IB DP Maths Topic 8.1 De Morgan’s laws: distributive, associative and commutative laws (for union and intersection) HL Paper 3

 

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Question

Given the sets \(A\) and \(B\), use the properties of sets to prove that \(A \cup (B’ \cup A)’ = A \cup B\), justifying each step of the proof.

Answer/Explanation

Markscheme

\(A \cup (B’ \cup A)’ = A \cup (B \cap A’)\)     De Morgan     M1A1

\( = (A \cup B) \cap (A \cup A’)\)     Distributive property     M1A1

\( = (A \cup B) \cap U\)     (Union of set and its complement)     A1

\( = A \cup B\)     (Intersection with the universal set)     AG

Note:     Do not accept proofs using Venn diagrams unless the properties are clearly stated.

Note:     Accept double inclusion proofs: M1A1 for each inclusion, final A1 for conclusion of equality of sets.

[5 marks]

Examiners report

[N/A]

Question

Consider the sets A = {1, 3, 5, 7, 9} , B = {2, 3, 5, 7, 11} and C = {1, 3, 7, 15, 31} .

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

[3]
a.i.

Verify that A \ C ≠ A.

[2]
a.ii.

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

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

[3]
b.
Answer/Explanation

Markscheme

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.i.

A \ C = {5, 9}     A1

= {15, 31}    A1

so A \ C ≠ A     AG

Note: Accept a Venn diagram representation.

[2 marks]

a.ii.

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]

b.

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

b.

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