(a)     Given a set \(U\), and two of its subsets \(A\) and \(B\), prove that

\[(A\backslash B) \cup (B\backslash A) = (A \cup B)\backslash (A \cap B),{\text{ where }}A\backslash B = A \cap B'.\]

(b)     Let \(S = \{ A,{\text{ }}B,{\text{ }}C,{\text{ }}D\} \) where \(A = \emptyset ,{\text{ }}B = \{ 0\} ,{\text{ }}C = \{ 0,{\text{ }}1\} \) and \(D = \{ {\text{0, 1, 2}}\} \).

State, with reasons, whether or not each of the following statements is true.

(i)     The operation \ is closed in \(S\).

(ii)     The operation \( \cap \) has an identity element in \(S\) but not all elements have an inverse.

(iii)     Given \(Y \in S\), the equation \(X \cup Y = Y\) always has a unique solution for \(X\) in \(S\).

(a)     \((A\backslash B) \cup (B\backslash A) = (A \cap B') \cup (B \cap A')\)     (M1)

\( = \left( {(A \cap B') \cup B} \right) \cap \left( {(A \cap B') \cup A'} \right)\)     M1

\( = \left( {(A \cup B) \cap \underbrace {(B' \cup B)}_U} \right) \cap \left( {\underbrace {(A \cup A')}_U \cap (B' \cup A')} \right)\)     A1

\( = (A \cup B') \cap (B' \cup A') = (A \cup B) \cap (A \cap B)'\)     A1

\( = (A \cup B)\backslash (A \cap B)\)     AG

[4 marks]

(b)     (i)     false     A1

counterexample

eg \(D\backslash C = \{ 2\}  \notin S\)     R1

(ii)     true     A1

as \(A \cap D = A,{\text{ }}B \cap D = B,{\text{ }}C \cap D = C\) and \(D \cap D = D\),

\(D\) is the identity     R1

\(A\) (or \(B\) or \(C\)) has no inverse as \(A \cap X = D\) is impossible     R1

(iii)     false     A1

when \(Y = D\) the equation has more than one solution (four solutions)     R1

[7 marks]

For part (a), candidates who chose to prove the given statement using the properties of Sets were often successful with the proof. Some candidates chose to use the definition of equality of sets, but made little to no progress. In a few cases candidates attempted to use Venn diagrams as a proof. Part (b) was challenging for most candidates, and few correct answers were seen.