Correct : B

Given:
A Boolean operator # with the following properties:
x # 0 = x
x # 1 = x'
x # x = 0
x # x' = 1
To find:
The equivalent expression for x # y.
Solution:
Let me check the standard Boolean operations that satisfy these conditions.
Looking at the first property, x # 0 = x.
Now looking at the second property, x # 1 = x'.
Next, the third property gives x # x = 0.
And the fourth property gives x # x' = 1.
Wait, this directly matches the standard XOR (Exclusive-OR) logic gate properties.
Let me verify with the XOR truth table and expressions:
x ⊕ 0 = x·0' + x'·0 = x·1 + 0 = x.
x ⊕ 1 = x·1' + x'·1 = x·0 + x' = x'.
x ⊕ x = x·x' + x'·x = 0 + 0 = 0.
x ⊕ x' = x·(x')' + x'·x' = x·x + x' = x + x' = 1.
All four properties are completely satisfied by the XOR operator.
So, x # y is simply x ⊕ y.
Writing the XOR operation in terms of basic Boolean operations, we get:
x ⊕ y = x'y + xy'
Thus, x # y = x'y + xy', which matches Option A.
Answer: (A) x'y + xy'