| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2021 |
| Session | October |
| 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 multi-part question involving differentiation (product rule with exponential), finding a normal line equation, and applying a given iterative formula. Part (a) is routine differentiation; part (b) requires algebraic manipulation but the rearrangement is shown; part (c) is straightforward calculator work applying the given iteration. All techniques are standard P3 material with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07m Tangents and normals: gradient and equations1.07q Product and quotient rules: differentiation1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(x) = (x^3 - 4x)e^{-\frac{1}{2}x} \Rightarrow f'(x) = (3x^2 - 4)e^{-\frac{1}{2}x} - \frac{1}{2}(x^3 - 4x)e^{-\frac{1}{2}x}\) | M1 A1 | M1: Uses valid method to differentiate - product rule giving form \(\pm A(x^3-4x)e^{-\frac{1}{2}x} \pm (Bx^2 \pm C)e^{-\frac{1}{2}x}\). Condone missing squared on \(Bx^2\) term. A1: Correct \(f'(x)\), may be unsimplified |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f'(0) = -4\) so equation of normal is \(y = -\frac{1}{-4}x\) | M1 | Full method to find equation of normal through \(O\); attempt at \(f'(0)\) followed by \(y = -\frac{1}{f'(0)}x\) |
| \(y = \frac{1}{4}x\) | B1 | Equation of normal is \(y = \frac{1}{4}x\) (seen or implied, may follow incorrect \(f'(x)\)) |
| Sets \(\frac{1}{4}x = x(x^2-4)e^{-\frac{1}{2}x} \Rightarrow x^2 - 4 = \frac{1}{4}e^{\frac{1}{2}x}\) | M1 | Equates \(y = \frac{1}{4}x\) (straight line through origin) with \(f(x) = x(x^2-4)e^{-\frac{1}{2}x}\), divides/factorises out \(x\) term, attempts to make \(x^2\) (or \(4x^2\)) the subject |
| \(\Rightarrow x^2 = \frac{16 + e^{\frac{1}{2}x}}{4} \Rightarrow x = -\frac{1}{2}\sqrt{16 + e^{\frac{1}{2}x}}\) * | A1* | Full proof showing all steps. No requirement to justify the \(-\) sign. Cannot be scored if A0 in part (a) unless restarted in (b) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x_2 = -\frac{1}{2}\sqrt{16 + e^{\frac{1}{2}(x-2)}} = -2.0229\) | M1 A1 | M1: Substitutes \(x = -2\) into iteration formula and finds \(x_2\). May be implied by \(-2.0228\) or \(-2.0229\). A1: awrt \(-2.0229\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((x =) -2.0226\) | A1 | Correct to 4 d.p. |
## Question 9:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) = (x^3 - 4x)e^{-\frac{1}{2}x} \Rightarrow f'(x) = (3x^2 - 4)e^{-\frac{1}{2}x} - \frac{1}{2}(x^3 - 4x)e^{-\frac{1}{2}x}$ | M1 A1 | M1: Uses valid method to differentiate - product rule giving form $\pm A(x^3-4x)e^{-\frac{1}{2}x} \pm (Bx^2 \pm C)e^{-\frac{1}{2}x}$. Condone missing squared on $Bx^2$ term. A1: Correct $f'(x)$, may be unsimplified |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f'(0) = -4$ so equation of normal is $y = -\frac{1}{-4}x$ | M1 | Full method to find equation of normal through $O$; attempt at $f'(0)$ followed by $y = -\frac{1}{f'(0)}x$ |
| $y = \frac{1}{4}x$ | B1 | Equation of normal is $y = \frac{1}{4}x$ (seen or implied, may follow incorrect $f'(x)$) |
| Sets $\frac{1}{4}x = x(x^2-4)e^{-\frac{1}{2}x} \Rightarrow x^2 - 4 = \frac{1}{4}e^{\frac{1}{2}x}$ | M1 | Equates $y = \frac{1}{4}x$ (straight line through origin) with $f(x) = x(x^2-4)e^{-\frac{1}{2}x}$, divides/factorises out $x$ term, attempts to make $x^2$ (or $4x^2$) the subject |
| $\Rightarrow x^2 = \frac{16 + e^{\frac{1}{2}x}}{4} \Rightarrow x = -\frac{1}{2}\sqrt{16 + e^{\frac{1}{2}x}}$ * | A1* | Full proof showing all steps. No requirement to justify the $-$ sign. Cannot be scored if A0 in part (a) unless restarted in (b) |
### Part (c)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x_2 = -\frac{1}{2}\sqrt{16 + e^{\frac{1}{2}(x-2)}} = -2.0229$ | M1 A1 | M1: Substitutes $x = -2$ into iteration formula and finds $x_2$. May be implied by $-2.0228$ or $-2.0229$. A1: awrt $-2.0229$ |
### Part (c)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(x =) -2.0226$ | A1 | Correct to 4 d.p. |
---9.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{9b0b8db0-79fd-4ad5-88c9-737447d9f894-26_698_744_255_593}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows a sketch of part of the curve with equation $y = \mathrm { f } ( x )$, where
$$f ( x ) = x \left( x ^ { 2 } - 4 \right) e ^ { - \frac { 1 } { 2 } x }$$
\begin{enumerate}[label=(\alph*)]
\item Find $f ^ { \prime } ( x )$.
The line $l$ is the normal to the curve at $O$ and meets the curve again at the point $P$. The point $P$ lies in the 3rd quadrant, as shown in Figure 3.
\item Show that the $x$ coordinate of $P$ is a solution of the equation
$$x = - \frac { 1 } { 2 } \sqrt { 16 + \mathrm { e } ^ { \frac { 1 } { 2 } x } }$$
\item Using the iterative formula
$$x _ { n + 1 } = - \frac { 1 } { 2 } \sqrt { 16 + \mathrm { e } ^ { \frac { 1 } { 2 } x _ { n } } } \quad \text { with } x _ { 1 } = - 2$$
find, to 4 decimal places,
\begin{enumerate}[label=(\roman*)]
\item the value of $x _ { 2 }$
\item the $x$ coordinate of $P$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel P3 2021 Q9 [9]}}