Higher November 2019 Unit 2 Q1
1. A solution of the equation
\[x^3 - 3x = 37\]lies between 3 and 4.
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 \(3 \leqslant x \leqslant 4\) | B1 | |
| 2 correct evaluations \(3{\cdot}55 \leqslant x \leqslant 3{\cdot}75\), one < 37, one > 37. | B1 | |
| 2 correct evaluations \(3{\cdot}55 \leqslant x \leqslant 3{\cdot}65\), one < 37, one > 37. | M1 | |
| \(x = 3{\cdot}6\) | A1 |
Correct evaluation regarded as enough to identify if <37 or >37. If evaluations not seen accept ‘too high’ or ‘too low’.
Look out for testing \(x^3 - 3x - 37 = 0\)
| \(x\) | \(x^3 - 3x\) | ||
|---|---|---|---|
| 3 | 18 | ||
| 3·1 | 20·491 | ||
| 3·2 | 23·168 | ||
| 3·3 | 26·037 | ||
| 3·4 | 29·104 | ||
| 3·5 | 32·375 | 3·55 | 34·08.... |
| 3·6 | 35·856 | 3·65 | 37·67... |
| 3·7 | 39·553 | 3·75 | 41·48... |
| 3·8 | 43·472 | ||
| 3·9 | 47·619 | ||
| 4 | 52 |