Higher November 2016 Unit 2 Q4
4. A solution to the equation
\[2x^3 - 3x - 17 = 0\]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 \(2{\cdot}15 \leqslant x \leqslant 2{\cdot}35\), ( one value < 0, one value > 0.) | B1 | |
| 2 correct evaluations \(2{\cdot}25 \leqslant x \leqslant 2{\cdot}35\), ( one value < 0, one value > 0.) | M1 | |
| \(x = 2{\cdot}3\) | A1 |
Correct evaluation regarded as enough to identify if negative or positive. If evaluations not seen accept ‘too high’ or ‘too low’.
| x | 2x3 − 3x − 17 | ||
|---|---|---|---|
| 2 | −7 | ||
| 2·1 | −4·778 | ||
| 2·2 | −2·304 | ||
| 2·3 | 0·434 | 2·15 | −3·573... |
| 2·4 | 3·448 | 2·25 | −0·968... |
| 2·5 | 6·75 | 2·26 | −0·693... |
| 2·6 | 10·352 | 2·27 | −0·415... |
| 2·7 | 14·266 | 2·28 | −0·135... |
| 2·8 | 18·504 | 2·29 | 0·147... |
| 2·9 | 23·078 | 2·35 | 1·905... |
| 3 | 28 |