Higher November 2020 Unit 2 Q3
3. A solution of the equation
\[x^3 - 5x - 350 = 0\]
lies between 7·2 and 7·3.
Use the method of trial and improvement to find this solution correct to 2 decimal places.
You must show all your working. [4]
| Scheme | Marks | Comments |
|---|---|---|
| One correct evaluation \(7{\cdot}2 \leqslant x \leqslant 7{\cdot}3\) | B1 | |
| 2 correct evaluations \(7{\cdot}275 \leqslant x \leqslant 7{\cdot}295\), one < 0, one > 0. | B1 | |
| 2 correct evaluations \(7{\cdot}275 \leqslant x \leqslant 7{\cdot}285\), one < 0, one > 0. | M1 | |
| \(x = 7{\cdot}28\) | 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 \(x^3 - 5x = 350\)
| \(x\) | \(x^3 - 5x - 350\) | ||
|---|---|---|---|
| 7·2 | −12·75(2) | ||
| 7·21 | −11(·2.. ) | ||
| 7·22 | − 9(·7…) | ||
| 7·23 | − 8(·2…) | ||
| 7·24 | − 6(·6…) | ||
| 7·25 | −5(·1...) | ||
| 7·26 | − 3(·6…) | 7·275 | −1(·3….) |
| 7·27 | − 2(·1…) | 7·284 | 0(·04..) |
| 7·28 | −0·5(7..) | 7·285 | 0·1(9..) |
| 7·29 | 0·9(7..) | 7·295 | 1(·7….) |
| 7·3 | 2·5(17) |