| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2021 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Derive stationary point equation |
| Difficulty | Standard +0.8 This is a multi-part question requiring product rule differentiation, algebraic manipulation involving trigonometric identities to derive an iteration formula, iterative calculations, and a change of sign method with interval bisection. While each individual step is standard P3 material, the algebraic rearrangement in part (b) requires insight to manipulate the derivative equation into the arctan form, and part (d) requires careful application of the change of sign method with explicit working. The combination of techniques and the non-routine algebraic manipulation elevates this above a typical textbook exercise. |
| Spec | 1.07q Product and quotient rules: differentiation1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.09a Sign change methods: locate roots1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(f'(x) = \cos\left(\frac{x}{3}\right) - \frac{1}{3}x\sin\left(\frac{x}{3}\right)\) | M1 A1 | Product rule giving form \(\alpha\cos\!\left(\frac{x}{3}\right) \pm \beta x\sin\!\left(\frac{x}{3}\right)\); may be unsimplified |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(f'(x)=0 \Rightarrow \cos\!\left(\frac{x}{3}\right) - \frac{1}{3}x\sin\!\left(\frac{x}{3}\right)=0 \Rightarrow 1 - \frac{1}{3}x\tan\!\left(\frac{x}{3}\right)=0\) | M1 | Sets \(f'(x)=0\), proceeds to equation involving \(\tan\!\left(\frac{x}{3}\right)\) |
| \(\tan\!\left(\frac{x}{3}\right) = \frac{3}{x} \Rightarrow x = 3\arctan\!\left(\frac{3}{x}\right)\) | A1* | CSO; requires intermediate line \(\tan\!\left(\frac{x}{3}\right)=\frac{3}{x}\) or \(\tan\!\left(\frac{x}{3}\right)=\frac{1}{(x/3)}\); do not condone \(\tan^{-1}\) unless correct notation also given |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x_2 = 3\arctan\!\left(\frac{3}{2.5}\right) =\) awrt \(2.6\) | M1 | Uses formula \(x = \alpha\arctan\!\left(\beta\times\frac{1}{x}\right)\) with \(2.5\) |
| \(x_2 =\) awrt \(2.628\) and \(x_6 =\) awrt \(2.586\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(f'(2.5815) = \cos\!\left(\frac{2.5815}{3}\right)-\frac{1}{3}(2.5815)\sin\!\left(\frac{2.5815}{3}\right) = -0.000345\ldots\) and \(f'(2.5805) = \ldots = 0.000346\ldots\) | M1 | Chooses suitable interval, attempts both values of their \(f'(x)=0\) function; need embedded values or one value correct to 1sf |
| Both values correct, sign change and continuous, therefore root | A1 | Requires: both values correct (rounded/truncated to 1sf); reference to sign change and continuity; minimal conclusion |
# Question 6:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $f'(x) = \cos\left(\frac{x}{3}\right) - \frac{1}{3}x\sin\left(\frac{x}{3}\right)$ | M1 A1 | Product rule giving form $\alpha\cos\!\left(\frac{x}{3}\right) \pm \beta x\sin\!\left(\frac{x}{3}\right)$; may be unsimplified |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $f'(x)=0 \Rightarrow \cos\!\left(\frac{x}{3}\right) - \frac{1}{3}x\sin\!\left(\frac{x}{3}\right)=0 \Rightarrow 1 - \frac{1}{3}x\tan\!\left(\frac{x}{3}\right)=0$ | M1 | Sets $f'(x)=0$, proceeds to equation involving $\tan\!\left(\frac{x}{3}\right)$ |
| $\tan\!\left(\frac{x}{3}\right) = \frac{3}{x} \Rightarrow x = 3\arctan\!\left(\frac{3}{x}\right)$ | A1* | CSO; requires intermediate line $\tan\!\left(\frac{x}{3}\right)=\frac{3}{x}$ or $\tan\!\left(\frac{x}{3}\right)=\frac{1}{(x/3)}$; do not condone $\tan^{-1}$ unless correct notation also given |
## Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $x_2 = 3\arctan\!\left(\frac{3}{2.5}\right) =$ awrt $2.6$ | M1 | Uses formula $x = \alpha\arctan\!\left(\beta\times\frac{1}{x}\right)$ with $2.5$ |
| $x_2 =$ awrt $2.628$ and $x_6 =$ awrt $2.586$ | A1 | |
## Part (d):
| Answer | Mark | Guidance |
|--------|------|----------|
| $f'(2.5815) = \cos\!\left(\frac{2.5815}{3}\right)-\frac{1}{3}(2.5815)\sin\!\left(\frac{2.5815}{3}\right) = -0.000345\ldots$ and $f'(2.5805) = \ldots = 0.000346\ldots$ | M1 | Chooses suitable interval, attempts both values of their $f'(x)=0$ function; need embedded values or one value correct to 1sf |
| Both values correct, sign change and continuous, therefore root | A1 | Requires: both values correct (rounded/truncated to 1sf); reference to sign change **and** continuity; minimal conclusion |6.
$$\mathrm { f } ( x ) = x \cos \left( \frac { x } { 3 } \right) \quad x > 0$$
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { f } ^ { \prime } ( x )$
\item Show that the equation $\mathrm { f } ^ { \prime } ( x ) = 0$ can be written as
$$x = k \arctan \left( \frac { k } { x } \right)$$
where $k$ is an integer to be found.
\item Starting with $x _ { 1 } = 2.5$ use the iteration formula
$$x _ { n + 1 } = k \arctan \left( \frac { k } { x _ { n } } \right)$$
with the value of $k$ found in part (b), to calculate the values of $x _ { 2 }$ and $x _ { 6 }$ giving your answers to 3 decimal places.
\item Using a suitable interval and a suitable function that should be stated, show that a root of $\mathrm { f } ^ { \prime } ( x ) = 0$ is 2.581 correct to 3 decimal places.\\
In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P3 2021 Q6 [8]}}