Let A and B be two disjoint subsets of a universal set U. Then `(AuuB )nnB'=`

Options

A. A
B. B
C. `phi`
D. None of these

Answer

To solve the problem, we need to find the expression \( (A \cup B) \cap B' \) where \( A \) and \( B \) are disjoint subsets of a universal set \( U \).

### Step-by-step Solution:

1. **Understand the components**:
- \( A \) and \( B \) are disjoint sets, meaning \( A \cap B = \emptyset \).
- \( B' \) (or \( B^c \)) is the complement of set \( B \) in the universal set \( U \), which includes all elements in \( U \) that are not in \( B \).

2. **Apply the Distributive Law**:
- We can use the distributive law of set theory, which states that \( X \cap (Y \cup Z) = (X \cap Y) \cup (X \cap Z) \).
- Here, we can rewrite the expression \( (A \cup B) \cap B' \) as:
\[ (A \cup B) \cap B' = (A \cap B') \cup (B \cap B') \]

3. **Evaluate \( B \cap B' \)**:
- Since \( B' \) contains all elements not in \( B \), the intersection \( B \cap B' \) is empty:
\[ B \cap B' = \emptyset \]

4. **Evaluate \( A \cap B' \)**:
- Since \( A \) and \( B \) are disjoint, all elements of \( A \) are also in \( B' \). Therefore:
\[ A \cap B' = A \]

5. **Combine the results**:
- Now substituting back into our expression:
\[ (A \cup B) \cap B' = (A \cap B') \cup (B \cap B') = A \cup \emptyset = A \]

### Final Result:
Thus, the expression \( (A \cup B) \cap B' \) simplifies to:
\[ \boxed{A} \]