C1 January 2008 Q7
7. A sequence is given by:\[\begin{aligned} x_1 &= 1, \\ x_{n+1} &= x_n(p + x_n), \end{aligned}\]where \(p\) is a constant \((p \neq 0)\).
Given that \(x_3 = 1\),
| Scheme | Marks |
|---|---|
| \(1(p + 1)\) or \(p + 1\) | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \(\left((a)\right)\left(p + (a)\right)\) [(a) must be a function of \(p\)]. \(\left[(p + 1)(p + p + 1)\right]\) | M1 |
| \(= 1 + 3p + 2p^2 \qquad (*)\) | A1cso |
| (2) |
Notes
M: Valid attempt to use the given recurrence relation to find \(x_3\).
Missing brackets, e.g. \(p + 1(p + p + 1)\) Condone for the M1, then if all terms in the expansion are correct, with working fully shown, M1 A1 is still allowed.Beware ‘working back from the answer’, e.g. \(1 + 3p + 2p^2 = (1 + p)(1 + 2p)\) scores no marks unless the recurrence relation is justified.
| Scheme | Marks |
|---|---|
| \(1 + 3p + 2p^2 = 1\) | M1 |
| \(p(2p + 3) = 0 \qquad p = \ldots\) | M1 |
| \(p = -\dfrac{3}{2}\) (ignore \(p = 0\), if seen, even if ‘chosen’ as the answer) | A1 |
| (3) |
Notes
2nd M: Attempt to solve a quadratic equation in \(p\) (e.g. quadratic formula or completing the square).
The equation must be based on \(x_3 = 1\).
The attempt must lead to a non-zero solution, so just stating the zero solution \(p = 0\) is M0.
A: The A mark is dependent on both M marks.
| Scheme | Marks |
|---|---|
| Noting that even terms are the same. This M mark can be implied by listing at least 4 terms, e.g. \(1,\ -\dfrac{1}{2},\ 1,\ -\dfrac{1}{2}, \ldots\) | M1 |
| \(x_{2008} = -\dfrac{1}{2}\) | A1 |
| (2) | |
| (8 marks) |
Notes
M: Can be implied by a correct answer for their \(p\) (answer is \(p + 1\)), and can also be implied if the working is ‘obscure’).
Trivialising, e.g. \(p = 0\), so every term = 1, is M0.
If the additional answer \(x_{2008} = 1\) (from \(p = 0\)) is seen, ignore this (isw).