Higher November 2017 Unit 2 Q2
2. A solution of the equation
\[x^3 + 2x = 91\]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 < 91, one > 91. | B1 | |
| 2 correct evaluations \(4{\cdot}25 \leqslant x \leqslant 4{\cdot}35\), one < 91, one > 91. | M1 | |
| \(x = 4{\cdot}3\) | A1 |
Correct evaluation regarded as enough to identify if <91 or >91. If evaluations not seen accept ‘too high’ or ‘too low’.
Look out for testing x3 + 2x – 91 = 0
| x | x3 + 2x | ||
|---|---|---|---|
| 4 | 72 | ||
| 4·1 | 77·121 | ||
| 4·2 | 82·488 | ||
| 4·3 | 88·107 | ||
| 4·4 | 93·984 | ||
| 4·5 | 100·125 | 4·25 | 85·26.... |
| 4·6 | 106·536 | 4·35 | 91·01... |
| 4·7 | 113·223 | 4·45 | 97·02... |
| 4·8 | 120·192 | ||
| 4·9 | 127·449 | ||
| 5 | 135 |