| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2017 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Rearrange to iterative form |
| Difficulty | Moderate -0.3 This is a multi-part question involving standard C3 techniques: finding a normal (differentiation + perpendicular gradient), algebraic manipulation to show an iterative form, and applying a given iteration formula. Part (b) requires rearranging the normal equation with the curve equation, which is routine algebra. Part (c) is straightforward calculator work. While multi-step, each component is a standard textbook exercise requiring no novel insight. |
| 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/Working | Mark | Guidance |
| At P: \(x = -2 \Rightarrow y = 3\) | B1 | \(y=3\) at point P, may be seen embedded in equation |
| \(\frac{dy}{dx} = \frac{4}{2x+5} - \frac{3}{2}\) | M1, A1 | M1: Differentiates \(\ln(2x+5) \to \frac{A}{2x+5}\); A1: correct unsimplified form |
| \(\frac{dy}{dx}\Big | _{x=-2} = \frac{5}{2} \Rightarrow\) normal equation: \(y - 3 = -\frac{2}{5}(x-(-2))\) | M1 |
| \(\Rightarrow 2x + 5y = 11\) | A1 | Must be in form \(ax + by = c\); score once seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Combines \(5y + 2x = 11\) and \(y = 2\ln(2x+5) - \frac{3x}{2}\): \(5\!\left(2\ln(2x+5) - \frac{3x}{2}\right) + 2x = 11\) | M1 | Condone slips on rearrangement of \(5y+2x=11\) |
| \(\Rightarrow x = \frac{20}{11}\ln(2x+5) - 2\) | dM1, A1* | dM1: collects \(x\) terms, proceeds to \(ax = b\ln(2x+5)+c\); A1*: given answer, all aspects correct including bracketing |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitutes \(x_1 = 2 \Rightarrow x_2 = \frac{20}{11}\ln(2\times2+5)-2\) | M1 | May be implied by \(x_2 = 1.99\ldots\) |
| \(x_2 = 1.9950\) and \(x_3 = 1.9929\) | A1 | Both correct awrt; condone \(x_2 = 1.995\); ignore subscripts, mark first two values given |
# Question 5:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| At P: $x = -2 \Rightarrow y = 3$ | B1 | $y=3$ at point P, may be seen embedded in equation |
| $\frac{dy}{dx} = \frac{4}{2x+5} - \frac{3}{2}$ | M1, A1 | M1: Differentiates $\ln(2x+5) \to \frac{A}{2x+5}$; A1: correct unsimplified form |
| $\frac{dy}{dx}\Big|_{x=-2} = \frac{5}{2} \Rightarrow$ normal equation: $y - 3 = -\frac{2}{5}(x-(-2))$ | M1 | Using $-\frac{dx}{dy}\Big|_{x=-2}$ as gradient; at least one bracket correct for $(-2,3)$ |
| $\Rightarrow 2x + 5y = 11$ | A1 | Must be in form $ax + by = c$; score once seen |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Combines $5y + 2x = 11$ and $y = 2\ln(2x+5) - \frac{3x}{2}$: $5\!\left(2\ln(2x+5) - \frac{3x}{2}\right) + 2x = 11$ | M1 | Condone slips on rearrangement of $5y+2x=11$ |
| $\Rightarrow x = \frac{20}{11}\ln(2x+5) - 2$ | dM1, A1* | dM1: collects $x$ terms, proceeds to $ax = b\ln(2x+5)+c$; A1*: given answer, all aspects correct including bracketing |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitutes $x_1 = 2 \Rightarrow x_2 = \frac{20}{11}\ln(2\times2+5)-2$ | M1 | May be implied by $x_2 = 1.99\ldots$ |
| $x_2 = 1.9950$ and $x_3 = 1.9929$ | A1 | Both correct awrt; condone $x_2 = 1.995$; ignore subscripts, mark first two values given |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f0a633e3-5c63-4d21-8ffa-d4e7dc43a536-14_549_958_221_493}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of part of the curve $C$ with equation
$$y = 2 \ln ( 2 x + 5 ) - \frac { 3 x } { 2 } , \quad x > - 2.5$$
The point $P$ with $x$ coordinate - 2 lies on $C$.
\begin{enumerate}[label=(\alph*)]
\item Find an equation of the normal to $C$ at $P$. Write your answer in the form $a x + b y = c$, where $a$, $b$ and $c$ are integers.
The normal to $C$ at $P$ cuts the curve again at the point $Q$, as shown in Figure 2
\item Show that the $x$ coordinate of $Q$ is a solution of the equation
$$x = \frac { 20 } { 11 } \ln ( 2 x + 5 ) - 2$$
The iteration formula
$$x _ { n + 1 } = \frac { 20 } { 11 } \ln \left( 2 x _ { n } + 5 \right) - 2$$
can be used to find an approximation for the $x$ coordinate of $Q$.
\item Taking $x _ { 1 } = 2$, find the values of $x _ { 2 }$ and $x _ { 3 }$, giving each answer to 4 decimal places.
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2017 Q5 [10]}}