## Question

The universal set contains all the positive integers less than 30. The set *A* contains all prime numbers less than 30 and the set *B* contains all positive integers of the form \(3 + 5n{\text{ }}(n \in \mathbb{N})\) that are less than 30. Determine the elements of

*A* \ *B* ;

\(A\Delta B\) .

**Answer/Explanation**

## Markscheme

*A* = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29} *(A1)*

*B* = {3, 8, 13, 18, 23, 28} *(A1)*

Note: *FT* on their *A* and *B*

* *

*A* \ *B* = {elements in *A* that are not in *B*} *(M1)*

= {2, 5, 7, 11, 17, 19, 29} *A1*

*[4 marks]*

\(B\backslash A\) = {8, 18, 28} *(A1)*

\(A\Delta B = (A\backslash B) \cup (B\backslash A)\) *(M1)*

= {2, 5, 7, 8, 11, 17, 18, 19, 28, 29} *A1*

*[3 marks]*

## Examiners report

It was disappointing to find that many candidates wrote the elements of A and B incorrectly. The most common errors were the inclusion of 1 as a prime number and the exclusion of 3 in B. It has been suggested that some candidates use *N* to denote the positive integers. If this is the case, then it is important to emphasise that the IB notation is that *N* denotes the positive integers and zero and IB candidates should all be aware of that. Most candidates solved the remaining parts of the question correctly and follow through ensured that those candidates with incorrect A and/or B were not penalised any further.

It was disappointing to find that many candidates wrote the elements of A and B incorrectly. The most common errors were the inclusion of 1 as a prime number and the exclusion of 3 in B. It has been suggested that some candidates use *N *to denote the positive integers. If this is the case, then it is important to emphasise that the IB notation is that *N* denotes the positive integers and zero and IB candidates should all be aware of that. Most candidates solved the remaining parts of the question correctly and follow through ensured that those candidates with incorrect A and/or B were not penalised any further.

## Question

Determine, using Venn diagrams, whether the following statements are true.

(i) \(A’ \cup B’ = (A \cup B)’\)

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

Prove, without using a Venn diagram, that \(A\backslash B\) and \(B\backslash A\) are disjoint sets.

**Answer/Explanation**

## Markscheme

(a) (i)

*A1**A1*

since the shaded regions are different, \(A’ \cup B’ \ne (A \cup B)’\) *R1*

\( \Rightarrow \) not true

(ii)

* A1*

*A1*

since the shaded regions are the same \((A\backslash B) \cup (B\backslash A) = (A \cup B)\backslash (A \cap B)\) *R1*

\( \Rightarrow \) true* *

*[6 marks]*

\(A\backslash B = A \cup B’\) and \(B\backslash A = B \cap A’\) *(A1)*

consider \(A \cap B’ \cap B \cap A’\) *M1*

now \(A \cap B’ \cap B \cap A’ = \emptyset \) *A1*

since this is the empty set, they are disjoint *R1*

**Note:** Accept alternative valid proofs.

* *

*[4 marks]*

## Examiners report

Part (a) was accessible to most candidates, but a number drew incorrect Venn diagrams. In some cases the clarity of the diagram made it difficult to follow what the candidate intended. Candidates found (b) harder, although the majority made a reasonable start to the proof. Once again a number of candidates were let down by poor explanation.

Part (a) was accessible to most candidates, but a number drew incorrect Venn diagrams. In some cases the clarity of the diagram made it difficult to follow what the candidate intended. Candidates found (b) harder, although the majority made a reasonable start to the proof. Once again a number of candidates were let down by poor explanation.

## Question

Determine, using Venn diagrams, whether the following statements are true.

(i) \(A’ \cup B’ = (A \cup B)’\)

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

Prove, without using a Venn diagram, that \(A\backslash B\) and \(B\backslash A\) are disjoint sets.

**Answer/Explanation**

## Markscheme

(a) (i)

*A1**A1*

since the shaded regions are different, \(A’ \cup B’ \ne (A \cup B)’\) *R1*

\( \Rightarrow \) not true

(ii)

* A1*

*A1*

since the shaded regions are the same \((A\backslash B) \cup (B\backslash A) = (A \cup B)\backslash (A \cap B)\) *R1*

\( \Rightarrow \) true* *

*[6 marks]*

\(A\backslash B = A \cup B’\) and \(B\backslash A = B \cap A’\) *(A1)*

consider \(A \cap B’ \cap B \cap A’\) *M1*

now \(A \cap B’ \cap B \cap A’ = \emptyset \) *A1*

since this is the empty set, they are disjoint *R1*

**Note:** Accept alternative valid proofs.

* *

*[4 marks]*

## Examiners report

Part (a) was accessible to most candidates, but a number drew incorrect Venn diagrams. In some cases the clarity of the diagram made it difficult to follow what the candidate intended. Candidates found (b) harder, although the majority made a reasonable start to the proof. Once again a number of candidates were let down by poor explanation.

## Question

The elements of sets *P *and *Q *are taken from the universal set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. *P *= {1, 2, 3} and *Q *= {2, 4, 6, 8, 10}.

Given that \(R = (P \cap Q’)’\) , list the elements of *R *.

For a set *S *, let \({S^ * }\) denote the set of all subsets of *S *,

(i) find \({P^ * }\) ;

(ii) find \(n({R^ * })\) .

**Answer/Explanation**

## Markscheme

\(P = \{ 1,{\text{ }}2,{\text{ }}3\} \)

\(Q’ = \{ 1,{\text{ }}3,{\text{ }}5,{\text{ }}7,{\text{ }}9\} \)

so \(P \cap Q’ = \{ 1,{\text{ }}3\} \) *(M1)(A1)*

so \((P \cap Q’)’ = \{ 2,{\text{ }}4,{\text{ }}5,{\text{ }}6,{\text{ }}7,{\text{ }}8,{\text{ }}9,{\text{ }}10\} \) *A1*

*[3 marks]*

(i) \({P^ * } = \left\{ {\{ 1\} ,{\text{ }}\{ 2\} ,{\text{ }}\{ 3\} ,{\text{ }}\{ 1,{\text{ }}2\} ,{\text{ }}\{ 2,{\text{ }}3\} ,{\text{ }}\{ 3,{\text{ }}1\} ,{\text{ }}\{ 1,{\text{ }}2,{\text{ }}3),{\text{ }}\emptyset } \right\}\) *A2*** **

**Note: **Award ** A1 **if one error,

**for two or more.**

*A0*(ii) \({R^ * }\) contains: the empty set (1 element); sets containing one element (8 elements); sets containing two elements *(M1)*

\( = \left( {\begin{array}{*{20}{c}}

8 \\

0

\end{array}} \right) + \left( {\begin{array}{*{20}{c}}

8 \\

1

\end{array}} \right) + \left( {\begin{array}{*{20}{c}}

8 \\

2

\end{array}} \right) + …\left( {\begin{array}{*{20}{c}}

8 \\

8

\end{array}} \right)\) ** (A1)**

\( = {2^8}{\text{ }}( = 256)\) ** A1**

**Note: FT **in (ii) applies if no empty set included in (i) and (ii).

*[5 marks]*

## Examiners report

This was also a well answered question with many candidates obtaining full marks on both parts of the problem. Some candidates attempted to use a factorial rather than a sum of combinations to solve part (b) (ii) and this led to incorrect answers.

This was also a well answered question with many candidates obtaining full marks on both parts of the problem. Some candidates attempted to use a factorial rather than a sum of combinations to solve part (b) (ii) and this led to incorrect answers.

## Question

The elements of sets *P *and *Q *are taken from the universal set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. *P *= {1, 2, 3} and *Q *= {2, 4, 6, 8, 10}.

Given that \(R = (P \cap Q’)’\) , list the elements of *R *.

For a set *S *, let \({S^ * }\) denote the set of all subsets of *S *,

(i) find \({P^ * }\) ;

(ii) find \(n({R^ * })\) .

**Answer/Explanation**

## Markscheme

\(P = \{ 1,{\text{ }}2,{\text{ }}3\} \)

\(Q’ = \{ 1,{\text{ }}3,{\text{ }}5,{\text{ }}7,{\text{ }}9\} \)

so \(P \cap Q’ = \{ 1,{\text{ }}3\} \) *(M1)(A1)*

so \((P \cap Q’)’ = \{ 2,{\text{ }}4,{\text{ }}5,{\text{ }}6,{\text{ }}7,{\text{ }}8,{\text{ }}9,{\text{ }}10\} \) *A1*

*[3 marks]*

(i) \({P^ * } = \left\{ {\{ 1\} ,{\text{ }}\{ 2\} ,{\text{ }}\{ 3\} ,{\text{ }}\{ 1,{\text{ }}2\} ,{\text{ }}\{ 2,{\text{ }}3\} ,{\text{ }}\{ 3,{\text{ }}1\} ,{\text{ }}\{ 1,{\text{ }}2,{\text{ }}3),{\text{ }}\emptyset } \right\}\) *A2*** **

**Note: **Award ** A1 **if one error,

**for two or more.**

*A0*(ii) \({R^ * }\) contains: the empty set (1 element); sets containing one element (8 elements); sets containing two elements *(M1)*

\( = \left( {\begin{array}{*{20}{c}}

8 \\

0

\end{array}} \right) + \left( {\begin{array}{*{20}{c}}

8 \\

1

\end{array}} \right) + \left( {\begin{array}{*{20}{c}}

8 \\

2

\end{array}} \right) + …\left( {\begin{array}{*{20}{c}}

8 \\

8

\end{array}} \right)\) ** (A1)**

\( = {2^8}{\text{ }}( = 256)\) ** A1**

**Note: FT **in (ii) applies if no empty set included in (i) and (ii).

*[5 marks]*

## Examiners report

This was also a well answered question with many candidates obtaining full marks on both parts of the problem. Some candidates attempted to use a factorial rather than a sum of combinations to solve part (b) (ii) and this led to incorrect answers.

## Question

Let \(A = \left\{ {a,{\text{ }}b} \right\}\).

Let the set of all these subsets be denoted by \(P(A)\) . The binary operation symmetric difference, \(\Delta\) , is defined on \(P(A)\) by \(X\Delta Y = (X\backslash Y) \cup (Y\backslash X)\) where \(X\) , \(Y \in P(A)\).

Let \({\mathbb{Z}_4} = \left\{ {0,{\text{ }}1,{\text{ }}2,{\text{ }}3} \right\}\) and \({ + _4}\) denote addition modulo \(4\).

Let \(S\) be any non-empty set. Let \(P(S)\) be the set of all subsets of \(S\) . For the following parts, you are allowed to assume that \(\Delta\), \( \cup \) and \( \cap \) are associative.

Write down all four subsets of *A *.

Construct the Cayley table for \(P(A)\) under \(\Delta \) .

Prove that \(\left\{ {P(A),{\text{ }}\Delta } \right\}\) is a group. You are allowed to assume that \(\Delta \) is associative.

Is \(\{ P(A){\text{, }}\Delta \} \) isomorphic to \(\{ {\mathbb{Z}_4},{\text{ }}{ + _4}\} \) ? Justify your answer.

(i) State the identity element for \(\{ P(S){\text{, }}\Delta \} \).

(ii) Write down \({X^{ – 1}}\) for \(X \in P(S)\) .

(iii) Hence prove that \(\{ P(S){\text{, }}\Delta \} \) is a group.

Explain why \(\{ P(S){\text{, }} \cup \} \) is not a group.

Explain why \(\{ P(S){\text{, }} \cap \} \) is not a group.

**Answer/Explanation**

## Markscheme

\(\emptyset {\text{, \{ a\} , \{ b\} , \{ a, b\} }}\) *A1*

*[1 mark]*

**A3**

**Note: **Award ** A2 **for one error,

**for two errors,**

*A1***for more than two errors.**

*A0** *

*[3 marks]*

closure is seen from the table above *A1*

\(\emptyset \) is the identity *A1*

each element is self-inverse *A1*

**Note: **Showing each element has an inverse is sufficient.

associativity is assumed so we have a group *AG*

*[3 marks]*

not isomorphic as in the above group all elements are self-inverse whereas in \(({\mathbb{Z}_4},{\text{ }}{ + _4})\) there is an element of order 4 (*e.g. *1) *R2*

*[2 marks]*

(i) \(\emptyset \) is the identity *A1*

* *

(ii) \({X^{ – 1}} = X\) *A1*

(iii) if *X *and *Y *are subsets of *S *then \(X\Delta Y\) (the set of elements that belong to *X *or *Y *but not both) is also a subset of *S*, hence closure is proved *R1*

\(\{ P(S){\text{, }}\Delta \} \) is a group because it is closed, has an identity, all elements have inverses (and \(\Delta \) is associative) *R1AG*

*[4 marks]*

not a group because although the identity is \(\emptyset {\text{, if }}X \ne \emptyset \) it is impossible to find a set *Y *such that \(X \cup Y = \emptyset \), so there are elements without an inverse *R1AG*

*[1 mark]*

not a group because although the identity is *S*, if \(X \ne S\) is impossible to find a set *Y *such that \(X \cap Y = S\), so there are elements without an inverse *R1AG*

*[1 mark]*

## Examiners report

A surprising number of candidates were unable to answer part (a) and consequently were unable to access much of the rest of the question. Most candidates however, were successful in parts (a), (b) and (c), and it was pleasing to see the preparedness of candidates in these parts. There were also many good answers for parts (d) and (e) although the third part of (e) caused the most problems with candidates failing to provide sufficient reasoning. Few candidates managed good responses to parts (f) and (g).

A surprising number of candidates were unable to answer part (a) and consequently were unable to access much of the rest of the question. Most candidates however, were successful in parts (a), (b) and (c), and it was pleasing to see the preparedness of candidates in these parts. There were also many good answers for parts (d) and (e) although the third part of (e) caused the most problems with candidates failing to provide sufficient reasoning. Few candidates managed good responses to parts (f) and (g).

A surprising number of candidates were unable to answer part (a) and consequently were unable to access much of the rest of the question. Most candidates however, were successful in parts (a), (b) and (c), and it was pleasing to see the preparedness of candidates in these parts. There were also many good answers for parts (d) and (e) although the third part of (e) caused the most problems with candidates failing to provide sufficient reasoning. Few candidates managed good responses to parts (f) and (g).

## Question

(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 \(S\).

**Answer/Explanation**

## Markscheme

(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]*

## Examiners report

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.

## Question

Sets *X *and *Y *are defined by \({\text{ }}X = \left] {0,{\text{ }}1} \right[;{\text{ }}Y = \{ 0,{\text{ }}1,{\text{ }}2,{\text{ }}3,{\text{ }}4,{\text{ }}5\} \).

(i) Sketch the set \(X \times Y\) in the Cartesian plane.

(ii) Sketch the set \(Y \times X\) in the Cartesian plane.

(iii) State \((X \times Y) \cap (Y \times X)\).

Consider the function \(f:X \times Y \to \mathbb{R}\) defined by \(f(x,{\text{ }}y) = x + y\) and the function \(g:X \times Y \to \mathbb{R}\) defined by \(g(x,{\text{ }}y) = xy\).

(i) Find the range of the function *f*.

(ii) Find the range of the function *g*.

(iii) Show that \(f\) is an injection.

(iv) Find \({f^{ – 1}}(\pi )\), expressing your answer in exact form.

(v) Find all solutions to \(g(x,{\text{ }}y) = \frac{1}{2}\).

**Answer/Explanation**

## Markscheme

(i)

correct horizontal lines *A1*

correctly labelled axes *A1*

clear indication that the endpoints are not included *A1*

(ii)

fully correct diagram *A1*

**Note:** Do not penalize the inclusion of endpoints twice.

(iii) the intersection is empty *A1*

*[5 marks]*

(i) range \((f) = \left] {0,{\text{ 1}}} \right[ \cup \left] {1,{\text{ 2}}} \right[ \cup \rm{L} \cup \left] {5,{\text{ 6}}} \right[\) *A1A1*

**Note: A1 **for six intervals and

**for fully correct notation.**

*A1*Accept \(0 < x < 6,{\text{ }}x \ne 0{\text{, 1, 2, 3, 4, 5, 6}}\).

(ii) range \((g) = \left[ {0,{\text{ 5}}} \right[\) *A1*

(iii) Attempt at solving

\(f({x_1},{\text{ }}{y_1}) = f({x_2},{\text{ }}{y_2})\) **M1**

\(f(x,{\text{ }}y) \in \left] {y,{\text{ }}y + 1} \right[ \Rightarrow {y_1} = {y_2}\) **M1**

and then \({x_1} = {x_2}\) *A1*

so \(f\) is injective *AG*

(iv) \({f^{ – 1}}(\pi ) = (\pi – 3,{\text{ }}3)\) **A1A1**

(v) solutions: (0.5, 1), (0.25, 2), \(\left( {\frac{1}{6},{\text{ 3}}} \right)\), (0.125, 4), (0.1, 5) *A2*

**Note:** ** A2 **for all correct,

**for 2 correct.**

*A1*

*[10 marks]*

## Examiners report

[N/A]

[N/A]

## Question

\(A\), \(B\) and \(C\) are three subsets of a universal set.

Consider the sets \(P = \{ 1,{\text{ }}2,{\text{ }}3\} ,{\text{ }}Q = \{ 2,{\text{ }}3,{\text{ }}4\} \) and \(R = \{ 1,{\text{ }}3,{\text{ }}5\} \).

Represent the following set on a Venn diagram,

\(A\Delta B\), the symmetric difference of the sets \(A\) and \(B\);

Represent the following set on a Venn diagram,

\(A \cap (B \cup C)\).

For sets \(P\), \(Q\) and \(R\), verify that \(P \cup (Q\Delta R) \ne (P \cup Q)\Delta (P \cup R)\).

In the context of the distributive law, describe what the result in part (b)(i) illustrates.

**Answer/Explanation**

## Markscheme

*A1*

*[1 mark]*

*Note: **Accept alternative set configurations*

*A1*

**Note:** Accept alternative set configurations.

*[1 mark]*

LHS:

\(Q\Delta R = \{ 1,{\text{ }}2,{\text{ }}4,{\text{ }}5\} \) *(A1)*

\(P \cup (Q\Delta R) = \{ 1,{\text{ }}2,{\text{ }}3,{\text{ }}4,{\text{ }}5\} \) *A1*

RHS:

\(P \cup Q = \{ 1,{\text{ }}2,{\text{ }}3,{\text{ }}4\} \) and \(P \cup R = \{ 1,{\text{ }}2,{\text{ }}3,{\text{ }}5\} \) *(A1)*

\((P \cup Q)\Delta (P \cup R) = \{ 4,{\text{ }}5\} \) *A1*

hence \(P \cup (Q\Delta R) \ne (P \cup Q)\Delta (P \cup R)\) *AG*

*[4 marks]*

the result shows that union is not distributive over symmetric difference *A1R1*

**Notes:** Award ** A1 **for “union is not distributive” and

**for “over symmetric difference”. Condone use of \( \cup \) and \(\Delta \).**

*R1**[2 marks]*

## Examiners report

[N/A]

[N/A]

[N/A]

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

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.

**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* \ *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 2^{0} = 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 \(n = 0\),

*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).*

**R1****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]**

## Examiners report

[N/A]

[N/A]

[N/A]