| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Derive stationary point equation |
| Difficulty | Standard +0.3 This is a structured multi-part question with clear scaffolding. Part (a) is routine differentiation and algebraic manipulation. Parts (b-c) involve standard curve sketching and graphical interpretation of roots. Parts (d-e) are straightforward iteration calculations requiring only calculator work. All techniques are standard C3 content with no novel problem-solving required—slightly easier than average due to the extensive guidance provided. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.06a Exponential function: a^x and e^x graphs and properties1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07n Stationary points: find maxima, minima using derivatives1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = 4e^{4x} + 4x^3 + 8\) | M1, A1 | M1: Two of the four terms differentiated correctly. A1: All correct |
| Sets \(\frac{dy}{dx} = 0\) to give \(x^3 = -2 - e^{4x}\) | A1* | States or sets \(\frac{dy}{dx}=0\) and proceeds correctly to achieve printed answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct shape and position for \(y = x^3\) | B1 | Must appear to go through origin, only in Quadrants 1 and 3, gradient always \(\geq 0\) |
| Correct shape for \(y = -2 - e^{4x}\) | B1 | Gradient approximately zero at left end, increasing negatively left to right |
| \(y = -2 - e^{4x}\) cuts \(y\)-axis at \((0,-3)\) | B1 | Accept intention of \((0,-3)\); also accept \(-3\) marked on \(y\)-axis. Do not accept 3 on negative \(y\)-axis |
| \(y = -2 - e^{4x}\) has asymptote at \(y = -2\) | B1 | Dependent on curve appearing to have asymptote. Do not accept '\(-2\)' or \(x=-2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Only one crossing point | B1 | Accept minimal statements such as 'one intersection'. Must have a diagram. Do not award if diagram shows more than one intersection |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(-1.26376,\ -1.26126\) | M1 A1 | M1: applying iteration formula once; sight of \(\sqrt[3]{-2-e^{-4}}\), \((-2-e^{4x-1})^{\frac{1}{3}}\) or awrt \(-1.264\). A1: Both values correct awrt 5dp |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\alpha = -1.26\) and so turning point is at \((-1.26, -2.55)\) | M1 A1cao | M1: rounding part (d) value to 2dp OR substituting \(x\) from (d) into \(y = e^{4x}+x^4+8x+5\). A1: correct solution only, cannot accept appearance of correct answer for two marks |
## Question 2:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = 4e^{4x} + 4x^3 + 8$ | M1, A1 | M1: Two of the four terms differentiated correctly. A1: All correct |
| Sets $\frac{dy}{dx} = 0$ to give $x^3 = -2 - e^{4x}$ | A1* | States or sets $\frac{dy}{dx}=0$ and proceeds correctly to achieve printed answer |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct shape and position for $y = x^3$ | B1 | Must appear to go through origin, only in Quadrants 1 and 3, gradient always $\geq 0$ |
| Correct shape for $y = -2 - e^{4x}$ | B1 | Gradient approximately zero at left end, increasing negatively left to right |
| $y = -2 - e^{4x}$ cuts $y$-axis at $(0,-3)$ | B1 | Accept intention of $(0,-3)$; also accept $-3$ marked on $y$-axis. Do not accept 3 on negative $y$-axis |
| $y = -2 - e^{4x}$ has asymptote at $y = -2$ | B1 | Dependent on curve appearing to have asymptote. Do not accept '$-2$' or $x=-2$ |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Only one crossing point | B1 | Accept minimal statements such as 'one intersection'. Must have a diagram. Do not award if diagram shows more than one intersection |
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $-1.26376,\ -1.26126$ | M1 A1 | M1: applying iteration formula once; sight of $\sqrt[3]{-2-e^{-4}}$, $(-2-e^{4x-1})^{\frac{1}{3}}$ or awrt $-1.264$. A1: Both values correct awrt 5dp |
### Part (e):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\alpha = -1.26$ and so turning point is at $(-1.26, -2.55)$ | M1 A1cao | M1: rounding part (d) value to 2dp OR substituting $x$ from (d) into $y = e^{4x}+x^4+8x+5$. A1: correct solution only, cannot accept appearance of correct answer for two marks |
---2. A curve $C$ has equation $y = \mathrm { e } ^ { 4 x } + x ^ { 4 } + 8 x + 5$
\begin{enumerate}[label=(\alph*)]
\item Show that the $x$ coordinate of any turning point of $C$ satisfies the equation
$$x ^ { 3 } = - 2 - \mathrm { e } ^ { 4 x }$$
\item On the axes given on page 5, sketch, on a single diagram, the curves with equations
\begin{enumerate}[label=(\roman*)]
\item $y = x ^ { 3 }$,
\item $y = - 2 - e ^ { 4 x }$
On your diagram give the coordinates of the points where each curve crosses the $y$-axis and state the equation of any asymptotes.
\end{enumerate}\item Explain how your diagram illustrates that the equation $x ^ { 3 } = - 2 - e ^ { 4 x }$ has only one root.
The iteration formula
$$x _ { n + 1 } = \left( - 2 - \mathrm { e } ^ { 4 x _ { n } } \right) ^ { \frac { 1 } { 3 } } , \quad x _ { 0 } = - 1$$
can be used to find an approximate value for this root.
\item Calculate the values of $x _ { 1 }$ and $x _ { 2 }$, giving your answers to 5 decimal places.
\item Hence deduce the coordinates, to 2 decimal places, of the turning point of the curve $C$.\\
\includegraphics[max width=\textwidth, alt={}, center]{be00fdaa-2fe3-4f06-a710-08ec67fb911e-04_1285_1294_308_331}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2014 Q2 [12]}}