Class 11 RD Sharma Solutions – Chapter 1 Sets – Exercise 1.5
Question 1. If A and B are two sets such that A ⊂ B, then Find:
(i) A ⋂ B
(ii) A ⋃ B
Solution:
(i) A ∩ B
A ∩ B denotes A intersection B, that is the common elements of both the sets.
Given A ⊂ B, every element of A is contained in B.
∴ A ∩ B = A
(ii) A ⋃ B
A ∪ B denotes A union B, that is it contains elements of either of the set.
Given A ⊂ B, B is having all elements including elements of A.
∴ A ∪ B = B
Question 2. If A = {1, 2, 3, 4, 5}, B = {4, 5, 6, 7, 8}, C = {7, 8, 9, 10, 11} and D = {10, 11, 12, 13, 14}. 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
(viii) A ∩ (B ∪ C)
(ix) (A ∩ B) ∩ (B ∩ C)
(x) (A ∪ D) ∩ (B ∪ C).
Solution:
We know,
X ∪ Y = {a: a ∈ X or a ∈ Y}
X ∩ Y = {a: a ∈ X and a ∈ Y}
(i) A = {1, 2, 3, 4, 5}
B = {4, 5, 6, 7, 8}
A ∪ B = Union of two sets A and B = {x: x ∈ A or x ∈ B}
= {1, 2, 3, 4, 5, 6, 7, 8}
(ii) A = {1, 2, 3, 4, 5}
C = {7, 8, 9, 10, 11}
A ∪ C = Union of two sets A and C = {x: x ∈ A or x ∈ C}
= {1, 2, 3, 4, 5, 7, 8, 9, 10, 11}
(iii) B = {4, 5, 6, 7, 8}
C = {7, 8, 9, 10, 11}
B ∪ C = {x: x ∈ B or x ∈ C}
= {4, 5, 6, 7, 8, 9, 10, 11}
(iv) B = {4, 5, 6, 7, 8}
D = {10, 11, 12, 13, 14}
B ∪ D = {x: x ∈ B or x ∈ D}
= {4, 5, 6, 7, 8, 10, 11, 12, 13, 14}
(v) A = {1, 2, 3, 4, 5}
B = {4, 5, 6, 7, 8}
C = {7, 8, 9, 10, 11}
A ∪ B = {x: x ∈ A or x ∈ B}
= {1, 2, 3, 4, 5, 6, 7, 8}
A ∪ B ∪ C = {x: x ∈ A ∪ B or x ∈ C}
= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}
(vi) A = {1, 2, 3, 4, 5}
B = {4, 5, 6, 7, 8}
D = {10, 11, 12, 13, 14}
A ∪ B = {x: x ∈ A or x ∈ B}
= {1, 2, 3, 4, 5, 6, 7, 8}
A ∪ B ∪ D = {x: x ∈ A ∪ B or x ∈ D}
= {1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14}
(vii) B = {4, 5, 6, 7, 8}
C = {7, 8, 9, 10, 11}
D = {10, 11, 12, 13, 14}
B ∪ C = {x: x ∈ B or x ∈ C}
= {4, 5, 6, 7, 8, 9, 10, 11}
B ∪ C ∪ D = {x: x ∈ B ∪ C or x ∈ D}
= {4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14}
(viii) A = {1, 2, 3, 4, 5}
B = {4, 5, 6, 7, 8}
C = {7, 8, 9, 10, 11}
B ∪ C = {x: x ∈ B or x ∈ C}
= {4, 5, 6, 7, 8, 9, 10, 11}
A ∩ B ∪ C = {x: x ∈ A and x ∈ B ∪ C}
= {4, 5}
(ix) A = {1, 2, 3, 4, 5}
B = {4, 5, 6, 7, 8}
C = {7, 8, 9, 10, 11}
(A ∩ B) = {x: x ∈ A and x ∈ B}
= {4, 5}
(B ∩ C) = {x: x ∈ B and x ∈ C}
= {7, 8}
(A ∩ B) ∩ (B ∩ C) = {x: x ∈ (A ∩ B) and x ∈ (B ∩ C)}
= ϕ
(x) A = {1, 2, 3, 4, 5}
B = {4, 5, 6, 7, 8}
C = {7, 8, 9, 10, 11}
D = {10, 11, 12, 13, 14}
A ∪ D = {x: x ∈ A or x ∈ D}
= {1, 2, 3, 4, 5, 10, 11, 12, 13, 14}
B ∪ C = {x: x ∈ B or x ∈ C}
= {4, 5, 6, 7, 8, 9, 10, 11}
(A ∪ D) ∩ (B ∪ C) = {x: x ∈ (A ∪ D) and x ∈ (B ∪ C)}
= {4, 5, 10, 11}
Question 3. Let A = {x: x ∈ N}, B = {x: x = 2n, n ∈ N), C = {x: x = 2n – 1, n ∈ N} and, D = {x: x is a prime natural number} Find:
(i) A ∩ B
(ii) A ∩ C
(iii) A ∩ D
(iv) B ∩ C
(v) B ∩ D
(vi) C ∩ D
Solution:
Let us assume,
A = All natural numbers i.e. {1, 2, 3…..}
B = All even natural numbers i.e. {2, 4, 6, 8…}
C = All odd natural numbers i.e. {1, 3, 5, 7……}
D = All prime natural numbers i.e. {1, 2, 3, 5, 7, 11, …}
(i) A ∩ B
A contains all elements of the set B.
∴ B ⊂ A = {2, 4, 6, 8…}
∴ A ∩ B = B
(ii) A ∩ C
A contains all elements of the set C.
∴ C ⊂ A = {1, 3, 5…}
∴ A ∩ C = C
(iii) A ∩ D
A contains all elements of the set D.
∴ D ⊂ A = {2, 3, 5, 7..}
∴ A ∩ D = D
(iv) B ∩ C
B ∩ C = ϕ
There cannot be any natural number which is both even and odd at same time.
(v) B ∩ D
B ∩ D = 2
{2} is the only natural number possible which is even and a prime number.
(vi) C ∩ D
C ∩ D = {1, 3, 5, 7…}
= D – {2}
Therefore, Every prime number is odd except {2}.
Question 4. Let A = {3,6,12,15,18,21}, 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
Solution:
For any two sets A and B, A-B is the set of elements belonging in A and not in B.
that is, A-B = {x : x ∈ A and x ∉ B}
(i) A-B = {x : x ∈ A and x ∉ B} = {3,6,15,18,21}
(ii) A-C = {x : x ∈ A and x ∉ C} = {3,15,18,21}
(iii) A-D = {x : x ∈ A and x ∉ D} = {3,6,12,18,21}
(iv) B-A = {x : x ∈ B and x ∉ A} = {4,8,16,20}
(v) C-A = {x : x ∈ C and x ∉ A} = {2,3,8,10,14,16}
(vi) D-A = {x : x ∈ D and x ∉ A} = {5,10,20}
(vii) B-C = {x : x ∈ B and x ∉ C} = {20}
(viii) B-D = {x : x ∈ B and x ∉ D} = {4,8,12,16}
Question 5. 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 U B)’
(v) (A’)’
(vi) (B-C)’
Solution:
(i) A’ ={x : x ∈ U and x ∉ A}
= {5,6,7,8,9}
(ii) B’ ={x : x ∈ U and x ∉ B}
= {1,3,5,7,9}
(iii) (A ∩ C)’ = A’ U C’ = {x : x ∈ U and x ∉ C and x ∉ A}
= {1,2,5,6,7,8,9}
(iv) (A U B)’ = {x : x ∈ U and x ∉ A ∩ B}
= {5,7,9}
(vi) (A’)’ = {x : x ∈ A} because, complement of complement cancels with each other
= {1,2,3,4}
(vi) (B-C)’ = {x : x ∈ U and x ∉ B-C}
={1,3,4,5,6,7,9}
Question 6. Let U = {1,2,3,4,5,6,7,8,9}, A = {2,4,6,8}, B = {2,3,5,7}. Verify that:
(i) (A U B)’ = A’ ∩ B’
(ii) (A ∩ B)’ = A’ U B’
Solution:
(i) We have, LHS = (A U B)’
Computing (A U B) = {2,3,4,5,6,7,8}
Now (A U B)’ = U – (A U B)
= {x : x ∈ U and x ∉ A U B}
= {1,9}
RHS = A’ ∩ B’
Now, A’ = {1,3,5,7,9}
B’ = {1,4,6,8,9}
A’ ∩ B’ = {x : x ∈ A’ and x ∈ B’}
={1,9}
Therefore, LHS = RHS
(ii) We have, LHS = (A ∩ B)’
Computing (A ∩ B) = {2}
Now, (A ∩ B)’ = U – (A ∩ B)
={x : x ∈ U and x ∉ A ∩ B}
={1,3,4,5,6,7,8,9}
RHS = A’ U B’
Computing A’ = {1,3,5,7,9}
B’ = {1,4,6,8,9}
Now A’ U B’ = {1,3,4,5,7,8,9}
Therefore, LHS= RHS.
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