| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2010 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Rearrange to iterative form |
| Difficulty | Moderate -0.3 This is a standard C3 fixed-point iteration question requiring routine algebraic rearrangement, calculator work to iterate three times, and verification of convergence to 3dp. The rearrangement is straightforward, and the iteration/verification follow textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.09b Sign change methods: understand failure cases1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(f(x)=0 \Rightarrow x^3+2x^2-3x-11=0\) | M1 | Sets \(f(x)=0\) (can be implied) and takes out a factor of \(x^2\) from \(x^3+2x^2\), or \(x\) from \(x^3+2x\) (slip) |
| \(\Rightarrow x^2(x+2)-3x-11=0\) | ||
| \(\Rightarrow x^2(x+2) = 3x+11\) | ||
| \(\Rightarrow x^2 = \frac{3x+11}{x+2}\) | ||
| \(\Rightarrow x = \sqrt{\left(\frac{3x+11}{x+2}\right)}\) | A1 AG | Then rearranges to give the quoted result on the question paper |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(x_2 = \sqrt{\left(\frac{3(0)+11}{(0)+2}\right)}\) | 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 |
| \(x_2 = 2.34520788...\) | A1 | Both \(x_2=\) awrt 2.345 and \(x_3=\) awrt 2.037 |
| \(x_3 = 2.037324945...\) | A1 | |
| \(x_4 = 2.058748112...\) | A1 | \(x_4=\) awrt 2.059 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Let \(f(x) = x^3+2x^2-3x-11=0\) | M1 | Choose suitable interval for \(x\), e.g. [2.0565, 2.0575] or tighter |
| \(f(2.0565) = -0.013781637...\) | dM1 | Any one value awrt 1 sf |
| \(f(2.0575) = 0.0041401094...\) | A1 | Both values correct awrt 1sf, sign change and conclusion |
| Sign change (and \(f(x)\) is continuous) therefore a root \(\alpha\) is such that \(\alpha \in (2.0565, 2.0575) \Rightarrow \alpha = 2.057\) (3 dp) | As a minimum, both values must be correct to 1 sf, candidate states "change of sign, hence root" |
## Question 2:
### Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $f(x)=0 \Rightarrow x^3+2x^2-3x-11=0$ | M1 | Sets $f(x)=0$ (can be implied) and takes out a factor of $x^2$ from $x^3+2x^2$, or $x$ from $x^3+2x$ (slip) |
| $\Rightarrow x^2(x+2)-3x-11=0$ | | |
| $\Rightarrow x^2(x+2) = 3x+11$ | | |
| $\Rightarrow x^2 = \frac{3x+11}{x+2}$ | | |
| $\Rightarrow x = \sqrt{\left(\frac{3x+11}{x+2}\right)}$ | A1 **AG** | Then rearranges to give the quoted result on the question paper |
**Total: (2)**
### Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $x_2 = \sqrt{\left(\frac{3(0)+11}{(0)+2}\right)}$ | 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 |
| $x_2 = 2.34520788...$ | A1 | Both $x_2=$ awrt 2.345 and $x_3=$ awrt 2.037 |
| $x_3 = 2.037324945...$ | A1 | |
| $x_4 = 2.058748112...$ | A1 | $x_4=$ awrt 2.059 |
**Total: (3)**
### Part (c):
| Working | Mark | Guidance |
|---------|------|----------|
| Let $f(x) = x^3+2x^2-3x-11=0$ | M1 | Choose suitable interval for $x$, e.g. [2.0565, 2.0575] or tighter |
| $f(2.0565) = -0.013781637...$ | dM1 | Any one value awrt 1 sf |
| $f(2.0575) = 0.0041401094...$ | A1 | Both values correct awrt 1sf, sign change and conclusion |
| Sign change (and $f(x)$ is continuous) therefore a root $\alpha$ is such that $\alpha \in (2.0565, 2.0575) \Rightarrow \alpha = 2.057$ (3 dp) | | As a minimum, both values must be correct to 1 sf, candidate states "change of sign, hence root" |
**Total: (3)**
**[8]**
---2.
$$f ( x ) = x ^ { 3 } + 2 x ^ { 2 } - 3 x - 11$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { f } ( x ) = 0$ can be rearranged as
$$x = \sqrt { } \left( \frac { 3 x + 11 } { x + 2 } \right) , \quad x \neq - 2 .$$
The equation $\mathrm { f } ( x ) = 0$ has one positive root $\alpha$.
The iterative formula $x _ { n + 1 } = \sqrt { } \left( \frac { 3 x _ { n } + 11 } { x _ { n } + 2 } \right)$ is used to find an approximation to $\alpha$.
\item Taking $x _ { 1 } = 0$, find, to 3 decimal places, the values of $x _ { 2 } , x _ { 3 }$ and $x _ { 4 }$.
\item Show that $\alpha = 2.057$ correct to 3 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2010 Q2 [8]}}