Higher June 2018 Paper 2 Q18
18.
(a) Show that \(x = 13 - \dfrac{9}{x}\) is a rearrangement of \(x^2 - 13x + 9 = 0\).
You must show each stage of your working. [1]
You must show each stage of your working. [1]
(b) Use the iteration formula \[x_{n+1} = 13 - \frac{9}{x_n} \quad\text{and}\quad x_1 = 12\] to find a solution of \(x^2 - 13x + 9 = 0\) 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 \(x = 13 - 9/x\) or starting with \(x^2 - 13x + 9 = 0\), 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 |
|---|---|---|
| Sight of \(x_2 = 12.25\) | M1 | |
| Sight of \(x_4 = 12.26(62229\ldots)\) and \(x_5 = 12.26(62778\ldots)\) | m1 | Allow for sight of \(x_3 = 12.26(5\ldots)\) and \(x_4 = 12.26(6\ldots)\) |
| Solution to 2 d.p. is 12.27 from sight of \(x_4 = 12.26(62229\ldots)\) and \(x_5 = 12.26(62778\ldots)\) | A1 | Ignore any further calculations |
| (4) |