C3 June 2007 Q4
4.\[\mathrm{f}(x) = -x^3 + 3x^2 - 1.\]
(a) Show that the equation \(\mathrm{f}(x) = 0\) can be rewritten as\[x = \sqrt{\left(\frac{1}{3 - x}\right)}.\] (2)
(b) Starting with \(x_1 = 0.6\), use the iteration\[x_{n+1} = \sqrt{\left(\frac{1}{3 - x_n}\right)}\]to calculate the values of \(x_2\), \(x_3\) and \(x_4\), giving all your answers to 4 decimal places. (2)
(c) Show that \(x = 0.653\) is a root of \(\mathrm{f}(x) = 0\) correct to 3 decimal places. (3)
| Scheme | Marks |
|---|---|
| \(x^2(3 - x) - 1 = 0\) o.e. (e.g. \(x^2(-x + 3) = 1\)) | M1 |
| \(x = \sqrt{\dfrac{1}{3 - x}}\) (✱) | A1 (cso) |
| (2) |
Notes
Note(✱), answer is given: need to see appropriate working and A1 is cso
[Reverse process: Squaring and non-fractional equation M1, form \(\mathrm{f}(x)\) A1]
| Scheme | Marks |
|---|---|
| \(x_2 = 0.6455\), \(x_3 = 0.6517\), \(x_4 = 0.6526\) | B1; B1 |
| (2) |
Notes
1st B1 is for one correct, 2nd B1 for other two correct
If all three are to greater accuracy, award B0 B1
| Scheme | Marks |
|---|---|
| Choose values in interval \((0.6525, 0.6535)\) or tighter and evaluate both | M1 |
| \(\mathrm{f}(0.6525) = -0.0005\ (372\ldots\) \(\mathrm{f}(0.6535) = 0.002\ (101\ldots\) At least one correct “up to bracket”, i.e. \(-0.0005\) or \(0.002\) | A1 |
| Change of sign, \(\therefore x = 0.653\) is a root (correct) to 3 d.p. Requires both correct “up to bracket” and conclusion as above | A1 |
| (3) | |
| (7 marks) |
Alternative (i)
| Continued iterations at least as far as \(x_6\) | M1 |
| \(x_5 = 0.6527\), \(x_6 = 0.6527\), \(x_7 = \ldots\) two correct to at least 4 s.f. | A1 |
| Conclusion: Two values correct to 4 d.p., so 0.653 is root to 3 d.p. | A1 |
Alternative (ii)
| If use \(\mathrm{g}(0.6525) = 0.6527.. > 0.6525\) and \(\mathrm{g}(0.6535) = 0.6528.. < 0.6535\) | M1A1 |
| Conclusion: Both results correct, so 0.653 is root to 3 d.p. | A1 |