Sets

Q1: Multiple Choice Type

i. If A = {3, 5, 7} and B = {5, 7, 9}, then \(n\left(A\cap B\right)\) is:

Step: \(A \cap B = \{5, 7\}\), so number of elements = 2
Answer: b. 2

ii. If n(universal set) = 80 and n(A) = 50, then n(complement of set A) is:

Step: \(n(A’) = 80 – 50 = 30\)
Answer: c. 30

iii. If A = {5, 8, 10} and empty set \(\phi\). Then \(\phi – A\) is equal to:

Step: Empty set minus any set is still empty.
Answer: b. \(\phi\)

iv. If set A = \({x:x \in W \text{ and } 0 < x \le 4}\), then set A is equal to:

Step: Whole numbers > 0 and ≤ 4 are: {1, 2, 3, 4}
Answer: c. {1, 2, 3, 4}

v. If set P = {factors of 8}, then set P is equal to:

Step: Factors of 8 = {1, 2, 4, 8}
Answer: c. {1, 2, 4, 8}

vi. The elements of the set \(\left\{x:x \in Z \text{ and } x^2 \le 9\right\}\) are:

Step: \(x^2 \le 9 \Rightarrow x = -3 \text{ to } 3\)
Answer: b. {-3, -2, -1, 0, 1, 2, 3}

vii. If A and B are equal sets, then A − B is eqaul to:

Step: If A = B, then A − B = \(\phi\)
Answer: d. { }

viii. If n(A) = n(B) then:

Step: Equal number of elements doesn’t imply equal sets
Answer: d. none of these

ix. A set has 5 elements, then number of its subsets is:

Step: Formula: \(2^n\), so \(2^5 = 32\)
Answer: a. \(2^5\)

x. Let M = {factors of 12}, N = {factors of 24}, then {24} is equal to:

Step: {24} ∈ N but not in M → {24} = N – M
Answer: d. \(N – M\)

xi. Statement 1: The number of subsets of {{1,{0}}, 2} is 8.
Statement 2: A set with ‘n’ elements has \(2^{n-1}\) proper subsets.

Analysis of Statement 1:Step 1: Let’s count the number of elements in the set \(\left\{\{1, \{0\}\}, 2\right\}\).
This set has two elements:
– One element is a set: \(\{1, \{0\}\}\)
– Another element is: \(2\)
So total elements = 2.
Step 2: Number of subsets of a set with \(n\) elements is \(2^n\).
Here, \(n = 2\), so number of subsets = \(2^2 = 4\)
But Statement 1 says 8, which is incorrect.
Analysis of Statement 2:
Step 1: A set with \(n\) elements has total subsets = \(2^n\)
Step 2: Proper subsets = \(2^n – 1\)
Statement 2 says: number of proper subsets = \(2^{n-1}\), which is false.
For example, if \(n = 3\):
– total subsets = \(2^3 = 8\)
– proper subsets = \(8 – 1 = 7\), not \(2^{3-1} = 4\)
Answer: b. Both the statements are false.

xii. Assertion (A): Let A = \(\left\{1,\ {\varphi}\right\}\), then each of \(\varphi,\ \left\{1\right\},\ \left\{\left\{\varphi \right\}\right\}\), is a proper subset of A.
Reason (R): The empty set has no proper subset.

Analysis:
ϕ (empty set) ⊂ A ✅
{1} ⊂ A ✅
{{ϕ}} ⊂ A ✅
So Assertion is True.
Reason is also True because ϕ has one subset (ϕ itself), but no proper subset ✅
Answer: b. Both A and R are correct, and R is not the correct explanation for A.

xiii. Assertion (A): Let A = {factors of 12) and B = {factors of 16}. Then B – A = {8, 16}.
Reason (R): B – A = \(\left\{x|x\ \in\ A\ but\ x\ \notin B\right\}\)

Analysis:
Factors of 12: {1, 2, 3, 4, 6, 12}
Factors of 16: {1, 2, 4, 8, 16}
B − A = {8, 16} ✅
But Reason says B − A = elements in A not in B ❌
Answer: c. A is true, but R is false.

xiv. Assertion (A): Let A = {1, 2, 3, 4, 5, 6}, and B = {1, 3, 5, 7, 9} then \(A\cap B\subseteq A\) and \(A\cap B\subseteq B\), always true for every pair of two sets.
Reason (R): For any sets A and B, we have \(A\cap B\subseteq A\) and \(A\cap B\subseteq B\).

Analysis:
A ∩ B = {1, 3, 5} ⊆ A and ⊆ B ✅
The Reason is the general rule and explains Assertion ✅
Answer: a. Both A and R are correct, and R is the correct explanation for A.

xv. Assertion (A): Let A = \(\left\{x|x+3=0,x\in N\right\}\), B = \(\left\{x|x\le3,x\in W\right\}\) then \(A\cap B=B\)
Reason (R): For any set A, \(A\cap\varphi=\varphi\).

Analysis:
A = ∅ because x = -3 ∉ N ⇒ A = ϕ
B = {0, 1, 2, 3}
A ∩ B = ϕ ≠ B ❌
Assertion is false
Reason is a correct general property ✅
Answer: d. A is false, but R is true.

Q2: If universal set = {all digits in our number system} and set A = {2, 3, 7, 9}. Write the complement of set A.

Step 1: Identify the universal set.
Digits in our number system = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
Step 2: Set A = {2, 3, 7, 9}
Step 3: Complement of set A (denoted by A′) = Elements in universal set but not in A
A′ = {0, 1, 4, 5, 6, 8}
Answer: A′ = {0, 1, 4, 5, 6, 8}

Q3: If A = {factors of 36} and B = {factors of 48}, find:

i. \(A \cup B\)

Step 1: List all the factors of 36.
A = {1, 2, 3, 4, 6, 9, 12, 18, 36}
Step 2: List all the factors of 48.
B = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}
Step 3: Union of A and B means combining all elements without repetition.
A ∪ B = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48}
Answer: A ∪ B = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48}

ii. \(A \cap B\)

Step 1: Intersection means common elements of A and B.
A ∩ B = {1, 2, 3, 4, 6, 12}
Answer: A ∩ B = {1, 2, 3, 4, 6, 12}

iii. \(A – B\)

Step 1: A – B means elements in A but not in B.
A – B = {9, 18, 36}
Answer: A – B = {9, 18, 36}

iv. \(B – A\)

Step 1: B – A means elements in B but not in A.
B – A = {8, 16, 24, 48}
Answer: B – A = {8, 16, 24, 48}

Q4: By taking the sets of your own, verify that:

i. \(n(A – B) = n(A \cup B) – n(B)\)

Step 1: Let A = {2, 4, 6, 8}, B = {4, 8, 10}
A ∪ B = {2, 4, 6, 8, 10} → n(A ∪ B) = 5
B = {4, 8, 10} → n(B) = 3
A – B = {2, 6} → n(A – B) = 2
Step 2: Now verify:
LHS = n(A – B) = 2
RHS = n(A ∪ B) – n(B) = 5 – 3 = 2
Answer: Verified: LHS = RHS = 2

ii. \(n(A \cap B) + n(A \cup B) = n(A) + n(B)\)

Step 1: Use same sets:
A = {2, 4, 6, 8} → n(A) = 4
B = {4, 8, 10} → n(B) = 3
A ∪ B = {2, 4, 6, 8, 10} → n(A ∪ B) = 5
A ∩ B = {4, 8} → n(A ∩ B) = 2
Step 2: Now verify:
LHS = n(A ∩ B) + n(A ∪ B) = 2 + 5 = 7
RHS = n(A) + n(B) = 4 + 3 = 7
Answer: Verified: LHS = RHS = 7

Q5: If n(A − B) = 24, n(B − A) = 32 and \(n(A \cap B) = 10\); find \(n(A \cup B)\)

Step 1: Use the identity: \[ n(A \cup B) = n(A – B) + n(B – A) + n(A \cap B) \]Step 2: Substitute the values: \[ n(A \cup B) = 24 + 32 + 10 = 66 \]Answer: \(n(A \cup B) = 66\)

Q6: If \(\xi=\left\{x:x\in\mathbb{N},x\le10\right\}\), A = \(\left\{x:x\geq5\right\}\) and B = \(\left\{x:3 < x < 6\right\}\), then find:

Universal Set: \[ \xi = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \]Set A: \[ A = \{5, 6, 7, 8, 9, 10\} \]Set B: \[ B = \{4, 5\} \]

i. \((A \cup B)’\):

\[ A \cup B = \{4, 5, 6, 7, 8, 9, 10\} \\ (A \cup B)’ = \xi – (A \cup B) = \{1, 2, 3\} \]Answer: \((A \cup B)’ = \{1, 2, 3\}\)

ii. \(A’ \cap B’\):

\[ A’ = \xi – A = \{1, 2, 3, 4\} \\ B’ = \xi – B = \{1, 2, 3, 6, 7, 8, 9, 10\} \\ A’ \cap B’ = \{1, 2, 3, 4\} \cap \{1, 2, 3, 6, 7, 8, 9, 10\} = \{1, 2, 3\} \]Answer: \(A’ \cap B’ = \{1, 2, 3\}\)

Are \(\left(A\cup B\right)’\) and \(A’\cap B’\) equal?

\[ (A \cup B)’ = A’ \cap B’ \]Answer: Yes, they are equal.

Q7: Write the elements of the set \(\left\{x:x=3y-1,\ y\in N\ and\ 8 < y \leq 12\right\}\).

Step 1: Identify values of \(y\) in natural numbers such that \(8 < y \leq 12\).
So, \(y = 9, 10, 11, 12\).
Step 2: Calculate \(x = 3y – 1\) for each \(y\).
For \(y=9:\quad x = 3(9) – 1 = 27 – 1 = 26\)
For \(y=10:\quad x = 3(10) – 1 = 30 – 1 = 29\)
For \(y=11:\quad x = 3(11) – 1 = 33 – 1 = 32\)
For \(y=12:\quad x = 3(12) – 1 = 36 – 1 = 35\)
Step 3: Write the set elements.
The set is \(\{26, 29, 32, 35\}\).
Answer: {26, 29, 32, 35}

Q8: If universal set = \(\left\{x : x \in \mathbb{Z}, -2 \leq x < 4 \right\}\),
A = \(\left\{x : -1 \leq x < 3 \right\}\),
B = \(\left\{x : 0 < x < 4 \right\}\),
C = \(\left\{x : -2 \leq x \leq 0 \right\}\);
Show that: \(A – \left(B \cup C\right) = (A – B) \cap (A – C)\).

Step 1: Write the universal set \(\xi\)
Universal set: \(\xi = \{-2, -1, 0, 1, 2, 3\}\)
Step 2: Find the sets A, B, and C explicitly \[ (A = \{x : -1 \leq x < 3\} = \{-1, 0, 1, 2\} \\ B = \{x : 0 < x < 4\} = \{1, 2, 3\} \\ C = \{x : -2 \leq x \leq 0\} = \{-2, -1, 0\} \]Step 3: Find \(B \cup C\) \[ B \cup C = \{1, 2, 3\} \cup \{-2, -1, 0\} = \{-2, -1, 0, 1, 2, 3\} \]Step 4: Find \(A – (B \cup C)\) \[ A – (B \cup C) = \{x \in A : x \notin B \cup C \} = \{-1, 0, 1, 2\} – \{-2, -1, 0, 1, 2, 3\} \]Step 5: Find \(A – B\) and \(A – C\) \[ A – B = \{-1, 0, 1, 2\} – \{1, 2, 3\} = \{-1, 0\} \\ A – C = \{-1, 0, 1, 2\} – \{-2, -1, 0\} = \{1, 2\} \]Step 6: Find \((A – B) \cap (A – C)\) \[ (A – B) \cap (A – C) = \{-1, 0\} \cap \{1, 2\} \]Step 7: Compare both sides \[ A – (B \cup C) = (A – B) \cap (A – C) \]Both sides are equal.
Answer: \(A – (B \cup C) = (A – B) \cap (A – C)\) is true.

Q9: Let set A = {x : x ∈ ℤ and x² – 9 = 0}, B = {x : x ∈ W and x² – 16 < 0}; Find:

i. A ∪ B

Step 1: Solve for set A.
We are given:
  x² – 9 = 0
 ⇒ x² = 9
 ⇒ x = ±3
Since x ∈ ℤ (integers),
So,
  A = {–3, 3}
Step 2: Solve for set B.
We are given:
  x² – 16 < 0
 ⇒ x² < 16
Also x ∈ W (Whole numbers)
So we find all natural numbers x such that x² < 16
Whole numbers: 0, 1, 2, 3, 4, 5…
Test squares:
0² = 0 ✅
1² = 1 ✅
2² = 4 ✅
3² = 9 ✅
4² = 16 ❌ (not < 16)
So, B = {0, 1, 2, 3}
Step 3: Find A ∪ B
A = {–3, 3}, B = {1, 2, 3}
Union means combine all elements without repetition.
 ⇒ A ∪ B = {–3, 1, 2, 3}
Answer: A ∪ B = {–3, 1, 2, 3}

ii. A ∩ B

Step 1: From previous part:
A = {–3, 3}
B = {1, 2, 3}
Step 2: Find common elements in A and B. Only common element = 3
Answer: A ∩ B = {3}