| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2018 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Rearrange to iterative form |
| Difficulty | Moderate -0.3 This is a standard fixed-point iteration question with routine algebraic manipulation in part (a), straightforward calculator work in part (b), and a textbook change-of-sign verification in part (c). While it requires understanding of iteration and logarithms, all steps follow predictable patterns with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.06d Natural logarithm: ln(x) function and properties1.09a Sign change methods: locate roots1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \(f(x) = \frac{x^2}{4} + \ln(2x) = 0\) so \(\ln(2x) = -\frac{x^2}{4}\) and so \(2x = e^{-\frac{x^2}{4}}\) and \(x = \frac{1}{2}e^{-\frac{x^2}{4}}\) | M1, A1* | M1: Sets \(f(x)=0\), makes \(2x\) the subject via exponential; A1*: Completely correct work ignoring bracketing on \(\ln(2x)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \(x_2 = \frac{1}{2}e^{-\frac{(0.5)^2}{4}}\) | M1 | Substitute \(x_1 = 0.5\) into iterative formula |
| \(x_2 = \text{awrt } 0.4697\) | A1 | |
| \(x_3 = \text{awrt } 0.4732\) and \(x_4 = \text{awrt } 0.4728\) | A1 | Ignore subscripts, mark in order given |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \(f(0.4725) = -0.000756\ldots < 0\), \(f(0.4735) = 0.001594\ldots > 0\); sign change and \(f(x)\) continuous therefore root lies in \([0.4725, 0.4735] \Rightarrow root = 0.473\) (3 dp) | M1, A1 | M1: Choose suitable interval e.g. \([0.4725, 0.4735]\), attempt \(f(x)\) at each endpoint; A1: Both evaluations correct to 1sf, sign change stated, conclusion \(root = 0.473\) |
# Question 3(a):
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $f(x) = \frac{x^2}{4} + \ln(2x) = 0$ so $\ln(2x) = -\frac{x^2}{4}$ and so $2x = e^{-\frac{x^2}{4}}$ and $x = \frac{1}{2}e^{-\frac{x^2}{4}}$ | M1, A1* | M1: Sets $f(x)=0$, makes $2x$ the subject via exponential; A1*: Completely correct work ignoring bracketing on $\ln(2x)$ |
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# Question 3(b):
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $x_2 = \frac{1}{2}e^{-\frac{(0.5)^2}{4}}$ | M1 | Substitute $x_1 = 0.5$ into iterative formula |
| $x_2 = \text{awrt } 0.4697$ | A1 | |
| $x_3 = \text{awrt } 0.4732$ and $x_4 = \text{awrt } 0.4728$ | A1 | Ignore subscripts, mark in order given |
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# Question 3(c):
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $f(0.4725) = -0.000756\ldots < 0$, $f(0.4735) = 0.001594\ldots > 0$; sign change and $f(x)$ continuous therefore root lies in $[0.4725, 0.4735] \Rightarrow root = 0.473$ (3 dp) | M1, A1 | M1: Choose suitable interval e.g. $[0.4725, 0.4735]$, attempt $f(x)$ at each endpoint; A1: Both evaluations correct to 1sf, sign change stated, conclusion $root = 0.473$ |3.
$$\mathrm { f } ( x ) = \frac { x ^ { 2 } } { 4 } + \ln ( 2 x ) , \quad x > 0$$
\begin{enumerate}[label=(\alph*)]
\item Show that the equation $\mathrm { f } ( x ) = 0$ can be rewritten as
$$x = \frac { 1 } { 2 } \mathrm { e } ^ { - \frac { 1 } { 4 } x ^ { 2 } }$$
The equation $\mathrm { f } ( x ) = 0$ has a root near 0.5
\item Starting with $x _ { 1 } = 0.5$ use the iterative formula
$$x _ { n + 1 } = \frac { 1 } { 2 } \mathrm { e } ^ { - \frac { 1 } { 4 } x _ { n } ^ { 2 } }$$
to calculate the values of $x _ { 2 } , x _ { 3 }$ and $x _ { 4 }$, giving your answers to 4 decimal places.
\item Using a suitable interval, show that 0.473 is a root of $\mathrm { f } ( x ) = 0$ correct to 3 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2018 Q3 [7]}}