Higher June 2023 Unit 2 Q3
3. A solution of the equation
\[x^3 - 8x + 3 = 0\]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 \(2{\cdot}55 \leqslant x \leqslant 2{\cdot}75\), (one evaluation < 0, one evaluation > 0) | B1 | |
| 2 correct evaluations \(2{\cdot}55 \leqslant x \leqslant 2{\cdot}65\), (one evaluation < 0, one evaluation > 0) | M1 | |
| \(x = 2{\cdot}6\) | A1 |
Comments
Correct evaluation regarded as enough to identify if negative or positive.If evaluations not seen accept ‘too high’ or ‘too low’.
Look out for equating \(x^3 - 8x = -3\)
| \(x\) | \(x^3 - 8x + 3\) | ||
|---|---|---|---|
| 2 | −5 | ||
| 2·1 | −4·539 | 2·55 | −0·818…. |
| 2·2 | − 3·952 | 2·61 | −0·1004… |
| 2·3 | −3·233 | 2·62 | 0·0247… |
| 2·4 | −2·376 | 2.63 | 0·1514… |
| 2·5 | −1·375 | 2.64 | 0·2797… |
| 2·6 | −0·224 | 2·65 | 0·409… |
| 2·7 | 1·083 | 2·75 | 1·796.. |
| 2·8 | 2·552 | ||
| 2·9 | 4·189 | ||
| 3 | 6 |
The rows 2·6 and 2·7 and the entry 2·65, 0·409… were shaded and in bold in the original.
Unsupported \(x = 2{\cdot}6\) is awarded B0B0M0A0.
An answer of \(x = 2{\cdot}6\) can only be awarded M1A1, following sight of 2 correct evaluations
\(2{\cdot}55 \leqslant x \leqslant 2{\cdot}65\)
(one evaluation < 0, one evaluation > 0).