If `A={1,3,5,7,9,11,13,15}, B and N={2,4,……,16}` is the universal set, then `A'uu((AuuB)nnB')` is (where, N is the set of natural numbers)

Options

A. A
B. N
C. B
D. None of these

Answer

To solve the problem, we need to find \( A' \cup ((A \cup B) \cap B') \).

### Step 1: Identify the Universal Set
The universal set \( N \) is given as \( N = \{2, 4, 6, 8, 10, 12, 14, 16\} \).

### Step 2: Identify Set A
Set \( A \) is given as:
\[ A = \{1, 3, 5, 7, 9, 11, 13, 15\} \]

### Step 3: Find the Complement of Set A
The complement of \( A \) (denoted as \( A' \)) is the set of elements in the universal set \( N \) that are not in \( A \).

Thus,
\[ A' = N - A = \{2, 4, 6, 8, 10, 12, 14, 16\} - \{1, 3, 5, 7, 9, 11, 13, 15\} = \{2, 4, 6, 8, 10, 12, 14, 16\} \]

### Step 4: Identify Set B
Set \( B \) is not explicitly defined in the problem, but we assume it can be any subset of \( N \). For this solution, we will assume \( B = \{2, 4, 6, 8, 10, 12, 14, 16\} \).

### Step 5: Find the Complement of Set B
The complement of \( B \) (denoted as \( B' \)) is the set of elements in the universal set that are not in \( B \):
\[ B' = N - B = \{1, 3, 5, 7, 9, 11, 13, 15\} \]

### Step 6: Calculate \( A \cup B \)
Now we need to find \( A \cup B \):
\[ A \cup B = \{1, 3, 5, 7, 9, 11, 13, 15\} \cup \{2, 4, 6, 8, 10, 12, 14, 16\} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16\} \]

### Step 7: Calculate \( (A \cup B) \cap B' \)
Now we find the intersection of \( (A \cup B) \) with \( B' \):
\[ (A \cup B) \cap B' = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16\} \cap \{1, 3, 5, 7, 9, 11, 13, 15\} = \{1, 3, 5, 7, 9, 11, 13, 15\} \]

### Step 8: Calculate \( A' \cup ((A \cup B) \cap B') \)
Finally, we find:
\[ A' \cup ((A \cup B) \cap B') = \{2, 4, 6, 8, 10, 12, 14, 16\} \cup \{1, 3, 5, 7, 9, 11, 13, 15\} \]
This gives us:
\[ A' \cup ((A \cup B) \cap B') = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16\} \]

### Final Answer
The final answer is:
\[ A' \cup ((A \cup B) \cap B') = N \]