AP DSC School Assistant Math Model Test 1

Concept:This question tests the application of De Morgan’s law and the distributive property of sets. De Morgan’s law states: $(A \cup B)^c = A^c \cap B^c$ and $(A \cap B)^c = A^c \cup B^c$. Also, for any set $X$, $X \cap X^c = \phi$ (empty set).Explanation:We are given two sets $A$ and $B$. We need to simplify $A \cap (B \cup A)^c$.Step 1: Apply De Morgan’s law to the complement of $B \cup A$.$(B \cup A)^c = B^c \cap A^c$ (since the order in union does not matter).Step 2: Substitute this into the original expression.$A \cap (B \cup A)^c = A \cap (B^c \cap A^c)$.Step 3: Using the associative property of intersection, we can rewrite:$A \cap (B^c \cap A^c) = (A \cap B^c) \cap A^c$.Step 4: Now regroup using the associative law (or simply note that $A \cap A^c = \phi$).$(A \cap B^c) \cap A^c = A \cap (B^c \cap A^c)$ is the same; a more direct approach is:$A \cap (B^c \cap A^c) = (A \cap A^c) \cap B^c$ (by commutativity and associativity).Step 5: Since $A \cap A^c = \phi$, we get $\phi \cap B^c = \phi$.Thus the whole expression simplifies to the empty set.Answer:$\phi$ (Option C).