Let A and B be two sets . The `(Acup B)'cup(A'capB)=`

Options

A. A'
B. A
C. B'
D. none of these

Explanation

To solve the problem \( (A \cup B)' \cup (A' \cap B) \), we will use set theory concepts and properties. Here’s a step-by-step solution:

### Step 1: Understand the components of the expression
- \( A \cup B \) is the union of sets A and B, which includes all elements that are in A, in B, or in both.
- The complement of a set, denoted by \( ' \), refers to all elements not in that set. Thus, \( (A \cup B)' \) represents all elements not in the union of A and B.
- \( A' \) is the complement of set A, which includes all elements not in A.
- \( A' \cap B \) is the intersection of the complement of A and set B, which includes all elements that are in B but not in A.

### Step 2: Find \( (A \cup B)' \)
Using the properties of complements:
\[ (A \cup B)' = U - (A \cup B) \]
where \( U \) is the universal set containing all possible elements.

### Step 3: Find \( A' \cap B \)
This is the set of elements that are in B but not in A:
\[ A' \cap B = \{ x \in U : x \notin A \text{ and } x \in B \} \]

### Step 4: Combine the two parts
Now we need to find the union of the two results:
\[ (A \cup B)' \cup (A' \cap B) \]
This means we take all elements that are not in \( A \cup B \) and combine them with all elements that are in B but not in A.

### Step 5: Simplify the expression
To simplify, we can analyze the resulting sets:
1. The elements in \( (A \cup B)' \) are those not in A or B.
2. The elements in \( A' \cap B \) are those in B but not in A.

Thus, the union will include:
- All elements outside of A and B.
- All elements in B that are not in A.

### Final Result
So, the expression simplifies to:
\[ (A \cup B)' \cup (A' \cap B) = A' \cap B \cup (U - (A \cup B)) \]
This means the final answer is:
\[ A' \cup (B - A) \]

### Conclusion
The final result of the expression \( (A \cup B)' \cup (A' \cap B) \) is \( A' \).