| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2007 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Exponential growth/decay model setup |
| Difficulty | Moderate -0.8 This is a straightforward exponential decay question requiring only direct substitution into a given formula (part a), verification of a calculation (part b), and solving a simple exponential equation using logarithms (part c). All techniques are routine C3 content with no problem-solving insight required, making it easier than average. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.06i Exponential growth/decay: in modelling context |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(D = 10, t = 5\), \(x = 10e^{-\frac{1}{2 \cdot 5}}\) \(= 5.353\) | M1 | |
| A1 | (2 marks) | |
| awrt | ||
| (b) \(D = 10 + 10e^{-x}\), \(t = 1\), \(x = 15.3526...xe^{-1}\) | M1 | |
| \(x = 13.549\) (✱) | A1 cso | (2 marks) |
| Alt.(b) \(x = 10e^{-1 \times 6} + 10e^{-1 \times 1}\) | M1 | |
| \(x = 13.549\) (✱) | A1 cso | |
| (c) \(15.3526...e^{-1/T} = 3\) | M1 | |
| \(e^{-1/T} = \frac{3}{15.3526...} = 0.1954...\) | ||
| \(-\frac{1}{T} = \ln 0.1954...\) | M1 | |
| \(T = 13.06...\) or 13.1 or 13 | A1 | (3 marks) |
| (7 marks) |
**(a)** $D = 10, t = 5$, $x = 10e^{-\frac{1}{2 \cdot 5}}$ $= 5.353$ | M1 | |
| A1 | (2 marks) |
awrt | | |
**(b)** $D = 10 + 10e^{-x}$, $t = 1$, $x = 15.3526...xe^{-1}$ | M1 | |
$x = 13.549$ (**✱**) | A1 cso | (2 marks) |
**Alt.(b)** $x = 10e^{-1 \times 6} + 10e^{-1 \times 1}$ | M1 |
$x = 13.549$ (**✱**) | A1 cso | |
**(c)** $15.3526...e^{-1/T} = 3$ | M1 | |
$e^{-1/T} = \frac{3}{15.3526...} = 0.1954...$ | | |
$-\frac{1}{T} = \ln 0.1954...$ | M1 | |
$T = 13.06...$ or 13.1 or 13 | A1 | (3 marks) |
| (7 marks) |
**Notes:**
- (b) (main scheme) M1 is for $(10 + 10e^{-x})e^{-1}$, or $\{10 +$ their $(a)\}e^{-(1/8)}$
- 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)
- (c) 1st M is for $(10+10e^{-x})e^{-T/8} = 3$
2nd M is for converting $e^{-T/8} = k$ ($k > 0$) to $-\frac{T}{8} = \ln k$. 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 scheme8. 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.
\begin{enumerate}[label=(\alph*)]
\item Find the amount of the drug in the bloodstream 5 hours after the dose is given. Give your answer in mg to 3 decimal places.
A second dose of 10 mg is given after 5 hours.
\item Show that the amount of the drug in the bloodstream 1 hour after the second dose is 13.549 mg to 3 decimal places.
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 .
\item Find the value of $T$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2007 Q8 [7]}}