| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2013 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Solve exponential equation via iteration |
| Difficulty | Standard +0.3 This is a standard C3 iteration question requiring algebraic rearrangement, calculator work to iterate three times, and interval verification using change of sign. All techniques are routine textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.06g Equations with exponentials: solve a^x = b1.09a Sign change methods: locate roots1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| \(0 = e^{x-1} + x - 6 \Rightarrow x = \ln(6-x) + 1\) | M1A1* | Sets \(g(x)=0\), and using correct \(\ln\) work, makes the \(x\) of the \(e^{x-1}\) term the subject of the formula. Look for \(e^{x-1} = \pm 6 \pm x \Rightarrow x = \ln(\pm 6 \pm x) \pm 1\). Do not accept \(e^{x-1} = 6 - x\) without firstly seeing \(e^{x-1} + x - 6 = 0\) or a statement that \(g(x)=0 \Rightarrow\). cso. \(x = \ln(6-x) + 1\) Note that this is a given answer (and a proof). 'Invisible' brackets are allowed for the M but not the A. Do not accept recovery from earlier errors for the A mark. The solution below scores 0 marks. \(0 = e^{x-1} + x - 6 \Rightarrow 0 = x - 1 + \ln(6-x) \Rightarrow x = \ln(6-x) + 1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Sub \(x_0 = 2\) into \(x_{n+1} = \ln(6-x_n) + 1 \Rightarrow x_1 = 2.3863\) | M1, A1 | Evidence for the award could be any of \(\ln(6-2) + 1\), \(\ln 4 + 1\), \(2.3...\)... or awrt 2.4 |
| AWRT 4 dp. \(x_2 = 2.2847\) \(x_3 = 2.3125\) | A1 | Awrt 4 dp. \(x_2 = 2.2847\) and \(x_3 = 2.3125\). The subscripts are not important. Mark as the second and third values given/found. |
| Answer | Marks | Guidance |
|---|---|---|
| Chooses interval \([2.3065, 2.3075]\) | M1 | Chooses the interval \([2.3065, 2.3075]\) or smaller containing the root 2.306558641 |
| \(g(2.3065)=−0.0002(7)\), \(g(2.3075)=0.004(4)\) | dM1 | Calculates \(g(2.3065)\) and \(g(2.3075)\) with at least one of these correct to 1sf. The answers can be rounded or truncated \(g(2.3065) = -0.0003\) rounded, \(g(2.3065)= -0.0002\) truncated \(g(2.3075) = (+) 0.004\) rounded and truncated |
| Sign change, hence root (correct to 3dp) | A1 | Both values correct (rounded or truncated). A reason which could include change of sign, \(>0 <0\), \(g(2.3065) \times g(2.3075) < 0\) AND a minimal conclusion such as hence root, \(α=2.307\) or ▭. Do not accept continued iteration as question demands an interval to be chosen. |
| (8 marks) |
| Answer | Marks |
|---|---|
| M1 | Proceeds from \(x = \ln(6-x) + 1\) using correct exp work to \(......=0\) |
| A1 | Arrives correctly at \(e^{x-1} + x - 6 = 0\) and makes a statement to the effect that this is \(g(x)=0\) |
| Answer | Marks |
|---|---|
| M1 | Chooses the interval \([2.3065, 2.3075]\) or smaller containing the root 2.306558641 |
| dM1 | Calculates \(f(2.3065)\) and \(f(2.3075)\) with at least 1 correct rounded or truncated \(f(2.3065) = 0.000074\). Accept \(0.00007\) rounded or truncated. Also accept \(0.0001\) |
| \(f(2.3075) = - 0.0011\).. Accept \(-0.001\) rounded or truncated |
**(a)**
$0 = e^{x-1} + x - 6 \Rightarrow x = \ln(6-x) + 1$ | M1A1* | Sets $g(x)=0$, and using correct $\ln$ work, makes the $x$ of the $e^{x-1}$ term the subject of the formula. Look for $e^{x-1} = \pm 6 \pm x \Rightarrow x = \ln(\pm 6 \pm x) \pm 1$. **Do not accept** $e^{x-1} = 6 - x$ **without firstly seeing** $e^{x-1} + x - 6 = 0$ **or a statement that** $g(x)=0 \Rightarrow$. **cso.** $x = \ln(6-x) + 1$ **Note that this is a given answer (and a proof). 'Invisible' brackets are allowed for the M but not the A. Do not accept recovery from earlier errors for the A mark. The solution below scores 0 marks.** $0 = e^{x-1} + x - 6 \Rightarrow 0 = x - 1 + \ln(6-x) \Rightarrow x = \ln(6-x) + 1$ | (2 marks)
**(b)**
Sub $x_0 = 2$ into $x_{n+1} = \ln(6-x_n) + 1 \Rightarrow x_1 = 2.3863$ | M1, A1 | Evidence for the award could be any of $\ln(6-2) + 1$, $\ln 4 + 1$, $2.3...$... or awrt 2.4 |
AWRT 4 dp. $x_2 = 2.2847$ $x_3 = 2.3125$ | A1 | Awrt 4 dp. $x_2 = 2.2847$ **and** $x_3 = 2.3125$. The subscripts are not important. Mark as the second and third values given/found. | (3 marks)
**(c)**
Chooses interval $[2.3065, 2.3075]$ | M1 | Chooses the interval $[2.3065, 2.3075]$ or smaller containing the root 2.306558641 |
$g(2.3065)=−0.0002(7)$, $g(2.3075)=0.004(4)$ | dM1 | Calculates $g(2.3065)$ and $g(2.3075)$ with at least one of these correct to 1sf. The answers can be rounded or truncated $g(2.3065) = -0.0003$ rounded, $g(2.3065)= -0.0002$ truncated $g(2.3075) = (+) 0.004$ rounded and truncated |
Sign change, hence root (correct to 3dp) | A1 | Both values correct (rounded or truncated). A reason which could include change of sign, $>0 <0$, $g(2.3065) \times g(2.3075) < 0$ **AND a minimal conclusion such as hence root,** $α=2.307$ or ▭. **Do not accept continued iteration as question demands an interval to be chosen.** | (3 marks)
| (8 marks) |
**Alternative solution to (a) working backwards**
M1 | Proceeds from $x = \ln(6-x) + 1$ using correct exp work to $......=0$ |
A1 | **Arrives correctly at** $e^{x-1} + x - 6 = 0$ **and makes a statement to the effect that this is** $g(x)=0$ |
**Alternative solution to (c) using** $f(x) = \ln(6-x) + 1 - x$ **{Similarly** $h(x) = x - 1 - \ln(6-x)$ **}**
M1 | Chooses the interval $[2.3065, 2.3075]$ or smaller containing the root 2.306558641 |
dM1 | Calculates $f(2.3065)$ and $f(2.3075)$ with at least 1 correct rounded or truncated $f(2.3065) = 0.000074$. Accept $0.00007$ rounded or truncated. Also accept $0.0001$ |
| | $f(2.3075) = - 0.0011$.. Accept $-0.001$ rounded or truncated |
---2.
$$\mathrm { g } ( x ) = \mathrm { e } ^ { x - 1 } + x - 6$$
\begin{enumerate}[label=(\alph*)]
\item Show that the equation $\mathrm { g } ( x ) = 0$ can be written as
$$x = \ln ( 6 - x ) + 1 , \quad x < 6$$
The root of $\mathrm { g } ( x ) = 0$ is $\alpha$.\\
The iterative formula
$$x _ { n + 1 } = \ln \left( 6 - x _ { n } \right) + 1 , \quad x _ { 0 } = 2$$
is used to find an approximate value for $\alpha$.
\item Calculate the values of $x _ { 1 } , x _ { 2 }$ and $x _ { 3 }$ to 4 decimal places.
\item By choosing a suitable interval, show that $\alpha = 2.307$ correct to 3 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2013 Q2 [8]}}