| Exam Board | Edexcel |
|---|---|
| Module | FP2 AS (Further Pure 2 AS) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and Series |
| Type | Modelling with Recurrence Relations |
| Difficulty | Standard +0.3 This is a straightforward recurrence relation question requiring explanation of a given model, solving for constants using standard substitution methods, and applying an inequality condition. While it involves multiple parts and real-world context, the mathematical techniques are routine for FP2 students (substituting into general solution, finding particular solution, checking convergence). The recurrence relation is already provided and the solution form is given, reducing problem-solving demands. |
| Spec | 1.04e Sequences: nth term and recurrence relations |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Immediately after first tablets taken there is 20 mg, so \(u_0=20\) | B1 | 2.4 |
| Reduction by 60% through day means just before next tablet there is \(0.4u_n\) mg; next tablet adds 10 mg giving \(u_{n+1}=0.4u_n+10\) | B1 | 3.3 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(u_1=0.4\times20+10=18\) | B1 | 3.1a |
| \(20=a+b\) and \(18=\frac{2}{5}a+b \Rightarrow 20-18=a\!\left(1-\frac{2}{5}\right)\Rightarrow a=\ldots\) | M1 | 1.1b |
| \(a=\dfrac{10}{3}\) and \(b=\dfrac{50}{3}\) | A1, A1 | 1.1b, 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| In long term \(u_n=\dfrac{10}{3}(0.4)^n+\dfrac{50}{3}\to\dfrac{50}{3}\) as \((0.4)^n\to 0\) | M1 | 3.4 |
| Minimum amount of vitamin occurs just before a tablet is taken: \(\dfrac{50}{3}-10=\dfrac{20}{3}=6\tfrac{2}{3}\) mg | M1 | 3.1b |
| This is greater than 6 mg so there is always at least 6 mg of the vitamin in the person. The course of vitamin will be effective. | A1 | 3.2a |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(n=0 \Rightarrow 20 = a+b\) | B1 | Uses \(n=0\) to form a correct relation relating \(a\) and \(b\) |
| \(u_{n+1} = 0.4u_n + 10 \Rightarrow a(0.4)^{n+1}+b = 0.4\left(a(0.4)^n+b\right)+10 \Rightarrow b = 0.4b+10\) | M1 | Substitutes the given general form into the recurrence relation and extracts an equation in \(b\) only |
| \(b = \dfrac{50}{3}\) | A1 | Correct value for \(b\) |
| \(a = \dfrac{10}{3}\) | A1 | Both \(a\) and \(b\) correct |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Particular solution: \(k = 0.4k+10 \Rightarrow k = \dfrac{50}{3}\) | B1 | Uses particular solution is constant to find \(b\) |
| \(u_n = a(0.4)^n + \dfrac{50}{3} \Rightarrow 20 = a(0.4)^0 + \dfrac{50}{3}\) | M1 A1ft | Substitutes \(n=0\) or \(n=1\) and sets equal to 20; correct equation following through their \(b\) |
| \(a = \dfrac{10}{3}\) | A1 | \(a\) correct |
| (4) |
## Question 5(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Immediately after first tablets taken there is 20 mg, so $u_0=20$ | B1 | 2.4 |
| Reduction by 60% through day means just before next tablet there is $0.4u_n$ mg; next tablet adds 10 mg giving $u_{n+1}=0.4u_n+10$ | B1 | 3.3 |
---
## Question 5(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $u_1=0.4\times20+10=18$ | B1 | 3.1a |
| $20=a+b$ and $18=\frac{2}{5}a+b \Rightarrow 20-18=a\!\left(1-\frac{2}{5}\right)\Rightarrow a=\ldots$ | M1 | 1.1b |
| $a=\dfrac{10}{3}$ and $b=\dfrac{50}{3}$ | A1, A1 | 1.1b, 1.1b |
---
## Question 5(c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| In long term $u_n=\dfrac{10}{3}(0.4)^n+\dfrac{50}{3}\to\dfrac{50}{3}$ as $(0.4)^n\to 0$ | M1 | 3.4 |
| Minimum amount of vitamin occurs just before a tablet is taken: $\dfrac{50}{3}-10=\dfrac{20}{3}=6\tfrac{2}{3}$ mg | M1 | 3.1b |
| This is greater than 6 mg so there is always at least 6 mg of the vitamin in the person. The course of vitamin will be effective. | A1 | 3.2a |
## Question (Part b):
### Alt (b) – Method 1:
| Working | Mark | Guidance |
|---------|------|----------|
| $n=0 \Rightarrow 20 = a+b$ | **B1** | Uses $n=0$ to form a correct relation relating $a$ and $b$ |
| $u_{n+1} = 0.4u_n + 10 \Rightarrow a(0.4)^{n+1}+b = 0.4\left(a(0.4)^n+b\right)+10 \Rightarrow b = 0.4b+10$ | **M1** | Substitutes the given general form into the recurrence relation and extracts an equation in $b$ only |
| $b = \dfrac{50}{3}$ | **A1** | Correct value for $b$ |
| $a = \dfrac{10}{3}$ | **A1** | Both $a$ and $b$ correct |
| | **(4)** | |
---
### Alt 2 (b) – Method 2:
| Working | Mark | Guidance |
|---------|------|----------|
| Particular solution: $k = 0.4k+10 \Rightarrow k = \dfrac{50}{3}$ | **B1** | Uses particular solution is constant to find $b$ |
| $u_n = a(0.4)^n + \dfrac{50}{3} \Rightarrow 20 = a(0.4)^0 + \dfrac{50}{3}$ | **M1 A1ft** | Substitutes $n=0$ or $n=1$ and sets equal to 20; correct equation following through their $b$ |
| $a = \dfrac{10}{3}$ | **A1** | $a$ correct |
| | **(4)** | |\begin{enumerate}
\item A person takes a course of a particular vitamin.
\end{enumerate}
Before the course there was none of the vitamin in the person's body.\\
During the course, vitamin tablets are taken at the same time each day.\\
Initially two tablets are taken and on each following day only one tablet is taken.\\
Each tablet contains 10 mg of the vitamin.\\
Between doses the amount of the vitamin in the person's body decreases naturally by 60\%
Let $u _ { n } \mathrm { mg }$ be the amount of the vitamin in the person's body immediately after a tablet is taken, $n$ days after the initial two tablets were taken.\\
(a) Explain why $u _ { n }$ satisfies the recurrence relation
$$u _ { 0 } = 20 \quad u _ { n + 1 } = 0.4 u _ { n } + 10$$
The general solution to this recurrence relation has the form $u _ { n } = a ( 0.4 ) ^ { n } + b$\\
(b) Determine the value of $a$ and the value of $b$.
The course is only effective if the amount of the vitamin in the person's body remains above 6 mg at all times throughout the course.\\
(c) Determine whether this course of the vitamin will be effective for this person, giving a reason for your answer.
\hfill \mbox{\textit{Edexcel FP2 AS 2022 Q5 [9]}}