Higher June 2023 Paper 2 Q22
22.
(a) The first 5 terms of a sequence are shown below.
5 8 11 14 17
Complete the term-to-term rule for this sequence. [1]\(t_1 = 5\)
\(t_{n+1} = t_n\) ……………… , \(n \geqslant 1\)
(b)
(i) Show that \(x^3 - 2x - 5 = 0\) can be rearranged as \(x = \sqrt{\dfrac{2x + 5}{x}}\). [1]
(ii) Use the iterative formula \(x_{n+1} = \sqrt{\dfrac{2x_n + 5}{x_n}}\) with \(x_1 = 2\)
to find a solution of \(x^3 - 2x - 5 = 0\) correct to 1 decimal place.
Write down all your values of \(x_{n+1}\). [3]
to find a solution of \(x^3 - 2x - 5 = 0\) correct to 1 decimal place.
Write down all your values of \(x_{n+1}\). [3]
| Scheme | Marks | Comments |
|---|---|---|
| +3 | B1 | Allow ‘Plus 3’ but not 3 |
| Scheme | Marks | Comments |
|---|---|---|
| (b)(i) Correct rearrangement to \(x = \sqrt{\dfrac{2x+5}{x}}\) | B1 | Allow working back from \(x = \sqrt{\dfrac{2x+5}{x}}\) to \(x^3 - 2x - 5 = 0\) |
| (b)(ii) \(x_2 = \sqrt{\dfrac{9}{2}}\) or 2.12(...) | M1 | |
| \(x_3 = 2.08(\ldots)\) \(x_4 = 2.09(\ldots)\) \(x_5 = 2.09(\ldots)\) | m1 | Allow omission of \(x_5 = 2.09\ldots\); allow for \(x_3 = 2.09 \quad x_4 = 2.1(0) \quad x_5 = 2.09\) (from rounding consistently to 2dp) |
| (\(x\) = ) 2.1 | A1 | Allow omission of \(x_5 = 2.09\ldots\), ignore extra iterations if listed; answer without/with incorrect working does not imply the method marks |
| (5) |