if `A={1,2,5,7,9,11,13,15,17,}={2,4,. . . . ,18}`and N is the universal set , then `A cup ((Acup B)cap B')`is

Options

A. A
B. N
C. B
D. `phi`

Answer

To solve the problem, we need to find the expression \( A \cup (A \cup B) \cap B' \) given the sets:

- \( A = \{1, 3, 5, 7, 9, 11, 13, 15, 17\} \) (the set of odd numbers from 1 to 17)
- \( B = \{2, 4, 6, 8, 10, 12, 14, 16, 18\} \) (the set of even numbers from 2 to 18)
- \( N \) is the universal set, which we can assume to be \( N = \{1, 2, 3, \ldots, 18\} \)

### Step 1: Find \( B' \) (the complement of B)
The complement of \( B \) in the universal set \( N \) is:
\[ B' = N - B = \{1, 3, 5, 7, 9, 11, 13, 15, 17\} \]
Thus, \( B' = A \).

### Step 2: Find \( A \cup B \)
Now, we find the union of sets \( A \) and \( B \):
\[ A \cup B = \{1, 3, 5, 7, 9, 11, 13, 15, 17\} \cup \{2, 4, 6, 8, 10, 12, 14, 16, 18\} \]
This results in:
\[ A \cup B = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18\} \]
This is the entire universal set \( N \).

### Step 3: Find \( (A \cup B) \cap B' \)
Next, we find the intersection of \( A \cup B \) and \( B' \):
\[ (A \cup B) \cap B' = N \cap B' = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18\} \cap \{1, 3, 5, 7, 9, 11, 13, 15, 17\} \]
The result is:
\[ (A \cup B) \cap B' = \{1, 3, 5, 7, 9, 11, 13, 15, 17\} = A \]

### Step 4: Find \( A \cup ((A \cup B) \cap B') \)
Finally, we find:
\[ A \cup ((A \cup B) \cap B') = A \cup A = A \]
Thus, the final result is:
\[ A \cup ((A \cup B) \cap B') = A = \{1, 3, 5, 7, 9, 11, 13, 15, 17\} \]

### Final Answer
The answer is:
\[ A = \{1, 3, 5, 7, 9, 11, 13, 15, 17\} \]