| Exam Board | Edexcel |
|---|---|
| Module | PMT Mocks (PMT Mocks) |
| Marks | 8 |
| 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 involving standard calculus techniques (finding roots, differentiation, product rule) and a straightforward iteration method. Part (c) is algebraic manipulation shown as 'show that', and part (d) is routine calculator work. While it requires multiple skills, each step follows standard A-level procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.06d Natural logarithm: ln(x) function and properties1.07l Derivative of ln(x): and related functions1.07n Stationary points: find maxima, minima using derivatives1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(A = 16\) | B1 |
| (b) | \(\frac{dy}{dx} = (0.5x-8) \times \frac{1}{(x+1)} + 0.5\ln(x+1)\), or equivalent such as \(\frac{dy}{dx} = \frac{0.5x}{(x+1)} + 0.5\ln(x+1) - \frac{8}{(x+1)}\) | A1 |
| (c) | \(x = \frac{17}{\ln(x+1)+1} - 1\) | A1 |
| (d) | \(x_1 = \text{awrt } 5.089\), \(x_6 = 5.066\) | A1 |
**(a)** | $A = 16$ | B1 | Solving $0.5x - 8 = 0 \Rightarrow x = \frac{8}{0.5} = 16$ |
**(b)** | $\frac{dy}{dx} = (0.5x-8) \times \frac{1}{(x+1)} + 0.5\ln(x+1)$, or equivalent such as $\frac{dy}{dx} = \frac{0.5x}{(x+1)} + 0.5\ln(x+1) - \frac{8}{(x+1)}$ | A1 | M1 Attempts to differentiate using the product rule. Look for $(0.5x-8) \times \frac{1}{(x+1)} \pm k\ln(x+1)$, where $k$ is a constant. You will see attempts from $f(x) = 0.5x\ln(x+1) - 8\ln(x+1)$ which can be similarly marked. In this case look for $\pm \frac{0.5x}{(x+1)} \pm b\ln(x+1) - \frac{c}{(x+1)}$ |
**(c)** | $x = \frac{17}{\ln(x+1)+1} - 1$ | A1 | M1 Score for setting their $\frac{dy}{dx} = 0$ and proceeding to an equation where the variable $x$ occurs only once. E.g. $(0.5x-8) \times \frac{1}{(x+1)} + 0.5\ln(x+1) = 0$ gives $0.5\ln(x+1) = (8-0.5x) \times \frac{1}{(x+1)}$, $\ln(x+1) = \frac{16-x}{(x+1)}$ (Dividing $(16-x)$ by $(x+1)$), $(x+1)(-1) + 16$ gives $-x - 1$ and $17$, $(x+1) \ln(x+1) + 1 = \frac{17}{(x+1)}$ or e.g. $\frac{0.5x}{(x+1)} + 0.5\ln(x+1) - \frac{8}{(x+1)} = 0$, $0.5\ln(x+1) = (8-0.5x) \times \frac{1}{(x+1)}$, $0.5(x+1)\ln(x+1) = 8 - 0.5x$, $0.5x\ln(x+1) + 0.5\ln(x+1) = 8 - 0.5x$, $0.5x\ln(x+1) + 0.5x = 8 - 0.5\ln(x+1)$, $0.5x(\ln(x+1)+1) = 8 - 0.5\ln(x+1)$. A1 Correctly proceeds to the given answer of $x = \frac{17}{\ln(x+1)+1} - 1$ showing all key steps. E.g. $\ln(x+1)+1 = \frac{17}{(x+1)}$, $(x+1)(\ln(x+1)+1) = 17$, $x+1 = \frac{17}{\ln(x+1)+1}$, $x = \frac{17}{\ln(x+1)+1} - 1$. Or e.g. $x(\ln(x+1)+1) = 16 - \ln(x+1)$, $x = \frac{16-\ln(x+1)}{\ln(x+1)+1}$ dividing $16 - \ln(x+1)$ by $\ln(x+1)+1$ gives $\frac{-1}{\ln(x+1)+16}$ and $-\ln(x+1) - 1$ and $17$, $x = \frac{17}{\ln(x+1)+1} - 1$ |
**(d)** | $x_1 = \text{awrt } 5.089$, $x_6 = 5.066$ | A1 | M1 Attempts to use the iteration formula at least once. Usually to find $x_1 = \frac{17}{\ln(5)+1} - 1$ which may be implied by awrt 5.089. A1 $x_1 = \text{awrt } 5.089$. A1 $x_6 = 5.066$ |
---8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{48f9a252-61a2-491d-94d0-8470aee96942-10_689_1011_294_486}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of part of the curve with equation $y = \mathrm { f } ( x )$, where $x \in R$, $x > 0$
$$\mathrm { f } ( x ) = ( 0.5 x - 8 ) \ln ( x + 1 ) \quad 0 \leq x \leq A$$
a. Find the value of $A$.\\
b. Find $\mathrm { f } ^ { \prime } ( x )$
The curve has a minimum turning point at $B$.\\
c. Show that the $x$-coordinate of $B$ is a solution of the equation
$$x = \frac { 17 } { \ln ( x + 1 ) + 1 } - 1$$
d. Use the iteration formula
$$x _ { n + 1 } = \frac { 17 } { \ln \left( x _ { n } + 1 \right) + 1 } - 1$$
with $x _ { 0 } = 5$ to find the values of $x _ { 1 }$ and the value of $x _ { 6 }$ giving your answers to three decimal places.\\
\hfill \mbox{\textit{Edexcel PMT Mocks Q8 [8]}}