Intermediate June 2023 Unit 2 Q13
13. A solution of the equation
\[x^3 - 8x + 3 = 0\]lies between 2 and 3.
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 2 ≤ \(x\) ≤ 3 | B1 | 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 \(x^3 - 8x = -3\)
| ||||||||||||||||||||||||||||||||||||||||||||||||
| 2 correct evaluations 2·55 ≤ \(x\) ≤ 2·75, (one evaluation < 0, one evaluation > 0) | B1 | |||||||||||||||||||||||||||||||||||||||||||||||||
| 2 correct evaluations 2·55 ≤ \(x\) ≤ 2·65, (one evaluation < 0, one evaluation > 0) | M1 | |||||||||||||||||||||||||||||||||||||||||||||||||
| \(x = 2{\cdot}6\) | A1 | Unsupported \(x = 2{\cdot}6\) is awarded B0B0M0A0. An answer of \(x = 2{\cdot}6\) can only be awarded M1A1, following sight of 2 correct evaluations 2·55 ≤ \(x\) ≤ 2·65 (one evaluation < 0, one evaluation > 0). |