| Exam Board | AQA |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2021 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof |
| Type | Identifying errors in proofs |
| Difficulty | Moderate -0.5 This is a straightforward proof-checking question requiring students to identify two clear errors (missing case n=3m+2 and algebraic mistake in Step 3) and complete a standard proof by exhaustion. The topic is routine A-level content with no novel insight required, making it slightly easier than average. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.01c Disproof by counter example |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Kai has not expanded the brackets correctly | E1 (2.3) | Identifies the algebraic mistake |
| Kai has not considered numbers of the form \(3m+2\) | E1 (2.3) | Identifies that all cases have not been exhausted |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((3m+1)(9m^2+6m+1-1)=(3m+1)(9m^2+6m)=3(3m+1)(3m^2+2m)\) which is a multiple of 3 | B1 (2.1) | Completes manipulation correctly in Step 4 to obtain \((3m+1)(9m^2+6m)\); OE |
| When \(n=3m+2\): \(n^3-n=(3m+2)((3m+2)^2-1)=(3m+2)(9m^2+12m+3)=3(3m+2)(3m^2+4m+1)\) which is a multiple of 3 | M1 (1.1a) | Manipulates expression with third substitution using \(n=3m+2\) or \(n=3m-1\) |
| Equivalent working shown convincingly as multiple of 3 | A1 (1.1b) | |
| \(n^3-n\) is always a multiple of 3 for all positive integer values of \(n\) | R1 (2.1) | Completes rigorous argument by clearly showing factor of 3 in both cases and concludes appropriately |
## Question 8(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Kai has not expanded the brackets correctly | E1 (2.3) | Identifies the algebraic mistake |
| Kai has not considered numbers of the form $3m+2$ | E1 (2.3) | Identifies that all cases have not been exhausted |
## Question 8(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(3m+1)(9m^2+6m+1-1)=(3m+1)(9m^2+6m)=3(3m+1)(3m^2+2m)$ which is a multiple of 3 | B1 (2.1) | Completes manipulation correctly in Step 4 to obtain $(3m+1)(9m^2+6m)$; OE |
| When $n=3m+2$: $n^3-n=(3m+2)((3m+2)^2-1)=(3m+2)(9m^2+12m+3)=3(3m+2)(3m^2+4m+1)$ which is a multiple of 3 | M1 (1.1a) | Manipulates expression with third substitution using $n=3m+2$ or $n=3m-1$ |
| Equivalent working shown convincingly as multiple of 3 | A1 (1.1b) | |
| $n^3-n$ is always a multiple of 3 for all positive integer values of $n$ | R1 (2.1) | Completes rigorous argument by clearly showing factor of 3 in both cases and concludes appropriately |8 Kai is proving that $n ^ { 3 } - n$ is a multiple of 3 for all positive integer values of $n$.
Kai begins a proof by exhaustion.\\
Step 1
$$n ^ { 3 } - n = n \left( n ^ { 2 } - 1 \right)$$
Step 2 When $n = 3 m$, where $m$ is a $n ^ { 3 } - n = 3 m \left( 9 m ^ { 2 } - 1 \right)$ non-negative integer which is a multiple of 3
Step 3 When $n = 3 m + 1$,
$$\begin{aligned}
& n ^ { 3 } - n = ( 3 m + 1 ) \left( ( 3 m + 1 ) ^ { 2 } - 1 \right) \\
& = ( 3 m + 1 ) \left( 9 m ^ { 2 } \right) \\
& = 3 ( 3 m + 1 ) \left( 3 m ^ { 2 } \right)
\end{aligned}$$
Step 5 Therefore $n ^ { 3 } - n$ is a multiple of 3 for all positive integer values of $n$
8
\begin{enumerate}[label=(\alph*)]
\item Explain the two mistakes that Kai has made after Step 3.
Step 4
\section*{which is a multiple of 3 \\
which is a multiple of 3}
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\end{center}
别\\
all positive integer values of $n$
\section*{a}
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$\_\_\_\_$ " $\_\_\_\_$\\
8
\item Correct Kai's argument from Step 4 onwards.
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 2 2021 Q8 [6]}}