AQA AS Paper 2 2021 June — Question 9 4 marks

Exam BoardAQA
ModuleAS Paper 2 (AS Paper 2)
Year2021
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProof
TypeDivisibility proof for all integers
DifficultyStandard +0.3 This is a standard divisibility proof requiring factorisation of a cubic expression and showing divisibility by 2 and 3. Part (a) is straightforward factorisation, and part (b) requires recognising that consecutive integers guarantee factors of 2 and 3, which is a common A-level proof technique but slightly above routine recall.
Spec1.01a Proof: structure of mathematical proof and logical steps

9
  1. Express \(n ^ { 3 } - n\) as a product of three factors. 9
  2. Given that \(n\) is a positive integer, prove that \(n ^ { 3 } - n\) is a multiple of 6

Question 9(a):
AnswerMarks Guidance
\(n^3 - n = n(n^2-1) = n(n-1)(n+1)\)B1 States correct factorisation
Question 9(b):
AnswerMarks Guidance
\((n-1)\), \(n\), \((n+1)\) are 3 consecutive integersE1 States three consecutive integers; or all integers are multiple of 3 or 1 more/less
So one must be a multiple of 3, and at least one must be a multiple of 2E1 Deduces one factor multiple of 3 and one multiple of 2
So product has factors 2 and 3, hence is a multiple of \(2 \times 3 = 6\)R1 States at least one multiple of 2, one of 3, draws correct conclusion
## Question 9(a):

$n^3 - n = n(n^2-1) = n(n-1)(n+1)$ | B1 | States correct factorisation |

## Question 9(b):

$(n-1)$, $n$, $(n+1)$ are 3 consecutive integers | E1 | States three consecutive integers; or all integers are multiple of 3 or 1 more/less |
So one must be a multiple of 3, and at least one must be a multiple of 2 | E1 | Deduces one factor multiple of 3 and one multiple of 2 |
So product has factors 2 and 3, hence is a multiple of $2 \times 3 = 6$ | R1 | States at least one multiple of 2, one of 3, draws correct conclusion |
9
\begin{enumerate}[label=(\alph*)]
\item Express $n ^ { 3 } - n$ as a product of three factors.

9
\item Given that $n$ is a positive integer, prove that $n ^ { 3 } - n$ is a multiple of 6
\end{enumerate}

\hfill \mbox{\textit{AQA AS Paper 2 2021 Q9 [4]}}