| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2006 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Derive stationary point equation |
| Difficulty | Standard +0.3 Part (a) requires product rule differentiation and a trigonometric identity (standard C3 techniques), while parts (b) and (c) involve straightforward calculator iteration. The derivation is routine for students who have practiced stationary points with trigonometric functions, making this slightly easier than average. |
| Spec | 1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.07n 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 |
|---|---|
| Using product rule: \(\frac{dy}{dx} = 2\tan 2x + 2(2x-1)\sec^2 2x\) | M1A1A1 |
| Use of \(\tan 2x = \frac{\sin 2x}{\cos 2x}\) and \(\sec 2x = \frac{1}{\cos 2x}\) | M1 |
| \(\left[= 2\frac{\sin 2x}{\cos 2x} + 2(2x-1)\frac{1}{\cos^2 2x}\right]\) | |
| Setting \(\frac{dy}{dx} = 0\) and multiplying through to eliminate fractions: \([{\Rightarrow} 2\sin 2x\cos 2x + 2(2x-1) = 0]\) | M1 |
| Completion: producing \(4k + \sin 4k - 2 = 0\) with no wrong working seen and at least previous line seen | A1* (6) |
| Answer | Marks | Guidance |
|---|---|---|
| \(x_1 = 0.2670\), \(x_2 = 0.2809\), \(x_3 = 0.2746\), \(x_4 = 0.2774\) | M1A1A1 (3) | M1 for first correct application; first A1 for two correct; second A1 for all four correct. Max \(-1\) if ALL correct to \(> 4\) d.p. |
| Answer | Marks |
|---|---|
| Choose suitable interval for \(k\): e.g. \([0.2765, 0.2775]\) and evaluate \(f(x)\) at these values | M1 |
| Show that \(4k + \sin 4k - 2\) changes sign and deduction | A1 (2) |
| \([f(0.2765) = -0.000087...,\ f(0.2775) = +0.0057]\) |
# Question 5:
## Part (a)
Using product rule: $\frac{dy}{dx} = 2\tan 2x + 2(2x-1)\sec^2 2x$ | M1A1A1 |
Use of $\tan 2x = \frac{\sin 2x}{\cos 2x}$ and $\sec 2x = \frac{1}{\cos 2x}$ | M1 |
$\left[= 2\frac{\sin 2x}{\cos 2x} + 2(2x-1)\frac{1}{\cos^2 2x}\right]$ | |
Setting $\frac{dy}{dx} = 0$ and multiplying through to eliminate fractions: $[{\Rightarrow} 2\sin 2x\cos 2x + 2(2x-1) = 0]$ | M1 |
Completion: producing $4k + \sin 4k - 2 = 0$ with no wrong working seen and at least previous line seen | A1* (6) |
## Part (b)
$x_1 = 0.2670$, $x_2 = 0.2809$, $x_3 = 0.2746$, $x_4 = 0.2774$ | M1A1A1 (3) | M1 for first correct application; first A1 for two correct; second A1 for all four correct. Max $-1$ if ALL correct to $> 4$ d.p.
## Part (c)
Choose suitable interval for $k$: e.g. $[0.2765, 0.2775]$ and evaluate $f(x)$ at these values | M1 |
Show that $4k + \sin 4k - 2$ changes sign and deduction | A1 (2) |
$[f(0.2765) = -0.000087...,\ f(0.2775) = +0.0057]$ | |
---5.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\includegraphics[alt={},max width=\textwidth]{f0f328ed-3550-4b8d-8b80-016df8773b21-07_465_565_296_701}
\end{center}
\end{figure}
Figure 2 shows part of the curve with equation
$$y = ( 2 x - 1 ) \tan 2 x , \quad 0 \leqslant x < \frac { \pi } { 4 }$$
The curve has a minimum at the point $P$. The $x$-coordinate of $P$ is $k$.
\begin{enumerate}[label=(\alph*)]
\item Show that $k$ satisfies the equation
$$4 k + \sin 4 k - 2 = 0$$
The iterative formula
$$x _ { n + 1 } = \frac { 1 } { 4 } \left( 2 - \sin 4 x _ { n } \right) , x _ { 0 } = 0.3$$
is used to find an approximate value for $k$.
\item Calculate the values of $x _ { 1 } , x _ { 2 } , x _ { 3 }$ and $x _ { 4 }$, giving your answers to 4 decimal places.
\item Show that $k = 0.277$, correct to 3 significant figures.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2006 Q5 [11]}}