RS AGGARWAL CLASS 11 MATHS SOLUTION SET THEORY
RS AGGARWAL CLASS 11 MATHS FREE SOLUTION SET THEORY (EXERCISE-1A, 1B, 1C)
R. S. AGGARWAL
CHAPTER 1 -SETS
Exercise 1A
1. Which of the following are sets? Justify your answer.
(i) The collection of all whole numbers less than 10.
Answer: One can definitely identify a number that belongs to this collection. So The collection of all whole numbers less than 10 is a well-defined collection. Hence, this collection is a set.
(ii) The collection of good hockey players in India.
Answer: Good hockey players in India may vary from person to person. So, it is not well-defined. This collection is not a set
(iii) The collection of all questions in this chapter.
Answer: The collection of all questions in this chapter is known. So this collection is a well-defined. Hence, the collection is a set.
(iv) The collection of all difficult chapters in this book.
Answer: Difficult chapters may vary from student to student. So the collection of all questions in this chapter is not a well-defined. Hence, this collection is not a set.
(v) A collection of Hindi novels written by Munshi Prem Chand.
Answer: A collection of Hindi novels written by Munshi Prem Chand is a well-defined collection, because one can definitely identify a book which written by Munshi Prem Chand. Hence, this collection is a set.
(vi) A team of 11 best cricket players of India.
Answer: The criteria for determining the best cricket players of India can vary from person to person. So the team is not a well-defined collection Hence, this collection is not a set
(vii) The collection of all the months of the year whose names begin with the letter M.
Answer: One can definitely identify a month that begin with the letter M. So The collection of all months of a year begin with the letter M is a well-defined. Hence, this collection is a set.
(viii) The collection of all interesting books.
Answer: The collection of all interesting books because it may vary from person to person. So the collection is not a well-defined collection Hence, this collection is not a set.
(ix)The collection of all short boys of your class.
Answer: As there is no specific height mentioned to guess whether the boy in your class is short or not. Here, the set is not well defined. Hence, this collection is not a set.
(x) The collection of all those students of your class whose ages exceed 15 years.
Answer: You can definitely identify a boy of your class whose ages exceed 15 years belongs to this collection. So this collection is a well-defined collection. Hence, this collection is a set.
(xi) The collection of all rich persons of Kolkata.
Answer: As the collection of all rich persons of Kolkata may vary from person to person. So the collection is not a well-defined collection Hence, this collection is not a set.
(xii) The collection of all persons of Kolkata whose assessed annual incomes exceed (say) ₹ 20 lakh in the financial year 2016-17.
Answer: The collection of all persons of Kolkata whose assessed annual incomes exceed (say) ₹ 20 lakh in the financial year 2016-17 is a well-defined collection. Hence, this collection is a set.
(xiii) The collection of all interesting dramas written by Shakespeare.
Answer: The collection of all interesting drams written by Shakespeare is not a well-defined collection because the criteria to determine a drama which is interesting may vary from person to person. Hence, this collection is not a set.
RS AGGARWAL CLASS 11 MATHS SOLUTION SET THEORY
Q. 2. Let A be the set of all even whole numbers less than 10.
(i) Write A in the roster from.
(ii) Fill in the blanks with the approximate symbol ∉ or ∈ :
(a) 0 ……. A
(b) 10 ……. A
(c) 3 ……. A
(d) 6 ……. A
Answer : (i) A = {0, 2, 4, 6, 8}
(a) 0 ∈ A
(b) 10 ∉ A
(c) 3 ∉ A
(d) 6 ∈ A
3. Write the following sets in roster form:
(i) A = {x: x is a natural number, 30 ≤ x < 36}
Answer: A = {30, 31, 32, 33, 34, 35}
3. Write the following sets in roster form:
(ii) B = {x : x is an integer and –4 < x < 6}.
Answer : B = {-3, -2, -1, 0, 1, 2, 3, 4, 5}
3 C. Write the following sets in roster from:
C = {x : x is a two-digit number such that the sum of its digits is 9}.
Answer: C = {18, 27, 36, 45, 54, 63, 72, 81, 90}
3 D. Write the following sets in roster from:
D = {x : x is an integer, x2 ≤ 9}.
Answer: D = {-3, -2, -1, 0, 1, 2, 3}
3 E. Write the following sets in roster from:
E = {x : x is a prime number, which is a divisor of 42}.
Answer: E = {2, 3, 7}
3 F. Write the following sets in roster from:
F = {x : x is a letter in the word’ MATHEMATICS’}.
Answer : F = {M, A, T, H, E, I, C, S}
3 G. Write the following sets in roster from:
G = {x : x is a prime number and 80 < x < 100}.
Answer: Prime numbers between 80 and 100 are 83, 89, 97.
G = {83, 89, 97}
[Prime number = Those number which is divisible by 1 and the number itself.]
3 H. Write the following sets in roster from:
H = {x : x is a perfect square and x < 50}.
Answer: H = {1, 4, 9, 16, 25, 36, 49}
3 I. Write the following sets in roster from:
J = {x : ∈ R and x2 + x – 12 = 0}.
Answer: x2 + x – 12 = 0
⇒ x2 + 4x – 3x – 12 = 0
⇒ x(x + 4) – 3(x + 4) = 0
or (x – 3)(x + 4) = 0
⇒ x – 3 = 0 or x + 4 = 0
⇒ x = 3 or x = -4
J = {-4, 3}
3 J. Write the following sets in roster from:
K = {x : ∈ N, x is a multiple of 5 and x2 < 400}.
Answer : Multiple of 5 are 5, 10, 15, 20, 25, …
So, 52 = 25; 102 = 100;
152 = 225; 202 = 400
252 = 625 > 400
K = {5, 10, 15}
RS AGGARWAL CLASS 11 MATHS SOLUTION SET THEORY
4 A. List all the elements of each of the sets given below.
A = {x : x = 2n, n ∈ N and n ≤ 5}.
Answer: Given: x = 2n and n ≤ 5
⇒ n = 1, 2, 3, 4 and 5
∴ A = {2, 4, 6, 8, 10}
4 B. List all the elements of each of the sets given below.
B = {x : x = 2n + 1, n ∈ W and n < 5}.
Answer: Given: x = 2n + 1 and n < 5
⇒ n = 0, 1, 2, 3 and 4
∴ B = {1, 3, 5, 7, 9}
4 C. List all the elements of each of the sets given below.
C= {x: x=1/n, n ∈ N and n < 6}
Answer: Given x = 1/n, n ∈ N and n < 6
⇒ n = 1, 2, 3, 4 and 5
∴ C = {1, 1/2, 1/3, 1/4, 1/5}
4 D. List all the elements of each of the sets given below.
D = {x : x = n2, n ∈ N and 2 ≤ n ≤ 5}.
Answer: Given x = n2 and 2 ≤ n ≤ 5
∴ n = 2, 3, 4, 5
∴ D = {4, 9, 16, 25}
4 E. List all the elements of each of the sets given below.
E = {x : x ∈ Z and x2 = x}.
Answer: Given: x ∈ Z and x2 = x
(0)2 = 0, (1)2 = 1
∴ E = {0, 1}
4 F. List all the elements of each of the sets given below.
F = {x : x = x ∈ Z and –1/2 < x < 13/2}.
Answer: Given x ∈ Z and –1/2 < x < 13/2
or -0.5 < x < 6.5
∴ F = {0, 1, 2, 3, 4, 5, 6}
4 G. List all the elements of each of the sets given below.
G = {x : x = 1/2n-1, x ∈ N and 1 ≤ n ≤ 5}.
Answer : Given: x ∈ N and 1 ≤ n ≤ 5
So, n = 1, 2, 3, 4, 5
∴ G = {1, 1/3, 1/5, 1/7, 1/9}
4 H. List all the elements of each of the sets given below.
H = {x : x ∈ Z, |x| ≤ 2}.
Answer : Given x ∈ Z and |x| ≤ 2
⇒ x ∈ Z and -2 ≤ x ≤ 2
∴ x = -2, -1, 0, 1, 2
∴ H = {-2, -1, 0, 1, 2}
5. Write each of the sets given below in set builder from:
(i) A = {1, 1/4, 1/9, 1/16, 1/25, 1/36, 1/49}
Answer :Hence, A = {1, 1/4, 1/9, 1/16, 1/25, 1/36, 1/49}
= {1/12, 1/22, 1/32, 1/42, 1/52, 1/62, 1/72}
Set builder from of A:
A = {x: x= 1/n2, x ∈ N and 1 ≤ n ≤ 7}
RS AGGARWAL CLASS 11 MATHS SOLUTION SET THEORY
5. Write each of the sets given below in set builder from:
(ii) B = {1/2, 2/5, 3/10, 4/17, 5/26, 6/37, 7/50}
Answer: Hence, B = {1/2, 2/5, 3/10, 4/17, 5/26, 6/37, 7/50}
B = {1/12+1, 2/22+1, 3/32+1, 4/42+1, 5/52+1, 6/62+1, 7/72+1}
Set builder from of B = {x: x= n/n2+1, x ∈ N and 1 ≤ n ≤ 7}
5. Write each of the sets given below in set builder from:
(iii) C = {53, 59, 61, 67, 71, 73, 79}.
Answer: C = {53, 59, 61, 67, 71, 73, 79}
Clearly the element of C are prime numbers.
So C = {x: x is a prime number and 50 < x < 80}
[Prime Number: prime numbers are those numbers which are divisible by 1 and the number itself.]
5. Write each of the sets given below in set builder from:
(iv) D = {–1, 1}.
Answer: Here, D = {–1, 1}
The given set can be write as:
D = {x: x= |x|}
5. Write each of the sets given below in set builder from:
(v) E = {14, 21, 28, 35, 42, …., 98}.
Answer : E = {14, 21, 28, 35, 42, …., 98}
E = {2×7, 3×7, 4×7, 5×7, 6×7, …., 14×7}
The given set can be write as:
E = {x: x = 7n, x ∈ N and 2 ≤ n ≤ 14 }
6. Match each of the sets on the left described in the roster from with the same set on the right described in the set-builder from:
(i) {–5, 5} (a) {x : x ϵ Z and x2 < 16}
(ii) {1, 2, 3, 6, 9, 18} (b) {x : x ϵ N and x2 = x}
(iii) {–3, –2, –1, 0, 1, 2, 3} (c) {x : x = ϵ Z and x2 =25}
(iv) {P, R, I, N, C, A, L} (d) {x : x ϵ N and x is a factor of 18}
(v) {1} (e) {x : x is a letter in the word ‘PRINCIPAL’}
Answer : (i) ⇒ (c)
[x2 = 25 ∴ x = ± 5 ∴ {-5, 5}]
(ii) ⇒ (d)
[ Divisor of 18 are 1, 2, 3, 6, 9, 18
∴ {x : x ϵ N and x is a factor of 18}
(iii) ⇒ (a)
[{–3, –2, –1, 0, 1, 2, 3}
(-3)2 = 9 < 16; (-2)2 = 4 < 16;
(-1)2 = 1 < 16; (-0)2 = 0 < 16;
(1)2 = 1 < 16; (2)2 = 4 < 16;
(3)2 = 9 < 16]
(iv) ⇒ (e)
(v) ⇒ (b)
[∵ (1)2 = 1]RS AGGARWAL CLASS 11 MATHS FREE SOLUTION SET THEORY (EXERCISE-1A,1B,1C,1D)
R. S. AGGARWAL
CHAPTER 1 -SETS
Click here to visit our Facebook
Exercise 1B
Q. 1(i) Which of the following are examples of the null set?
Set of odd natural numbers divisible by 2.
Answer: Odd numbers are not divisible by 2.
∴ No elements in this set.
It is a null set.
Therefore It is not a null set.
Q. 1 (iii) Which of the following are examples of the null set?
A = {x : x ∈ N, 1 < x ≤ 2}.
Answer: A = {x : x ∈ N, 1 < x ≤ 2}.
A = {2}
∴ It is not a null set.
Q. 1 (iv) Which of the following are examples of the null set?
B = {x : x ∈ N, 2x + 3 = 4}.
Answer : Given 2x + 3 = 4
⇒ 2x = 1
⇒ x = 1/2
1/2 ∉ N
∴ No elements in the set B.
∴ It is a null set.
Q. 1 (v) Which of the following are examples of the null set?
C = {x : x is prime, 90 < x < 96}.
Answer: There are not any prime numbers between 90 and 96
∴ The set is empty
∴ It is a null set.
Q. 1 (vi) Which of the following are examples of the null set?
D = {x : x ∈ N, x2 + 1 = 0}.
Answer: x2 + 1 = 0
⇒ x2 = -1
⇒ x = ±√-1
√-1 is an imaginary number.
∴ No elements in the set F.
∴ It is a null set.
Q. 1 (vii) Which of the following are examples of the null set?
E = {x : x ∈ W, x + 3 ≤ 3}.
Answer : Given: x + 3 ≤ 3
⇒ x ≤ 0
∴ x = 0 which is a whole number.
∴ 0 is the element of set E.
So It is not a null set.
Q. 1 (viii) Which of the following are examples of the null set?
F = {x : x ∈ Q, 1 < x < 2}.
Answer: Here, x ∈ Q i.e. x is a rational number.
There are infinitely rational numbers between any two distinct real numbers.
So, the set is not empty
Therefore It is not a null set.
Q. 1 (ix) Which of the following are examples of the null set?
G = {0}
Answer: G = {0}
∴ 0 ∈ G
∴ The set is not empty
G is not a null set
RS AGGARWAL CLASS 11 MATHS SOLUTION SET THEORY
Q. 2. Which of the following are examples of the singleton set?
(i) {x : x ∈ Z, x2 = 4}.
Answer: (i) Given equation:
x2 = 4
⇒ x = ±√4
⇒ x = ± 2
So there are two elements in a set.
∴ It is not a singleton set.
Q. 2. Which of the following are examples of the singleton set?
(ii) {x : x ∈ Z, x + 5 = 0}.
Answer: Given equations:
x + 5 = 0
⇒ x = – 5
So, there is only 1 element in a given set.
∴ It is a singleton set.
Q. 2. Which of the following are examples of the singleton set?
(iii) {x : x ∈ Z, |x| = 1}.
Answer: Given equation:
|x| = 1
If x < 0, then |x| = -x
∴ -x = 1 ⇒ x = -1
If x > 0, then |x| = x ∴ x = 1
So {x : x ∈ Z, |x| = 1} = {-1, 1}
∴ It is not a singleton set.
Q. 2. Which of the following are examples of the singleton set?
(iv) {x : x ∈ N, x2 = 16}.
Answer: Given equation:
x2= 16
⇒ x = ± 4
⇒ x = -4, 4
but -x ∉ N
So, there is only 1 element in a set.
∴ It is a singleton set.
Q. 2. Which of the following are examples of the singleton set?
(v) {x : x is an even prime number}
Answer: Even Prime number is only 2.
∴ It is a singleton set.
Q. 3 (i) Which of the following are pairs of equal sets?
A = set of letters in the word, ‘ALLOY.’
B = set of letters in the word, ‘LOYAL.’
[Equal Sets = Two sets A and B are said to be equal if they have exactly the same elements.]
Answer: A = set of letters in the word, ALLOY
= {A, L, O, Y}
B = set of letters in the word, LOYAL
= {L, O, Y, A}
Every element of A = Every element of B and also every element of B = Every element of A.
Thus, A and B are equal sets.
RS AGGARWAL CLASS 11 MATHS SOLUTION SET THEORY
Q. 3 (ii) Which of the following are pairs of equal sets?
C = set of letters in the word, ‘CATARACT.’
D = set of letters in the word, ‘TRACT.’
Answer :
C = set of letters in the word, ‘CATARACT.’
= {C, A, T, R}
D = set of letters in the word, ‘TRACT.’
= {T, R, A, C}
Every element of C = Every element of D and also every element of D = Every element of C.
Thus, C and D are equal sets.
Q. 3 (iii) Which of the following are pairs of equal sets?
E = {x : x ∈ Z, x2 ≤ 4}
F = {x : x ∈ Z, x2 = 4}.
Answer : E = {x : x ∈ Z, x2 ≤ 4}
= { -2, -1, 0, 1, 2}
F = {x : x ∈ Z, x2= 4}
= {-2, 2}
∴ E ≠ F
Therefore E and F are not equal sets.
Q. 3 (iv) Which of the following are pairs of equal sets?
G = {–1, 1}
H = {x : x ∈ Z, x2 – 1 = 0}.
Answer: G = {-1, 1} and
H = {x : x ∈ Z, x2 – 1 = 0}
Here, x ∈ Z and x2 – 1 = 0
Given equation:
x2 – 1 = 0
⇒ x2 = 1
⇒ x = ± 1
Therefore x = -1 and 1
∴ H = {-1, 1}
⇒ G = H
∴ G and H are equal sets.
Q. 3 (v) Which of the following are pairs of equal sets?
J = {2, 3} K = {x : x ∈ Z, (x2 + 5x + 6) = 0}
Answer: J = {2, 3}
K = {x : x ∈ Z, (x2 + 5x + 6) = 0}
Given equation :
x2 + 5x + 6 = 0
⇒ x2 + 2x + 3x + 6 = 0
⇒ x(x + 2) + 3(x + 2) = 0
or (x + 2)(x + 3) = 0
⇒ x = -2 and -3
∴ K = {-2, -3}
∴ J ≠ K
So J and K are not equal sets.
RS AGGARWAL CLASS 11 MATHS SOLUTION SET THEORY
Q. 4. Which of the following are pairs of equivalent sets?
(i) A = {–1, –2, 0} and
B = { 1, 2, 3,}
[Equivalent sets are sets that contain the same number of elements, although the elements themselves may be different i.e. Equivalent sets have the same cardinality]
Answer :
A = {–1, –2, 0} and
B = {1, 2, 3,}
Both have 3 elements.
∴ A and B are equivalent sets.
(ii) C = {x : x ∈ N, x < 3} and
D ={x : x ∈ W, x < 3}
Answer :
C = {x : x ∈ N, x < 3} = {1, 2} and
D ={x : x ∈ W, x < 3} = {0, 1, 2}
∴ C and D are not equivalent sets because their cardinality is not same.
(iii) E = {a, e, i, o, u} and
F = {p, q, r, s, t}
Answer :
E = {a, e, i, o, u} and
F = {p, q, r, s, t}
∴ n(E) = n(F) = 5
E and F are equivalent sets.
RS AGGARWAL CLASS 11 MATHS SOLUTION SET THEORY
Q. 5 (i) State whether any given set is finite or infinite:
A = Set of all triangles in a plane.
Answer: in a plane there is an infinite number of triangles.
The set of all triangles in a plane is an infinite set.
Q. 5 (ii) State whether any given set is finite or infinite:
B = Set of all points on the circumference of a circle.
Answer: There are infinite numbers of points on the circumference of the circle.
So the set will have infinite elements.
∴ the given set is infinite.
Q. 5 (iii) State whether any given set is finite or infinite:
C = set of all lines parallel to the y-axis
Answer: There are infinite lines parallel to y-axis,
so the set will have infinite elements.
∴ the given set is infinite.
Q. 5 (iv) State whether any given set is finite or infinite:
D = set of all leaves on a tree.
Answer: A tree has a limited number of leaves, which can be counted.
Therefore, the set of all leaves on a tree is a finite set.
Q. 5 (v) State whether any given set is finite or infinite:
E = set of all positive integers greater than 500
Answer: Positive Integers greater than 500
= 501, 502, 503, …
There are infinite positive integers which are greater than 500.
So the given set is infinite.
Q. 5 (vi) State whether any given set is finite or infinite:
F = {x ∈ R: 0 < x < 1].
Answer: There are infinitely real numbers between 0 and 1.
So the given set is infinite.
Q. 5 (vii) State whether any given set is finite or infinite:
G = {x ∈ Z: x < 1].
Answer: Integers less than 1
= …… -4, -3, -2, -1, 0 There are infinite integers less than 1.
∴ The given set is infinite.
Q. 5 (viii) State whether any given set is finite or infinite:
H = {x ∈ Z: –15 < x < 15].
Answer: H = {x ∈ Z: –15 < x < 15]
= {-3, -2, -1, 0, 1, 2, 3, …..}
The integers lies between -15 and 15 are finite.
∴ The given set is finite.
Q. 5 (ix) State whether any given set is finite or infinite:
j = {x : x ∈ N and x is prime}.
Answer : The set of prime numbers are infinite.
Hence, the given set is infinite.
Q. 5 (x) State whether any given set is finite or infinite:
K = {x : x ∈ N and x is odd}.
Answer: The set of odd natural numbers are infinite.
Hence, the given set is infinite.
Q. 5 (xi) State whether any given set is finite or infinite:
K = set of all circles passing through the origin (0, 0)
Answer: Infinite number of circles can pass through the origin, so the set will have infinite elements.
So the given set is infinite
RS AGGARWAL CLASS 11 MATHS SOLUTION SET THEORY
Q. 6. Rewrite the following statements using the set notation:
(i) a is an element of set A.
Answer: a ∈ A
(ii) b is not an element of A.
Answer: b ∉ A
(iii) A is an empty set and B is a nonempty set.
Answer: A = ɸ and B ≠ ɸ
(iv) A number of elements in A is 6.
Answer: n(A) = 6
(v) 0 is a whole number but not a natural number.
Answer: 0 ∈ W but 0 ∉ N
RS AGGARWAL CLASS 11 MATHS FREE SOLUTION SET THEORY (EXERCISE-1A,1B,1C,1D)
R. S. AGGARWAL
CHAPTER 1 -SETS
Exercise 1C
Q. 1 (i) State in each case whether A ⊂ B or A ⊄ B.
A = {0, 1, 2, 3}, B = {1, 2, 3, 4, 5}
Answer: A ⊄ B
[Since 0 ∈ A and 0 ∉ B.]
Q. 1 (ii) State in each case whether A ⊂ B or A ⊄ B.
A = ϕ, B = {0}
Answer: A ⊂ B
[A is a null set.
Null set is a subset of every set.
∴ A ⊂ B.]
Q. 1 (iii) State in each case whether A ⊂ B or A ⊄ B.
A = {1, 2, 3}, B = {1, 2, 4}
Answer: A ⊄ B
[Since 3 ∈ A and 3 ∉ B.]
Q. 1 (iv) State in each case whether A ⊂ B or A ⊄ B.
A = {x : x ∈ Z, x2 = 1},
B = {x : x ∈ N, x2 = 1}
Answer: A ⊄ B
[ Given x2 = 1
⇒ x = ± 1
-1 ∈ Z and -1 ∉ N
∴ A = { -1, 1} and
B = {1}
∴ A ⊄ B .]
Q. 1 (v) State in each case whether A ⊂ B or A ⊄ B.
A = {x : x is an even natural number,},
B = {x : x is an integer}
Answer: A ⊂ B
[ A = {2, 4, 6, 8,…} and
B = {. . ., –3, –2, –1, 0, 1, 2, 3, . . .}.
since, even natural numbers are also integers.
∴ A ⊂ B.]
Q. 1 (vi) State in each case whether A ⊂ B or A ⊄ B.
A = {x : x is an integer},
B = {x: x is a rational number}
Answer: A = {…. -3, -2, -1, 0, 1, 2, 3….}
B = {-∞, ……..0, ……∞ }
All integers are contained in rational numbers.
∴ A ⊂ B
Q. 1 (vii) State in each case whether A ⊂ B or A ⊄ B.
A = {x : x is a real number,},
B = {x : x is a complex number}
Answer: A ⊂ B
[ A = set of real numbers and
B = set of complex numbers.
Real number can be written as a complex number in the form of a+ib, where a and b are real, and i is imaginary.
So set of real numbers is a subset of set of complex numbers.]
Q. 1 (viii) State in each case whether A ⊂ B or A ⊄ B.
A = {x : x is an isosceles triangle in the plane},
B = {x : x is an equilateral triangle in the same plane}
Answer: A ⊄ B
[All isosceles triangles are not equilateral triangles.
Therefore set of isosceles triangle is not contained in the set of equilateral triangle.]
Q. 1 (ix) State in each case whether A ⊂ B or A ⊄ B.
A = {x : x is a square in a plane},
B = {x : x is a rectangle in the same plane}
Answer: A ⊂ B
[Since all squares are rectangles.
So set of squares is a subset of set of rectangles.]
Q. 1 (x) State in each case whether A ⊂ B or A ⊄ B.
A = {x : x is a triangle in a plane},
B = {x : x is a rectangle in the same plane}\
Answer: A ⊄ B
[ A = set of triangles and
B = set of rectangles.Since set of triangles does not include a set of rectangles.]
Q. 1 (xi) State in each case whether A ⊂ B or A ⊄ B.
A = {x : x is an even natural number less than 8},
B = {x : x is a natural number which divides 32}
Answer: A ⊄ B
[ A = {2,4,6} and
B = {1,2,4,8,16,32}.
6 ∈ A but 6 ∉ B.
∴ A ⊄ B]
RS AGGARWAL CLASS 11 MATHS SOLUTION SET THEORY
Q. 2. Examine whether the following statements are true of false:
(i) {a, b} ⊄ {b, c, a}
Answer: False
[Since elements of {a,b} is not an element of {b,c,a}.
It is a subset of {b, c, a}.
So {a, b} ⊂ {b, c, a}]
Q. 2. Examine whether the following statements are true of false:
(ii) {a} ∈ {a, b, c}
Answer: False
[{a} is not an element of {a, b, c}.
It is a subset of {a, b, c}.
So {a} ⊂ {a, b, c}]
Q. 2. Examine whether the following statements are true of false:
(iii) φ ⊂ {a, b, c}
Answer: True
[ Null set is a subset of every set.
φ is a null set.
So φ ⊂ {a, b, c}]
Q. 2. Examine whether the following statements are true of false:
(iv) {a, e} ⊂ {x : x is a vowel in the English alphabet}
Answer: True
[a, e are vowels of the English alphabet.
So {a, e} is the subset of the given set.]
Q. 2. Examine whether the following statements are true of false:
(v) {x : x ∈ W, x + 5 = 5} = φ
Answer: False
[ x + 5 = 5
⇒ x = 0 and 0 ∈ W
Hence, {0} ≠ φ
Therefore it is mot a null set,]
Q. 2. Examine whether the following statements are true of false:
(vi) a ∈ {{a}, b}
Answer: False
[ a is not an element of {{a}, b},
So a ∉ {{a}, b}]
Q. 2. Examine whether the following statements are true of false:
(vii) {a} ⊂ {{a}, b}
Answer: False
[{a} is an element of set {{a}, b},
So {a} ∈ {{a}, b} but {a} ⊄ {{a}, b}]
Q. 2. Examine whether the following statements are true of false:
(viii) {b, c} ⊂ {a, {b, c}}
Answer: False
[{b,c} is an element of {a, {b, c}} and an element cannot be a subset of a set.
So {b, c} ⊄ {a, {b, c}}]
Q. 2. Examine whether the following statements are true of false:
(ix) {a, a, b, b} = {a, b}
Answer: True
[ Repetition of elements in a set does not change a set.]
Q. 2. Examine whether the following statements are true of false:
(x) {a, b, a, b, a, b, ….} is an infinite set.
Answer: False
[In a set all the elements are unique.
So the given set will be {a,b} and it is a finite set.]
Q. 2. Examine whether the following statements are true of false:
(xi) If A = set of all circles of unit radius in a plane and B = set of all circles in the same plane then A⊂B.
Answer: True
[Circle in a plane with unit radius is a subset of circle in a plane of any Radius.]
RS AGGARWAL CLASS 11 MATHS SOLUTION SET THEORY
Q. 3. If A = {1} and B = {{1}, 2} then show that A ⊄ B.
Answer: 1 ∈ A but ∉ B So A ⊄ B.
Q. 4. Write down all subsets of each of the following sets:
(i) A = {a}
Answer: The subsets of A are –
φ and {a}
Q. 4. Write down all subsets of each of the following sets:
(ii) B = {a, b}
Answer:
The subsets of B are –
φ, {a}, {b}, and {a, b}
Q. 4. Write down all subsets of each of the following sets:
(iii) C = {–2, 3}
Answer: The subsets of C are –
φ, {-2}, {3}, and {-2, 3}
Q. 4. Write down all subsets of each of the following sets:
(iv) D = {–1, 0, 1}
Answer: The subsets of D are –
φ ,{-1}, {0}, {1}, {-1,0}, {0,1}, {1,-1}, {–1, 0, 1}
Q. 4. Write down all subsets of each of the following sets:
(v) E = φ
Answer: The subsets of E is φ
Q. 4. Write down all subsets of each of the following sets:
(vi) F = {2, {3}}
Answer: The subsets of F are –
φ , {2}, {{3}}, {2,{3}}
Q. 4. Write down all subsets of each of the following sets:
(vii) G = {3, 4, {5, 6}}
Answer: The subsets of G are –
φ , {3}, {4}, {{5,6}}, {3.4}. {4,{5,6}}, {{5,6}, 3}, {3,4,{5,6}}
RS AGGARWAL CLASS 11 MATHS SOLUTION SET THEORY
Q. 5. Express each of the following sets as an interval:
(i) A = {x : x ∈ R, –4 < x < 0}
(ii) B = {x : x ∈ R, 0 ≤ x < 3}
(iii) C = {x : x ∈ R, 2 < x ≤ 6}
[▶️ (a, b) is called an open interval i.e. all the points between a and b belong to this interval excluding a and b.
👉 [a, b] is a closed interval i.e. all the points between a and b belong to this interval including a and b.
▶️ (a, b] is a left side open or right side closed interval i.e. all the points between a and b belong to this interval excluding a but including b.
👉 [a, b) is a left side closed or right side open interval i.e. all the points between a and b belong to this interval including a but excluding b.]
Answer:
(i) A = (-4, 0)
(ii) B = [0, 3)
(iii) C = (2, 6]
Q. 5. Express each of the following sets as an interval:
(iv) D = {x : x ∈ R, –5 ≤ x ≤ 2}
(v) E = {x : x ∈ R, –3 ≤ x < 2}
(vi) F = {x : x ∈ R, –2 ≤ x < 0}
Answer:
(iv) D = [-5, 2]
(v) E = [-3, 2)
(vi) F = [-2, 0)
Q. 6. Write each of the following intervals in the set-builder form:
(i) A = (–2, 3)
(ii) B = [4, 10]
(iii) C = [–1, 8)
Answer:
(i) A = {x : x ∈ R, –2 < x < 3}
(ii) B = {x : x ∈ R, 4 ≤ x ≤ 10}
(iii) C= {x : x ∈ R, –1 ≤ x < 8}
Q. 6. Write each of the following intervals in the set-builder form:
(iv) D = (4, 9]
(v) E = [–10, 0)
(vi) F = (0, 5]
Answer:
(iv) D = {x : x ∈ R, 4 < x ≤ 9}
(v) E = {x : x ∈ R, –10 ≤ x < 0}
(vi) F = {x : x ∈ R, 0 < x ≤ 5}
RS AGGARWAL CLASS 11 MATHS SOLUTION SET THEORY
Q. 7. if A = {3, {4, 5}, 6} find which of the following statements are true.
(i) {4, 5} ⊆ A
Answer: False
[{4,5} is an element of A.
{4,5} ∈ A but not a subset of A]
Q. 7. if A = {3, {4, 5}, 6} find which of the following statements are true.
(ii) {4, 5} ∈ A
Answer: (ii) True
[{4,5} is an element of A.
{4,5} ∈ A]
Q. 7. if A = {3, {4, 5}, 6} find which of the following statements are true.
(iii) {{4, 5}} ⊆ A
Answer: True
[{4,5} is an element of the set A.
So {{4,5}} is a subset of A
Hence, {{4, 5}} ⊆ A]
Q. 7. if A = {3, {4, 5}, 6} find which of the following statements are true.
(iv) 4 ∈ A
Answer: False
[4 is not an element of A.
So 4 ∉ A]
Q. 7. if A = {3, {4, 5}, 6} find which of the following statements are true.
(v) {3} ⊆ A
Answer: True
[3 is an element of the set A.
So {3} is a subset of A
Hence, {3} ⊆ A]
Q. 7. if A = {3, {4, 5}, 6} find which of the following statements are true.
(vi) {φ} ⊆ A
Answer: False
[φ is an element in { φ} but not in A.
So {φ} ⊄ A]
Q. 7. if A = {3, {4, 5}, 6} find which of the following statements are true.
(vii) φ ⊆ A
Answer: True
[φ is a null set.
Null set is a subset of every set.
So φ ⊆ A]
Q. 7. if A = {3, {4, 5}, 6} find which of the following statements are true.
(viii) {3, 4, 5} ⊆ A
Answer: False
[A = {3, {4, 5}, 6}
{4,5} ∈ A but 5 ∉ A.
So {3, 4, 5} ⊄ A]
Q. 7. if A = {3, {4, 5}, 6} find which of the following statements are true.
(ix) {3, 6} ⊆ A
Answer: True
[3,6 is in {3,6} and also in A,
Thus {3, 6} ⊆A]
RS AGGARWAL CLASS 11 MATHS SOLUTION SET THEORY
Q. 8. If A = {a, b, c}, find P(A) and n{P(A)}.
[P(A) is a power set of A which is the collection of all subsets of a set A]
Answer: φ, {a}, {b}, {c}, {a,b}, {b,c}, {c, a}, {a, b, c} are subsets of {a, b, c}
∴ P(A) = { φ , {a}, {b}, {c}, {a,b}, {b,c}, {c, a}, {a, b, c}}
Here n(A) = 3
∴ n{P(A)} = 2n = 23 = 8
Q. 9. If A = {1, {2, 3}}, find P(A) and n {P(A)}
Answer: A = {1, {2, 3}}
Subsets of A are φ , {1} , {2, 3} , {1, {2, 3}}
∴ P(A) = φ , {1} , {2,3} , {1, {2,3}}
Here n(A) = 2
∴ n{P(A)} = 2n = 22 = = 4
Q. 10. If A = φ then find n{P(A)}
Answer: A = φ , i.e. A is a null set.
So n(A)= 0
∴ n{P(A)} = 2n = 20 = 1.
Q. 11. If A = {1, 3, 5} B = {2, 4, 6} and C = {0, 2, 4, 6, 8} then find the universal set.
Answer: Elements of A + B + C
= {1, 3, 5, 2, 4, 6, 0, 8}
So the universal set of A, B and C
= {0, 1, 2, 3, 4, 5, 6, 8}
Q. 12. Prove that A ⊆ B, B ⊆ C and C ⊆ A ⇒ A = C.
Answer: Let x ∈ A x ∈ A ⇒ x ∈ B. . . [∵ A ⊆ B]
Again x ∈ B ⇒ x ∈ C. . . [∵ B ⊆ C]
and x ∈ C ⇒ x ∈ A. . . [∵ C ⊆ A]
Hence, A = C. – – (Proved)
Q. 13. For any set A, prove that A ⊆ φ ⇔ A = φ
Answer: Given A ⊆ φ
A is a subset of the null set ,
So A is also an empty set.
∴ A = φ
RS AGGARWAL CLASS 11 MATHS SOLUTION SET THEORY
Q. 14. State whether the given statement is true false:
(i) If A ⊂ B and x ∉ B then x ∉ A.
Answer: True
[Given A ⊂ B
A is a subset of B
So all elements of A should be in B.
Let A = {a, b} and B = {a, b, c}
Here d ∉ B so d ∉ A
Hence, If A ⊂ B and x ∉ B than x ∉ A.
Q. 14. State whether the given statement is true false:
(ii) If A ⊆ φ then A = φ
Answer: True
[Given A ⊆ φ
A is a subset of a null set , this implies A is also an empty set.
∴ A = φ]
Q. 14. State whether the given statement is true false:
(iii) If A, B and C are three sets such than A ∈ B and B ⊂ C then A ⊂ C.
Answer: False
[Let A = {1}, B = {{1}, 2} and C = {{1}, 2, 3}.
Here , {1} ∈ B i.e. A ∈ B
Since, {1}, 2 ∈ B so B ⊂ C.
Element of A is 1 but {1} i.e A is an element of C.
The element of a set cannot be a subset of a set.
Hence,A ⊄ C.]
(iv) If A, B and C are three sets such than A ⊂ B and B ∈ C then A ∈ C.
Answer: False
[Given A ⊂ B and B ∈ C.
Let A = {1},B = {1, 2} and C = {{1, 2}, 3}
{1} is not an element of C.
So A ∉ C ]
Q. 14. State whether the given statement is true false:
(v) If A, B and C are three sets such that A ⊄ B and B ⊄ C then A ⊄ C.
Answer: False.
[Given A ⊄ B and B ⊄ C
Let A = {1}, B = {2, 3} and C = {1, 3}.
Since 1 ∈ A and 1 ∈ C.Then,
∴ A ⊂ C]
Q. 14. State whether the given statement is true false:
(vi) If A and B are sets such that x ∈ A and A ∈ B then x ∈ B.
Answer: False.
[x ∈ A and A ∈ B
Let A = {x}, B = {{x}, y}
{x} is an element of B and x is not an element of B.
Thus, x ∉ B.]
- RS AGGARWAL CLASS 11 MATHS SOLUTION FUNCTIONS-2
- RS AGGARWAL CLASS 11 MATHS SOLUTION FUNCTIONS-1
- RS AGGARWAL CLASS 11 MATHS SOLUTION RELATIONS-2
- RS AGGARWAL CLASS 11 MATHS SOLUTION RELATIONS-1
- RS AGGARWAL CLASS 11 MATHS SOLUTION SET THEORY-3
- RS AGGARWAL CLASS 11 MATHS SOLUTION SET THEORY-2
- RS AGGARWAL CLASS 11 MATHS SOLUTION SET THEORY-1
Leave a Reply