Higher November 2017 Paper 2 Q19
19.
(a) Show that \(x = 8 - \dfrac{10}{x}\) is a rearrangement of \(x^2 - 8x + 10 = 0\).
You must show each stage of your working. [1]
You must show each stage of your working. [1]
(b) Use the iteration formula \[x_{n+1} = 8 - \frac{10}{x_n} \ \text{ and } \ x_1 = 5\] to find a solution of \(x^2 - 8x + 10 = 0\) correct to 2 decimal places.
You must give all your calculated values of \(x_{n+1}\). [3]
You must give all your calculated values of \(x_{n+1}\). [3]
| Scheme | Marks | Comments |
|---|---|---|
| Either starting \(x = 8 - 10/x\) or starting with \(x^2 - 8x + 10 = 0\), showing the 2 stages of rearrangement | B1 | 2 stages required either multiplication by \(x\) and ‘= 0’, or division by \(x\) and isolating the original ‘\(x^2\)’ term |
| Scheme | Marks | Comments |
|---|---|---|
| (\(x_1 = 5\) gives) \(x_2 = 6\) | M1 | |
| Sight of \(x_7 = 6.449\ldots\) OR \(x_8 = 6.449\ldots\) | m1 | Must be shown to at least 3 d.p. |
| Solution to 2 d.p. is 6.45 | A1 | Do not allow FT from M0 \(x_2 = 6\) \(x_3 = 6.333\ldots\) \(x_4 = 6.421\ldots\) \(x_5 = 6.4426\ldots\) \(x_6 = 6.4478\ldots\) \(x_7 = 6.449\ldots\) \(x_8 = 6.449\ldots\) |
| (4) |