Edexcel FD2 2022 June — Question 8 9 marks

Exam BoardEdexcel
ModuleFD2 (Further Decision 2)
Year2022
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDynamic Programming
TypeRecurrence relation solution
DifficultyStandard +0.8 This is a standard Further Maths recurrence relation question requiring students to derive the general solution to a first-order linear recurrence (involving geometric series summation) and then apply logarithms to solve an inequality. While it requires multiple techniques and careful algebraic manipulation, it follows a well-established template taught in FP2/FD2 courses with no novel problem-solving insight needed.
Spec1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form4.04e Line intersections: parallel, skew, or intersecting4.04f Line-plane intersection: find point
8. The owner of a new company models the number of customers that the company will have at the end of each month. The owner assumes that
  • a constant proportion, \(p\) (where \(0 < p < 1\) ), of the previous month's customers will be retained for the next month
  • a constant number of new customers, \(k\), will be added each month.
Let \(u _ { n }\) (where \(n \geqslant 1\) ) represent the number of customers that the company will have at the end of \(n\) months. The company has 5000 customers at the end of the first month.
  1. By setting up a first order recurrence relation for \(u _ { n + 1 }\) in terms of \(u _ { n }\), determine an expression for \(u _ { n }\) in terms of \(n , p\) and \(k\). The owner believes that \(95 \%\) of the previous month's customers will be retained each month and that there will be 10000 new customers each month. According to the model, the company will first have at least 135000 customers by the end of the \(m\) th month.
  2. Using logarithms, determine the value of \(m\). Please check the examination details below before entering your candidate information \section*{Further Mathematics} Advanced
    PAPER 4D: Decision Mathematics 2 \section*{Answer Book} Do not return the question paper with the answer book.
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    Minimum cost: \(\_\_\_\_\) 7. Player A \begin{table}[h]
    \captionsetup{labelformat=empty} \caption{Player B}
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    \end{table} The tableau for (d)(ii) can be found at the bottom of page 16
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    8.
Question 8:
Part (a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(u_{n+1} = pu_n + k\)B1 CAO
Auxiliary equation \(m - p = 0 \Rightarrow\) complementary function is \(A(p)^n\)B1 CAO
Consider trial solution \(u_n = \lambda\) so \(\lambda - p\lambda = k\)M1 Substituting trial solution into recurrence relation to find \(\lambda\) (which if correct is \(\frac{k}{1-p}\))
General solution is \(u_n = A(p)^n + \frac{k}{1-p}\)A1 CAO for general solution
\(u_1 = 5000 \Rightarrow 5000 = A(p) + \frac{k}{1-p}\) and solve for \(A\)M1 Using conditions to calculate \(A\) (which if correct is \(p^{-1}\left(5000 - \frac{k}{1-p}\right)\))
\(u_n = \left(5000 - \frac{k}{1-p}\right)p^{n-1} + \frac{k}{1-p}\)A1 CAO for particular solution (oe)
(6 marks)
Part (b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Set \(k = 10000\), \(p = 0.95\) and \(u_m \ldots 135000\)B1 Applying \(u_m \ldots 135000\) (equality or strict inequality) to general solution with correct \(k\) and \(p\) (dependent on both M marks in (a))
\(\left(5000 - \frac{10000}{1-0.95}\right)(0.95)^{m-1} + \frac{10000}{1-0.95} \ldots 135000\)
\((0.95)^{m-1} \ldots \frac{1}{3} \Rightarrow (m-1)\log(0.95) \ldots \log\left(\frac{1}{3}\right) \Rightarrow m \ldots\)M1 Dependent on previous B mark — solving equation using logarithms
\(m \ldots 22.418\ldots\) so 23 months after the company was first set upA1 CAO — must be rounded to 23
(3 marks)
Special case: A common misread is 500 for 5000. Mark as a misread so final A marks in (a) and (b) deducted.
## Question 8:

### Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $u_{n+1} = pu_n + k$ | B1 | CAO |
| Auxiliary equation $m - p = 0 \Rightarrow$ complementary function is $A(p)^n$ | B1 | CAO |
| Consider trial solution $u_n = \lambda$ so $\lambda - p\lambda = k$ | M1 | Substituting trial solution into recurrence relation to find $\lambda$ (which if correct is $\frac{k}{1-p}$) |
| General solution is $u_n = A(p)^n + \frac{k}{1-p}$ | A1 | CAO for general solution |
| $u_1 = 5000 \Rightarrow 5000 = A(p) + \frac{k}{1-p}$ and solve for $A$ | M1 | Using conditions to calculate $A$ (which if correct is $p^{-1}\left(5000 - \frac{k}{1-p}\right)$) |
| $u_n = \left(5000 - \frac{k}{1-p}\right)p^{n-1} + \frac{k}{1-p}$ | A1 | CAO for particular solution (oe) |

**(6 marks)**

---

### Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Set $k = 10000$, $p = 0.95$ and $u_m \ldots 135000$ | B1 | Applying $u_m \ldots 135000$ (equality or strict inequality) to general solution with correct $k$ and $p$ (dependent on both M marks in **(a)**) |
| $\left(5000 - \frac{10000}{1-0.95}\right)(0.95)^{m-1} + \frac{10000}{1-0.95} \ldots 135000$ | | |
| $(0.95)^{m-1} \ldots \frac{1}{3} \Rightarrow (m-1)\log(0.95) \ldots \log\left(\frac{1}{3}\right) \Rightarrow m \ldots$ | M1 | Dependent on previous B mark — solving equation using logarithms |
| $m \ldots 22.418\ldots$ so **23 months** after the company was first set up | A1 | CAO — must be rounded to 23 |

**(3 marks)**

**Special case:** A common misread is 500 for 5000. Mark as a misread so final A marks in (a) and (b) deducted.
8. The owner of a new company models the number of customers that the company will have at the end of each month. The owner assumes that

\begin{itemize}
  \item a constant proportion, $p$ (where $0 < p < 1$ ), of the previous month's customers will be retained for the next month
  \item a constant number of new customers, $k$, will be added each month.
\end{itemize}

Let $u _ { n }$ (where $n \geqslant 1$ ) represent the number of customers that the company will have at the end of $n$ months.

The company has 5000 customers at the end of the first month.
\begin{enumerate}[label=(\alph*)]
\item By setting up a first order recurrence relation for $u _ { n + 1 }$ in terms of $u _ { n }$, determine an expression for $u _ { n }$ in terms of $n , p$ and $k$.

The owner believes that $95 \%$ of the previous month's customers will be retained each month and that there will be 10000 new customers each month.

According to the model, the company will first have at least 135000 customers by the end of the $m$ th month.
\item Using logarithms, determine the value of $m$.

Please check the examination details below before entering your candidate information

\section*{Further Mathematics}
Advanced\\
PAPER 4D: Decision Mathematics 2

\section*{Answer Book}
Do not return the question paper with the answer book.\\
1.

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Minimum cost: $\_\_\_\_$\\
7.

Player A

\begin{table}[h]
\begin{center}
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\caption{Player B}
\begin{tabular}{|l|l|l|l|l|}
\hline
 & Option W & Option X & Option Y & Option Z \\
\hline
Option Q & 4 & 3 & -1 & -2 \\
\hline
Option R & -3 & 5 & -4 & $k$ \\
\hline
Option S & -1 & 6 & 3 & -3 \\
\hline
\end{tabular}
\end{center}
\end{table}

The tableau for (d)(ii) can be found at the bottom of page 16

\begin{center}
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b.v. &  &  &  & □ & □ & □ & □ &  & Value \\
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8.
\end{enumerate}

\hfill \mbox{\textit{Edexcel FD2 2022 Q8 [9]}}
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