AS June 2022 Q5
5. A person takes a course of a particular vitamin.
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 \(u_n\) 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.
The general solution to this recurrence relation has the form \(u_n = a(0.4)^n + b\)
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.
| Scheme | Marks | AO |
|---|---|---|
| B1 B1 | 2.4 3.3 |
| (2) |
Notes
B1: For explaining any two of the three aspects in the scheme.
B1: All three aspects explained.
| Scheme | Marks | AO |
|---|---|---|
| \(u_1 = 0.4 \times 20 + 10 = 18\) | B1 | 3.1a |
| So \(\left.\begin{matrix} 20 = a + b \\ 18 = \frac{2}{5}a + b \end{matrix}\right\} \Rightarrow 20 - 18 = a\left(1 - \dfrac{2}{5}\right) \Rightarrow a = \ldots\) | M1 | 1.1b |
| \(a = \dfrac{10}{3}\) and \(b = \dfrac{50}{3}\) | A1 A1 | 1.1b 1.1b |
| (4) |
Notes
B1: For \(u_1 = 18\) found in order to be able to form and solve simultaneous equations.
M1: Setting up and solving equations for \(a\) and \(b\) using their \(u_1\) and 20, with at least one value found.
A1: Either \(a\) or \(b\) correct.
A1: Both \(a\) and \(b\) correct.
Alt (b)
| Scheme | Marks | AO |
|---|---|---|
| \(n = 0 \Rightarrow 20 = a + b\) | B1 | 3.1a |
| \(u_{n+1} = 0.4u_n + 10 \Rightarrow a(0.4)^{n+1} + b = 0.4\left(a(0.4)^n + b\right) + 10\) \(\Rightarrow b = 0.4b + 10\) | M1 | 1.1b |
| \(b = \dfrac{50}{3}\) and \(a = \dfrac{10}{3}\) | A1 A1 | 1.1b 1.1b |
| (4) |
B1: Uses \(n = 0\) to form a correct relation relating \(a\) and \(b\).
M1: Substitutes the given general form into the recurrence relation and extracts an equation in \(b\) only.
A1: Correct value for \(b\)
A1: Both \(a\) and \(b\) correct.
Alt 2 (b)
| Scheme | Marks | AO |
|---|---|---|
| Particular solution is \(k \Rightarrow k = 0.4k + 10 \Rightarrow k = \dfrac{50}{3}\) | B1 | 1.ba |
| \(u_n = a(0.4)^n + \dfrac{50}{3} \Rightarrow 20 = a(0.4)^0 + \dfrac{50}{3}\) | M1 A1ft | 3.1a 1.1b |
| \(a = \dfrac{10}{3}\) | A1 | 1.1b |
| (4) |
B1: Uses particular solution is constant to find \(b\)
M1: Substitutes for \(n = 0\) or \(n = 1\) and sets equal to 20 in the general form to form an equation in \(a\)
A1: Correct equation follow through their \(b\).
A1: \(a\) correct.
| Scheme | Marks | AO |
|---|---|---|
| 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, so is \(\dfrac{50}{3} - 10 = \dfrac{20}{3} = 6\tfrac{2}{3}\) mg | M1 | 3.1b |
| This is greater than 6 mg and so there is always at least 6 mg of the vitamin in the person. The course of vitamin will be effective. | A1 | 3.2a |
| (3) | ||
| (9 marks) |
Notes
M1: Uses the model to work out the long-term behaviour or the amount of vitamin in the person. Accept identifying \(\dfrac{50}{3}\) as a lower limit for \(u_n\).
M1: Finds the lower bound for the amount of vitamin in the person, subtracting 10 from their long-term value or multiplying the \(\dfrac{50}{3}\) limit by 0.4 to find amount just before next tablet.
A1: Concludes the course of vitamin will be effective following correct work and reasoning.
NB: Attempts that set \(u_n = 6\) and try to solve for \(n\) score no marks. Attempts that set \(u_n - 10 = 6\) and try to solve for \(n\) can score the second M only. Long term behaviour needs to be considered for first M.