Higher November 2021 Unit 2 Q3
3. A solution of the equation
\[x^3 + 3x = 20\]
lies between 2 and 3.
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 \(2 \leqslant x \leqslant 3\) | B1 | |
| 2 correct evaluations \(2{\cdot}25 \leqslant x \leqslant 2{\cdot}45\), one < 20, one > 20. | B1 | |
| 2 correct evaluations \(2{\cdot}25 \leqslant x \leqslant 2{\cdot}35\), one < 20, one > 20. | M1 | |
| \(x = 2{\cdot}3\) | A1 |
Correct evaluation regarded as enough to identify if <20 or >20. If evaluations not seen accept ‘too high’ or ‘too low’.
Look out for testing \(x^3 + 3x - 20 = 0\)
| \(x\) | \(x^3 + 3x\) | ||
|---|---|---|---|
| 2 | 14 | ||
| 2·1 | 15·561 | ||
| 2·2 | 17·248 | 2·25 | 18·140…. |
| 2·3 | 19·067 | 2·35 | 20·027…. |
| 2·4 | 21·024 | 2·45 | 22·056…. |
| 2·5 | 23·125 | ||
| 2·6 | 25·376 | ||
| 2·7 | 27·783 | ||
| 2·8 | 30·352 | ||
| 2·9 | 33·089 | ||
| 3 | 36 |
Note
Evidence for M1 must be seen before A1 can be awarded.