Higher November 2020 Paper 2 Q22
22.
(a) Show that \(x=\sqrt{x+7}\) is a rearrangement of \(x^2-x-7=0\). [1]
(b) Use the iteration formula
You must give all your calculated values of \(x_{n+1}\). [3]
\(x_{n+1}=\sqrt{x_n+7}\) starting with \(x_1=3\)
to find a solution of \(x^2-x-7=0\). Give your answer correct to 2 decimal places.You must give all your calculated values of \(x_{n+1}\). [3]
| Scheme | Marks | Comments |
|---|---|---|
| Starting with either form, show the two stages of rearrangement | B1 | \(x=\sqrt{x+7}\) \(x^2=x+7\) \(x^2-x-7=0\) or \(x^2-x-7=0\) \(x^2=x+7\) \(x=\sqrt{x+7}\) |
| Scheme | Marks | Comments |
|---|---|---|
| Sight of \(x_2\)= 3.16(22…) | M1 | |
| Sight of both \(x_4\) = 3.19(18……) and \(x_5\) = 3.19(24……) | m1 | Allow for sight of \(x_3\) = 3.18(78……) or 3.19 and \(x_4\) = 3.19(18……) |
| Solution to 2 d.p. is 3.19 from sight of both \(x_4\) = 3.191(8……) and \(x_5\) = 3.192(4……) | A1 | Ignore any further calculations |
| (4) |