C3 January 2006 Q5
5. \[\mathrm{f}(x) = 2x^3 - x - 4.\]
(a) Show that the equation \(\mathrm{f}(x) = 0\) can be written as\[x = \sqrt{\left(\frac{2}{x} + \frac{1}{2}\right)}.\] (3)
The equation \(2x^3 - x - 4 = 0\) has a root between 1.35 and 1.4.
(b) Use the iteration formula\[x_{n+1} = \sqrt{\left(\frac{2}{x_n} + \frac{1}{2}\right)},\]with \(x_0 = 1.35\), to find, to 2 decimal places, the values of \(x_1\), \(x_2\) and \(x_3\). (3)
The only real root of \(\mathrm{f}(x) = 0\) is \(\alpha\).
(c) By choosing a suitable interval, prove that \(\alpha = 1.392\), to 3 decimal places. (3)
| Scheme | Marks |
|---|---|
| \(2x^2 - 1 - \dfrac{4}{x} = 0\) Dividing equation by \(x\) | M1 |
| \(x^2 = \dfrac{1}{2} + \dfrac{4}{2x}\) Obtaining \(x^2 = \ldots\) | M1 |
| \(x = \sqrt{\left(\dfrac{2}{x} + \dfrac{1}{2}\right)}\) * cso | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(x_1 = 1.41,\ x_2 = 1.39,\ x_3 = 1.39\) | B1, B1, B1 |
| (3) |
Notes
If answers given to more than 2 dp, penalise first time then accept awrt above.
| Scheme | Marks |
|---|---|
| Choosing \((1.3915, 1.3925)\) or a tighter interval | M1 |
| \(\mathrm{f}(1.3915) \approx -3 \times 10^{-3},\ \mathrm{f}(1.3925) \approx 7 \times 10^{-3}\) Both, awrt | A1 |
| Change of sign (and continuity) \(\Rightarrow \alpha \in (1.3915, 1.3925)\) \(\Rightarrow \alpha = 1.392\) to 3 decimal places * cso | A1 |
| (3) | |
| (9 marks) |