| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2008 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Solve exponential equation via iteration |
| Difficulty | Moderate -0.3 This is a standard C3 iteration question requiring routine application of change of sign test, iterative formula calculation, and verification of root accuracy. All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average but still requiring careful execution across multiple parts. |
| Spec | 1.09a Sign change methods: locate roots1.09b Sign change methods: understand failure cases1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(2) = 0.38\ldots\), \(f(3) = -0.39\ldots\) | M1 | |
| Change of sign (and continuity) \(\Rightarrow\) root in \((2,3)\) | A1 | cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x_1 = \ln 4.5 + 1 \approx 2.50408\) | M1 | |
| \(x_2 \approx 2.50498\) | A1 | |
| \(x_3 \approx 2.50518\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Selecting \([2.5045, 2.5055]\) or appropriate tighter range, evaluating at both ends | M1 | |
| \(f(2.5045) \approx 6\times10^{-4}\), \(f(2.5055) \approx -2\times10^{-4}\) | ||
| Change of sign \(\Rightarrow\) root \(= 2.505\) to 3 d.p. | A1 | cso; note root correct to 5 d.p. is 2.50524 |
# Question 3:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(2) = 0.38\ldots$, $f(3) = -0.39\ldots$ | M1 | |
| Change of sign (and continuity) $\Rightarrow$ root in $(2,3)$ | A1 | cso |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x_1 = \ln 4.5 + 1 \approx 2.50408$ | M1 | |
| $x_2 \approx 2.50498$ | A1 | |
| $x_3 \approx 2.50518$ | A1 | |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Selecting $[2.5045, 2.5055]$ or appropriate tighter range, evaluating at both ends | M1 | |
| $f(2.5045) \approx 6\times10^{-4}$, $f(2.5055) \approx -2\times10^{-4}$ | | |
| Change of sign $\Rightarrow$ root $= 2.505$ to 3 d.p. | A1 | cso; note root correct to 5 d.p. is 2.50524 |
---3.
$$\mathrm { f } ( x ) = \ln ( x + 2 ) - x + 1 , \quad x > - 2 , x \in \mathbb { R } .$$
\begin{enumerate}[label=(\alph*)]
\item Show that there is a root of $\mathrm { f } ( x ) = 0$ in the interval $2 < x < 3$.
\item Use the iterative formula
$$x _ { n + 1 } = \ln \left( x _ { n } + 2 \right) + 1 , x _ { 0 } = 2.5$$
to calculate the values of $x _ { 1 } , x _ { 2 }$ and $x _ { 3 }$ giving your answers to 5 decimal places.
\item Show that $x = 2.505$ is a root of $\mathrm { f } ( x ) = 0$ correct to 3 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2008 Q3 [7]}}