| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2015 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Find intersection point coordinates |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question requiring routine algebraic manipulation to derive the iteration formula, mechanical application of the iterative formula (calculator work), and simple substitution to find coordinates. While it involves logarithms and iteration, these are standard C3 techniques with no novel problem-solving required—slightly easier than average due to its procedural nature. |
| Spec | 1.02q Use intersection points: of graphs to solve equations1.06a Exponential function: a^x and e^x graphs and properties1.06d Natural logarithm: ln(x) function and properties1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(2^{x+1} - 3 = 17 - x \Rightarrow 2^{x+1} = 20 - x\) | M1 | Setting equations equal and making \(2^{x+1}\) subject |
| \((x+1)\ln 2 = \ln(20-x) \Rightarrow x = \ldots\) | dM1 | Take ln of both sides, use power law, proceed to \(x=\) |
| \(x = \frac{\ln(20-x)}{\ln 2} - 1\) | A1* | Given answer; bracketing on \((x+1)\) and \(\ln(20-x)\) must be correct; ln or \(\log_e\) required not \(\log_{10}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Sub \(x_0 = 3\) into \(x_{n+1} = \frac{\ln(20-x_n)}{\ln 2} - 1 \Rightarrow x_1 = 3.087\) (awrt) | M1A1 | M1: Correct substitution. Note: 4.087 alone without working is M0 |
| \(x_2 = 3.080\), \(x_3 = 3.081\) (awrt) | A1 | Tolerate 3.08 for 3.080; subscripts not important |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(A = (3.1, 13.9)\) cao | M1, A1 | M1: Sight of 3.1. A1: Can also score by substituting \(x\) value from (b) into \(2^{x+1}-3\) or \(17-x\) |
## Question 6:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $2^{x+1} - 3 = 17 - x \Rightarrow 2^{x+1} = 20 - x$ | M1 | Setting equations equal and making $2^{x+1}$ subject |
| $(x+1)\ln 2 = \ln(20-x) \Rightarrow x = \ldots$ | dM1 | Take ln of both sides, use power law, proceed to $x=$ |
| $x = \frac{\ln(20-x)}{\ln 2} - 1$ | A1* | Given answer; bracketing on $(x+1)$ and $\ln(20-x)$ must be correct; ln or $\log_e$ required not $\log_{10}$ |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Sub $x_0 = 3$ into $x_{n+1} = \frac{\ln(20-x_n)}{\ln 2} - 1 \Rightarrow x_1 = 3.087$ (awrt) | M1A1 | M1: Correct substitution. Note: 4.087 alone without working is M0 |
| $x_2 = 3.080$, $x_3 = 3.081$ (awrt) | A1 | Tolerate 3.08 for 3.080; subscripts not important |
### Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $A = (3.1, 13.9)$ cao | M1, A1 | M1: Sight of 3.1. A1: Can also score by substituting $x$ value from (b) into $2^{x+1}-3$ or $17-x$ |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{57ea7a94-6939-4c12-a6fd-01cd6af73310-10_1004_1120_260_420}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 is a sketch showing part of the curve with equation $y = 2 ^ { x + 1 } - 3$ and part of the line with equation $y = 17 - x$.
The curve and the line intersect at the point $A$.
\begin{enumerate}[label=(\alph*)]
\item Show that the $x$ coordinate of $A$ satisfies the equation
$$x = \frac { \ln ( 20 - x ) } { \ln 2 } - 1$$
\item Use the iterative formula
$$x _ { n + 1 } = \frac { \ln \left( 20 - x _ { n } \right) } { \ln 2 } - 1 , \quad x _ { 0 } = 3$$
to calculate the values of $x _ { 1 } , x _ { 2 }$ and $x _ { 3 }$, giving your answers to 3 decimal places.
\item Use your answer to part (b) to deduce the coordinates of the point $A$, giving your answers to one decimal place.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2015 Q6 [8]}}