| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof by induction |
| Type | Prove polynomial divisibility property |
| Difficulty | Standard +0.3 Part (i) is a simple logic/counterexample question requiring minimal work (x=-4 disproves it). Part (ii) is a standard induction proof with straightforward algebra showing divisibility by 6, typical of AS-level induction questions. The combination is slightly easier than average due to the trivial first part and routine nature of the induction. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.01c Disproof by counter example |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| The statement is not true because e.g. when \(x = -4\), \(x^2 = 16\) (which is \(> 9\) but \(x < 3\)) | B1 | Identifies the error by giving a counter example and a reason e.g. \(x = -4\) with \(x^2 = 16\), e.g. \(x = -4\) with \((-4)^2 > 9\), and concludes not true. No errors in brackets. Do not accept "sometimes true". Alternatively explains why statement is not true: e.g. it is not true as when \(x < -3\) then \(x^2 > 9\) so \(x\) does not have to be greater than 3. E.g. \(x^2 > 9 \Rightarrow x < -3\) or \(x > 3\) so not true |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(n^3 + 3n^2 + 2n = n(n^2 + 3n + 2) = n(n+1)(n+2)\) | M1 | Takes out a factor of \(n\) and attempts to factorise the resulting quadratic |
| \(n(n+1)(n+2)\) is the product of 3 consecutive integers | A1 | Deduces that the expression is the product of 3 consecutive integers |
| As \(n(n+1)(n+2)\) is a multiple of 2 and a multiple of 3 it must be a multiple of 6 and so \(n^3 + 3n^2 + 2n\) is divisible by 6 for all integers \(n\) | A1 | Explains that as the expression is a multiple of 3 and 2, it must be a multiple of 6 and so is divisible by 6 |
## Question 14:
### Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| The statement is **not true** because e.g. when $x = -4$, $x^2 = 16$ (which is $> 9$ but $x < 3$) | B1 | Identifies the error by giving a counter example and a reason e.g. $x = -4$ with $x^2 = 16$, e.g. $x = -4$ with $(-4)^2 > 9$, and concludes **not true**. No errors in brackets. Do not accept "sometimes true". Alternatively explains why statement is **not true**: e.g. it is not true as when $x < -3$ then $x^2 > 9$ so $x$ does not have to be greater than 3. E.g. $x^2 > 9 \Rightarrow x < -3$ or $x > 3$ so not true |
**(1 mark)**
---
### Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $n^3 + 3n^2 + 2n = n(n^2 + 3n + 2) = n(n+1)(n+2)$ | M1 | Takes out a factor of $n$ and attempts to factorise the resulting quadratic |
| $n(n+1)(n+2)$ is the product of 3 consecutive integers | A1 | Deduces that the expression is the product of 3 consecutive integers |
| As $n(n+1)(n+2)$ is a multiple of 2 **and** a multiple of 3 it must be a multiple of 6 and so $n^3 + 3n^2 + 2n$ is divisible by 6 for all integers $n$ | A1 | Explains that as the expression is a multiple of 3 **and** 2, it must be a multiple of 6 and so is divisible by 6 |
**(3 marks)**
**Total: 4 marks**\begin{enumerate}
\item (i) A student states\\
"if $x ^ { 2 }$ is greater than 9 then $x$ must be greater than 3 "
\end{enumerate}
Determine whether or not this statement is true, giving a reason for your answer.\\
(ii) Prove that for all positive integers $n$,
$$n ^ { 3 } + 3 n ^ { 2 } + 2 n$$
is divisible by 6
\hfill \mbox{\textit{Edexcel AS Paper 1 2022 Q14 [4]}}