C3 June 2007 Q8
8. The amount of a certain type of drug in the bloodstream \(t\) hours after it has been taken is given by the formula\[x = D\mathrm{e}^{-\frac{1}{8}t},\]where \(x\) is the amount of the drug in the bloodstream in milligrams and \(D\) is the dose given in milligrams.
A dose of 10 mg of the drug is given.
Give your answer in mg to 3 decimal places. (2)
A second dose of 10 mg is given after 5 hours.
No more doses of the drug are given. At time \(T\) hours after the second dose is given, the amount of the drug in the bloodstream is 3 mg.
| Scheme | Marks |
|---|---|
| \(D = 10,\ t = 5,\ \ x = 10\mathrm{e}^{-\frac{1}{8}\times 5}\) | M1 |
| \(= 5.353\) awrt | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(D = 10 + 10\mathrm{e}^{-\frac{5}{8}},\ t = 1,\ \ x = 15.3526\ldots\times\mathrm{e}^{-\frac{1}{8}}\) | M1 |
| \(x = 13.549\) (✱) | A1 cso |
| (2) |
Alternative
| \(x = 10\mathrm{e}^{-\frac{1}{8}\times 6} + 10\mathrm{e}^{-\frac{1}{8}\times 1}\) | M1 |
| \(x = 13.549\) (✱) | A1 cso |
Notes
(main scheme) M1 is for \(\left(10 + 10\mathrm{e}^{-\frac{5}{8}}\right)\mathrm{e}^{-\frac{1}{8}}\), or \(\{10 + \text{their(a)}\}\mathrm{e}^{-(1/8)}\)
N.B. The answer is given. There are many correct answers seen which deserve M0A0 or M1A0(If adding two values, these should be 4.724 and 8.825)
| Scheme | Marks |
|---|---|
| \(15.3526\ldots\mathrm{e}^{-\frac{1}{8}T} = 3\) | M1 |
| \(\mathrm{e}^{-\frac{1}{8}T} = \dfrac{3}{15.3526\ldots} = 0.1954\ldots\) | |
| \(-\dfrac{1}{8}T = \ln 0.1954\ldots\) | M1 |
| \(T = 13.06\ldots\) or 13.1 or 13 | A1 |
| (3) | |
| (7 marks) |
Notes
1st M is for \(\left(10 + 10\mathrm{e}^{-\frac{5}{8}}\right)e^{-\frac{T}{8}} = 3\)
2nd M is for converting \(e^{-\frac{T}{8}} = k\ (k > 0)\) to \(-\dfrac{T}{8} = \ln k\). This is independent of 1st M.
Trial and improvement: M1 as scheme,
M1 correct process for their equation (two equal to 3 s.f.)
A1 as scheme