Explanation
Step 1: Find the elements of Set X
The set X contains ordered pairs of integers (x,y) that satisfy the equation:
x2+2y2=3
Since x and y must be integers (Z), we can analyze the possible values for y2:
If y2=0⟹y=0, then x2+2(0)=3⟹x2=3 (No integer solution for x).
If y2=1⟹y=±1, then:
x2+2(1)=3
x2=1⟹x=±1
If y2≥4, then 2y2≥8, which makes x2 negative (x^2 = 3 - 2y^2 < 0), yielding no real solutions.
By combining the valid integer values x=±1 and y=±1, we get the complete set of elements for X:
X={(1,1),(1,−1),(−1,1),(−1,−1)}
Step 2: Apply the condition for Set Y (x > y)
The intersection set X∩Y will consist only of those elements from set X that satisfy the condition from set Y, which is x > y.
Let's test each ordered pair from set X:
For (1,1): Here 1 > 1 is False (1=1).
For (1,−1): Here 1 > -1 is True.
For (−1,1): Here -1 > 1 is False.
For (−1,−1): Here -1 > -1 is False (−1=−1).
Step 3: Determine the elements in X∩Y
The only ordered pair that satisfies both conditions is:
X∩Y={(1,−1)}
Counting the elements in this intersection set gives:
n(X∩Y)=1
Conclusion
The number of elements in X∩Y is 1.
Therefore, the correct option is B.
