Intermediate June 2018 Unit 2 Q13
13. A solution of the equation
\[x^3 - 7x - 51 = 0\]lies between 4 and 5.
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 \(4 \leqslant x \leqslant 5\) | B1 | |
| 2 correct evaluations \(4{\cdot}25 \leqslant x \leqslant 4{\cdot}45\), one < 0, one > 0. | B1 | |
| 2 correct evaluations \(4{\cdot}25 \leqslant x \leqslant 4{\cdot}35\), one < 0, one > 0. | M1 | |
| ( x = ) 4·3 | A1 |
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 x3 – 7x = 51
| x | x3 – 7x – 51 | ||
|---|---|---|---|
| 4 | −15 | ||
| 4·1 | −10·779 | ||
| 4·2 | − 6·312 | 4·25 | −3·984... |
| 4·3 | − 1·593 | 4·34 | 0·366... |
| 4·4 | 3·384 | 4·35 | 0·862... |
| 4·5 | 8·625 | 4·45 | 5·971... |
| 4·6 | 14·136 | ||
| 4·7 | 19·923 | ||
| 4·8 | 25·992 | ||
| 4·9 | 32·349 | ||
| 5 | 39 |
For this question A1 can only be awarded if M1 given.