| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Derive stationary point equation |
| Difficulty | Standard +0.3 This is a structured multi-part question that guides students through standard techniques: finding roots (trivial), showing a derivative result equals a given form (routine calculus), sign change verification (direct substitution), and applying a given iterative formula. All steps are scaffolded with no novel problem-solving required, making it slightly easier than average despite multiple parts. |
| Spec | 1.06d Natural logarithm: ln(x) function and properties1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams1.09d Newton-Raphson method |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(x) = (8-x)\ln x,\ x > 0\); crosses \(x\)-axis \(\Rightarrow f(x) = 0 \Rightarrow (8-x)\ln x = 0\); \(x\) coordinates are \(1\) and \(8\) | B1 | Either 1 and 8, or marks 1 next to \(A\) and 8 next to \(B\) on figure |
| (1 mark) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Complete strategy of setting \(f'(x) = 0\) and rearranging to make \(x = \ldots\) | M1 | Recognises \(Q\) is a stationary point (not a root); AO 3.1a |
| Applies product rule \(vu' + uv'\) where \(u = 8-x\), \(v = \ln x\); \(\dfrac{du}{dx} = -1\), \(\dfrac{dv}{dx} = \dfrac{1}{x}\) | M1 | This mark can be recovered in part (c); AO 1.1b |
| \(f'(x) = -\ln x + \dfrac{8-x}{x}\) | A1 | \((8-x)\ln x \to -\ln x + \dfrac{8-x}{x}\), or equivalent. Can be recovered in part (c); AO 1.1b |
| \(-\ln x + \dfrac{8-x}{x} = 0 \Rightarrow -\ln x + \dfrac{8}{x} - 1 = 0 \Rightarrow \dfrac{8}{x} = 1 + \ln x \Rightarrow x = \dfrac{8}{1 + \ln x}\) | A1* | Correct proof with no errors seen in working; AO 2.1 |
| (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Evaluates both \(f'(3.5)\) and \(f'(3.6)\) | M1 | AO 1.1b |
| \(f'(3.5) = 0.032951317\ldots\) and \(f'(3.6) = -0.058711623\ldots\); sign change and as \(f'(x)\) is continuous, \(x\) coordinate of \(Q\) lies between \(x = 3.5\) and \(x = 3.6\) | A1 | \(f'(3.5) =\) awrt \(0.03\) and \(f'(3.6) =\) awrt \(-0.06\) or \(-0.05\) (truncated), and a correct conclusion; AO 2.4 |
| (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x_5 = 3.5340\) | B1 | AO 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x_Q = 3.54\) (2 dp) | B1 | Deduces by further iterations that \(x\) coordinate of \(Q\) is \(3.54\) accurate to 2 dp. Note: \(3.5 \to 3.55119 \to 3.52845 \to 3.53848 \to 3.53404 \to 3.53600 \to 3.53514\ (\to 3.535518\ldots)\); AO 2.2a |
| (2 marks total for d(i) and d(ii)) |
# Question 6:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) = (8-x)\ln x,\ x > 0$; crosses $x$-axis $\Rightarrow f(x) = 0 \Rightarrow (8-x)\ln x = 0$; $x$ coordinates are $1$ and $8$ | B1 | Either 1 and 8, or marks 1 next to $A$ and 8 next to $B$ on figure |
| **(1 mark)** | | |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Complete strategy of setting $f'(x) = 0$ and rearranging to make $x = \ldots$ | M1 | Recognises $Q$ is a stationary point (not a root); AO 3.1a |
| Applies product rule $vu' + uv'$ where $u = 8-x$, $v = \ln x$; $\dfrac{du}{dx} = -1$, $\dfrac{dv}{dx} = \dfrac{1}{x}$ | M1 | This mark can be recovered in part (c); AO 1.1b |
| $f'(x) = -\ln x + \dfrac{8-x}{x}$ | A1 | $(8-x)\ln x \to -\ln x + \dfrac{8-x}{x}$, or equivalent. Can be recovered in part (c); AO 1.1b |
| $-\ln x + \dfrac{8-x}{x} = 0 \Rightarrow -\ln x + \dfrac{8}{x} - 1 = 0 \Rightarrow \dfrac{8}{x} = 1 + \ln x \Rightarrow x = \dfrac{8}{1 + \ln x}$ | A1* | Correct proof with no errors seen in working; AO 2.1 |
| **(4 marks)** | | |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Evaluates both $f'(3.5)$ and $f'(3.6)$ | M1 | AO 1.1b |
| $f'(3.5) = 0.032951317\ldots$ and $f'(3.6) = -0.058711623\ldots$; sign change and as $f'(x)$ is continuous, $x$ coordinate of $Q$ lies between $x = 3.5$ and $x = 3.6$ | A1 | $f'(3.5) =$ awrt $0.03$ and $f'(3.6) =$ awrt $-0.06$ or $-0.05$ (truncated), and a correct conclusion; AO 2.4 |
| **(2 marks)** | | |
## Part (d)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x_5 = 3.5340$ | B1 | AO 1.1b |
## Part (d)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x_Q = 3.54$ (2 dp) | B1 | Deduces by further iterations that $x$ coordinate of $Q$ is $3.54$ accurate to 2 dp. Note: $3.5 \to 3.55119 \to 3.52845 \to 3.53848 \to 3.53404 \to 3.53600 \to 3.53514\ (\to 3.535518\ldots)$; AO 2.2a |
| **(2 marks total for d(i) and d(ii))** | | |
**Total: 9 marks**
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{659a0479-c8c6-418b-b8a9-67ad68474023-12_624_1057_258_504}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of the curve with equation $y = \mathrm { f } ( x )$, where
$$\mathrm { f } ( x ) = ( 8 - x ) \ln x , \quad x > 0$$
The curve cuts the $x$-axis at the points $A$ and $B$ and has a maximum turning point at $Q$, as shown in Figure 2.
\begin{enumerate}[label=(\alph*)]
\item Find the $x$ coordinate of $A$ and the $x$ coordinate of $B$.
\item Show that the $x$ coordinate of $Q$ satisfies
$$x = \frac { 8 } { 1 + \ln x }$$
\item Show that the $x$ coordinate of $Q$ lies between 3.5 and 3.6
\item Use the iterative formula
$$x _ { n + 1 } = \frac { 8 } { 1 + \ln x _ { n } } \quad n \in \mathbb { N }$$
with $x _ { 1 } = 3.5$ to
\begin{enumerate}[label=(\roman*)]
\item find the value of $x _ { 5 }$ to 4 decimal places,
\item find the $x$ coordinate of $Q$ accurate to 2 decimal places.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 2 Q6 [9]}}