C1 June 2008 Q5
5. A sequence \(x_1, x_2, x_3, \ldots\) is defined by\[\begin{aligned} x_1 &= 1, \\ x_{n+1} &= ax_n - 3, \quad n \geqslant 1, \end{aligned}\]where \(a\) is a constant.
Given that \(x_3 = 7\),
| Scheme | Marks |
|---|---|
| \(\left[x_2 =\right]\ a - 3\) | B1 |
| (1) |
Notes
B1: for \(a\times 1 - 3\) or better. Give for \(a - 3\) in part (a) or if it appears in (b) they must state \(x_2 = a - 3\)
This must be seen in (a) or before the \(a(a - 3) - 3\) step.
| Scheme | Marks |
|---|---|
| \(\left[x_3 =\right]\ ax_2 - 3\) or \(a(a - 3) - 3\) | M1 |
| \(= a(a - 3) - 3\) \(= a^2 - 3a - 3 \quad (*)\) (both lines needed for A1) | A1cso |
| (2) |
Notes
M1: for clear show that. Usually for \(a(a - 3) - 3\) but can follow through their \(x_2\) and even allow \(ax_2 - 3\)
A1: for correct processing leading to printed answer. Both lines needed and no incorrect working seen.
| Scheme | Marks |
|---|---|
| \(a^2 - 3a - 3 = 7\) | |
| \(a^2 - 3a - 10 = 0\) or \(a^2 - 3a = 10\) | M1 |
| \((a - 5)(a + 2) = 0\) | dM1 |
| \(\underline{a = 5 \text{ or } -2}\) | A1 |
| (3) | |
| (6 marks) |
Notes
1st M1: for attempt to form a correct equation and start to collect terms. It must be a quadratic but need not lead to a 3TQ=0
2nd dM1: This mark is dependent upon the first M1.
for attempt to factorize their 3TQ=0 or to solve their 3TQ=0. The “=0” can be implied.
\((x \pm p)(x \pm q) = 0\), where \(pq = 10\) or \(\left(x \pm \tfrac{3}{2}\right)^2 \pm \tfrac{9}{4} - 10 = 0\) or correct use of quadratic formula with \(\pm\)
They must have a form that leads directly to 2 values for \(a\).
Trial and Improvement that leads to only one answer gets M0 here.
A1: for both correct answers. Allow \(x = \ldots\)
Give 3/3 for correct answers with no working or trial and improvement that gives both values for \(a\)