Class 11 Sets NCERT solution|NCERT Solutions for Class 11 Maths Chapter 1 Sets| Sets class 11 1.4 exercise solutions|VECTOR STUDIES
NCERT Solutions for Class 11 Maths Chapter 1- Sets
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 …}
Hence, 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}
Hence, 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:
It is given that
A = {a, b} and B = {a, b, c}
Yes, A ⊂ B
So the union of the pairs of set can be written as
A∪ B = {a, b, c} = B
Q3. If A and B are two sets such that A ⊂ B, then what is A ∪ B?
Solution:
If A and B are two sets such that A ⊂ B, then 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:
It is given that
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:
(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} = {2, 3, 4, 5, 6}
B = {x: x is a natural number and 6 < x < 10} = {7, 8, 9}
A ∩ B = Φ
(v) A = {1, 2, 3}, B = Φ
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 ∩ B = {7, 9, 11}
(ii) B ∩ C = {11, 13}
(iii) A ∩ C ∩ D = {A ∩ C} ∩ D
= {11} ∩ {15, 17}= Φ
(iv) A ∩ C = {11}
(v) B ∩ D = Φ
(vi) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
= {7, 9, 11} ∪ {11}= {7, 9, 11}
(vii) A ∩ D = Φ
(viii) A ∩ (B ∪ D) = (A ∩ B) ∪ (A ∩ D)
= {7, 9, 11} ∪ Φ= {7, 9, 11}
(ix) (A ∩ B) ∩ (B ∪ C) = {7, 9, 11} ∩ {7, 9, 11, 13, 15}
= {7, 9, 11}
(x) (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:
It can be written as
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 …}
(i) A ∩B = {x: x is a even natural number} = B
(ii) A ∩ C = {x: x is an odd natural number} = C
(iii) A ∩ D = {x: x is a prime number} = D
(iv) B ∩ C = Φ
(v) B ∩ D = {2}
(vi) 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}
Solution:
(i) {1, 2, 3, 4}
{x: x is a natural number and 4 ≤ x ≤ 6} = {4, 5, 6}
So we get
{1, 2, 3, 4} ∩ {4, 5, 6} = {4}
Hence, this pair of sets is not disjoint.
(ii) {a, e, i, o, u} ∩ (c, d, e, f} = {e}
Hence, {a, e, i, o, u} and (c, d, e, f} are not disjoint.
(iii) {x: x is an even integer} ∩ {x: x is an odd integer} = Φ
Hence, 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:
We know that
R – Set of real numbers
Q – Set of rational numbers
Hence, 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.
(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
If 3 ∈ {2, 3, 4, 5}, 3 ∈ {3, 6}
So we get {2, 3, 4, 5} ∩ {3, 6} = {3}
(ii) False
If a ∈ {a, e, i, o, u}, a ∈ {a, b, c, d}
So we get {a, e, i, o, u} ∩ {a, b, c, d} = {a}
(iii) True
Here {2, 6, 10, 14} ∩ {3, 7, 11, 15} = Φ
(iv) True
Here {2, 6, 10} ∩ {3, 7, 11} = Φ
For notes of this chapter visit-https://vectorstudiesindia.blogspot.com/2022/07/class-11-set-chapter-notesclass-11.html
