Higher November 2018 Unit 2 Q5
5. A solution of the equation
\[x^3 - 13x = 107\]lies between 5 and 6.
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 \(5 \leqslant x \leqslant 6\) | B1 | |
| 2 correct evaluations \(5{\cdot}55 \leqslant x \leqslant 5{\cdot}75\), one < 107, one > 107. | B1 | |
| 2 correct evaluations \(5{\cdot}65 \leqslant x \leqslant 5{\cdot}75\), one < 107, one > 107. | M1 | |
| \(x = 5{\cdot}7\) | A1 |
Correct evaluation regarded as enough to identify if ‘too high’ or ‘too low’. If evaluations not seen accept ‘too high’ or ‘too low’.
| \(x\) | \(x^3 - 13x\) (or check \(x^3 - 13x - 107 = 0\)) | ||
|---|---|---|---|
| 5 | 60 | ||
| 5·1 | 66·351 | ||
| 5·2 | 73·008 | ||
| 5·3 | 79·977 | ||
| 5·4 | 87·264 | ||
| 5·5 | 94·875 | ||
| 5·6 | 102·816 | 5·55 | 98·803... |
| 5·7 | 111·093 | 5·65 | 106·912... |
| 5·8 | 119·712 | 5·655 | 107·326... |
| 5·9 | 128·679 | 5·75 | 115·359... |
| 6 | 138 |