Higher November 2024 Unit 2 Q2
2. A solution of the equation
\[x^3 + 5x - 4 = 0\]lies between 0 and 1.
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 \(0 \leqslant x \leqslant 1\) | B1 | Correct evaluation regarded as enough to identify if < 0 or > 0. If evaluations not seen accept ‘too high’ or ‘too low’. |
| 2 correct evaluations \(0{\cdot}65 \leqslant x \leqslant 0{\cdot}85\), (one value < 0, one value > 0) | B1 | |
| 2 correct evaluations \(0{\cdot}65 \leqslant x \leqslant 0{\cdot}75\), (one value < 0, one value > 0) | M1 | |
| \(x = 0{\cdot}7\) | A1 | Unsupported \(x = 0{\cdot}7\) is awarded B0B0M0A0. An answer of \(x = 0{\cdot}7\) can only be awarded B1B1M1A1, following sight of 2 correct evaluations \(0{\cdot}65 \leqslant x \leqslant 0{\cdot}75\) with one evaluation < 0 and one evaluation > 0. |
Table of evaluations (printed in the Comments column)
| \(x\) | \(x^3 + 5x - 4\) (check \(x^3 + 5x = 4\)) |
|---|---|
| 0 | -4 |
| 0.1 | -3.499 |
| 0.2 | -2.992 |
| 0.3 | -2.473 |
| 0.4 | -1.936 |
| 0.5 | -1.375 |
| 0.6 | -0.784 |
| 0.7 | -0.157 |
| 0.8 | 0.512 |
| 0.9 | 1.229 |
| 1 | 2 |
| 0.65 | -0.475 |
| 0.73 | 0.039 |
| 0.74 | 0.105 |
| 0.75 | 0.172 |
| 0.85 | 0.864 |
The rows for 0.7, 0.8 and 0.75 were shaded (and bold) in the original.