Miscellaneous Exercise on Chapter 1 - Sets - Complete Solutions

📚 Miscellaneous Exercise — Chapter 1

Sets — Complete Step-by-Step Solutions

1 Decide which sets are subsets of one and another
A = {x : x ∈ ℝ and x satisfies x² − 8x + 12 = 0}
B = {2, 4, 6},   C = {2, 4, 6, 8, …},   D = {6}
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1 Solve for Set A
x² − 8x + 12 = 0
(x − 2)(x − 6) = 0
A = {2, 6}
2 List All Sets
A = {2, 6},   B = {2, 4, 6},   C = {2, 4, 6, 8, …},   D = {6}
3 Determine Subset Relations
A ⊂ B    (2, 6 both in B)
A ⊂ C    (2, 6 both in C)
B ⊂ C    (2, 4, 6 all in C)
D ⊂ A    (6 ∈ A)
D ⊂ B    (6 ∈ B)
D ⊂ C    (6 ∈ C)
A ⊂ B, A ⊂ C, B ⊂ C, D ⊂ A, D ⊂ B, D ⊂ C
2 Determine whether each statement is True or False. Prove or give a counterexample.
(i) If x ∈ A and A ∈ B, then x ∈ B
(ii) If A ⊂ B and B ∈ C, then A ∈ C
(iii) If A ⊂ B and B ⊂ C, then A ⊂ C
(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
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i If x ∈ A and A ∈ B, then x ∈ B
FALSE
Counterexample: Let A = {1}, B = {{1}, 2}
Clearly 1 ∈ A and A ∈ B, but 1 ∉ B.
∴ x ∈ A and A ∈ B does not imply x ∈ B.
ii If A ⊂ B and B ∈ C, then A ∈ C
FALSE
Counterexample: Let A = {1}, B = {1,2}, C = {{1,2}, 3}
A ⊂ B and B ∈ C but A ∉ C.
iii If A ⊂ B and B ⊂ C, then A ⊂ C
TRUE
Let x ∈ A
⇒ x ∈ B    [∵ A ⊂ B]
⇒ x ∈ C    [∵ B ⊂ C]
∴ x ∈ A ⇒ x ∈ C for all x ∈ A, therefore A ⊂ C.
iv If A ⊄ B and B ⊄ C, then A ⊄ C
FALSE
Counterexample: A = {1,2}, B = {2,3}, C = {1,2,5}
A ⊄ B (since 1 ∈ A, 1 ∉ B) and B ⊄ C (since 3 ∈ B, 3 ∉ C), but A ⊂ C.
v If x ∈ A and A ⊄ B, then x ∈ B
FALSE
Counterexample: A = {1,2}, B = {2,3}
1 ∈ A and A ⊄ B, but 1 ∉ B.
vi If A ⊂ B and x ∉ B, then x ∉ A
TRUE
Let A ⊂ B and x ∉ B. Suppose x ∈ A.
Then x ∈ A and A ⊂ B ⇒ x ∈ B — contradiction!
∴ Our supposition is wrong. Hence x ∉ A.
3 Let A ∪ B = A ∪ C and A ∩ B = A ∩ C. Show that B = C.
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1 Prove B ⊂ C
Let x ∈ B
⇒ x ∈ A ∪ B    [∵ B ⊂ A ∪ B always]
⇒ x ∈ A ∪ C    [∵ A ∪ B = A ∪ C]
⇒ x ∈ A or x ∈ C

Case I: x ∈ A
Then x ∈ A and x ∈ B ⇒ x ∈ A ∩ B ⇒ x ∈ A ∩ C ⇒ x ∈ C

Case II: x ∈ C
In each case x ∈ C. ∴ B ⊂ C    ...(i)
2 Prove C ⊂ B (Similarly)
By symmetry (interchanging B and C in the argument above),
C ⊂ B    ...(ii)
From (i) and (ii): B = C
4 Show that the following four conditions are equivalent: (i) A ⊂ B   (ii) A − B = φ   (iii) A ∪ B = B   (iv) A ∩ B = A
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1 (i) ⟹ (ii): A ⊂ B ⟹ A − B = φ
Suppose A − B ≠ φ. Let x ∈ A − B ⇒ x ∈ A and x ∉ B.
But A ⊂ B ⇒ x ∈ B — contradiction!
∴ A − B = φ
2 (i) ⟹ (iii): A ⊂ B ⟹ A ∪ B = B
Let x ∈ A ∪ B ⇒ x ∈ A or x ∈ B ⇒ x ∈ B or x ∈ B (since A ⊂ B)
⇒ x ∈ B ⇒ A ∪ B ⊂ B
Also B ⊂ A ∪ B always. ∴ A ∪ B = B
3 (i) ⟹ (iv): A ⊂ B ⟹ A ∩ B = A
We know A ∩ B ⊂ A always.
Now let x ∈ A ⇒ x ∈ B [∵ A ⊂ B] ⇒ x ∈ A ∩ B
∴ A ⊂ A ∩ B. Therefore A ∩ B = A
💡 Key Insight: All four conditions are logically equivalent — proving any one of them from A ⊂ B establishes all four.
5 Show that if A ⊂ B, then C − B ⊂ C − A.
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1 Start with an element of C − B
Given: A ⊂ B
Let x ∈ (C − B)
⇒ x ∈ C and x ∉ B
2 Use A ⊂ B to conclude
Since A ⊂ B and x ∉ B ⇒ x ∉ A
∴ x ∈ C and x ∉ A ⇒ x ∈ (C − A)
∴ (C − B) ⊂ (C − A)   ■
6 Assume that P(A) = P(B). Show that A = B.
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1 Prove A ⊂ B
Let x ∈ A
⇒ {x} ⊂ A ⇒ {x} ∈ P(A)
⇒ {x} ∈ P(B)    [∵ P(A) = P(B)]
⇒ {x} ⊂ B ⇒ x ∈ B
∴ A ⊂ B
2 Prove B ⊂ A (Similarly)
By identical argument (interchanging A and B):
B ⊂ A
Since A ⊂ B and B ⊂ A, we have A = B   ■
7 Is it true that for any sets A and B, P(A) ∪ P(B) = P(A ∪ B)? Justify your answer.
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1 Take a Counterexample
Let A = {1}, B = {2}
P(A) = {φ, {1}} and P(B) = {φ, {2}}
P(A) ∪ P(B) = {φ, {1}, {2}}
2 Compute P(A ∪ B)
A ∪ B = {1, 2}
P(A ∪ B) = {φ, {1}, {2}, {1,2}}
3 Compare
{1, 2} ∈ P(A ∪ B) but {1, 2} ∉ P(A) ∪ P(B)
∴ P(A) ∪ P(B) ≠ P(A ∪ B)
No, it is NOT true in general. P(A) ∪ P(B) ≠ P(A ∪ B)
8 Show that for any sets A and B: A = (A ∩ B) ∪ (A − B) and A ∪ (B − A) = A ∪ B
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1 Prove A = (A ∩ B) ∪ (A − B)
(A ∩ B) ∪ (A − B) = (A ∩ B) ∪ (A ∩ B′)
= A ∩ (B ∪ B′)    [Distributive Law]
= A ∩ U = A
2 Prove A ∪ (B − A) = A ∪ B
A ∪ (B − A) = A ∪ (B ∩ A′)
= (A ∪ B) ∩ (A ∪ A′)    [Distributive Law]
= (A ∪ B) ∩ U = A ∪ B
Both identities are proved using Distributive Law ■
9 Using properties of sets, show that: (i) A ∪ (A ∩ B) = A   (ii) A ∩ (A ∪ B) = A
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1 Prove A ∪ (A ∩ B) = A
We know: X ⊂ Y ⇒ X ∪ Y = Y (the superset)
Here A ∩ B ⊂ A
∴ A ∪ (A ∩ B) = A
2 Prove A ∩ (A ∪ B) = A
We know: X ⊂ Y ⇒ X ∩ Y = X (the subset)
Here A ⊂ A ∪ B
∴ A ∩ (A ∪ B) = A
💡 Key Insight: These are called Absorption Laws — A "absorbs" any set obtained by intersecting or unioning with itself.
10 Show that A ∩ B = A ∩ C need not imply B = C.
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1 Counterexample
Let A = {1, 2, 3}, B = {2, 4}, C = {2, 5}
A ∩ B = {2} and A ∩ C = {2}
∴ A ∩ B = A ∩ C = {2}
But B = {2, 4} ≠ {2, 5} = C
A ∩ B = A ∩ C does NOT necessarily imply B = C ■
11 If A ∩ X = B ∩ X = φ and A ∪ X = B ∪ X for some set X, show that A = B.
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1 Express A using given conditions
A ∩ (A ∪ X) = A    [∵ A ⊂ A ∪ X always]
= A ∩ (B ∪ X)    [∵ A ∪ X = B ∪ X]
= (A ∩ B) ∪ (A ∩ X)    [Distributive Law]
= (A ∩ B) ∪ φ    [∵ A ∩ X = φ]
= A ∩ B
∴ A = A ∩ B    ...(i)
2 Interchange A and B
By the same argument (interchanging A and B):
B = B ∩ A = A ∩ B    ...(ii) [Commutative Law]
From (i) and (ii): A = A ∩ B = B   ∴ A = B   ■
12 Find sets A, B and C such that A ∩ B, B ∩ C and A ∩ C are non-empty but A ∩ B ∩ C = φ.
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1 Choose the Sets
Let A = {x, y},   B = {x, z},   C = {y, z}
2 Verify Pairwise Intersections
A ∩ B = {x} ≠ φ ✓
B ∩ C = {z} ≠ φ ✓
A ∩ C = {y} ≠ φ ✓
3 Verify Triple Intersection
A ∩ B ∩ C = {x,y} ∩ {x,z} ∩ {y,z} = φ ✓
A = {x, y}, B = {x, z}, C = {y, z} satisfies all conditions ■
13 In a survey of 600 students, 150 drink tea, 225 drink coffee, 100 drink both. Find how many drink neither.
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1 Define the Sets
Total students = 600
n(T) = 150,   n(C) = 225,   n(T ∩ C) = 100
2 Apply Union Formula
n(T ∪ C) = n(T) + n(C) − n(T ∩ C)
= 150 + 225 − 100 = 275
3 Find Neither
Neither = Total − n(T ∪ C)
= 600 − 275 = 325
325 students drink neither tea nor coffee
14 In a group, 100 know Hindi, 50 know English, 25 know both. Each knows at least one. How many in the group?
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1 Define and Apply Formula
n(H) = 100,   n(E) = 50,   n(H ∩ E) = 25
n(H ∪ E) = n(H) + n(E) − n(H ∩ E)
= 100 + 50 − 25 = 125
There are 125 students in the group
15 Survey of 60 people: H=25, T=26, I=26, H∩I=9, H∩T=11, T∩I=8, H∩T∩I=3. Find (i) at least one newspaper (ii) exactly one newspaper.
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1 Given Data
n(H)=25, n(T)=26, n(I)=26
n(H∩T)=11, n(H∩I)=9, n(T∩I)=8
n(H∩T∩I)=3
2 (i) At Least One Newspaper
n(H∪T∪I) = n(H)+n(T)+n(I) − n(H∩T) − n(H∩I) − n(T∩I) + n(H∩T∩I)
= 25+26+26 − 11 − 9 − 8 + 3 = 52
3 (ii) Exactly One Newspaper
Only H = n(H) − n(H∩T) − n(H∩I) + n(H∩T∩I) = 25−11−9+3 = 8
Only T = n(T) − n(T∩H) − n(T∩I) + n(H∩T∩I) = 26−11−8+3 = 10
Only I = n(I) − n(I∩H) − n(I∩T) + n(H∩T∩I) = 26−9−8+3 = 12
Total = 8 + 10 + 12 = 30
(i) 52 people read at least one newspaper
(ii) 30 people read exactly one newspaper
16 21 liked A, 26 liked B, 29 liked C. 14 liked A&B, 12 liked C&A, 14 liked B&C, 8 liked all three. Find how many liked C only.
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1 Define Variables
Let d = people who liked all three = 8
b = people who liked B and C only (not A)
c = people who liked C and A only (not B)
2 Solve for b and c
b + d = 14 ⇒ b = 14 − 8 = 6
c + d = 12 ⇒ c = 12 − 8 = 4
3 Find C Only
n(Z) = 29 (total who liked C)
z + b + c + d = 29
z + 6 + 4 + 8 = 29
z = 29 − 18 = 11
11 people liked product C only
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