NCERT MATHS SOLUTION CLASS 11 SETS CHAPTER 1 EXERCISE 1.4
NCERT MATHS SOLUTION CLASS 11 CHAPTER 1 SETS EXERCISE 1.4-1.6
EXERCISE 1.4
Q.1.(i) Find the union of each of the following pairs of sets: X = {1, 3, 5}, Y = {1, 2, 3}
Solution: Given: X = {1, 3, 5} Y = {1, 2, 3}
∴ X U Y = {1, 2, 3, 5}
Q.1.(ii) Find the union of each of the following pairs of sets: A = {a, e, i, o, u} and B = {a, b, c}
Solution: Given: A = {a, e, i, o, u} and B = {a, b, c}
∴ A U B = {a, b, c, e, i, o, u}
Q.1.(iii) Find the union of each of the following pairs of sets:
A = {x: x is a natural number and multiple of 3} and
B = {x: x is a natural number less than 6}
Solution: Given:
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}
⇒ B = {1, 2, 3, 4, 5}
∴ A U B = {3, 6, 9, 12….}
= {x: x is a natural number and multiple of 3}
Q.1.(iv) Find the union of each of the following pairs of sets:
A = {x: x is a natural number and 1 < x ≤ 6} and
B = {x: x is a natural number and 6 < x < 10}
Solution: Given: A = {x: x is a natural number and 1 < x ≤ 6}
⇒ A = {2, 3, 4, 5, 6} and
B = {x: x is a natural number and 6 < x < 10}
⇒ B = {7, 8, 9}
∴ A U B = {2, 3, 4, 5, 6, 7, 8, 9}
= {x: x is a natural number and 1 < x < 10}
Q.1.(v) Find the union of each of the following pairs of sets: A = {1, 2, 3} and B = Φ
Solution: Given: A = {1, 2, 3} and B = Φ
∴ A U B = {1, 2, 3, 5}
Q.2. Let A = {a, b} and B = {a, b, c} Is A ⊂ B?
What is A ∪ B?
Solution: Given: A = {a, b} and B = {a, b, c}
Yes,
A ⊂ B
A ∪ B
= {a, b, c}
⇒ B
Q.3. If A and B are two sets such that A ⊂ B, then what is A ∪ B?
Solution: Here A ⊂ B,
∴ A ∪ B = B
Q.4. If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8} and D = {7, 8, 9, 10} Find the following:
(i) A ∪ B
(ii) A ∪ C
(iii) B ∪ C
(iv) B ∪ D
Solution:
(i) Given: A = {1, 2, 3, 4}, B = {3, 4, 5, 6}
∴ A ∪ B
= {1, 2, 3, 4, 5, 6}
(ii) Given: A = {1, 2, 3, 4}, C = {5, 6, 7, 8}
∴ A ∪ C
= {1, 2, 3, 4, 5, 6, 7, 8}
(iii) Given: B = {3, 4, 5, 6}, C = {5, 6, 7, 8}
∴ B ∪ C
= {3, 4, 5, 6, 7, 8}
(iv) Given: B = {3, 4, 5, 6} D = {7, 8, 9, 10}
∴ B ∪ D
= {3, 4, 5, 6, 7, 8, 9, 10}
Q.4. If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8} and D = {7, 8, 9, 10} Find the following:
(v) A ∪ B ∪ C
(vi) A ∪ B ∪ D
(vii) B ∪ C ∪ D
Solution: Given: A = {1, 2, 3, 4}, B = {3, 4, 5, 6} and C = {5, 6, 7, 8}
∴ A ∪ B ∪ C
= {1, 2, 3, 4, 5, 6, 7, 8}
(vi) Given: A = {1, 2, 3, 4}, B = {3, 4, 5, 6} and D = {7, 8, 9, 10}
∴ A ∪ B ∪ D
= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(vii) Given: B = {3, 4, 5, 6}, C = {5, 6, 7, 8} and D = {7, 8, 9, 10}
∴ B ∪ C ∪ D
= {3, 4, 5, 6, 7, 8, 9, 10}
Q.5. Find the intersection of each pairs of sets of question above.
(i) A ∩ B
(ii) A ∩ C
(iii) B ∩ C
(iv) B ∩ D
Solution:
(i) Given: A = {1, 2, 3, 4}, B = {3, 4, 5, 6},
∴ A ∩ B = {3, 4}
(ii) Given: A = {1, 2, 3, 4}, C = {5, 6, 7, 8}
∴ A ∩ C = ϕ
(iii) Given: B = {3, 4, 5, 6}, C = {5, 6, 7, 8}
∴ B ∩ C = {5, 6}
(iv) Given: B = {3, 4, 5, 6}, D = {7, 8, 9, 10}
∴ B ∩ D = ϕ
Q.5. Find the intersection of each pairs of sets of question above.
(v) A ∩ B ∩ C
(vi) A ∩ B ∩ D
(vii) B ∩ C ∩ D
(v) Given: A = {1, 2, 3, 4}, B = {3, 4, 5, 6} and C = {5, 6, 7, 8}
∴ A ∩ B ∩ C = ϕ
(vi) Given: A = {1, 2, 3, 4}, B = {3, 4, 5, 6} and D = {7, 8, 9, 10}
∴ A ∩ B ∩ D = ϕ
(vii) Given: B = {3, 4, 5, 6} and C = {5, 6, 7, 8} and D = {7, 8, 9, 10}
∴ B ∩ C ∩ D = ϕ
Q.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
Solution:
(i) Given: A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13}
∴ A ∩ B
= {9, 11}
(ii) Given: B = {7, 9, 11, 13}, C = {11, 13, 15}
∴ B ∩ C
= {11, 13}
(iii) Given: A = {3, 5, 7, 9, 11}, C = {11, 13, 15}, D = {15, 17}∴ A ∩ C ∩ D
= {11} ∩ {15, 17}
= ϕ
(iv)
∴ A ∩ C
= {11}
Q.6. If A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13}, C = {11, 13, 15} and D = {15, 17}; Find
(v) B ∩ D
(vi) A ∩ (B ∪ C)
(vii) A ∩ D
Solution:
(v) Given: B = {7, 9, 11, 13}, D = {15, 17}
∴ B ∩ D
= ϕ
(vi) Given: A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13}, C = {11, 13, 15}
B ∪ C
= {7, 9, 11, 13, 15}
∴ A ∩ (B ∪ C)
= {3, 5, 7, 9, 11} ∩ {7, 9, 11, 13, 15}
= {7, 9, 11}
(vii) Given: A = {3, 5, 7, 9, 11}, D = {15, 17}
∴ A ∩ D
= ϕ
Q.6. If A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13}, C = {11, 13, 15} and D = {15, 17}; Find
(viii) A ∩ (B ∪ D)
(ix) (A ∩ B) ∩ (B ∪ C)
(ix) (A ∪ D) ∩ (B ∪ C)
Solution:
(viii) Given: A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13}, D = {15, 17}
(B ∪ D)
= {7, 9, 11, 13, 15, 17}
∴ A ∩ (B ∪ D)
= {3, 5, 7, 9, 11} ∩ {7, 9, 11, 13, 15, 17}
= {7, 9, 11}
(ix) Given: A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13}, C = {11, 13, 15}
(A ∩ B)
= {7, 9, 11, 13}
(B ∪ C)
= {7, 9, 11, 13, 15}
∴ (A ∩ B) ∩ (B ∪ C)
= {7, 9, 11}
(ix) Given: A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13}, C = {11, 13, 15}, D = {15, 17}
(A ∪ D)
= {3, 5, 7, 9, 11, 15, 17}
(B ∪ C)
{7, 9, 11, 13, 15}
∴ (A ∪ D) ∩ (B ∪ C)
= {7, 9, 11, 15}
Q.7. 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 the following:
(i) A ∩ B
(ii) A ∩ C
(iii) A ∩ D
Solution:
(i) Given, A = {x: x is a natural number}
= {1, 2, 3, 4, 5, . . . . . . . },
B = {x: x is an even natural number}
⇒ {2, 4, 6, 8, . . . . . . . }
∴ A ∩ B
⇒ {2, 4, 6, 8, . . . . . . . }
= {x: x is an even natural number}
⇒ {2, 4, 6, 8, . . . . . . . }
= B
(ii) Given, A = {x: x is a natural number}
= {1, 2, 3, 4, 5, . . . . . . . },
C = {x: x is an odd natural number}
= {1, 3, 5, 7, . . . . . . . }
∴ A ∩ C
= {1, 3, 5, 7, . . . . . . . }
⇒ {x: x is an odd natural number}
= C
(iii) Given, A = {x: x is a natural number}
= {1, 2, 3, 4, 5, . . . . . . . }
D = {x: x is a prime number}
⇒ {2, 3, 5, 7, . . . . . . . }
∴ A ∩ D
= {2, 3, 5, 7, . . . . . . . }
⇒ {x: x is a prime number}
= D
Q.7. 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 the following:
(iv) B ∩ C
(v) B ∩ D
(vi) C ∩ D
Solution:
(iv) Given, B = {x: x is an even natural number}
= {2, 4, 6, 8, . . . . . . . },
C = {x: x is an odd natural number}
= {1, 3, 5, 7, . . . . . . . }
∴ B ∩ C = ϕ
(v) Given, B = {x: x is an even natural number}
= {2, 4, 6, 8, . . . . . . . }
D = {x: x is a prime number}
= {2, 3, 5, 7, . . . . . . . }
∴ B ∩ D = {2}
(vi) Given, C = {x: x is an odd natural number}
= {1, 3, 5, 7, . . . . . . . }, and
D = {x: x is a prime number}
= {2, 3, 5, 7, . . . . . . . }
∴ C ∩ D
= {3, 5, 7, . . . . . . . }
= {x: x is a odd prime number}
Q.8. Which of the given pairs of sets are disjoint?
(i) A = {1, 2, 3, 4} and B = {x: x is a natural number and 4 ≤ x ≤ 6}
(ii) A = {a, e, i, o, u} and B = {c, d, e, f}
(iii) A = {x: x is an even integer} and B = {x: x is an odd integer}
Solution:
(i) A = {1, 2, 3, 4}
B = {x: x is a natural number and 4 ≤ x ≤ 6}
= {4, 5, 6}
∴ A ∩ B = {4}
Hence, A and B are not disjoint.
(ii) A = {a, e, i, o, u} and B = {c, d, e, f}
∴ A ∩ B = {e}
Hence, A and B are not disjoint.
(iii) A = {x: x is an even integer}
= {2, 4, 6, 8, . . . . . . . }
and B = {x: x is an odd integer}
= {1, 3, 5, 7, . . . . . . . }
∴ A ∩ B = ϕ
Hence, A and B are disjoint.
Q.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 the following:
(i) A – B
(ii) A – C
(iii) A – D
(iv) B – A
Solution:
(i) Given, A = {3, 6, 9, 12, 15, 18, 21}, B = {4, 8, 12, 16, 20}
∴ A – B
= {3, 6, 9, 15, 18, 21}
(ii) Given, A = {3, 6, 9, 12, 15, 18, 21}, C = {2, 4, 6, 8, 10, 12, 14, 16}
∴ A – C
= {3, 9, 15, 18, 21}
(iii) Given, A = {3, 6, 9, 12, 15, 18, 21}, D = {5, 10, 15, 20}
∴ A – D
= {3, 6, 9, 12, 18, 21}
(iv) Given, A = {3, 6, 9, 12, 15, 18, 21}, B = {4, 8, 12, 16, 20}
∴ B – A
= {4, 8, 16, 20},
Q.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 the following:
(v) C – A
(vi) D – A
(vii) B – C
(viii) B – D
Solution:
(v) Given, A = {3, 6, 9, 12, 15, 18, 21}, C = {2, 4, 6, 8, 10, 12, 14, 16}
∴ C – A
= {2, 4, 8, 10, 14, 16}
(vi) Given, A = {3, 6, 9, 12, 15, 18, 21}, D = {5, 10, 15, 20}
∴ D – A
= {5, 10, 20}
(vii) Given, B = {4, 8, 12, 16, 20}, C = {2, 4, 6, 8, 10, 12, 14, 16}
∴ B – C = {20}
(viii) Given, B = {4, 8, 12, 16, 20} and D = {5, 10, 15, 20}
∴ B – D
= {4, 8, 12, 16}
Q.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 the following:(v) C – A
(vi) D – A
(vii) B – C
(viii) B – D
Solution:
(v) Given, A = {3, 6, 9, 12, 15, 18, 21}, C = {2, 4, 6, 8, 10, 12, 14, 16}
∴ C – A
= {2, 4, 8, 10, 14, 16}
(vi) Given, A = {3, 6, 9, 12, 15, 18, 21}, D = {5, 10, 15, 20}
∴ D – A
= {5, 10, 20}
(vii) Given, B = {4, 8, 12, 16, 20}, C = {2, 4, 6, 8, 10, 12, 14, 16}
∴ B – C = {20}
(viii) Given, B = {4, 8, 12, 16, 20} and D = {5, 10, 15, 20}
∴ B – D
= {4, 8, 12, 16}
Q.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 the following:
(ix) C – B
(x) D – B
(xi) C – D
(xii) D – C
Solution:
(ix) Given, B = {4, 8, 12, 16, 20}, C = {2, 4, 6, 8, 10, 12, 14, 16}
∴ C – B
= {2, 6, 10, 14}
(x) Given, B = {4, 8, 12, 16, 20} and D = {5, 10, 15, 20}
∴ D – B
= {5, 10, 15}
(xi) Given, C = {2, 4, 6, 8, 10, 12, 14, 16} and D = {5, 10, 15, 20}
∴ C – D
= {2, 4, 6, 8, 12, 14, 16}
(xii) Given, C = {2, 4, 6, 8, 10, 12, 14, 16} and D = {5, 10, 15, 20}
∴ D – C
= {5, 15, 20}
10: If X = {a, b, c, d} and Y = {f, b, d, g}
Find the following:
(i) X – Y
(ii) Y – X
(iii) X ∩ Y
Solution: (i) Given, X = {a, b, c, d} and Y = {f, b, d, g}
∴ X – Y = {a, c}
(ii) Given, X = {a, b, c, d} and Y = {f, b, d, g}
∴ Y – X = {f, g}
(iii) Given, X = {a, b, c, d} and Y = {f, b, d, g}
∴ X ∩ Y = {b, d}
Q.11. If R is the set of real numbers and Q is the set of rational numbers then what is R – Q?
Solution: R is the set of real numbers and Q is the set of rational numbers. ∴ R – Q = Set of irrational numbers.
Q.12: State whether each of the following statements are true or false. Give reason.
(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.
Solution:
(i) False.
{2, 3, 4, 5} ∩ {3, 6}
= {3}
∴ {2, 3, 4, 5} and {3, 6} are not disjoint sets.
(ii) False.
{a, e, i, o, u } ∩ {a, b, c, d}
= {a}
∴ {a, e, i, o, u } and {a, b, c, d} are not disjoint sets.
(iii) True.
{2, 6, 10, 14} ∩ {3, 7, 11, 15}
= ϕ
∴ {2, 6, 10, 14} and {3, 7, 11, 15} are disjoint sets.
(iv) True.
{2, 6, 10} and {3, 7, 11}
= ϕ
∴ {2, 6, 10} and {3, 7, 11} are disjoint sets.
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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 U C)’
Solution:
(i) Given, U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4}
∴ A’ = U – A
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4}
= {5, 6, 7, 8, 9}
(ii) Given, U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, B = {2, 4, 6, 8}
∴ B’= U – B
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 4, 6, 8}
= {1, 3, 5, 7, 9}
(iii) Given, U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4}
and C = {3, 4, 5, 6}
∴ (A U C) = {1, 2, 3, 4, 5, 6}
∴ (A U C)’
= U – (A U C)
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4, 5, 6}= {7, 8, 9}
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
(iv) ( A U B)’
(v) (A’)’
(vi) (B – C)
Solution:
(iv) Given, U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4}, B = {2, 4, 6, 8}
∴ (A U B) = {1, 2, 3, 4, 6, 8}
∴ ( A U B)’
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4, 6, 8}= {5, 7, 9}
(v) Given, U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4}
∴ A’ = U – A
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4}
= {5, 6, 7, 8, 9}
∴ (A’)’= U – A’
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {5, 6, 7, 8, 9}
= {1, 2, 3, 4}
(vi) Given, U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, B = {2, 4, 6, 8} and C = {3, 4, 5, 6}
∴ (B – C) = {2, 8}
∴ (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}
Solution:
(i) Given, U = {a, b, c, d, e, f, g, h}, A = {a, b, c}
∴ A’ = {a, b, c, d, e, f, g, h} – {a, b, c}
= {d, e, f, g, h}
(ii) Given, U = {a, b, c, d, e, f, g, h}, B = {d, e, f, g}
∴ B’ = {a, b, c, d, e, f, g, h} – {d, e, f, g}
= {a, b, c, h}
(iii) Given, U = {a, b, c, d, e, f, g, h}, C = {a, c, e, g}
∴ C’ = {a, b, c, d, e, f, g, h} – {a, c, e, g}
= {b, d, f, g}
(iv) Given, U = {a, b, c, d, e, f, g, h}, D = {f, g, h, a}
∴ D’ = {a, b, c, d, e, f, g, h} – {f, g, h, a}
= {b, c, d, e}
3. 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 numbers}.
(ii) {x: x is an odd natural numbers}.
(iii) {x: x is a positive multiple of 3}
(iv) {x: x is a prime number}
Solution:
(i) U = {x: x is a natural numbers}
∴ complements of the set {x: x is an even natural numbers} is
{x: x is an odd natural numbers}.
(ii) U = {x: x is a natural numbers}
∴ complements of the set {x: x is an odd natural numbers} is
{x: x is an even natural numbers}
(iii) U = {x: x is a natural numbers}
∴ complements of the set {x: x is a positive multiple of 3}is
{x: x ∈ N and x is not a multiple of 3}
(iv) U = {x: x is a natural numbers}
∴ complements of the set {x: x is a prime number} is
{x: x ∈ N and x = 1 and x is a composite number}
3. Taking the set of natural numbers as the universal set, write down the complements of the following sets:
(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}
Solution:
(v) U = {x: x is a natural numbers}
∴ complements of the set {x: x is a natural number divisible by 3 and 5} is
{x: x ∈ N and x is not divisible by 3 and 5}
(vi) U = {x: x is a natural numbers}
∴ complements of the set {x: x is a perfect square} is
{x: x ∈ N and x is not a perfect square}
(vii) U = {x: x is a natural numbers}
∴ complements of the set {x: x is a perfect cube} is
{x: x ∈ N and x is not a perfect cube}
(viii) U = {x: x is a natural numbers}
{x: x + 5 = 8}
Given, x + 5 = 8
⇒ x = 3
∴ complements of the given set is
{x: x ∈ N and x ≠ 3}
3. Taking the set of natural numbers as the universal set, write down the complements of the following sets:
(ix) {x: 2x + 5 = 9}
(x) {x: x ≥ 7}
(xi) {x: x ∈ N ; and 2x + 1 > 10}
Solution:
(ix) U = {x: x is a natural numbers}
{x: 2x + 5 = 9}
Given, 2x + 5 = 9
⇒ 2x = 4
⇒ x = 2
∴ complements of the given set is
{x: x ∈ N and x ≠ 2}
(x) U = {x: x is a natural numbers}
{x: x ≥ 7}
∴ complements of the given set is
{x: x ∈ N and x < 7}
(xi) U = {x: x is a natural numbers}
{x: x ∈ N ; and 2x + 1 > 10}
Given, 2x + 1 > 10
⇒ 2x > 9
⇒ x > 9/2
∴ complements of the given set is
{x: x ∈ N and x ≤ 9/2}
4.(i) If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 4, 6, 8}, B = {2, 3, 5, 7} verify that (A U B)’ = A’ ∩ B’
Solution: Given, U = {1, 2, 3, 4, 5, 6, 7, 8, 9},
A = {2, 4, 6, 8}, B = {2, 3, 5, 7}
A U B
= {2, 4, 6, 8} U {2, 3, 5, 7}
= {2, 3, 4, 5, 6, 7, 8}
L.H.S.
(A U B)’ = U – (A U B)
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 3, 4, 5, 6, 7, 8}
= {1, 9}
A’ = U – A
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 4, 6, 8}
= {1, 3, 5, 7, 9}
B’ = U – B
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 3, 5, 7}
= {1, 4, 6, 8, 9}
R. H.S.
= A’ ∩ B’
⇒ {1, 3, 5, 7, 9} ∩ {1, 4, 6, 8, 9}
= {1, 9} = L.H.S.
∴ (A U B)’ = A’ ∩ B’ (Verified).
4.(ii) If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 4, 6, 8}, B = {2, 3, 5, 7}
verify that (A ∩ B)’ = A’ U B’
Solution: Given, U = {1, 2, 3, 4, 5, 6, 7, 8, 9},
A = {2, 4, 6, 8}, B = {2, 3, 5, 7}
A ∩ B
= {2, 4, 6, 8} ∩ {2, 3, 5, 7}
= {2}
L.H.S.
(A ∩ B)’
= U – (A ∩ B)
⇒ {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2}
= {1, 3, 4, 5, 6, 7, 8, 9}
A’ = U – A
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 4, 6, 8}
= {1, 3, 5, 7, 9}
B’ = U – B
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 3, 5, 7}
= {1, 4, 6, 8, 9}
R. H.S. = A’ U B’
⇒ {1, 3, 5, 7, 9} U {1, 4, 6, 8, 9}
= {1, 3, 4, 5, 6, 7, 8, 9}
= L.H.S.
∴ (A ∩ B)’ = A’ U B’ (Verified)
5. Draw appropriate Venn diagram for each of the following:
(i) (A U B)’.
(ii) A’ ∩ B’
(i) Solution:
(A ∪ B)’ denoted by green region.
Venn diagrams:
Two sets A and B are shown with the following Venn Diagrams.
A ∪ B denoted by the blue, orange and yellow region
(A ∪ B)’ denoted by green region.
(ii) Solution:
(A’ ∩ B’) = denoted by blue region.
Venn diagrams:
A’ denoted by whole region except A circle.
B’ denoted by whole region except B circle.
A’ ∩ B’ denoted by blue region
5. Draw appropriate Venn diagram for each of the following:
(iii) (A ∩ B)’
(iv) A’ U B’
(iii) Solution:
(A ∩ B)’ = denoted by green region.
Venn diagrams:
(A ∩ B) denoted by green region.
(A ∩ B)’ = denoted by whole region except the overlapping region.
(iv) Solution:
A’ ∪ B’ = denoted by whole region except green region.
Venn diagrams:
A’ denoted by whole region except A circle.
B’ denoted by whole region except B circle.
A’ ∪ B’ denoted by whole region except the green region.
6. 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: U = Set of all triangles in a plane. = {x: x is a triangle}
A = Set of all triangles with at least one angle different from 60°
= {x: x is a triangle with at least one angle different from 60°}
∴ A’ = U – A
= {x: x is a triangle} – {x: x is a triangle with at least one angle different from 60°}
⇒ {x: x is a triangle in which all angles are equal to 60°}
⇒ {x: x is a triangle whose all angle is 60°}
= {x: x is a equilateral triangle.
Thus, Thus A’ is the set of all equilateral triangles.
7. Fill in the blanks to make each of the following a true statement:
(i) A U A’ = ……… (ii) ϕ’ ∩ A = ………
(iii) A ∩ A’ = ………(iv) U’ ∩ A = …
Solution:
(i) A ∪ A′ = U
(ii) Ø′ ∩ A = U ∩ A = A
(iii) A ∩ A′ = Ø
(iv) U′ ∩ A = Ø ∩ A = Ø
Miscellaneous Exercise on Chapter 1
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}.
Solution: 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
∴ A = {2, 6}
B = {2, 4, 6},
C = {2, 4, 6, 8 ,…},
D = {6}
Hence A ⊂ B, A ⊂ C, B ⊂ C, D ⊂ A, D ⊂ B, D ⊂ C,
2. 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
(ii) A ⊂ B and B ∈ C then A ∈ C
(iii) If A ⊂ B and B ⊂ C then A ⊂ C
Solution:
(i) False;
Let A = {a, b}, B = {a, {a, b}, c}
Here, b ∈ A and {a, b} ∈ B i.e. A ∈ B but 2 ∉ B
(ii) False;
Let A = {1}, B = {1, 2} and C = {{1, 2}, 3}
Here, A ⊂ B and B ∈ C but {1} ∉ C
(iii) True;
Let x ∈ A
⇒ x ∈ B . . . [∴ A ⊂ B]
⇒ x ∈ C . . . [∴ B ⊂ C]
∴ A ⊂ C
2. 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.
(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
Solution:
(iv) False;
Let A = {1,2}, B = {2, 3}, C = {1, 3}
Here, 1 ∈ A but 1 ∉ B
∴ A ⊄ B
Again 2 ∈ B but 2 ∉ C
∴ B ⊄ C
However A ⊄ C
∴ The statement is false.
(v) False;
Let A = {1,2}, B = {2, 3}
Here, 1 ∈ A
1 ∈ A but 1 ∉ B
∴ A ⊄ B
However, 1 ∉ B
(vi) True;
If A ⊂ B all elements of A are also in B i.e., if any element is not in B then it will not also be in A
∴ x ∉ B then x ∉ A
Therefore, if A ⊂ B and x ∉ B , then x ∉ A is true
3. Let A, B, and C be the sets such that A U B = A U C and A ∩ B = A ∩ C. Show that B = C
Solution: A U B = A U C
Taking intersection with B on both sides,
(A U B) ∩ B = (A U C) ∩ B
⇒ (A ∩ B) U (B ∩ B) = (A ∩ B) U (C ∩ B)… [Apply distributive property]
= (A ∩ B) U B = (A ∩ B) U (B ∩ C) . . . . [∵ B ∩ B = B]
⇒ B = (A ∩ B) U (B ∩ C) . . . . [∵ B ⊆ (A ∩ B)] . . . (i)
A U B = A U C
Taking intersection with C on both sides,
(A U B) ∩ C = (A U C) ∩ C
⇒ (A ∩ C) U (B ∩ C) = (A ∩ C) U (C ∩ C)… [Apply distributive property]
⇒ (A ∩ C) U (B ∩ C) = (A ∩ C) U C . . . . [∵ C ∩ C = C] ⇒ (A ∩ B) U (B ∩ C) = (A ∩ C) U C . . . . [∵ A ∩ B = A ∩ C] . . . . (i)
⇒ (A ∩ B) U (B ∩ C) = C . . . . (ii)
From (i) and (ii)
we get B = C
4. Show that the following four conditions are equivalent:
(i) A ⊂ B (ii) A – B = ϕ
(iii) A U B = B (iv) A ∩ B = A
Solution: A ⊂ B It means that every elements of A are in B.
A – B = Φ
Since A – B = Φ
∴ There is no element in A which is not in B.
Thus, (i) is equivalent to (ii)
A – B = Φ This implies that there are no element in A which are not in B.
So A ⊂ B
Since A ⊂ B Therefore A U B = B
Thus, (ii) is equivalent to (iii)
Since A U B = B
So A ⊂ B
Since A ⊂ B Therefore A ∩ B = A
Thus, (iii) is equivalent to (iv)
Since A ∩ B = A
So A ⊂ B
Thus, (iv) is equivalent to (i)
We prove (i) ⇔ (ii) ⇔ (iii) ⇔ (iv) ⇔ (i)
Therefore four conditions are equivalent.
5. Show that if A ⊂ B then C – B ⊂ C – A.
Solution: Let us consider x ∈ (C – B)
⇒ x ∈ C and x ∉ B
⇒ x ∈ C and x ∉ A . . . . [∵ A ⊂ B]
⇒ x ∈ C and x ∉ A
⇒ x ∈ (C – A)
Therefore, C – B ⊂ C – A
6. Show that for any sets A and B, A = (A ∩ B) U (A – B) and A U (B – A) = (A U B)
L.H.S.
= A ∪ (B – A)
= A ∪ (B ∩ Ac)
⇒ (A ∪ B) ∩ (A ∪ Ac) . . . [Apply distributive property]
= (A ∪ B) ∩ S
= (A ∪ B) = R.H.S.(Proved)
7. Using properties of sets, show that
(i) A U (A ∩ B) = A
(ii) A ∩ (A U B)= A.
Solution:
(i) L.H.S.
=A U (A ∩ B)
= (A U A) ∩ (A U B) . . . . [Use Distributive law]
= A ∩ (A U B) = A . . . . [∵ A ⊆ A U B]
∴ A U (A ∩ B) = A (Proved)
(ii) L.H.S.
= A ∩ (A U B)
= (A ∩ A) U (A ∩ B) . . . . [Use Distributive law]
= A U (A ∩ B) = A . . . . [∵ A ∩ B ⊂ A]
∴ A ∩ (A U B) = A (Proved)
8. Show A ∩ B = A ∩ C need not imply B = C
Solution: Let A = {a, b}, B = {b, c, d} and C = {b, e}
∴ A ∩ B = {a, b} ∩ B = {b, c, d} = {b} and
A ∩ C = {a, b} ∩ {b, e} = {b}
Here, A ∩ B = A ∩ C = {b}
However,
B ≠ C [Since e ∉ B but e ∈ C]
So A ∩ B = A ∩ C need not imply B = C
9. Let A and B be sets. If A ∩ X = B ∩ X = ϕ and A U X = B U X for some set X, show that A = B
Solution:
A = A ∩ (A U X)
= A ∩ (B U X) . . . . [∵ A U X = B U X]
= (A ∩ B) U (A ∩ X) . . . . [Use Distributive law]
= (A ∩ B) U ϕ . . . . [∵ A ∩ X = ϕ]
= (A ∩ B) ϕ . . . . (i)
B = B ∩ (B U X)
= B ∩ (A U X) . . . . [∵ A U X = B U X]
= (B ∩ A) U (B ∩ X) . . . . [Use Distributive law]
= (B ∩ A) U ϕ . . . . [∵ B ∩ X = ϕ]
= (B ∩ A) ϕ . . . . (ii)
From (ii) and (ii), we get
A = B (Proved)
10. 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 = {a, b}; B = {b, c } and C = {a, c}
We can see that A ∩ B = {b}, B ∩ C = {c} and A ∩ C = {a}
Hence, three sets are non-empty sets and A ∩ B ∩ C = ϕ .
However, A ∩ B ∩ C = {} = φ
Ans: Therefore, A = {a, b}; B = {b, c } and C = {a, c}.
- NCERT MATHS SOLUTION CLASS 11 SETS CHAPTER 1 EXERCISE 1.4
- NCERT MATHS SOLUTION CLASS 11 CHAPTER 1 SETS EXERCISE 1.1
- 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
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