| Exam Board | Edexcel |
|---|---|
| Module | FD2 AS (Further Decision 2 AS) |
| Year | 2018 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Dynamic Programming |
| Type | Recurrence relation solution |
| Difficulty | Moderate -0.8 This is a straightforward recurrence relation question requiring standard techniques: identifying parameters from the model, solving a first-order linear recurrence (standard formula application), and solving an inequality using logarithms. All steps are routine for Further Maths students with no novel problem-solving required, making it easier than average but not trivial due to the algebraic manipulation involved. |
| Spec | 8.01f First-order recurrence: solve using auxiliary equation and complementary function8.01g Second-order recurrence: solve with distinct, repeated, or complex roots |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(r = 2\) | B1 | cao |
| \(N = 50\) | B1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Auxiliary equation \(m - 1.02 = 0\), complementary function is \(A(1.02)^n\) | B1 | cao |
| Trial solution \(u_n = \lambda\), so \(\lambda - 1.02\lambda = 50 \Rightarrow \lambda = \ldots\) | M1 | Substituting trial solution into recurrence relation to find \(\lambda\) |
| General solution \(u_n = A(1.02)^n - 2500\) | A1 | cao |
| \(n=1,\ u_1 = 560 \Rightarrow A = \ldots\) | M1 | Using conditions in model to calculate \(A\) |
| \(u_n = 3000(1.02)^n - 2500\) | A1 | cao for particular solution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(3000(1.02)^n - 2500 > 3000\) | M1 | Sets particular solution greater than 3000 (condone equals); solution must be of correct form \(u_n = c(1.02)^n \pm d\) |
| \((1.02)^n > \dfrac{11}{6} \Rightarrow n\log(1.02) > \log\!\left(\dfrac{11}{6}\right)\) | M1 | Re-arranging and correctly applying logs to their particular solution |
| \(n > 30.6088\ldots \Rightarrow n = 31\) | A1 | cao (allow correct answer to 3 s.f. or 31) |
# Question 4:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $r = 2$ | B1 | cao |
| $N = 50$ | B1 | cao |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| Auxiliary equation $m - 1.02 = 0$, complementary function is $A(1.02)^n$ | B1 | cao |
| Trial solution $u_n = \lambda$, so $\lambda - 1.02\lambda = 50 \Rightarrow \lambda = \ldots$ | M1 | Substituting trial solution into recurrence relation to find $\lambda$ |
| General solution $u_n = A(1.02)^n - 2500$ | A1 | cao |
| $n=1,\ u_1 = 560 \Rightarrow A = \ldots$ | M1 | Using conditions in model to calculate $A$ |
| $u_n = 3000(1.02)^n - 2500$ | A1 | cao for particular solution |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $3000(1.02)^n - 2500 > 3000$ | M1 | Sets particular solution greater than 3000 (condone equals); solution must be of correct form $u_n = c(1.02)^n \pm d$ |
| $(1.02)^n > \dfrac{11}{6} \Rightarrow n\log(1.02) > \log\!\left(\dfrac{11}{6}\right)$ | M1 | Re-arranging and correctly applying logs to their particular solution |
| $n > 30.6088\ldots \Rightarrow n = 31$ | A1 | cao (allow correct answer to 3 s.f. or 31) |
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There is **no mark scheme content visible** on this page to extract. This appears to be a blank reverse/back page of an exam document.
Please share the pages that contain the actual mark scheme content and I'll be happy to extract and format it for you.4. A village has an expected population growth rate (birth rate minus death rate) of $r \%$ per year. In addition, $N$ people are expected to move into the village each year. The expected population of the village is modelled by
$$u _ { n + 1 } = 1.02 u _ { n } + 50$$
where $u _ { n }$ is the expected population of the village $n$ years from now.
\begin{enumerate}[label=(\alph*)]
\item State
\begin{enumerate}[label=(\roman*)]
\item the value of $r$,
\item the value of $N$.
Given that the population 1 year from now is expected to be 560
\end{enumerate}\item solve the recurrence relation for $u _ { n }$
\item Hence determine, using algebra, the number of years from now when the model predicts that the population of the village will first be greater than 3000\\
(Total for Question 4 is 10 marks)\\
TOTAL FOR DECISION MATHEMATICS 2 IS 40 MARKS
END
\end{enumerate}
\hfill \mbox{\textit{Edexcel FD2 AS 2018 Q4 [10]}}