| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2022 |
| Session | October |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Rearrange to iterative form |
| Difficulty | Standard +0.3 This is a straightforward multi-part question testing standard P3 techniques: substitution to verify a value, sign change to locate a root, algebraic rearrangement of logarithms (using log laws), and applying a given iteration formula. All steps are routine with clear guidance, requiring only careful execution rather than problem-solving insight. Slightly easier than average due to the scaffolded structure. |
| Spec | 1.06f Laws of logarithms: addition, subtraction, power rules1.09a Sign change methods: locate roots1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(1)=\frac{4-1}{10}+\frac{3}{4}\ln\!\left(\frac{2}{3^2}\right)\;(=-0.828...)\) | M1 | Attempts to substitute \(t=1\); allow slip but attempt at substitution seen, or sight of awrt \(-0.828\) |
| \(P(1)=-0.828...\) (loss of £0.828 million), so approximately £830 000 loss | A1* | Correct value to at least 2 s.f. with substitution shown, or 3 s.f. without; must mention "loss" and include units |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(6)=-0.08799...\) and \(P(7)=0.1975...\) | M1 | Attempts both \(P(6)\) and \(P(7)\); at least one correct to 1 s.f. |
| Sign change and \(P\) continuous on \([6,7]\), so root for \(t\) lies in \([6,7]\) | A1 | Must state sign change and continuity conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P=0\Rightarrow\frac{4t-1}{10}=-\frac{3}{4}\ln\!\left(\frac{t+1}{(2t+1)^2}\right)\Rightarrow t=...\) | M1 | Attempt to rearrange to iterative formula |
| \(t=\frac{1}{4}+\frac{15}{8}\ln\!\left(\frac{(2t+1)^2}{t+1}\right)\) | A1* | Correct rearrangement (shown) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t_2=\frac{1}{4}+\frac{15}{8}\ln\!\left(\frac{13^2}{7}\right)=...\,(=6.219978...)\) | M1 | Correct substitution into formula |
| \(t_2=\text{awrt } 6.220\) | A1 | |
| \(t_6=6.314\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \("6.3..."\times12=...\) months, or repeated iteration to root \(6.31487\) gives \(75.7785...\), or \("0.314..."\times12="3.768..."\) | M1 | |
| So it will take 76 months (accept 75 or awrt 76 months) | A1 |
# Question 5:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(1)=\frac{4-1}{10}+\frac{3}{4}\ln\!\left(\frac{2}{3^2}\right)\;(=-0.828...)$ | M1 | Attempts to substitute $t=1$; allow slip but attempt at substitution seen, or sight of awrt $-0.828$ |
| $P(1)=-0.828...$ (loss of £0.828 million), so approximately £830 000 loss | A1* | Correct value to at least 2 s.f. with substitution shown, or 3 s.f. without; must mention "loss" and include units |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(6)=-0.08799...$ and $P(7)=0.1975...$ | M1 | Attempts both $P(6)$ and $P(7)$; at least one correct to 1 s.f. |
| Sign change and $P$ continuous on $[6,7]$, so root for $t$ lies in $[6,7]$ | A1 | Must state sign change and continuity conclusion |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P=0\Rightarrow\frac{4t-1}{10}=-\frac{3}{4}\ln\!\left(\frac{t+1}{(2t+1)^2}\right)\Rightarrow t=...$ | M1 | Attempt to rearrange to iterative formula |
| $t=\frac{1}{4}+\frac{15}{8}\ln\!\left(\frac{(2t+1)^2}{t+1}\right)$ | A1* | Correct rearrangement (shown) |
## Part (d)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t_2=\frac{1}{4}+\frac{15}{8}\ln\!\left(\frac{13^2}{7}\right)=...\,(=6.219978...)$ | M1 | Correct substitution into formula |
| $t_2=\text{awrt } 6.220$ | A1 | |
| $t_6=6.314$ | A1 | |
## Part (e)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $"6.3..."\times12=...$ months, or repeated iteration to root $6.31487$ gives $75.7785...$, or $"0.314..."\times12="3.768..."$ | M1 | |
| So it will take 76 months (accept 75 or awrt 76 months) | A1 | |5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{83e12fa4-1abb-4bea-bff4-8d36757bd9c3-12_479_551_214_699}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
The profit made by a company, $\pounds P$ million, $t$ years after the company started trading, is modelled by the equation
$$P = \frac { 4 t - 1 } { 10 } + \frac { 3 } { 4 } \ln \left[ \frac { t + 1 } { ( 2 t + 1 ) ^ { 2 } } \right]$$
The graph of $P$ against $t$ is shown in Figure 2.
According to the model,
\begin{enumerate}[label=(\alph*)]
\item show that exactly one year after it started trading, the company had made a loss of approximately £ 830000
A manager of the company wants to know the value of $t$ for which $P = 0$
\item Show that this value of $t$ occurs in the interval [6,7]
\item Show that the equation $P = 0$ can be expressed in the form
$$t = \frac { 1 } { 4 } + \frac { 15 } { 8 } \ln \left[ \frac { ( 2 t + 1 ) ^ { 2 } } { t + 1 } \right]$$
\item Using the iteration formula
$$t _ { n + 1 } = \frac { 1 } { 4 } + \frac { 15 } { 8 } \ln \left[ \frac { \left( 2 t _ { n } + 1 \right) ^ { 2 } } { t _ { n } + 1 } \right] \text { with } t _ { 1 } = 6$$
find the value of $t _ { 2 }$ and the value of $t _ { 6 }$, giving your answers to 3 decimal places.
\item Hence find, according to the model, how many months it takes in total, from when the company started trading, for it to make a profit.\\
(2)
\end{enumerate}
\hfill \mbox{\textit{Edexcel P3 2022 Q5 [11]}}