Question 1:
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
Answer:
(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 …}
As 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 …}
∴ 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}
∴ A∪ B = {x: x ∈ N and 1 < x < 10}
(v) A = {1, 2, 3}, B = Φ
A∪ B = {1, 2, 3}
Question 2:
Let A = {a, b}, B = {a, b, c}. Is A ⊂ B? What is A ∪ B?
Answer:
Here, A = {a, b} and B = {a, b, c}
Yes, A ⊂ B.
A∪ B = {a, b, c} = B
Question 3:
If A and B are two sets such that A ⊂ B, then what is A ∪ B?
Answer:
If A and B are two sets such that A ⊂ B, then A ∪ B = B.
Question 4:
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
Answer:
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}
Page No 18:
Question 5:
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
Answer:
(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
Question 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
(v) B ∩ D
(vi) A ∩ (B ∪ C)
(vii) A ∩ D
(viii) A ∩ (B ∪ D)
(ix) (A ∩ B) ∩ (B ∪ C)
(x) (A ∪ D) ∩ (B ∪ C)
Answer:
(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}
Question 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
(i) A ∩ B
(ii) A ∩ C
(iii) A ∩ D
(iv) B ∩ C
(v) B ∩ D
(vi) C ∩ D
Answer:
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}
Question 8:
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}
Answer:
(i) {1, 2, 3, 4}
{x: x is a natural number and 4 ≤ x ≤ 6} = {4, 5, 6}
Now, {1, 2, 3, 4} ∩ {4, 5, 6} = {4}
Therefore, this pair of sets is not disjoint.
(ii) {a, e, i, o, u} ∩ (c, d, e, f} = {e}
Therefore, {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
Therefore, this pair of sets is disjoint.
Question 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}, 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
Answer:
(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}
Question 10:
If X = {a, b, c, d} and Y = {f, b, d, g}, find
(i) X – Y
(ii) Y – X
(iii) X ∩ Y
Answer:
(i) X – Y = {a, c}
(ii) Y – X = {f, g}
(iii) X ∩ Y = {b, d}
Question 11:
If R is the set of real numbers and Q is the set of rational numbers, then what is R – Q?
Answer:
R: set of real numbers
Q: set of rational numbers
Therefore, R – Q is a set of irrational numbers.
Question 12:
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.
Answer:
(i) False
As 3 ∈ {2, 3, 4, 5}, 3 ∈ {3, 6}
⇒ {2, 3, 4, 5} ∩ {3, 6} = {3}
(ii) False
As a ∈ {a, e, i, o, u}, a ∈ {a, b, c, d}
⇒ {a, e, i, o, u } ∩ {a, b, c, d} = {a}
(iii) True
As {2, 6, 10, 14} ∩ {3, 7, 11, 15} = Φ
(iv) True
As {2, 6, 10} ∩ {3, 7, 11} = Φ
1.5 exercise
Ex 1.5 Class 11 Maths Question 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 ∪ C)’
(iv) (A ∪B)’
(v) (A’)’
(vi) (B – C)’
Solution.
Here 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}
(i) A’=U – A
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4}
= {5, 6, 7, 8, 9}
6, 7, 8, 9}
(ii) B’=U – B
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 4, 6, 8}
= {1, 3, 5, 7, 9}
(iii) A ∪ C = {1, 2, 3, 4} ∪ {3, 4, 5, 6}
= (1, 2, 3, 4, 5, 6}
(A∪C)’=U-(A∪C)
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4, 5, 6}
= {7, 8, 9}
(iv) A ∪ B = {1, 2, 3,4} ∪ {2, 4, 6, 8}
= {1, 2, 3, 4, 6, 8}
(A∪B)’ = U – (A∪B)
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4, 6, 8}
= {5, 7, 9}
(v) We know that A’ = {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) B – C = {2, 4, 6, 8} – {3, 4, 5, 6} = {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}.
Ex 1.5 Class 11 Maths Question 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) A’ = U – A = {a, b, c, d, e, f, g, h} – {a, b, c}
= {d, e,f, g, h}
(ii) B’ = U – B = {a, b, c, d, e,f, g, h} – {d, e, f, g}
= {a, b, c, h}
(iii) C’ = U – C = {a, b, c, d, e, f, g, h} – {a, c, e, g}
= {b, d, f, h}
(iv) D’ = U – D = {a, b, c, d, e, f, g, h} – {f, g, h, a}
= {b, c, d, e}.
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 number}
(ii) {x: x is an odd natural number}
(iii) {x: x is a positive multiple of 3}
(iv) {x: x is a prime number}
(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}
(ix) (x: 2x + 5 = 9)
(x) {x: x ≥ 7}
(xi) {x: x ∈ W and 2x + 1 > 10}
Ex 1.5 Class 11 Maths Question 4.
If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 4, 6, 8} and B = {2, 3, 5, 7}. Verify that
(i) (A ∪ B)’ = A’∩B’
(ii) (A ∩ B)’ = A’∪B’
Ex 1.5 Class 11 Maths Question 5.
Draw appropriate Venn diagram for each of the following:
(i) (A ∪ B)’
(ii) A’∩B’
(iii) (A ∩ B)’
(iv) A’ ∪ B’
Ex 1.5 Class 11 Maths Question 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.
Here U = {x : x is a triangle}
A = {x: x is a triangle and has at least one angle different from 60°}
∴ A’ = U – A = {x : x is a triangle} – {x : x is a triangle and has atleast one angle different from 60°}
= {x : x is a triangle and has all angles equal to 60°)
= Set of all equilateral triangles.
1.5 Class 11 Maths Question 7.
Fill in the blanks to make each of the following a true statement:
(i) A ∪ A’ = …….
(ii) φ’ ∩ A = .…….
(iii) A ∩ A’ = …….
(iv) U’ ∩ A = .…….
Solution.
(i) A ∪ A’= U
(ii) φ’ ∩ A = U ∩ A = A
(iii) A ∩ A’ = φ
(iv) U’ ∩ A = φ ∩ A = φ
1.7 exercise
1. For any two sets A and B, prove that: A‘ – B‘ = B – A
Solution:
To prove, A’ – B’ = B – A
Firstly we need to show
A’ – B’ ⊆ B – A
Let, x ∈ A’ – B’
⇒ x ∈ A’ and x ∉ B’
⇒ x ∉ A and x ∈ B (since, A ∩ A’ = ϕ )
⇒ x ∈ B – A
It is true for all x ∈ A’ – B’
∴ A’ – B’ = B – A
Hence Proved.
2. For any two sets A and B, prove the following:
(i) A ∩ (A‘ ∪ B) = A ∩ B
(ii) A – (A – B) = A ∩ B
(iii) A ∩ (A ∪ B’) = ϕ
(iv) A – B = A Δ (A ∩ B)
Solution:
(i) A ∩ (A’ ∪ B) = A ∩ B
Let us consider LHS A ∩ (A’ ∪ B)
Expanding
(A ∩ A’) ∪ (A ∩ B)
We know, (A ∩ A’) =ϕ
⇒ ϕ ∪ (A∩ B)
⇒ (A ∩ B)
∴ LHS = RHS
Hence proved.
(ii) A – (A – B) = A ∩ B
For any sets A and B we have De-Morgan’s law
(A ∪ B)’ = A’ ∩ B’, (A ∩ B) ‘ = A’ ∪ B’
Consider LHS
= A – (A–B)
= A ∩ (A–B)’
= A ∩ (A∩B’)’
= A ∩ (A’ ∪ B’)’) (since, (B’)’ = B)
= A ∩ (A’ ∪ B)
= (A ∩ A’) ∪ (A ∩ B)
= ϕ ∪ (A ∩ B) (since, A ∩ A’ = ϕ)
= (A ∩ B) (since, ϕ ∪ x = x, for any set)
= RHS
∴ LHS=RHS
Hence proved.
(iii) A ∩ (A ∪ B’) = ϕ
Let us consider LHS A ∩ (A ∪ B’)
= A ∩ (A ∪ B’)
= A ∩ (A’∩ B’) (By De–Morgan’s law)
= (A ∩ A’) ∩ B’ (since, A ∩ A’ = ϕ)
= ϕ ∩ B’
= ϕ (since, ϕ ∩ B’ = ϕ)
= RHS
∴ LHS=RHS
Hence proved.
(iv) A – B = A Δ (A ∩ B)
Let us consider RHS A Δ (A ∩ B)
A Δ (A ∩ B) (since, E Δ F = (E–F) ∪ (F–E))
= (A – (A ∩ B)) ∪ (A ∩ B –A) (since, E – F = E ∩ F’)
= (A ∩ (A ∩ B)’) ∪ (A ∩ B ∩ A’)
= (A ∩ (A’ ∪ B’)) ∪ (A ∩ A’ ∩ B) (by using De-Morgan’s law and associative law)
= (A ∩ A’) ∪ (A ∩ B’) ∪ (ϕ ∩ B) (by using distributive law)
= ϕ ∪ (A ∩ B’) ∪ ϕ
= A ∩ B’ (since, A ∩ B’ = A–B)
= A – B
= LHS
∴ LHS=RHS
Hence Proved
3. If A, B, C are three sets such that A ⊂ B, then prove that C – B ⊂ C – A.
Solution:
Given, ACB
To prove: C – B ⊂ C – A
Let us consider, x ∈ C–B
⇒ x ∈ C and x ∉ B
⇒ x ∈ C and x ∉ A
⇒ x ∈ C – A
Thus, x ∈ C–B ⇒ x ∈ C – A
This is true for all x ∈ C–B
∴ C – B ⊂ C – A
Hence proved.
4. For any two sets A and B, prove that
(i) (A ∪ B) – B = A – B
(ii) A – (A ∩ B) = A – B
(iii) A – (A – B) = A ∩ B
(iv) A ∪ (B – A) = A ∪ B
(v) (A – B) ∪ (A ∩ B) = A
Solution:
(i) (A ∪ B) – B = A – B
Let us consider LHS (A ∪ B) – B
= (A–B) ∪ (B–B)
= (A–B) ∪ ϕ (since, B–B = ϕ)
= A–B (since, x ∪ ϕ = x for any set)
= RHS
Hence proved.
(ii) A – (A ∩ B) = A – B
Let us consider LHS A – (A ∩ B)
= (A–A) ∩ (A–B)
= ϕ ∩ (A – B) (since, A-A = ϕ)
= A – B
= RHS
Hence proved.
(iii) A – (A – B) = A ∩ B
Let us consider LHS A – (A – B)
Let, x ∈ A – (A–B) = x ∈ A and x ∉ (A–B)
x ∈ A and x ∉ (A ∩ B)
= x ∈ A ∩ (A ∩ B)
= x ∈ (A ∩ B)
= (A ∩ B)
= RHS
Hence proved.
(iv) A ∪ (B – A) = A ∪ B
Let us consider LHS A ∪ (B – A)
Let, x ∈ A ∪ (B –A) ⇒ x ∈ A or x ∈ (B – A)
⇒ x ∈ A or x ∈ B and x ∉ A
⇒ x ∈ B
⇒ x ∈ (A ∪ B) (since, B ⊂ (A ∪ B))
This is true for all x ∈ A ∪ (B–A)
∴ A ∪ (B–A) ⊂ (A ∪ B)…… (1)
Conversely,
Let x ∈ (A ∪ B) ⇒ x ∈ A or x ∈ B
⇒ x ∈ A or x ∈ (B–A) (since, B ⊂ (A ∪ B))
⇒ x ∈ A ∪ (B–A)
∴ (A ∪ B) ⊂ A ∪ (B–A)…… (2)
From 1 and 2 we get,
A ∪ (B – A) = A ∪ B
Hence proved.
(v) (A – B) ∪ (A ∩ B) = A
Let us consider LHS (A – B) ∪ (A ∩ B)
Let, x ∈ A
Then either x ∈ (A–B) or x ∈ (A ∩ B)
⇒ x ∈ (A–B) ∪ (A ∩ B)
∴ A ⊂ (A – B) ∪ (A ∩ B)…. (1)
Conversely,
Let x ∈ (A–B) ∪ (A ∩ B)
⇒ x ∈ (A–B) or x ∈ (A ∩ B)
⇒ x ∈ A and x ∉ B or x ∈ B
⇒ x ∈ A
(A–B) ∪ (A ∩ B) ⊂ A………. (2)
∴ From (1) and (2), We get
(A–B) ∪ (A ∩ B) = A
Hence proved.
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