| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trig Proofs |
| Type | Algebraic proof about integers |
| Difficulty | Standard +0.3 Part (a) requires finding a simple counter-example (e.g., p=1, q=6 gives 215=5×43), which is straightforward trial. Part (b) involves substituting q=p+2, expanding (p+2)³-p³=6p²+12p+8=2(3p²+6p+4), then noting p is even so p²+2p is even, making the expression divisible by 8. This is a routine algebraic proof with standard expansion and factoring, slightly easier than average A-level questions. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.01c Disproof by counter example |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Provides a counter example with a reason, e.g. \(6^3 - 1^3 = 215\) which is a multiple of 5 | B1 | No need to state "not true". Here \(q\) must be greater than \(p\), both natural numbers, not 0 or negatives. Any pair of positive integers \(n\) and \(n+5k\) will work, but \(q^3-p^3\) must be evaluated correctly. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| States or uses e.g. \(2n\) and \(2n+2\) or \(2n+2\) and \(2n+4\) | M1 | Key step stating algebraic form of consecutive even numbers. |
| Attempts \((2n+2)^3 - (2n)^3 = 8n^3 + 24n^2 + 24n + 8 - 8n^3\) leading to a quadratic | dM1 | Condone slips but must lead to a quadratic. Alternatively \((2n+2)^3-(2n)^3 = 2^3\{(n+1)^3-n^3\}\) |
| \(= 24n^2 + 24n + 8\) | A1 | e.g. \((2n+2)^3-(2n)^3 = 24n^2+24n+8\) or \((2n+4)^3-(2n+2)^3=24n^2+72n+56\) |
| \(24n^2 + 24n + 8 = 8(3n^2 + 3n + 1)\), so \(q^3 - p^3\) is a multiple of 8 | A1 | Full and rigorous proof required including: correct quadratic, factoring out 8 (or showing all coefficients are multiples of 8), and minimal conclusion "hence true" |
## Question 17(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Provides a counter example with a reason, e.g. $6^3 - 1^3 = 215$ which is a multiple of 5 | B1 | No need to state "not true". Here $q$ must be greater than $p$, both natural numbers, not 0 or negatives. Any pair of positive integers $n$ and $n+5k$ will work, but $q^3-p^3$ must be evaluated correctly. |
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## Question 17(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| States or uses e.g. $2n$ and $2n+2$ **or** $2n+2$ and $2n+4$ | M1 | Key step stating algebraic form of consecutive even numbers. |
| Attempts $(2n+2)^3 - (2n)^3 = 8n^3 + 24n^2 + 24n + 8 - 8n^3$ leading to a quadratic | dM1 | Condone slips but must lead to a quadratic. Alternatively $(2n+2)^3-(2n)^3 = 2^3\{(n+1)^3-n^3\}$ |
| $= 24n^2 + 24n + 8$ | A1 | e.g. $(2n+2)^3-(2n)^3 = 24n^2+24n+8$ or $(2n+4)^3-(2n+2)^3=24n^2+72n+56$ |
| $24n^2 + 24n + 8 = 8(3n^2 + 3n + 1)$, so $q^3 - p^3$ is a multiple of 8 | A1 | Full and rigorous proof required including: correct quadratic, factoring out 8 (or showing all coefficients are multiples of 8), and minimal conclusion "hence true" |\begin{enumerate}
\item In this question $p$ and $q$ are positive integers with $q > p$
\end{enumerate}
Statement 1: $q ^ { 3 } - p ^ { 3 }$ is never a multiple of 5\\
(a) Show, by means of a counter example, that Statement 1 is not true.
Statement 2: When $p$ and $q$ are consecutive even integers $q ^ { 3 } - p ^ { 3 }$ is a multiple of 8\\
(b) Prove, using algebra, that Statement 2 is true.
\hfill \mbox{\textit{Edexcel AS Paper 1 2023 Q17 [5]}}