Intermediate June 2019 Unit 2 Q16
16. A solution of the equation
\[2x^3 + x - 10 = 0\]lies between 1 and 2.
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 \(1 \leqslant x \leqslant 2\) | B1 | |
| 2 correct evaluations \(1{\cdot}55 \leqslant x \leqslant 1{\cdot}75\), one < 0, one > 0. | B1 | |
| 2 correct evaluations \(1{\cdot}55 \leqslant x \leqslant 1{\cdot}65\), one < 0, one > 0. | M1 | |
| x = 1·6 | A1 |
Correct evaluation regarded as enough to identify if ‘too high’ or ‘too low’. If evaluations not seen accept ‘too high’ or ‘too low’.
| x | 2x3 + x − 10 (or check 2x3 + x =10) | ||
|---|---|---|---|
| 1 | −7 | ||
| 1·1 | −6·238 | ||
| 1·2 | −5·344 | ||
| 1·3 | −4·306 | ||
| 1·4 | −3·112 | 1·45 | − 2·452... |
| 1·5 | −1·75 | 1·55 | − 1·002... |
| 1·6 | −0·208 | 1·65 | 0·634... |
| 1·7 | 1·526 | 1·75 | 2·468... |
| 1·8 | 3·464 | (1·62 | 0·123..) |
| 1·9 | 5·618 | (1·63 | 0·291..) |
| 2 | 8 | (1·64 | 0·461..) |