Find \(\left( {A \cup B} \right) \cap \left( {A \cup C} \right)\).

Verify that A \ C ≠ C \ A.

Let S be a set containing \(n\) elements where \(n \in \mathbb{N}\).

Show that S has \({2^n}\) subsets.

EITHER

\(\left( {A \cup B} \right) \cap \left( {A \cup C} \right) = \left\{ {1,\,2,\,3,\,5,\,7,\,9,\,11} \right\} \cap \left\{ {1,\,3,\,5,\,7,\,9,\,15,\,31} \right\}\)    M1A1

OR

\(A \cup \left( {B \cap C} \right) = \left\{ {1,\,3,\,5,\,7,\,9,\,11} \right\} \cup \left\{ {3,\,7} \right\}\)     M1A1

OR

\({B \cap C}\) is contained within A     (M1)A1

THEN

= {1, 3, 5, 7, 9} (= A)     A1

Note: Accept a Venn diagram representation.

 [3 marks]

A \ C = {5, 9}     A1

C \ A = {15, 31}    A1

so A \ C ≠ C \ A     AG

Note: Accept a Venn diagram representation.

[2 marks]

METHOD 1

if \(S = \emptyset \) then \(n = 0\) and the number of subsets of S is given by 2⁰ = 1     A1

if \(n > 0\)

for every subset of S, there are 2 possibilities for each element \(x \in S\) either \(x\) will be in the subset or it will not     R1

so for all \(n\) elements there are \(\left( {2 \times 2 \times  \ldots  \times 2} \right){2^n}\) different choices in forming a subset of S      R1

so S has \({2^n}\) subsets      AG

Note: If candidates attempt induction, award A1 for case \(n = 0\), R1 for setting up the induction method (assume \(P\left( k \right)\) and consider \(P\left( {k + 1} \right)\) and R1 for showing how the \(P\left( k \right)\) true implies \(P\left( {k + 1} \right)\) true).

METHOD 2

\(\sum\limits_{k = 0}^n {\left( \begin{gathered}
n \hfill \\
k \hfill \\
\end{gathered} \right)} \) is the number of subsets of S (of all possible sizes from 0 to \(n\))     R1

\({\left( {1 + 1} \right)^n} = \sum\limits_{k = 0}^n {\left( \begin{gathered}
n \hfill \\
k \hfill \\
\end{gathered} \right)} \left( {{1^k}} \right)\left( {{1^{n - k}}} \right)\)      M1

\({2^n} = \sum\limits_{k = 0}^n {\left( \begin{gathered}
n \hfill \\
k \hfill \\
\end{gathered} \right)} \) (= number of subsets of S)     A1

so S has \({2^n}\) subsets      AG

[3 marks]