Intermediate November 2022 Unit 2 Q12
12. A solution to the equation
\[x^3+5x-8=0\]lies between 1 and 2.
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 1 ≤ \(x\) ≤ 2 | B1 | Correct evaluation regarded as enough to identify if < 0 or > 0. Look out for testing \(x^3 + 5x = 8\) or equivalent |
| 2 correct evaluations 1·15 ≤ \(x\) ≤ 1·35, (one value < 0, one value > 0) | B1 | If evaluations not seen accept ‘too high’ or ‘too low’. |
| 2 correct evaluations 1·15 ≤ \(x\) ≤ 1·25, (one value < 0, one value > 0) | M1 | |
| \(x\) = 1·2 | A1 |
| \(x\) | \(x^3+5x-8\) | |||
|---|---|---|---|---|
| 1 | -2 | |||
| 1.1 | -1.169 | |||
| 1.2 | -0.272 | |||
| 1.3 | 0.697 | |||
| 1.4 | 1.744 | 1.15 | -0.72913 | |
| 1.5 | 2.875 | 1.22 | -0.08415 | |
| 1.6 | 4.096 | 1.23 | 0.010867 | |
| 1.7 | 5.413 | 1.24 | 0.106624 | |
| 1.8 | 6.832 | 1.25 | 0.203125 | |
| 1.9 | 8.359 | 1.35 | 1.210375 | |
| 2 | 10 |
The highlighted cells were shaded in the original mark scheme.