| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2005 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Derive stationary point equation |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question testing standard C3 techniques: basic differentiation of exponential and logarithmic functions, finding stationary points by setting f'(x)=0 and rearranging, applying a given iterative formula (pure calculation), and verifying accuracy via sign change. All parts are routine applications of textbook methods with no problem-solving insight required, making it slightly easier than average. |
| Spec | 1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07l Derivative of ln(x): and related functions1.07n Stationary points: find maxima, minima using derivatives1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(f'(x) = 3e^x - \frac{1}{2x}\) | M1A1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(3e^x - \frac{1}{2x} = 0\) | M1 | |
| \(\Rightarrow 6\alpha e^\alpha = 1 \Rightarrow \alpha = \frac{1}{6}e^{-\alpha}\) * | A1 cso |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(x_1 = 0.0613\ldots,\ x_2 = 0.1568\ldots,\ x_3 = 0.1425\ldots,\ x_4 = 0.1445\ldots\) | M1 A1 | M1 at least \(x_1\) correct, A1 all correct to 4 d.p. |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Using \(f'(x) = 3e^x - \frac{1}{2x}\) with suitable interval, e.g. \(f'(0.14425) = -0.0007\), \(f'(0.14435) = +0.002(1)\) | M1 | |
| Accuracy (change of sign and correct values) | A1 |
## Question 4:
### Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $f'(x) = 3e^x - \frac{1}{2x}$ | M1A1A1 | |
### Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $3e^x - \frac{1}{2x} = 0$ | M1 | |
| $\Rightarrow 6\alpha e^\alpha = 1 \Rightarrow \alpha = \frac{1}{6}e^{-\alpha}$ * | A1 cso | |
### Part (c):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $x_1 = 0.0613\ldots,\ x_2 = 0.1568\ldots,\ x_3 = 0.1425\ldots,\ x_4 = 0.1445\ldots$ | M1 A1 | M1 at least $x_1$ correct, A1 all correct to 4 d.p. |
### Part (d):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Using $f'(x) = 3e^x - \frac{1}{2x}$ with suitable interval, e.g. $f'(0.14425) = -0.0007$, $f'(0.14435) = +0.002(1)$ | M1 | |
| Accuracy (change of sign and correct values) | A1 | |
---4.
$$\mathrm { f } ( x ) = 3 \mathrm { e } ^ { x } - \frac { 1 } { 2 } \ln x - 2 , \quad x > 0 .$$
\begin{enumerate}[label=(\alph*)]
\item Differentiate to find $\mathrm { f } ^ { \prime } ( x )$.
The curve with equation $y = \mathrm { f } ( x )$ has a turning point at $P$. The $x$-coordinate of $P$ is $\alpha$.
\item Show that $\alpha = \frac { 1 } { 6 } \mathrm { e } ^ { - \alpha }$.
The iterative formula
$$x _ { n + 1 } = \frac { 1 } { 6 } \mathrm { e } ^ { - x _ { n } } , x _ { 0 } = 1$$
is used to find an approximate value for $\alpha$.
\item Calculate the values of $x _ { 1 } , x _ { 2 } , x _ { 3 }$ and $x _ { 4 }$, giving your answers to 4 decimal places.
\item By considering the change of sign of $\mathrm { f } ^ { \prime } ( x )$ in a suitable interval, prove that $\alpha = 0.1443$ correct to 4 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2005 Q4 [9]}}