Higher November 2019 Paper 2 Q27
27 \(x_{n+1} = \sqrt[3]{3x_n + 7}\)
Use a starting value of \(\quad x_1 = 2 \quad\) to work out a solution to \(\quad x = \sqrt[3]{3x + 7}\)
Give your answer to 3 decimal places. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\sqrt[3]{13}\) or 2.35(1…) | M1 | \(\sqrt[3]{6 + 7}\) or \(\sqrt[3]{3 \times 2 + 7}\) |
| 2.413(…) or 2.4238… or 2.424 or 2.4256… or 2.4259… | M1dep | |
| 2.426 | A1 |
Additional guidance
| Answer 2.426 (eg from using starting value of 1) | M2A1 |
| Answer only 2.425 | M0M0A0 |
| \(\sqrt{13}\) | M0M0A0 |
| Condone \(2 = \sqrt[3]{13}\) etc |