| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2018 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Rearrange to iterative form |
| Difficulty | Standard +0.3 This is a standard C3 iteration question requiring routine differentiation to find a normal, algebraic manipulation to rearrange an equation into iterative form, and straightforward application of an iterative formula. All steps are procedural with no novel insight required, making it slightly easier than average. |
| 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 |
|---|---|
| \(\frac{dy}{dx} = -2e^{-2x} + 2x\) | M1A1 |
| At \(x = 0\): \(\frac{dy}{dx} = -2 \Rightarrow \frac{dx}{dy} = \frac{1}{2}\) | M1 |
| Equation of normal is \(y - (-2) = \frac{1}{2}(x-0) \Rightarrow y = \frac{1}{2}x - 2\) | M1 A1 |
| Answer | Marks |
|---|---|
| \(y = e^{-2x} + x^2 - 3\) meets \(y = \frac{1}{2}x - 2\) when \(e^{-2x} + x^2 - 3 = "\frac{1}{2}x - 2"\) | M1 |
| \(x^2 = 1 + \frac{1}{2}x - e^{-2x}\) | M1 |
| \(x = \sqrt{1 + \frac{1}{2}x - e^{-2x}}\) * | A1* |
| Answer | Marks |
|---|---|
| \(x_2 = \sqrt{1 + 0.5 - e^{-2}}\) | M1 |
| \(x_2 = 1.168, x_3 = 1.220\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| - M1: A correct method of finding the gradient of the normal at \(x = 0\). To score this the candidate must find the negative reciprocal of \(\left.\frac{dy}{dx}\right | _{x=0}\) | |
| - M1: An attempt at the equation of the normal at \((0,-2)\). To score this mark the candidate must be using the point \((0,-2)\) and a gradient that has been changed from \(\left.\frac{dy}{dx}\right | _{x=0}\). Look for \(y - (-2) = changed\left.\frac{dy}{dx}\right | _{x=0}(x-0)\) or \(y = mx - 2\) where \(m = changed\left.\frac{dy}{dx}\right |
**(a)**
| $\frac{dy}{dx} = -2e^{-2x} + 2x$ | M1A1 |
| At $x = 0$: $\frac{dy}{dx} = -2 \Rightarrow \frac{dx}{dy} = \frac{1}{2}$ | M1 |
| Equation of normal is $y - (-2) = \frac{1}{2}(x-0) \Rightarrow y = \frac{1}{2}x - 2$ | M1 A1 |
(5 marks)
**(b)**
| $y = e^{-2x} + x^2 - 3$ meets $y = \frac{1}{2}x - 2$ when $e^{-2x} + x^2 - 3 = "\frac{1}{2}x - 2"$ | M1 |
| $x^2 = 1 + \frac{1}{2}x - e^{-2x}$ | M1 |
| $x = \sqrt{1 + \frac{1}{2}x - e^{-2x}}$ * | A1* |
(2 marks)
**(c)**
| $x_2 = \sqrt{1 + 0.5 - e^{-2}}$ | M1 |
| $x_2 = 1.168, x_3 = 1.220$ | A1 |
(2 marks)
(9 marks total)
**Guidance notes:**
**(a)**
- M1: Attempts to differentiate with $e^{-2x} \rightarrow Ae^{-2x}$ with any non-zero $A$, even 1. Watch for $e^{-2x} \rightarrow Ae^{2x}$ which is M0 A0
- A1: $\frac{dy}{dx} = -2e^{-2x} + 2x$
- M1: A correct method of finding the gradient of the normal at $x = 0$. To score this the candidate must find the negative reciprocal of $\left.\frac{dy}{dx}\right|_{x=0}$
- M1: An attempt at the equation of the normal at $(0,-2)$. To score this mark the candidate must be using the point $(0,-2)$ and a gradient that has been changed from $\left.\frac{dy}{dx}\right|_{x=0}$. Look for $y - (-2) = changed\left.\frac{dy}{dx}\right|_{x=0}(x-0)$ or $y = mx - 2$ where $m = changed\left.\frac{dy}{dx}\right|_{x=0}$
- A1: $y = \frac{1}{2}x - 2$ cso with as well as showing the correct differentiation. So reaching $y = \frac{1}{2}x - 2$ from $\frac{dy}{dx} = -2e^{-2x} + 2x$ is A0. If it is not simplified (or written in the required form) you may award this if $y = \frac{1}{2}x - 2$ is seen in part (b).
**(b)**
- M1: Equates $y = e^{-2x} + x^2 - 3$ and their $y = mx + c, m \neq 0$ and proceeds to $x^2 = ...$. Condone an attempt for this M mark where the candidate uses an adapted $y = mx + c$ in an attempt to get the printed answer.
- A1*: Proceeds to $x = \sqrt{1 + \frac{1}{2}x - e^{-2x}}$. It is a printed answer but you may accept a different order. $x = \sqrt{1 - e^{-2x} + \frac{1}{2}x}$. For this mark, the candidate must start with a normal equation of $y = \frac{1}{2}x - 2$ oe found in (a). It can be awarded when the candidate finds the equation incorrectly, for example from $\frac{dy}{dx} = -2e^{2x} + 2x$.
**(c)**
- M1: Sub $x_1 = 1$ in $x = \sqrt{1 + \frac{1}{2}x - e^{-2x}}$ to find $x_2$. May be implied by $\sqrt{1 + 0.5 - e^{-2}}$ oe or awrt 1.17
- A1: $x_2 = awrt 1.168, x_3 = awrt 1.220$ 3dp. Condone 1.22 for $x_3$. Mark these in the order given, the subscripts are not required and incorrect ones may be ignored.
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{42aff260-e734-48ff-a92a-674032cb0377-12_595_930_219_603}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of part of the curve $C$ with equation
$$y = \mathrm { e } ^ { - 2 x } + x ^ { 2 } - 3$$
The curve $C$ crosses the $y$-axis at the point $A$.
The line $l$ is the normal to $C$ at the point $A$.
\begin{enumerate}[label=(\alph*)]
\item Find the equation of $l$, writing your answer in the form $y = m x + c$, where $m$ and $c$ are constants.
The line $l$ meets $C$ again at the point $B$, as shown in Figure 1 .
\item Show that the $x$ coordinate of $B$ is a solution of
$$x = \sqrt { 1 + \frac { 1 } { 2 } x - \mathrm { e } ^ { - 2 x } }$$
Using the iterative formula
$$x _ { n + 1 } = \sqrt { 1 + \frac { 1 } { 2 } x _ { n } - \mathrm { e } ^ { - 2 x _ { n } } }$$
with $x _ { 1 } = 1$
\item find $x _ { 2 }$ and $x _ { 3 }$ to 3 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2018 Q4 [9]}}