Intermediate November 2023 Unit 2 Q17
17. A solution of the equation
\[x^3 + 6x = 80\]lies between 3 and 4.
Use the method of trial and improvement to find this solution correct to 1 decimal place.
You must show all your working. [4]
| Scheme | Marks | Comments |
|---|---|---|
| One correct evaluation \(3 \leqslant x \leqslant 4\) | B1 | Correct evaluation regarded as enough to identify if < 80 or > 80. If evaluations not seen accept ‘too high’ or ‘too low’. Look out for \(x^3 + 6x - 80 = 0\) |
| 2 correct evaluations \(3{\cdot}75 \leqslant x \leqslant 3{\cdot}95\), (one value < 80, one value > 80) | B1 | |
| 2 correct evaluations \(3{\cdot}75 \leqslant x \leqslant 3{\cdot}85\), (one value < 80, one value > 80) | M1 | |
| \(x = 3{\cdot}8\) | A1 | Unsupported \(x = 3{\cdot}8\) is awarded B0B0M0A0. An answer of \(x = 3{\cdot}8\) can only be awarded M1A1, following sight of 2 correct evaluations \(3{\cdot}75 \leqslant x \leqslant 3{\cdot}85\) (one evaluation < 80, one evaluation > 80). If 3·85 is given as 80 (truncated) award M0 A0 unless ‘too high’ or equivalent is indicated. |
Table of evaluations (from the comments column)
| \(x\) | \(x^3 + 6x\) |
|---|---|
| 3 | 45 |
| 3.1 | 48·391 |
| 3.2 | 51·968 |
| 3.3 | 55·737 |
| 3.4 | 59·704 |
| 3.5 | 63·875 |
| 3.6 | 68·256 |
| 3.7 | 72·853 |
| 3.8 | 77·672 |
| 3.9 | 82·719 |
| 4 | 88 |
| 3.75 | 75·2343.. |
| 3.84 | 79·6631.. |
| 3.85 | 80·1666.. |
| 3.95 | 85·3298.. |
The highlighted cells were shaded in the original mark scheme.