Higher June 2017 Unit 2 Q3
3. A solution to the equation
\[x^3 - 2x - 45 = 0\]lies between 3 and 4.
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 3 ≤ x ≤ 4 | B1 | Correct evaluation regarded as enough to identify if negative or positive. Evaluations can be rounded or truncated. If evaluations not seen condone ‘too high’ or ‘too low’. Look out for testing for x3 − 2x = 45. |
| 2 correct evaluations 3·65 ≤ x ≤ 3·85, one < 0, one > 0. | B1 | |
| 2 correct evaluations 3·65 ≤ x ≤ 3·75, one < 0, one > 0. | M1 | |
| x = 3·7 | A1 | Mark final answer. |
Table of values printed in the Comments column for Q3:
| x | x3 − 2x − 45 | ||
|---|---|---|---|
| 3 | −24 | ||
| 3·1 | −21·409 | ||
| 3·2 | −18·632 | ||
| 3·3 | −15·663 | ||
| 3·4 | −12·496 | ||
| 3·5 | −9·125 | 3·55 | −7·361... |
| 3·6 | −5·544 | 3·65 | −3·672... |
| 3·7 | −1·747 | 3·74 | −0·166... |
| 3·8 | 2·272 | 3·75 | 0·234... |
| 3·9 | 6·519 | 3.85 | 4.366… |
| 4 | 11 |