| Exam Board | Edexcel |
|---|---|
| Module | FD2 AS (Further Decision 2 AS) |
| Year | 2021 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Dynamic Programming |
| Type | Recurrence relation solution |
| Difficulty | Standard +0.3 This is a standard recurrence relation problem requiring identification of the interest rate from the multiplier, solving a first-order linear recurrence (textbook method), and applying a boundary condition. While it involves multiple steps (9 marks), each component follows routine procedures taught in FD2 with no novel insight required—slightly easier than average A-level difficulty. |
| Spec | 1.04e Sequences: nth term and recurrence relations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(r = 0.25\) | B1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| (aux equation \(m - 1.0025 = 0 \Rightarrow\)) complementary function is \(A(1.0025)^n\) | B1 | cao |
| Consider a trial solution of the form \(u_n = \lambda\) so \(\lambda - 1.0025\lambda = -x \Rightarrow \lambda = \ldots\) | M1 | Substituting their trial solution into the recurrence relation in an attempt to find their \(\lambda\) (which if correct is \(400x\)) |
| General solution is \(u_n = A(1.0025)^n + 400x\) | A1 | cao for the general solution |
| \(n = 1,\ u_1 = 155000 \Rightarrow A = \ldots\) | M1 | Using the conditions in the model to calculate \(A\) (which if correct is \((155000 - 400x)(1.0025)^{-1}\)) |
| \(u_n = (155000 - 400x)(1.0025)^{n-1} + 400x\) | A1 | cao for the particular solution (oe) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Applying \(u_{360} = 0\) | B1 | Applying \(u_{360} = 0\) to their general solution (dependent on both M marks in (b)) |
| \(155000(1.0025)^{359} + 400x\left(1 - 1.0025^{359}\right) = 0 \Rightarrow x = \ldots\) | M1 | Dependent on previous B mark – solving their linear equation to find \(x\) |
| \(x = 654.61\) | A1 | cao |
## Question 4:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $r = 0.25$ | B1 | cao |
**Total: (1)**
---
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| (aux equation $m - 1.0025 = 0 \Rightarrow$) complementary function is $A(1.0025)^n$ | B1 | cao |
| Consider a trial solution of the form $u_n = \lambda$ so $\lambda - 1.0025\lambda = -x \Rightarrow \lambda = \ldots$ | M1 | Substituting their trial solution into the recurrence relation in an attempt to find their $\lambda$ (which if correct is $400x$) |
| General solution is $u_n = A(1.0025)^n + 400x$ | A1 | cao for the general solution |
| $n = 1,\ u_1 = 155000 \Rightarrow A = \ldots$ | M1 | Using the conditions in the model to calculate $A$ (which if correct is $(155000 - 400x)(1.0025)^{-1}$) |
| $u_n = (155000 - 400x)(1.0025)^{n-1} + 400x$ | A1 | cao for the particular solution (oe) |
**Total: (5)**
---
### Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Applying $u_{360} = 0$ | B1 | Applying $u_{360} = 0$ to their general solution (dependent on both M marks in **(b)**) |
| $155000(1.0025)^{359} + 400x\left(1 - 1.0025^{359}\right) = 0 \Rightarrow x = \ldots$ | M1 | Dependent on previous B mark – solving their linear equation to find $x$ |
| $x = 654.61$ | A1 | cao |
**Total: (3)**
---
**Overall Total: (9 marks)**4. Sarah takes out a mortgage of $\pounds 155000$ to buy a house. Interest is added each month on the outstanding balance at a constant rate of $r$ \% each month. Sarah makes fixed monthly repayments to reduce the amount owed.
Each month, interest is added, and then her monthly repayment is used to reduce the outstanding amount owed.
The recurrence relationship for the amount of the mortgage outstanding after $n + 1$ months is modelled by
$$u _ { n + 1 } = 1.0025 u _ { n } - x \quad n \geqslant 0$$
where $\pounds u _ { n }$ is the amount of the mortgage outstanding after $n$ months and $\pounds x$ is the monthly repayment.
\begin{enumerate}[label=(\alph*)]
\item State the value of $r$.
\item Solve the recurrence relation to find an expression for $u _ { n }$ in terms of $x$ and $n$.
Given that the mortgage will be paid off in exactly 30 years,
\item determine, to 2 decimal places, the least possible value of $x$.
\section*{(Total for Question 4 is 9 marks)}
TOTAL FOR DECISION MATHEMATICS 2 IS 40 MARKS\\
END
\end{enumerate}
\hfill \mbox{\textit{Edexcel FD2 AS 2021 Q4 [9]}}