Higher November 2022 Paper 2 Q12
12. The equation \(2x^3 + x^2 - 12 = 0\) has a solution between 1 and 2.
Use trial and improvement to find this solution correct to 1 decimal place. [4]
| Scheme | Marks | Comments |
|---|---|---|
| Correct evaluation enough to imply > 0 or < 0; If evaluations not seen then may be implied by e,g, ‘too big’ or ‘too small’; Mark solving \(2x^3 + x^2 = 12\) equivalently | ||
| One correct evaluation with \(1 \leqslant x \leqslant 2\) | B1 | |
| Two correct evaluations with \(1.55 \leqslant x \leqslant 1.75\) and one > 0, one < 0 si | B1 | |
| Two correct evaluations with \(1.65 \leqslant x \leqslant 1.75\) and one > 0, one < 0 si | M1 | |
| 1.7 | A1 | |
| (4) |
Notes
| \(x\) | \(2x^3 + x^2 - 12\) | or | \(2x^3 + x^2\) | |
|---|---|---|---|---|
| 1 | −9 | 3 | ||
| 1.1 | −8.128 | 3.872 | ||
| 1.2 | −7.104 | 4.896 | ||
| 1.3 | −5.916 | 6.084 | ||
| 1.4 | −4.552 | 7.448 | ||
| 1.5 | −3 | 9 | ||
| 1.55 | −2.149... | 9.850... | ||
| 1.6 | −1.248 | 10.752 | ||
| 1.65 | −0.293... | 11.706... | ||
| 1.7 | 0.716 | 12.716 | ||
| 1.75 | 1.781... | 13.781... | ||
| 1.8 | 2.904 | 14.904 | ||
| 1.9 | 5.328 | 17.328 | ||
| 2 | 8 | 20 | ||
If no marks, award SC2 for a complete method with a final answer of 1.1 from \((2x)^3 + x^2 - 12 = 0\) or \((2x)^3 + x^2 = 12\)