C3 January 2010 Q2
2.
\[\mathrm{f}(x) = x^3 + 2x^2 - 3x - 11\]The equation \(\mathrm{f}(x) = 0\) has one positive root \(\alpha\).
The iterative formula \(x_{n+1} = \sqrt{\left(\dfrac{3x_n + 11}{x_n + 2}\right)}\) is used to find an approximation to \(\alpha\).
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(x) = x^3 + 2x^2 - 3x - 11\) | |
| \(\mathrm{f}(x) = 0 \Rightarrow x^3 + 2x^2 - 3x - 11 = 0\) \(\Rightarrow x^2(x + 2) - 3x - 11 = 0\) | M1 |
| \(\Rightarrow x^2(x + 2) = 3x + 11\) \(\Rightarrow x^2 = \dfrac{3x + 11}{x + 2}\) | |
| \(\Rightarrow x = \sqrt{\left(\dfrac{3x + 11}{x + 2}\right)}\) | A1 AG |
| (2) |
Notes
M1: Sets \(\mathrm{f}(x) = 0\) (can be implied) and takes out a factor of \(x^2\) from \(x^3 + 2x^2\), or \(x\) from \(x^3 + 2x\) (slip).
A1 AG: then rearranges to give the quoted result on the question paper.
| Scheme | Marks |
|---|---|
| Iterative formula: \(x_{n+1} = \sqrt{\left(\dfrac{3x_n + 11}{x_n + 2}\right)}\), \(x_1 = 0\) | |
| \(x_2 = \sqrt{\left(\dfrac{3(0) + 11}{(0) + 2}\right)}\) | M1 |
| \(x_2 = 2.34520788\ldots\) \(x_3 = 2.037324945\ldots\) | A1 |
| \(x_4 = 2.058748112\ldots\) | A1 |
| (3) |
Notes
M1: An attempt to substitute \(x_1 = 0\) into the iterative formula. Can be implied by \(x_2 = \sqrt{5.5}\) or 2.35 or awrt 2.345
A1: Both \(x_2 = \text{awrt } 2.345\) and \(x_3 = \text{awrt } 2.037\)
A1: \(x_4 = \text{awrt } 2.059\)
| Scheme | Marks |
|---|---|
| Let \(\mathrm{f}(x) = x^3 + 2x^2 - 3x - 11 = 0\) | |
| \(\mathrm{f}(2.0565) = -0.013781637\ldots\) \(\mathrm{f}(2.0575) = 0.0041401094\ldots\) | M1 dM1 |
| Sign change (and \(\mathrm{f}(x)\) is continuous) therefore a root \(\alpha\) is such that \(\alpha \in (2.0565, 2.0575) \Rightarrow \alpha = 2.057\ (3\text{ dp})\) | A1 |
| (3) | |
| (8 marks) |
Notes
M1: Choose suitable interval for \(x\), e.g. [2.0565, 2.0575] or tighter
dM1: any one value awrt 1 sf
A1: both values correct awrt 1sf, sign change and conclusion
As a minimum, both values must be correct to 1 sf, candidate states “change of sign, hence root”.