Intermediate June 2017 Unit 2 Q13
13. 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 \(3 \leqslant x \leqslant 4\) | B1 | |
| 2 correct evaluations \(3{\cdot}65 \leqslant x \leqslant 3{\cdot}85\), one < 0, one > 0. | B1 | |
| 2 correct evaluations \(3{\cdot}65 \leqslant x \leqslant 3{\cdot}75\), one < 0, one > 0. | M1 | |
| \(x = 3{\cdot}7\) | A1 | Mark final answer. |
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 \(x^3 - 2x = 45\).
| \(x\) | \(x^3 - 2x - 45\) | ||
|---|---|---|---|
| 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 |