Higher June 2024 Paper 2 Q13
13. A solution to the equation \(2x^3 - 5x - 23 = 0\) lies between 2 and 3.
Use the method of trial and improvement to find the solution correct to 1 decimal place. [4]
| Scheme | Marks | Comments |
|---|---|---|
| One correct evaluation \(2 \leqslant x \leqslant 3\) | B1 | |
| 2 correct evaluations \(2{\cdot}55 \leqslant x \leqslant 2{\cdot}75\), one < 0, one > 0. | B1 | |
| 2 correct evaluations \(2{\cdot}55 \leqslant x \leqslant 2{\cdot}65\), one < 0, one > 0. | M1 | |
| \(x = 2{\cdot}6\) | A1 | |
| (4) |
Notes
Correct evaluation regarded as enough to identify if < 0 or > 0.
If evaluations not seen accept ‘too high’ or ‘too low’.
Note: Some candidates may test \(2x^3 - 5x = 23\)
| \(x\) | \(2x^3 - 5x - 23\) | ||
|---|---|---|---|
| 2 | −17 | 2.62 | −0.130… |
| 2·1 | −14.978 | 2.63 | 0.232… |
| 2·2 | −12.704 | 2.64 | 0.559…. |
| 2·3 | −10.166 | ||
| 2·4 | −7.352 | ||
| 2·5 | −4.25 | 2.55 | −2.587… |
| 2·6 | −0.848 | 2.65 | 0.969… |
| 2·7 | 2.866 | 2.75 | 4.843… |
| 2·8 | 6.904 | ||
| 2·9 | 11.278 | … | |
| 3 | 16 |
Note: Evidence for M1 must be seen before A1 can be awarded.