| Exam Board | AQA |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof |
| Type | Existence of greatest/smallest element |
| Difficulty | Moderate -0.8 This question tests understanding of open intervals and proof by contradiction at a basic level. Part (a) requires only recognizing that 3 is not in the open interval (1 mark, trivial). Part (b)(i) asks students to explain a given contradiction (1 mark, straightforward). Part (b)(ii) requires adapting the given proof structure to show no largest value exists (4 marks), but the template is provided and the adaptation is mechanical—students just need to construct y=(x+4)/2 and show x<y<4. While proof by contradiction appears sophisticated, the question scaffolds heavily and requires minimal independent reasoning or problem-solving beyond pattern recognition. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.01d Proof by contradiction |
| Answer | Marks | Guidance |
|---|---|---|
| 11(a) | Explains that 3 is not in 3 < k < 4 | |
| OE | 2.4 | E1 |
| Subtotal | 1 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 11(b)(i) | Explains that 3 < y < x which |
| Answer | Marks | Guidance |
|---|---|---|
| the smallest value OE | 2.4 | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Subtotal | 1 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 11(b)(ii) | Assumes there is a largest value |
| Answer | Marks | Guidance |
|---|---|---|
| OE | 2.1 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| before this step | 2.2a | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| OE | 2.4 | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| CSO | 2.1 | R1 |
| Subtotal | 4 | |
| Question 11 Total | 6 | |
| Q | Marking instructions | AO |
Question 11:
--- 11(a) ---
11(a) | Explains that 3 is not in 3 < k < 4
OE | 2.4 | E1 | 3 is not in the interval
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 11(b)(i) ---
11(b)(i) | Explains that 3 < y < x which
contradicts the definition of x as
the smallest value OE | 2.4 | E1 | y is between 3 and x which contradicts
the definition of x as the smallest
value in (3, 4)
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 11(b)(ii) ---
11(b)(ii) | Assumes there is a largest value
in (3, 4)
OE | 2.1 | B1 | Step 1: Assume there is a largest
number in the interval 3 < k < 4 and
let this largest number be x
x+4
Step 2: let y =
2
Step 3: x < y < 4 which is a
contradiction.
Step 4: Therefore, there is no largest
value in 3 < k < 4
Constructs a value “y” in (3, 4)
which is greater than their “x”
Must have referenced their “x”
before this step | 2.2a | B1
States that x < y < 4 which is a
contradiction
OE | 2.4 | E1
Concludes that there is no
largest value in (3, 4)
OE
CSO | 2.1 | R1
Subtotal | 4
Question 11 Total | 6
Q | Marking instructions | AO | Marks | Typical solution\begin{enumerate}[label=(\alph*)]
\item A student states that 3 is the smallest value of $k$ in the interval $3 < k < 4$
Explain the error in the student's statement.
[1 mark]
\item The student's teacher says there is no smallest value of $k$ in the interval $3 < k < 4$
The teacher gives the following correct proof:
Step 1: Assume there is a smallest number in the interval $3 < k < 4$ and let this smallest number be $x$
Step 2: let $y = \frac{3 + x}{2}$
Step 3: $3 < y < x$ which is a contradiction.
Step 4: Therefore, there is no smallest number in interval $3 < k < 4$
\begin{enumerate}[label=(\roman*)]
\item Explain the contradiction stated in Step 3
[1 mark]
\item Prove that there is no largest value of $k$ in the interval $3 < k < 4$
[4 marks]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 2 2024 Q11 [6]}}