# 01_Sets.pdf

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### 01_Sets.pdf

• 10. How to Avail Discount ? Apply Coupon Code: AVKPRO Visit: https://vdnt.in/YTPRO SpecialDiscount for this class Link in Description
• 11. Sets, Relations and Functions
• 12. Sets
• 13. Definition A set is a collectionof well defined objects. Definition and Description of Sets
• 14. (a) Sets are generally denotedby capital letters like A, B, X, Y etc. Elements of the set are denotedby small letters like a, b, x, y etc. (a) If a is an element of the set A, then it’s written as a ∈ A. (b) Elements are not repeated in the set. Note Definition A set is a collectionof well defined objects.
• 15. Description Sets are describedin two ways. (i) Roster form (ii) SetBuilderform Elements are listed using commas and brackets For example, {a, e, i, o, u} Elements are told by telling the property or rule they follow For example, {x : x is a vowel}
• 16. Match the following (a) {1, 2, 4} (i) {x : x is a prime divisor of 6} (b) {2, 3} (ii) {x : x ∈ N and x is a divisor of 4} (c) {A, C, E, H, I, M, S, T} (iii) {x : x is a letter of the word MATHEMATICS} (iv) {x : x ∈ N and x < 7} (v) {x : x is a letter of the word ALCHEMIST} Q Question!
• 17. Solution: A ➝ II; B ➝ I; C ➝ III
• 18. Ex Write the set {1, 2, 4, 8, 16, 32, 64} in set-builder form. Q Question!
• 19. Solution: Given set is A = {1, 2, 4, 8, 16, 32, 64} By observation, we can say A = {2n - 1 : n ≤ 7 and n ∈ N} Alternatively, we can say A = {2n : n ≤ 6 and n ∈ W}
• 20. Write the set in set-builder form. Q Question!
• 22. Write the following set in the roster form. C = (x : x2 + 7x - 8 = 0, x ∈ R} Q Question!
• 25. Nullset orEmptyset or Voidset A set which does not contain any element is called Empty set or Null set or Void set. It’s denoted by 𝜙 or { }. Eg. (i) (ii)
• 26. The set A = {x : x ∈ R, x2 = 16 and 2x = 6] equal- Q Question! 𝜙 {3} {4} A B C D {14, 3, 4}
• 27. The set A = {x : x ∈ R, x2 = 16 and 2x = 6] equal- Q Question! 𝜙 {3} {4} B C D {14, 3, 4} A
• 29. SingletonSet A set containingsingle element is called singleton set. Remark : Curly brackets ie { . } are use to write singletons. Eg. {z} , {0} , {𝜙} , { {1,2,3,4,5} } are singleton sets.
• 30. FiniteSet A set which is either empty or has finite number of elements is called a finite set. Number of elements in a finite set A is called the order or the cardinality of the set A, generally denoted by o(A) or (A). Remark
• 31. If A = {1, 2, {3, 4} }, then n(A) = _____. Q Question! 2 4 None of these A B C D 3 Recall Eg. {z} , {0} , {𝜙} , { {1,2,3,4,5} } are singleton sets.
• 32. If A = {1, 2, {3, 4} }, then n(A) = _____. Q Question! 2 4 None of these A C D 3 B
• 33. Given A = {1, 2, {3, 4} } Here, {3, 4} is an element of A So, n(A) = 3 Solution:
• 34. InfiniteSet A set having infinite number of elements is called infiniteset. Eg. (i) {1, 2, 3, 4, ...}, (ii) {..., -2, -1, 0, 1, 2, ...} etc.
• 35. Equal Sets Two sets are said to be equal if they have exactly the same elements. Eg. If A = {0, 1, 2} , B = {2, 1, 0} , C = {0, 1, 1, 2}, then A = B = C. Equivalent Sets Two finite sets are said to be equivalent if their cardinalities are equal. Eg. {1, 2, 3} is equivalent to {dog, cat, parrot}.
• 37. Definition Set A is said to be a subset of set B if every element of A is also an element of B. It is generally denotedas A ⊂ B. Ex. If A = {1, 2, 3} and B = {1, 2, 3, 4}, then A ⊂ B. Subsets
• 38. Note (a) Every set is a subset of itself. (b) ɸ is a subset of all sets. (c) A ⊂ B and B ⊂ A ⇒ A = B .
• 39. If A = { p , q , {2, 3} } , then try to observe that (i) { p , q } ⊂ A (ii) { { 2, 3} , p } ⊂ A Observations
• 40. Write the subsets of {1, 2} Q Question!
• 41. Solution: Let A = {1, 2} The subsets of A are: 𝜙 , {1}, {2}, {1,2}
• 43. Definition The collectionof all the subsets of a set A is called the power set of A, denoted by P(A). Power set
• 44. Write down the power sets of the following sets. (a) A = {1, 2} (b) A = {1, 2, 3} Q Question!
• 45. (a) Given A = {1, 2} Power set of A = {ɸ, {1}, {2}, {1, 2}} Solution: (b) Given A = {1, 2, 3} P(A) = {ɸ, {1}, {2}, {3}, {1, 2} {2, 3}, {1, 3}, {1, 2, 3}}
• 46. If n(A) = p, then the number of subsets of A is 2p. Clearly, same will be n(P(A)). Result
• 47. Two finite sets have p and q number of elements Respectively. The number of elements in the power set of the first set is 12 more than the number of elements in the power set of the second set. Find p and q. Q Question!
• 48. Here, let the two sets be A and B, such that n(A) = p and n(B) = q. Number of elements in power set of A = 2p Number of elements in power set of B = 2q Now, according to question, 2p − 2q = 12, that is 2q(2p - q - 1) = 22(22 -1) ∴ q = 2, p = 4 Solution:
• 49. To understand operations on sets, first we need to know the term “universal set”. Operations on Sets
• 50. UniversalSet If we talk about any particular context in sets then the set which is superset of all possible sets in the given context is called universal set. (Generally, it’s given in the question). VennDiagram Ex. A = {1, 2, 3} B = {3, 4, 5} U = {1, 2, 3, 4, 5, 6} Operations on Sets
• 51. Followingare the operationson sets that we are now goingto learn. (a) Unionof two Sets (b) Intersection of two Sets (c) Difference of two Sets (d) Symmetric Difference of two Sets (e) Complement of a Set Operations on Sets
• 52. Union of two Sets A ∪ B = {x : x ∈ A or x ∈ B} Ex. {1, 2, 3} ∪ {2, 3, 4} = {1, 2, 3, 4} A B Operations on Sets
• 53. Ex 1.5 Intersection of two Sets A ∩ B = {x : x ∈ A and x ∈ B simultaneously} Ex. {1, 2, 3} ∩ {2, 3, 4} = {2, 3} A B Operations on Sets
• 54. Note (a) If C ⊂ B, then (i) C ∪ B = B (ii) C ∩ B = C
• 55. Note (b) C ∩ (B ∪ A) = (C ∩ B) ∪ (C ∩ A); C ∪ (B ∩ A) = (C ∪ B) ∩ (C ∪ A)
• 56. Note (a) If C ⊂ B, then (i) C ∪ B = B (ii) C ∩ B = C (b) C ∩ (B ∪ A) = (C ∩ B) ∪ (C ∩ A); C ∪ (B ∩ A) = (C ∪ B) ∩ (C ∪ A) (c) If C ∩ B = ɸ, then C and B are called disjointsets.
• 57. Ac or A’ = {x : x ∉ A and x ∈ U} Eg. If U = {1, 2, 3, 4, 5, 6}, then complement of {1, 2, 3, 4} is {5, 6}. A Ac Complement of a Set Operations on Sets
• 59. Note (a) (Ac)c = A (b) A ∪ Ac = U
• 60. Note (a) (Ac)c = A (b) A ∪ Ac = U (c) A ∩ B = 𝜙 ⇒ A ⊂ Bc
• 61. De-Morgan’s Laws (i) (A ∪ B)c = Ac ∩ Bc (ii) (A ∩ B)c = Ac ∪ Bc
• 62. Difference of two Sets A – B (or) AB = {x : x ∈ A and x ∉ B} Eg. {1, 2, 3} – {2, 3, 4} = Eg. {1,2,3}{4,5} = B A Operations on Sets
• 63. AB = A ∩ Bc Observation
• 64. SymmetricDifference of two Sets B A Operations on Sets
• 66. (i) n(A ∪ B) = n(A) + n(B) - n(A ∩ B) Result Cardinality based Problems
• 67. (i) n(A ∪ B) = n(A) + n(B) - n(A ∩ B) (ii) n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(A ∩ C) - n(B ∩ C) + n(A ∩ B ∩ C) Result Cardinality based Problems
• 68. B A
• 69. (a) n(A) = x1 + x2 (b) n(only A) = x1 (c) n(B) = x2 + x3 (d) n(only B) = x3 (e) n(none of A or B) = x4 Remarks B A x1 x4 x2 x3
• 70. In a school, every teacher teaches either maths or physics. If 15 teach maths, 24 teach physics, while 6 teach both, then find (a) the number of teachers. (b) the number of teachers who teach physics only. Q Question!
• 71. In a school, every teacher teaches either maths or physics. If 15 teach maths, 24 teach physics, while 6 teach both, then find (a) the number of teachers. Q Question!
• 72. Solution: Let A be the set of teacher who teaches Maths B be the set of teacher who teaches Physics Then n(A) = 15, n(B) = 24 n(A ∩ B) = 6 ∴ Total number of teachers n(A ∪ B) = n(A) + n(B) - n(A ∩ B) = 15 + 24 – 6 = 33 A B
• 73. In a school, every teacher teaches either maths or physics. If 15 teach maths, 24 teach physics, while 6 teach both, then find (b) the number of teachers who teach physics only. Q Question!
• 74. Solution: Let x1 = num of teachers who teach physics only Let x2 = num of teachers who teach maths only Let x3 = num of teachers who teach both maths and physics According to the question, x1 + x3 = 24 and x2 + x3 = 15 and x3 = 6 Solving, we get x1 = 24 - 6 = 18 x1 x3 x2
• 75. In a class 60% passed their Physics examination and 58% passed in Mathematics. What percentage of students passed both their physics and Mathematics examination? Q A B C D Question! 18% 17% 16% 2%
• 76. In a class 60% passed their Physics examination and 58% passed in Mathematics. What percentage of students passed both their physics and Mathematics examination? Q B C D Question! 18% 17% 16% 2% A
• 77. Solution: n(P) = 60, n(M) = 58, n(P ∪ M) = 100, n(P ∩ M) = n(P) + n(M) - n(P ∪ M) = 60 + 58 - 100 = 18
• 78. In a survey of 600 students in a school, 150 students were found to be drinking Tea and 225 drinking Coffee, 100 were drinking both Tea and Coffee. Find how many students were drinking neither Tea nor Coffee. Q Question!
• 79. Solution: Given, Total number of students = 600 Number of students who were drinking Tea = n(T) = 150 Number of students who were drinking Coffee = n(C) = 225 Number of students who were drinking both Tea and Coffee = n(T ∩ C) = 100 n(T ∪ C) = n(T) + n(C) - n(T ∩ C) = 150 + 225 - 100 = 375 - 100 = 275 Hence, the number of students who are drinking neither Tea nor Coffee = 600 - 275 = 325
• 80. In a school, if 15 teachers teach maths and 25 teachers teach physics, and if total number of teachers is 30, then try to observe that minimum value of n(M ∩ P) is 10. Observation
• 81. (a) If n(A) + n(B) ≥ n(U) , then minimum value of n(A ∩ B) is given by n(A) + n(B) − n(U) (b) If n(A) + n(B) < n(U), then minimum n(A ∩ B) = 0 Remarks
• 82. If n(A) = 5 and n(B) = 10 , then find maximum and minimum possible values of n(A ∩ B) if (a) n(U) = 8 (b) n(U) = 20 Q Question!
• 83. Solution: (a) Given, n(A) = 5 and n(B) = 10 Here, maximum possible value of n(A ∩ B) = 5 Now, as n(U) = 8 and n(A) + n(B) = 15 Thus, minimum value of n(A ∩ B) = 7
• 84. Solution: (b) Given, n(A) = 5 and n(B) = 10 Here, maximum possible value of n(A ∩ B) = 5 Now, as n(U) = 20 and n(A) + n(B) = 15 Thus, minimum value of n(A ∩ B) = 0
• 85. There are 25 students. Every student reads 10 newspapers and each newspaper is read by (i.e., shared by) 5 students. Find the number of newspapers. Q A B C D Question!
• 86. There are 25 students. Every student reads 10 newspapers and each newspaper is read by (i.e., shared by) 5 students. Find the number of newspapers. Q B C D Question! A
• 87. Solution: Since every student reads 10 newspaper Thus, total number of newspaper read = 10 × 25 = 250 But each news paper is shared by 5 students Thus actual number of newspapers =
• 88. Suppose A1, A2, ….., A30 are 30 sets, each with 5 elements and B1, B2, …., Bn are n sets, each with 3 elements. and each element of S belongs to exactly 10 of Ai‘s and 9 of Bj’s. Find ‘n’. Q Question!
• 91. (a) n(A) = x1 + x4 + x6 + x7 (b) n(only A) = x1 (c) n(only A and B) = x1 + x4 + x2 (d) n(only one of A, B or C) = x1 + x2 + x3 (e) n(exactly two of A, B and C) = x4 + x5 + x6 (f) n(A, B or C) = x1 + x2 + x3 + x4 + x5 + x6 + x7 B A x1 x4 x2 x5 x6 x7 x3 Remarks Cardinality based Problems C
• 92. Of the number of three athletic teams in a school, 21 are in the basketball team, 26 in hockey team and 29 in the football team, 14 play hockey and basketball, 15 play hockey and football, 12 play football and basketball and 8 play all the games. The total number of members is Q Question! A B C D 42 43 45 None of these n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(A ∩ C) - n(B ∩ C) + n(A ∩ B ∩ C) Recall:
• 93. Of the number of three athletic teams in a school, 21 are in the basketball team, 26 in hockey team and 29 in the football team, 14 play hockey and basketball, 15 play hockey and football, 12 play football and basketball and 8 play all the games. The total number of members is Q Question! A C D 42 43 45 None of these B
• 94. Solution: Let B, H, F be the sets of the three teams respectively So n(B) = 21, n(H) = 26, n(F) = 29, n(H ∩ B) = 14, n(H ∩ F) = 15, n(F ∩ B) = 12, n(B ∩ H ∩ F) = 8 and n(B ∪ H ∪ F) = n(B) + n(H) + n(F) - n(H ∩ B) - n(H ∩ F) - n(F ∩ B) + n(B ∩ H ∩ F) = 21 + 26 + 29 - 14 - 15 - 12 + 8 = 43
• 95. In a class, every student plays either cricket or hockey or badminton. 10 play cricket, 15 play hockey, 20 play badminton, 5 play cricket and hockey, 2 play hockey and badminton, 5 play badminton and cricket and 2 play all the games. How many play exactly one of the games? Q Question!
• 96. Solution: n(C) = 10 = x1 + x4 + x6 + x7 n(H) = 15 = x2 + x4 + x7 + x5 n(B) = 20 = x3 + x5 + x6 + x7 n(C ∩ H) = 5 = x4 + x7 n(H ∩ B) = 2 = x5 + x7 n(B ∩ C) = 5 = x6 + x7 n(C ∩ H ∩ B) = 2 = x7 ∴ x4 = 3 ; x5 = 0 ; x6 = 3 x1 = 2 ; x2 = 10 ; x3 = 15 H C x1 x4 x2 x5 x6 x7 x3 B
• 97. Solution: Number of students playing exactly one game = x1 + x2 + x3 = 27 H C x1 x4 x2 x5 x6 x7 x3 B
• 105. How to Avail Discount ? Apply Coupon Code: AVKPRO Visit: https://vdnt.in/YTPRO SpecialDiscount for this class Link in Description
• 106. Join Vedantu JEE Telegram channel NOW! https://vdnt.in/jeepro Assignments Daily Update Notes
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