| Exam Board | OCR MEI |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2019 |
| Session | June |
| 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 straightforward application of differentiation followed by standard fixed-point iteration. Part (a) requires routine differentiation of polynomial and logarithmic terms, then setting equal to zero. Part (b) is a direct application of an iterative method (likely Newton-Raphson or rearrangement) which is a standard A-level technique. The question is slightly easier than average because it's well-scaffolded and uses familiar functions without requiring novel insight or complex algebraic manipulation. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Differentiate to obtain \(2x - 4\) | B1 (3.1a) | |
| \(+ 1 \times \ln x + x \times \frac{1}{x}\) oe | M1, A1 (2.1, 1.1) | Use of Product Rule; all correct; allow one error |
| Derivative \(= 0\) oe seen and terms combined; \(2x - 3 + \ln x = 0\) isw AG | A1 (2.4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Any rearrangement to obtain \(x = g(x)\) from given derivative \(= 0\) | M1* (2.1) | Allow sign error; e.g. \(x = e^{2x-3}\) |
| \(x = \frac{3 - \ln x}{2}\) | A1 (1.1) | Any correct rearrangement; need not see subscripts in iterative formula |
| Use of their \(g(x_n) = \frac{3 - \ln x_n}{2}\) to obtain at least two iterates e.g. 2, 1.1534, 1.4286… | M1dep* (1.1) | Must see iterates |
| 1.3500 cao | A1 (2.2a) | 0 for 1.3500 unsupported; trial and improvement does not score |
| Alternatively (Newton-Raphson): \(x_{n+1} = x_n - \frac{2x_n - 3 + \ln x_n}{\text{their}\left(2 + \frac{1}{x_n}\right)}\) | M1* (2.1) | Newton-Raphson iterative formula seen (not for solving \(f(x)=0\)); need not see subscripts |
| Use of their N-R formula to obtain \(x_1, x_2,...\) e.g. 1.5, 1.34795, 1.34996,…1.349962 | A1, M1dep* (1.1, 1.1) | Formula all correct; must see iterates |
| 1.3500 cao | A1 (2.2a) |
## Question 11(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Differentiate to obtain $2x - 4$ | B1 (3.1a) | |
| $+ 1 \times \ln x + x \times \frac{1}{x}$ oe | M1, A1 (2.1, 1.1) | Use of Product Rule; all correct; allow one error |
| Derivative $= 0$ oe seen and terms combined; $2x - 3 + \ln x = 0$ **isw AG** | A1 (2.4) | |
---
## Question 11(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Any rearrangement to obtain $x = g(x)$ from given derivative $= 0$ | M1* (2.1) | Allow sign error; e.g. $x = e^{2x-3}$ |
| $x = \frac{3 - \ln x}{2}$ | A1 (1.1) | Any correct rearrangement; need not see subscripts in iterative formula |
| Use of their $g(x_n) = \frac{3 - \ln x_n}{2}$ to obtain at least two iterates e.g. 2, 1.1534, 1.4286… | M1dep* (1.1) | Must see iterates |
| 1.3500 cao | A1 (2.2a) | **0** for 1.3500 unsupported; trial and improvement does not score |
| **Alternatively** (Newton-Raphson): $x_{n+1} = x_n - \frac{2x_n - 3 + \ln x_n}{\text{their}\left(2 + \frac{1}{x_n}\right)}$ | M1* (2.1) | Newton-Raphson iterative formula seen (not for solving $f(x)=0$); need not see subscripts |
| Use of their N-R formula to obtain $x_1, x_2,...$ e.g. 1.5, 1.34795, 1.34996,…1.349962 | A1, M1dep* (1.1, 1.1) | Formula all correct; must see iterates |
| 1.3500 cao | A1 (2.2a) | |
---11 Fig. 11 shows the graph of $y = x ^ { 2 } - 4 x + x \ln x$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{95eb3bcc-6d3c-4f7e-9b27-5e046ab57ec5-08_697_463_338_246}
\captionsetup{labelformat=empty}
\caption{Fig. 11}
\end{center}
\end{figure}
\begin{enumerate}[label=(\alph*)]
\item Show that the $x$-coordinate of the stationary point on the curve may be found from the equation $2 x - 3 + \ln x = 0$.
\item Use an iterative method to find the $x$-coordinate of the stationary point on the curve $y = x ^ { 2 } - 4 x + x \ln x$, giving your answer correct to 4 decimal places.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 2 2019 Q11 [8]}}