| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Session | Specimen |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof |
| Type | Counter example to disprove statement |
| Difficulty | Moderate -0.8 Part (i) requires only testing small values of n until finding one where n²-n-1 is composite (n=4 gives 11, n=5 gives 19, n=6 gives 29, n=7 gives 41, n=8 gives 53, n=9 gives 71, n=10 gives 89 - actually n=4 gives 11 which is prime, but the statement claims ALL values give primes, so any composite result disproves it). Part (ii) is a straightforward algebraic proof: represent odd number as 2k+1, expand (2k+1)³-(2k+1)², factor out 2. Both parts are routine AS-level proof techniques with minimal problem-solving required. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.01c Disproof by counter example |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Tries at least one value in the interval, e.g. \(4^2 - 4 - 1 = 11\) | M1 | Attempts any \(n^2 - n - 1\) for \(n\) in interval |
| When \(n=8\): \(8^2 - 8 - 1 = 55 = (11\times 5)\), hence NOT prime | A1 | Must state \(n=8\) is FALSE and show \(55=11\times5\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Knows odd number is of the form \(2n+1\) | B1 | |
| Attempts to simplify \((2n+1)^3 - (2n+1)^2\) | M1 | |
| Factorises: \(8n^3 + 8n^2 + 2n = 2(4n^3 + 4n^2 + n)\) | dM1 | |
| States \(2\times\ldots\) is always even | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Let odd number be \(n\), attempts \(n^3 - n^2\) | B1 | |
| Factorises \(n^3 - n^2 = n^2(n-1)\) | M1 | |
| \(n^2\) is odd (odd\(\times\)odd) and \((n-1)\) is even (odd \(-1\)) | dM1 | |
| Product is even (odd \(\times\) even) | A1 |
## Question 6:
**Part (i):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Tries at least one value in the interval, e.g. $4^2 - 4 - 1 = 11$ | M1 | Attempts any $n^2 - n - 1$ for $n$ in interval |
| When $n=8$: $8^2 - 8 - 1 = 55 = (11\times 5)$, hence NOT prime | A1 | Must state $n=8$ is FALSE and show $55=11\times5$ |
**Part (ii):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Knows odd number is of the form $2n+1$ | B1 | |
| Attempts to simplify $(2n+1)^3 - (2n+1)^2$ | M1 | |
| Factorises: $8n^3 + 8n^2 + 2n = 2(4n^3 + 4n^2 + n)$ | dM1 | |
| States $2\times\ldots$ is always even | A1 | |
**Part (ii) Alternative:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Let odd number be $n$, attempts $n^3 - n^2$ | B1 | |
| Factorises $n^3 - n^2 = n^2(n-1)$ | M1 | |
| $n^2$ is odd (odd$\times$odd) and $(n-1)$ is even (odd $-1$) | dM1 | |
| Product is even (odd $\times$ even) | A1 | |\begin{enumerate}
\item (i) Use a counter example to show that the following statement is false.
\end{enumerate}
$$" n ^ { 2 } - n - 1 \text { is a prime number, for } 3 \leqslant n \leqslant 10 \text {." }$$
(ii) Prove that the following statement is always true.\\
"The difference between the cube and the square of an odd number is even."\\
For example $5 ^ { 3 } - 5 ^ { 2 } = 100$ is even.\\
\includegraphics[max width=\textwidth, alt={}, center]{fa7abe9f-f5c0-4578-afd1-73176c717536-12_2255_51_314_1978}
\hfill \mbox{\textit{Edexcel AS Paper 1 Q6 [6]}}