| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Topic | Proof |
| Type | Counter example to disprove statement |
| Difficulty | Standard +0.3 Part (a) requires finding a simple counterexample (p=3 or p=5 work), which is straightforward trial. Part (b) is a routine algebraic proof using m=2k+1, n=2k+3, expanding mn+1=(2k+2)² - standard manipulation. Both parts are below average difficulty for A-level, requiring only basic techniques without novel insight. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.01c Disproof by counter example |
| Answer | Marks | Guidance |
|---|---|---|
| \(2^3 < 3^2\) | B1 | or eg \(2^3 \geq 3^2\), \(8 \geq 9\) Not true. Allow if poorly expressed. |
| [1] |
| Answer | Marks |
|---|---|
| \(n(n+2)+1\) \(\left(= n^2+2n+1\right) = (n+1)^2\) | M1, A1 |
| But \(n\) is odd so \(n+1\) is even, so this is (even number)\(^2\) | M1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Let \(m = 2k-1\) and \(n = 2k+1\) where \(k\) is a positive integer | M1 | Allow \(2n-1\) and \(2n+1\). Condone omission of "where \(k\) is a positive integer" |
| \((2k-1)(2k+1)+1\) \((= 4k^2) = (2k)^2\) or \(2^2\times k^2\) | M1, A1 | |
| This is (even number)\(^2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Let \(m = 2k+1\) and \(n = 2k+3\) where \(k\) is a positive integer | M1 | Allow \(2n+1\) and \(2n+3\). Condone omission of "where \(k\) is a positive integer" |
| \((2k+1)(2k+3)+1\) \((= 4k^2+8k+4 = 4(k^2+2k+1)) = [2(k+1)]^2\) or \(2^2\times(k+1)^2\) | M1, A1 | |
| This is (even number)\(^2\) | A1 | |
| [4] |
# Question 8:
## Part (a):
| $2^3 < 3^2$ | B1 | or eg $2^3 \geq 3^2$, $8 \geq 9$ Not true. Allow if poorly expressed. |
|---|---|---|
| | [1] | |
## Part (b):
| $n(n+2)+1$ $\left(= n^2+2n+1\right) = (n+1)^2$ | M1, A1 | |
|---|---|---|
| But $n$ is odd so $n+1$ is even, so this is (even number)$^2$ | M1, A1 | |
**Alternative Method 1:**
| Let $m = 2k-1$ and $n = 2k+1$ where $k$ is a positive integer | M1 | Allow $2n-1$ and $2n+1$. Condone omission of "where $k$ is a positive integer" |
|---|---|---|
| $(2k-1)(2k+1)+1$ $(= 4k^2) = (2k)^2$ or $2^2\times k^2$ | M1, A1 | |
| This is (even number)$^2$ | A1 | |
**Alternative Method 2:**
| Let $m = 2k+1$ and $n = 2k+3$ where $k$ is a positive integer | M1 | Allow $2n+1$ and $2n+3$. Condone omission of "where $k$ is a positive integer" |
|---|---|---|
| $(2k+1)(2k+3)+1$ $(= 4k^2+8k+4 = 4(k^2+2k+1)) = [2(k+1)]^2$ or $2^2\times(k+1)^2$ | M1, A1 | |
| This is (even number)$^2$ | A1 | |
| | [4] | |
---8
\begin{enumerate}[label=(\alph*)]
\item Prove that the following statement is not true.
$$p \text { is a positive integer } \Rightarrow 2 ^ { p } \geqslant p ^ { 2 }$$
\item Prove that the following statement is true.\\
$m$ and $n$ are consecutive positive odd numbers $\Rightarrow m n + 1$ is the square of an even number
\end{enumerate}
\hfill \mbox{\textit{OCR PURE Q8 [5]}}