| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2018 |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Verification of solutions |
| Difficulty | Moderate -0.3 This is a straightforward application of an exponential decay model requiring only substitution and basic algebraic manipulation. Part (a) is direct substitution, part (b) is verification requiring addition of two exponential terms, and part (c) involves solving an exponential equation using logarithms—all standard P3 techniques with no novel problem-solving required. Slightly easier than average due to the guided structure and verification format. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Subs \(D=15\) and \(t=4\): \(x = 15e^{-0.2\times4} = 6.740\) (mg) | M1 A1 | Condone slips on power e.g. \(-0.02\); note \(6.74\) (mg) is A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(15e^{-0.2\times7} + 15e^{-0.2\times2} = 13.754\) (mg) | M1 | Both \(D=15\) terms with \(t\) values of 2 and 7; award for sight of awrt 3.70 and awrt 10.05 |
| A1* | Both expression \(15e^{-0.2\times7}+15e^{-0.2\times2}\) and \(13.754\) (mg) required; sight of just numbers insufficient |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(15e^{-0.2\times T} + 15e^{-0.2\times(T+5)} = 7.5\) | M1 | Correct equation involving \(T\) or \(t\); accept with or without correct bracketing |
| \(15e^{-0.2T} + 15e^{-0.2T}e^{-1} = 7.5\) | ||
| \(15e^{-0.2T}(1+e^{-1}) = 7.5 \Rightarrow e^{-0.2T} = \dfrac{7.5}{15(1+e^{-1})}\) | dM1 | Proceed via index law \(x^{m+n}=x^m\times x^n\), factoring out \(e^{-0.2T}\) |
| \(T = -5\ln\!\left(\dfrac{7.5}{15(1+e^{-1})}\right)\) | A1 | Any correct form e.g. \(-5\ln\!\left(\dfrac{7.5}{15(1+e^{-1})}\right)\) |
| \(T = 5\ln\!\left(2+\dfrac{2}{e}\right)\) | A1 | Cso; condone \(t\) appearing for \(T\) throughout |
## Question 10:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Subs $D=15$ and $t=4$: $x = 15e^{-0.2\times4} = 6.740$ (mg) | M1 A1 | Condone slips on power e.g. $-0.02$; note $6.74$ (mg) is A0 |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $15e^{-0.2\times7} + 15e^{-0.2\times2} = 13.754$ (mg) | M1 | Both $D=15$ terms with $t$ values of 2 and 7; award for sight of awrt **3.70** and awrt **10.05** |
| | A1* | Both expression $15e^{-0.2\times7}+15e^{-0.2\times2}$ and $13.754$ (mg) required; sight of just numbers insufficient |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $15e^{-0.2\times T} + 15e^{-0.2\times(T+5)} = 7.5$ | M1 | Correct equation involving $T$ or $t$; accept with or without correct bracketing |
| $15e^{-0.2T} + 15e^{-0.2T}e^{-1} = 7.5$ | | |
| $15e^{-0.2T}(1+e^{-1}) = 7.5 \Rightarrow e^{-0.2T} = \dfrac{7.5}{15(1+e^{-1})}$ | dM1 | Proceed via index law $x^{m+n}=x^m\times x^n$, factoring out $e^{-0.2T}$ |
| $T = -5\ln\!\left(\dfrac{7.5}{15(1+e^{-1})}\right)$ | A1 | Any correct form e.g. $-5\ln\!\left(\dfrac{7.5}{15(1+e^{-1})}\right)$ |
| $T = 5\ln\!\left(2+\dfrac{2}{e}\right)$ | A1 | Cso; condone $t$ appearing for $T$ throughout |10. The amount of an antibiotic in the bloodstream, from a given dose, is modelled by the formula
$$x = D \mathrm { e } ^ { - 0.2 t }$$
where $x$ is the amount of the antibiotic in the bloodstream in milligrams, $D$ is the dose given in milligrams and $t$ is the time in hours after the antibiotic has been given.
A first dose of 15 mg of the antibiotic is given.
\begin{enumerate}[label=(\alph*)]
\item Use the model to find the amount of the antibiotic in the bloodstream 4 hours after the dose is given. Give your answer in mg to 3 decimal places.
A second dose of 15 mg is given 5 hours after the first dose has been given. Using the same model for the second dose,
\item show that the total amount of the antibiotic in the bloodstream 2 hours after the second dose is given is 13.754 mg to 3 decimal places.
No more doses of the antibiotic are given. At time $T$ hours after the second dose is given, the total amount of the antibiotic in the bloodstream is 7.5 mg .
\item Show that $T = a \ln \left( b + \frac { b } { \mathrm { e } } \right)$, where $a$ and $b$ are integers to be determined.\\
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel P3 2018 Q10 [8]}}