| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2014 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Derive stationary point equation |
| Difficulty | Standard +0.3 This is a standard multi-part calculus question requiring product rule differentiation, algebraic manipulation to derive a fixed-point equation from f'(x)=0, straightforward iteration calculations, and coordinate evaluation. While it involves several steps, each technique is routine for C3/C4 level with no novel insights required—slightly easier than average due to the guided structure. |
| Spec | 1.07q Product and quotient rules: differentiation1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams1.09d Newton-Raphson method |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Applies \(vu' + uv'\) with \(u = 2x + 2x^2\) and \(v = \ln x\) or vice versa | M1 | Must correctly identify u and v; product rule must be correct if quoted |
| \(f'(x) = \ln(x)(2+4x) + (2x+2x^2) \times \frac{1}{x}\) | A1 | Two out of four separate terms correct (unsimplified) |
| \(f'(x) = \ln(x)(2+4x) + (2x+2x^2) \times \frac{1}{x}\) | A1 | All four terms correct (unsimplified) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Sets \(\ln(x)(2+4x) + (2x+2x^2) \times \frac{1}{x} = 0\) and makes \(\ln x\) the subject | M1 | \(f'(x)\) need not be correct but must be of equal difficulty; look for more than one term in \(\ln x\) and two other unlike terms |
| \(\ln(x) = -\frac{1+x}{1+2x} \Rightarrow x = e^{-\frac{1+x}{1+2x}}\) | dM1 | Dependent on previous M1; moving from \(\ln x = \ldots \Rightarrow x = e^{\ldots}\) |
| \(x = e^{-\frac{1+x}{1+2x}}\) | A1* | CSO; all aspects correct including minus sign and bracketing |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Substitutes \(x_0 = 0.46\) into \(x = e^{-\frac{1+x}{1+2x}}\) | M1 | Attempt to find \(x_1\) from \(x_0\); accept sight of \(e^{-\frac{1+0.46}{1+2\times0.46}}\) or awrt 0.47 |
| \(x_1 =\) awrt \(0.4675\), \(x_2 =\) awrt \(0.4684\) | A1 | |
| \(x_3 =\) awrt \(0.4685\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(A = (0.47, -1.04)\) | M1 | For either \(x = 0.47\) or \(y = -1.04\) using \(x_3\); or substituting \(x_3\) into \(f(x)\); accept sight of \(2 \times x_3(1+x_3)\ln x_3\) |
| \(A = (0.47, -1.04)\) | A1 | Both coordinates correct |
## Question 7:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Applies $vu' + uv'$ with $u = 2x + 2x^2$ and $v = \ln x$ or vice versa | M1 | Must correctly identify u and v; product rule must be correct if quoted |
| $f'(x) = \ln(x)(2+4x) + (2x+2x^2) \times \frac{1}{x}$ | A1 | Two out of four separate terms correct (unsimplified) |
| $f'(x) = \ln(x)(2+4x) + (2x+2x^2) \times \frac{1}{x}$ | A1 | All four terms correct (unsimplified) |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Sets $\ln(x)(2+4x) + (2x+2x^2) \times \frac{1}{x} = 0$ and makes $\ln x$ the subject | M1 | $f'(x)$ need not be correct but must be of equal difficulty; look for more than one term in $\ln x$ and two other unlike terms |
| $\ln(x) = -\frac{1+x}{1+2x} \Rightarrow x = e^{-\frac{1+x}{1+2x}}$ | dM1 | Dependent on previous M1; moving from $\ln x = \ldots \Rightarrow x = e^{\ldots}$ |
| $x = e^{-\frac{1+x}{1+2x}}$ | A1* | CSO; all aspects correct including minus sign and bracketing |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Substitutes $x_0 = 0.46$ into $x = e^{-\frac{1+x}{1+2x}}$ | M1 | Attempt to find $x_1$ from $x_0$; accept sight of $e^{-\frac{1+0.46}{1+2\times0.46}}$ or awrt 0.47 |
| $x_1 =$ awrt $0.4675$, $x_2 =$ awrt $0.4684$ | A1 | |
| $x_3 =$ awrt $0.4685$ | A1 | |
### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $A = (0.47, -1.04)$ | M1 | For either $x = 0.47$ or $y = -1.04$ using $x_3$; or substituting $x_3$ into $f(x)$; accept sight of $2 \times x_3(1+x_3)\ln x_3$ |
| $A = (0.47, -1.04)$ | A1 | Both coordinates correct |
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5b698944-41ac-4072-b5e1-c580b7752c39-20_689_712_248_680}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows a sketch of part of the curve with equation $y = \mathrm { f } ( x )$, where
$$f ( x ) = 2 x ( 1 + x ) \ln x , \quad x > 0$$
The curve has a minimum turning point at $A$.
\begin{enumerate}[label=(\alph*)]
\item Find f'(x)
\item Hence show that the $x$ coordinate of $A$ is the solution of the equation
$$x = \mathrm { e } ^ { - \frac { 1 + x } { 1 + 2 x } }$$
\item Use the iteration formula
$$x _ { n + 1 } = \mathrm { e } ^ { - \frac { 1 + x _ { n } } { 1 + 2 x _ { n } } } , \quad x _ { 0 } = 0.46$$
to find the values of $x _ { 1 } , x _ { 2 }$ and $x _ { 3 }$ to 4 decimal places.
\item Use your answer to part (c) to estimate the coordinates of $A$ to 2 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2014 Q7 [11]}}