Intermediate November 2018 Unit 2 Q14
14. 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 ≤ x ≤ 6 | B1 | Correct evaluation regarded as enough to identify if ‘too high’ or ‘too low’. If evaluations not seen accept ‘too high’ or ‘too low’. |
| 2 correct evaluations 5·55 ≤ x ≤ 5·75, one < 107, one > 107. | B1 | |
| 2 correct evaluations 5·65 ≤ x ≤ 5·75, one < 107, one > 107. | M1 | |
| x = 5·7 | A1 |
(or check x3 – 13x – 107=0)
| x | x3 – 13x | ||
|---|---|---|---|
| 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 |
In the original this table of values is printed in the Comments column. The column headings are underlined and the bold values are bold in the original.