(T) = 32, n(S ∩ T) = 11
We know that: n (S ∪ T) = n (S) + n (T) – n (S ∩ T)
∴ n (S ∪ T) = 21 + 32 – 11 = 42
Thus, the set (S ∪ T) has 42 elements.
Q5. If X and Y are two sets such that X has 40 elements, X ∪ Y has 60 elements and X ∩ Y has 10 elements, how many elements does Y have?
Solution:
Given that: n(X) = 40, n(X ∪ Y) = 60, n(X ∩ Y) = 10
We know that:
n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)
∴ 60 = 40 + n(Y) – 10
∴ n(Y) = 60 – (40 – 10)
= 60 - 30
= 30
Thus, the set Y has 30 elements.
Q6. In a group of 70 people, 37 like coffee, 52 like tea and each person likes at least one of the two drinks. How many people like both coffee and tea?
Solution:
Let C denote the set of people who like coffee, and T denote the set of people who like tea then
n(C ∪ T) = 70, n(C) = 37, n(T) = 52
We know that: n(C ∪ T) = n(C) + n(T) – n(C ∩ T)
∴ 70 = 37 + 52 – n(C ∩ T)
⇒ 70 = 89 – n(C ∩ T)
⇒ n(C ∩ T) = 89 – 70 = 19
Thus, 19 people like both coffee and tea.
Q7. In a group of 65 people, 40 like cricket, 10 like both cricket and tennis. How many like tennis only and not cricket? How many like tennis?
Solutions:
Let C denote the set of people who like cricket, and T denote the set of people who like tennis 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)
⇒ n(T) = 65 – 30 = 35
Therefore, 35 people like tennis.
Now,
(T – C) ∪ (T ∩ C) = T
Also,
(T – C) ∩ (T ∩ C) = Φ
∴ n (T) = n (T – C) + n (T ∩ C)
⇒ 35 = n (T – C) + 10
⇒ n (T – C) = 35 – 10 = 25
Thus, 25 people like only tennis.
Q8. In a committee, 50 people speak French, 20 speak Spanish and 10 speak both Spanish and French. How many speak at least one of these two languages?
Solution:
Let F be the set of people in the committee who speak French, and
S be the set of people in the committee who speak Spanish then-
∴ n(F) = 50, n(S) = 20, n(S ∩ F) = 10
We know that: n(S ∪ F) = n(S) + n(F) – n(S ∩ F)
= 20 + 50 – 10
= 70 – 10 = 60
Thus, 60 people in the committee speak at least one of the two languages.
Miscellaneous
Miscellaneous Exercise on Chapter 1
Q1. 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 }.
Solution:
x2 - 8x + 12 = 0
x2 - 6x - 2x + 12 = 0
x(x - 6) -2 (x - 6) = 0
x - 6 = 0, x - 2 = 0;
x = 6 , x = 2
A = {2, 6} , B = { 2, 4, 6 }, C = { 2, 4, 6, 8, . . . }, D = { 6 }.
Every element of A is in B and C
∴ A ⊂ B, A ⊂ C,
Every Element of B is in C
∴ B ⊂ C ,
Every Element of D is in A, B and C
∴ D ⊂ A, D ⊂ B, D ⊂ C
Q2. In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example.
(i) If x ∈ A and A ∈ B , then x ∈ B
Solution:
False, Let be A = {1}, and B = {{1}, 2}
then 1 ∈ A and A ∈ B but 1 ∉ B
∴ x ∈ A and A ∈ B but not emply x ∉ B
(ii) If A ⊂ B and B ∈ C , then A ∈ C
Solution: False,
Let A = {1, 2} and B = {1, 2, 3, 4} and C = {{1, 2, 3, 4}, 5, 6}
According to condition,
A ⊂ B and B ∈ C but A ∉ C
A ∈ B but does not emply A ∉ C
(iii) If A ⊂ B and B ⊂ C , then A ⊂ C
Solution: True,
Let A = {a, b} , B = {a, b, c} and C = {a, b, c, d}
Here, A ⊂ B ⇒ x ∈ A and x ∈ B ------------- (i)
Similarily, B ⊂ C ⇒ x ∈ B and x ∈ C ------------------ (ii)
So, A ⊂ B and B ⊂ C ⇒ A ⊂ C
(iv) If A ⊄ B and B ⊄ C , then A ⊄ C
Solution: false;
Let A = {a, b}, B = {b, c} and C = {a, b, d, e}
Here, A ⊄ B and B ⊄ C but A ⊂ C
So, A ⊄ B and B ⊄ C do not emply A ⊄ C
(v) If x ∈ A and A ⊄ B , then x ∈ B
Solution: False;
Let A = {1,2}, B = {2, 3, 5}
1 ∈ A for all x, but 1 ∉ B
Therefore, A ⊄ B which does not imply x ∈ B.
(vi) If A ⊂ B and x ∉ B , then x ∉ A
Solution: True
A ⊂ B given
then x ∈ A and x ∈ B
While x ∉ B ⇒ x ∉ A
Q3. Let A, B, and C be the sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C. Show that B = C.
Solution:
A ∪ B = A ∪ C ..................................... (i)
A ∩ B = A ∩ C ................................................. (ii)
taking equation (i)
(A ∪ B) ∩ C = (A ∪ C) ∩ C
Using distributive law;
(A ∩ C) ∪ (B ∩ C) = C [ ∵ (A ∪ C) ∩ C = C ]
(A ∩ B) ∪ (B ∩ C) = C From equ (i) [ ∵ A ∩ B = A ∩ C ]
C = (A ∩ B) ∪ (B ∩ C) ................... (iii)
Now
(A ∪ B) ∩ B = (A ∪ C) ∩ B
using distributive law
(A ∩ B) ∪ (B ∩ B) = (A ∩ B) ∪ (C ∩ B)
(A ∩ B) ∪ B = (A ∩ B) ∪ ( B ∩ C) [∵ (B ∩ B) = B]
B = (A ∩ B) ∪ ( B ∩ C) ....................... (iv) [∵ (A ∩ B) ∪ B = B ]
From equation (iii) and (iv)
B = C proved
Q4. Show that the following four conditions are equivalent :
(i) A ⊂ B
(ii) A – B = φ
(iii) A ∪ B = B
(iv) A ∩ B = A
Solution:
First we try to prove
A ⊂ B ⇔ A - B = φ
Given: A ⊂ B
To prove: A - B = φ
A ⊂ B ⇒ x ∈ A and x ∈ B
then (A ∪ B) = B
(A ∪ B) ∩ B = (B ∩ B)
(A ∩ B ) ∪ (B ∩ B) = (B ∩ B)
(A ∩ B ) ∪ B = B
then (A ∩ B ) = B .................................(i)
Now also, x ∈ A ∩ B
which gives A ∩ B = A .................. (ii) [∵ A ⊂ B ]
L H S = A - B
= A - (A ∩ B ) from eqa .... (i)
= A - A using equa .. (ii)
= φ
∴ A - B = φ proved
(ii) ⇔ (i)
Given : A - B = φ
To prove : A ⊂ B
Solution:
x ∈ A - B ⇒ x ∈ A and x ∉ B
⇒ x ∈ A and x ∈ B'
⇒ A ∩ B' Here x is the element of A and B' both
∴ A ∩ B' = φ [∵ A - B = φ ] given
Again,
Let x ∈ A ⇒ x ∉ B' [∵ A ∩ B' = φ ]
There is no element that is common (A ∩ B') resulting φ.
⇒ x ∈ B
So, x ∈ A ⇒ x ∈ B
∴ A ⊂ B Proved
(i) ⇔ (iii)
Given : A ⊂ B
To prove : A ∪ B = B
Solution:
x ∈ A ⊂ B ⇔ x ∈ A and x ∈ B
clearly, A ∪ B = B All elements of A is in set B. [∵ A ⊂ B ]
(i) ⇔ (iv)
Given : A ⊂ B
To prove : A ∩ B = A
Solution:
x ∈ A ⊂ B ⇔ x ∈ A and x ∈ B
then x ∈ A ∩ B [∵ x is element of set A as well as set B ]
Clearly, which gives A ∩ B = A
Hence Proved
Q5. Show that if A ⊂ B, then C – B ⊂ C – A.
Solution:
Given: A ⊂ B
To show: C – B ⊂ C – A.
x ∈ C – B ⇒ x ∈ C and x ∉ B ................................... (i)
x ∈ C – A ⇒ x ∈ C and x ∉ A ................................... (ii)
while A ⊂ B
then A - B = φ
from equa. (i) and (ii)
x ∈ C – B
x ∈ C – A
∴ C – B ⊂ C – A Proved
Q6. Assume that P ( A ) = P ( B ). Show that A = B
Solution:
Let x is any element of set A and it is an also element of its subset say X
such that x ∈ X
then
X ⊂ A ⇒ X ∈ P(A)
⇒ X ∈ P(B) [∵ P(A) = P(B) ]
⇒ X ⊂ B
⇒ x ∈ B
x ∈ A and x ∈ B
A ⊂ B .............................. (i)
similarily,
y is any elemet of set B and it is an also element of its subset say Y such that y ∈ Y
then
Y ⊂ B ⇒ Y ∈ P(B)
⇒ Y ∈ P(A)
Y ⊂ B ⇒ y ∈ Y and y ∈ A
B ⊂ A .................................................. (ii)
From equation (i) and (ii)
A = B Proved
Q7. Is it true that for any sets A and B, P (A) ∪ P (B) = P (A ∪ B)? Justify your answer.
Solution : False Let A = {0, 1} and B = {1, 2}
∴ A ∪ B = {0, 1, 2}
P(A) = {Φ, {0}, {1}, {0, 1}}
P(B) = {Φ, {1}, {2}, {1, 2}}
P(A ∪ B) = {Φ, {0}, {1}, {2}, {0, 1}, {1, 2}, {0, 2}, {0, 1, 2}}
P(A) ∪ P(B) = {Φ, {0}, {1}, {0, 1}, {2}, {1, 2}}
∴ P(A) ∪ P(B) ≠ P(A ∪ B)
Q8. Show that for any sets A and B, A = (A ∩ B) ∪ (A – B) and A ∪ (B – A) = (A ∪ B)
Solution :
To show: A = (A ∩ B) ∪ (A – B)
Let x ∈ A
We have to show that x ∈ (A ∩ B) ∪ (A – B)
Case I x ∈ A ∩ B
Then, x ∈ (A ∩ B) ⊂ (A ∪ B) ∪ (A – B)
Case II x ∉ A ∩ B ⇒ x ∉ A or x ∉ B
∴ x ∉ B [x ∉ A] ∴ x ∉ A – B ⊂ (A ∪ B) ∪ (A – B)
∴ A ⊂ (A ∩ B) ∪ (A – B) … (1)
It is clear that A ∩ B ⊂ A and (A – B) ⊂ A
∴ (A ∩ B) ∪ (A – B) ⊂ A … (2)
From (1) and (2), we obtain
A = (A ∩ B) ∪ (A – B)
To prove: A ∪ (B – A) ⊂ A ∪ B
Let x ∈ A ∪ (B – A)
⇒ x ∈ A or x ∈ (B – A)
⇒ x ∈ A or (x ∈ B and x ∉ A)
⇒ (x ∈ A or x ∈ B) and (x ∈ A or x ∉ A)
⇒ x ∈ (A ∪ B)
∴ A ∪ (B – A) ⊂ (A ∪ B) … (3)
Next, we show that (A ∪ B) ⊂ A ∪ (B – A).
Let y ∈ A ∪ B ⇒ y ∈ A or y ∈ B
⇒ (y ∈ A or y ∈ B) and (y ∈ A or y ∉ A)
⇒ y ∈ A or (y ∈ B and y ∉ A)
⇒ y ∈ A ∪ (B – A)
∴ A ∪ B ⊂ A ∪ (B – A) … (4)
Hence, from (3) and (4), we obtain A ∪ (B – A) = A ∪B.
Q9. Using properties of sets show that (i) A ∪ (A ∩ B) = A (ii) A ∩ (A ∪ B) = A.
Solution : (i) To show: A ∪ (A ∩ B) = A
We know that A ⊂ A A ∩ B ⊂ A
∴ A ∪ (A ∩ B) ⊂ A … (1)
Also, A ⊂ A ∪ (A ∩ B) … (2)
∴ From (1) and (2), A ∪ (A ∩ B) = A
(ii) To show: A ∩ (A ∪ B) = A
A ∩ (A ∪ B) = (A ∩ A) ∪ (A ∩ B)
= A ∪ (A ∩ B)
Q10. Show that A ∩ B = A ∩ C need not imply B = C.
Solution : Let A = {0, 1}, B = {0, 2, 3}, and C = {0, 4, 5}
Thus, A ∩ B = {0} and A ∩ C = {0}
Now, A ∩ B = A ∩ C = {0}
However, B ≠ C [2 ∈ B and 2 ∉ C]
Q11. 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. (Hints A = A ∩ (A ∪ X), B = B ∩ (B ∪ X) and use distributive law)
Solution:
Let A and B be two sets such that A ∩ X = B ∩ X = f
and A ∪ X = B ∪ X for some set X.
To show: A = B
Here A = A ∩ (A ∪ X) = A ∩ (B ∪ X) [A ∪ X = B ∪ X]
= (A ∩ B) ∪ (A ∩ X) [Distributive law]
= (A ∩ B) ∪ Φ [A ∩ X = Φ] = A ∩ B ….. (1)
Now, B = B ∩ (B ∪ X)
= B ∩ (A ∪ X) [A ∪ X = B ∪ X]
= (B ∩ A) ∪ (B ∩ X) [Distributive law]
= (B ∩ A) ∪ Φ [B ∩ X = Φ]
= B ∩ A = A ∩ B …… (2)
Hence, from (1) and (2), we have that A = B.
Q12. 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:
Let A = {0, 1}, B = {1, 2}, and C = {2, 0}.
Thus, A ∩ B = {1}, B ∩ C = {2}, and A ∩ C = {0}.
∴ A ∩ B, B ∩ C, and A ∩ C are non-empty.
However, A ∩ B ∩ C = Φ
Q13. 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?
Solution:
Let U be the set of all students who took part in the survey.
Let T be the set of students taking tea.
Let C be the set of students taking coffee.
Then n(U) = 600, n(T) = 150, n(C) = 225, n(T ∩ C) = 100
To find: Number of student taking neither tea nor coffee.
Therefore, we have to find n(T' ∩ C').
n(T' ∩ C') = n(T ∪ C)'
= n(U) – n(T ∪ C)
= n(U) – [n(T) + n(C) – n(T ∩ C)]
= 600 – [150 + 225 – 100]
= 600 – 275
= 325
Hence, 325 students were taking neither tea nor coffee.
Q14. 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?
Solution:
Let U be the set of all students in the group.
Let E be the set of all students who know English.
Let H be the set of all students who know Hindi.
∴ H ∪ E = U
Then, n(H) = 100 and n(E) = 50
n( H U E ) = n(H) + n(E) – n(H ∩ E)
= 100 + 50 – 25
= 125
Hence, there are 125 students in the group.
Q15. 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 read all three newspapers. Find:
(i) the number of people who read at least one of the newspapers.
(ii) the number of people who read exactly one newspaper.
Solution:
Let A be the set of people who read newspaper H.
Let B be the set of people who read newspaper T.
Let C be the set of people who read newspaper I.
Then,
n(A) = 25, n(B) = 26, n(C) = 26 , n(A ∩ C) = 9, n(A ∩ B) = 11,
n(B ∩ C) = 8 , n(A ∩ B ∩ C) = 3
Let U be the set of people who took part in the survey.
(i) n(A U B U C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(C ∩ A) + n(A ∩ B ∩ C)
= 25 + 26 + 26 – 11 – 8 – 9 + 3 = 52
Hence, 52 people read at least one of the newspapers.
(ii) Let a be the number of people who read newspapers H and T only.
Let b denote the number of people who read newspapers I and H only.
Let c denote the number of people who read newspapers T and I only.
Let d denote the number of people who read all three newspapers.
Accordingly, d = n(A ∩ B ∩ C) = 3
Now, n(A ∩ B) = a + d
n(B ∩ C) = c + d
n(C ∩ A) = b + d
∴ a + d + c + d + b + d = 11 + 8 + 9 = 28
Q16. 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 products 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.
Solution:
Let A, B, and C be the set of people who like product A, product B, and product C respectively.
Then, n(A) = 21,
n(B) = 26,
n(C) = 29,
n(A ∩ B) = 14,
n(C ∩ A) = 12,
n(B ∩ C) = 14,
n(A ∩ B ∩ C) = 8
The Venn diagram for the given problem-
It can be seen that number of people who like product C only is
{29 – (4 + 8 + 6)} = 11
Select Class for NCERT Books Solutions
NCERT Solutions
sponder's Ads
