June 2022 Paper 3 Q9
9 Assume that \(a\) and \(b\) are integers such that
\[a^2 - 4b - 2 = 0\](a) Prove that \(a\) is even. [2 marks]
(b) Hence, prove that \(2b + 1\) is even and explain why this is a contradiction. [3 marks]
(c) Explain what can be deduced about the solutions of the equation\[a^2 - 4b - 2 = 0\] [1 mark]
| Scheme | Marks | AO |
|---|---|---|
Begins argument with either of:
begins proof by contradiction by assuming \(a\) is odd therefore \(a^2\) is odd | M1 | 2.1 |
| Completes reasoned argument by deducing that \(a^2\) must be even or has a factor of 2, which means that \(a\) must be even or Completes reasoned argument by deducing that \(a^2 = 4b + 2\) which is even because \(4b\) and 2 are both even hence \(a^2\) is even which is a contradiction OE | R1 | 2.2a |
| (2) |
Typical solution
\[a^2 - 4b - 2 = 0\]\[a^2 = 4b + 2\]\[a^2 = 2(2b + 1)\]Hence \(a^2\) must be even, which means that \(a\) must be even
| Scheme | Marks | AO |
|---|---|---|
| Uses \(2p\) and obtains \((2p)^2\) PI by \(4p^2\) Allow any letter for \(p\) except \(a\) and \(b\) | M1 | 1.1a |
| Obtains either \(4p^2 = 2(2b + 1)\) or \(4p^2 = 4b + 2\) and followed by \(2p^2 = 2b + 1\) or \(4p^2 = 2(2b + 1)\) or \(4p^2 = 4b + 2\) and followed by \(2 \times 2p^2 = 2(2b + 1)\) | A1 | 3.1a |
| 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 | R1 | 2.2a |
| (3) |
Typical solution
\[a = 2p \Rightarrow (2p)^2 = 4p^2\]\[4p^2 = 2(2b + 1)\]\[2p^2 = 2b + 1\]Hence \(2b + 1\) must be even
\(2b + 1\) is an odd number which is a contradiction
| Scheme | Marks | AO |
|---|---|---|
| Deduces that there are no solutions to \(a^2 - 4b - 2 = 0\) where \(a\) and \(b\) are integers | R1 | 2.2a |
| (1) | ||
| (6 marks) |
Typical solution
There are no solutions to \(a^2 - 4b - 2 = 0\) where \(a\) and \(b\) are integers