| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2018 |
| Session | October |
| Marks | 10 |
| 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 C3/C4 question involving routine techniques: (a) solving 8x = xe^(3x) by factoring, (b) differentiating a product and rearranging to the given form, (c) applying an iterative formula three times with a calculator. Each part follows textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.06b Gradient of e^(kx): derivative and exponential model1.07j Differentiate exponentials: e^(kx) and a^(kx)1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Sets \(8x - xe^{3x} = 0 \Rightarrow e^{3x} = 8 \Rightarrow 3x = \ln 8 \Rightarrow x = \frac{1}{3}\ln 8 = \ln 2\) | M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = 8 - (3xe^{3x} + e^{3x})\) | M1, A1 | |
| Sets \(\frac{dy}{dx} = 0 \Rightarrow (1+3x)e^{3x} = 8\) | M1 | |
| \(\Rightarrow e^{3x} = \frac{8}{(1+3x)} \Rightarrow x = \frac{1}{3}\ln\left(\frac{8}{1+3x}\right)\) | dM1, A1* |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x_1 = \frac{1}{3}\ln\left(\frac{8}{1+3\times 0.4}\right) =\) awrt \(0.430\) | M1A1 | |
| \(x_2 =\) awrt \(0.417\), \(x_3 =\) awrt \(0.423\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(e^{3x} = 8 \Rightarrow 3x = \ln 8\) | M1 | Attempts to solve using correct order of operations; allow \(3x = \ln 8\); condone slip on 8 |
| \(x = \ln 2\) | A1 | Note \(x = \frac{1}{3}\ln 8\) is M1 A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Differentiates \(-xe^{3x}\) term to \(\pm Axe^{3x} \pm Be^{3x}\) using product rule | M1 | If rule is quoted it must be correct |
| \(\frac{dy}{dx} = 8 - (3xe^{3x} + e^{3x})\) or \(\frac{dy}{dx} = 8 - 3xe^{3x} - e^{3x}\) | A1 | Correct derivative with correct bracketing |
| Sets \(\frac{dy}{dx} = 0\) and takes out common factor \(e^{3x}\) reaching \((\ldots\pm\ldots)e^{3x} = \ldots\) or \(e^{3x} = \frac{\ldots}{(\ldots\pm\ldots)}\) | M1 | May be implied |
| \((\ldots\pm\ldots)e^{3x} = \ldots \Rightarrow x =\) or \(e^{3x} = \frac{\ldots}{(\ldots\pm\ldots)} \Rightarrow x =\) | dM1 | Dependent on both previous M's; correct ln work |
| \(x = \frac{1}{3}\ln\left(\frac{8}{1+3x}\right)\), \(x = \frac{1}{3}\ln\left(\frac{8}{3x+1}\right)\) or \(x = \frac{1}{3}\ln\frac{8}{3x+1}\) | A1* | No errors or omissions |
| Answer | Marks | Guidance |
|---|---|---|
| Calculates \(x_1\) from iterative formula; \(\frac{1}{3}\ln\left(\frac{8}{1+3\times0.4}\right)\) or awrt 0.43 | M1 | May be implied |
| awrt 0.430 (allow 0.43) | A1 | |
| awrt \(x_2 = 0.417\), \(x_3 = 0.423\) | A1 | Subscripts not important |
# Question 4:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Sets $8x - xe^{3x} = 0 \Rightarrow e^{3x} = 8 \Rightarrow 3x = \ln 8 \Rightarrow x = \frac{1}{3}\ln 8 = \ln 2$ | M1A1 | |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = 8 - (3xe^{3x} + e^{3x})$ | M1, A1 | |
| Sets $\frac{dy}{dx} = 0 \Rightarrow (1+3x)e^{3x} = 8$ | M1 | |
| $\Rightarrow e^{3x} = \frac{8}{(1+3x)} \Rightarrow x = \frac{1}{3}\ln\left(\frac{8}{1+3x}\right)$ | dM1, A1* | |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x_1 = \frac{1}{3}\ln\left(\frac{8}{1+3\times 0.4}\right) =$ awrt $0.430$ | M1A1 | |
| $x_2 =$ awrt $0.417$, $x_3 =$ awrt $0.423$ | A1 | |
# Question (from first page - part a,b,c):
## Part (a):
| $e^{3x} = 8 \Rightarrow 3x = \ln 8$ | M1 | Attempts to solve using correct order of operations; allow $3x = \ln 8$; condone slip on 8 |
| $x = \ln 2$ | A1 | Note $x = \frac{1}{3}\ln 8$ is M1 A0 |
## Part (b):
| Differentiates $-xe^{3x}$ term to $\pm Axe^{3x} \pm Be^{3x}$ using product rule | M1 | If rule is quoted it must be correct |
| $\frac{dy}{dx} = 8 - (3xe^{3x} + e^{3x})$ or $\frac{dy}{dx} = 8 - 3xe^{3x} - e^{3x}$ | A1 | Correct derivative with correct bracketing |
| Sets $\frac{dy}{dx} = 0$ and takes out common factor $e^{3x}$ reaching $(\ldots\pm\ldots)e^{3x} = \ldots$ or $e^{3x} = \frac{\ldots}{(\ldots\pm\ldots)}$ | M1 | May be implied |
| $(\ldots\pm\ldots)e^{3x} = \ldots \Rightarrow x =$ or $e^{3x} = \frac{\ldots}{(\ldots\pm\ldots)} \Rightarrow x =$ | dM1 | Dependent on both previous M's; correct ln work |
| $x = \frac{1}{3}\ln\left(\frac{8}{1+3x}\right)$, $x = \frac{1}{3}\ln\left(\frac{8}{3x+1}\right)$ or $x = \frac{1}{3}\ln\frac{8}{3x+1}$ | A1* | No errors or omissions |
## Part (c):
| Calculates $x_1$ from iterative formula; $\frac{1}{3}\ln\left(\frac{8}{1+3\times0.4}\right)$ or awrt 0.43 | M1 | May be implied |
| awrt 0.430 (allow 0.43) | A1 | |
| awrt $x_2 = 0.417$, $x_3 = 0.423$ | A1 | Subscripts not important |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c6bde466-61ec-437d-a3b4-84511a98d788-08_510_783_260_584}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of part of the curve with equation $y = 8 x - x \mathrm { e } ^ { 3 x } , x \geqslant 0$ The curve meets the $x$-axis at the origin and cuts the $x$-axis at the point $A$.
\begin{enumerate}[label=(\alph*)]
\item Find the exact $x$ coordinate of $A$, giving your answer in its simplest form.
The curve has a maximum turning point at the point $M$.
\item Show, by using calculus, that the $x$ coordinate of $M$ is a solution of
$$x = \frac { 1 } { 3 } \ln \left( \frac { 8 } { 1 + 3 x } \right)$$
\item Use the iterative formula
$$x _ { n + 1 } = \frac { 1 } { 3 } \ln \left( \frac { 8 } { 1 + 3 x _ { n } } \right)$$
with $x _ { 0 } = 0.4$ to calculate the values of $x _ { 1 } , x _ { 2 }$ and $x _ { 3 }$, giving your answers to 3 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2018 Q4 [10]}}