| Exam Board | AQA |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof |
| Type | Contradiction proof about integers |
| Difficulty | Standard +0.8 This is a proof by contradiction involving parity arguments and integer properties. Part (a) requires rearranging to show a² is even, hence a is even. Part (b) requires substituting a=2k and manipulating to show 2b+1 is even (contradiction since odd=even). While the individual steps are accessible, the question requires careful logical reasoning about parity, proof structure, and recognizing the contradiction, making it moderately challenging but within reach of strong A-level students. |
| Spec | 1.01d Proof by contradiction |
| Answer | Marks |
|---|---|
| 9(a) | Begins argument with either of: |
| Answer | Marks | Guidance |
|---|---|---|
| is odd | 2.1 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| which is a contradiction OE | 2.2a | R1 |
| Subtotal | 2 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 9(b) | Uses 2p and obtains (2p)2 |
| Answer | Marks | Guidance |
|---|---|---|
| and b | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| followed by 2 ×2p2 = 2(2b+1) | 3.1a | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| be an integer | 2.2a | R1 |
| Subtotal | 3 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 9(c) | Deduces that there are no |
| Answer | Marks | Guidance |
|---|---|---|
| where a and b are integers | 2.2a | R1 |
| Answer | Marks | Guidance |
|---|---|---|
| Subtotal | 1 | |
| Question 9 Total | 6 | |
| Q | Marking instructions | AO |
Question 9:
--- 9(a) ---
9(a) | Begins argument with either of:
• Factorises 4b 2 or 4b + 2
correctly to 2b+1 as a factor
• States 4b a − nd 2 − are both even
or
begins proof by contradiction by
assuming a is odd thereforea2
is odd | 2.1 | M1 | a2 −4b−2=0
a2 =4b+2
a2 =2 ( 2b+1 )
Hence a2 must be even, which
means that a must be even
Completes reasoned argument
by deducing that a2 must be
even or has a factor of 2, which
means that a must be even
or
Completes reasoned argument
by deducing that a2 = 4b+2
which is even because 4b and 2
are both even hence a2is even
which is a contradiction OE | 2.2a | R1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 9(b) ---
9(b) | Uses 2p and obtains (2p)2
PI by 4p2
Allow any letter for p except a
and b | 1.1a | M1 | a =2p⇒( 2p )2 =4p2
4p2 =2 ( 2b+1 )
2p2 =2b+1
Hence 2b+1 must be even
2b+1 is an odd number which is a
contradiction
Obtains either
4p2 = 2(2b+1) or 4p2 = 4b+2 and
followed by 2p2 = 2b+1
or
4p2 = 2(2b+1) or 4p2 = 4b+2 and
followed by 2 ×2p2 = 2(2b+1) | 3.1a | A1
Complete reasoned argument
by deducing that 2b + 1 is even
hence contradiction as 2b+1 is
an odd number
or
Complete reasoned argument
by deducing that 2b + 1 is even
hence contradiction as b cannot
be an integer | 2.2a | R1
Subtotal | 3
Q | Marking instructions | AO | Marks | Typical solution
--- 9(c) ---
9(c) | Deduces that there are no
solutions to a2 −4b−2=0
where a and b are integers | 2.2a | R1 | There are no solutions to
a2 −4b−2=0 where a and b are
integers
Subtotal | 1
Question 9 Total | 6
Q | Marking instructions | AO | Marks | Typical solutionAssume that $a$ and $b$ are integers such that
$$a^2 - 4b - 2 = 0$$
\begin{enumerate}[label=(\alph*)]
\item Prove that $a$ is even.
[2 marks]
\item Hence, prove that $2b + 1$ is even and explain why this is a contradiction.
[3 marks]
\item Explain what can be deduced about the solutions of the equation
$$a^2 - 4b - 2 = 0$$
[1 mark]
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 3 2022 Q9 [6]}}