| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2007 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Exponential model with shifted asymptote |
| Difficulty | Moderate -0.8 This is a straightforward application question requiring substitution into a given exponential model, basic rearrangement to solve for t, and differentiation of an exponential function. All parts follow standard procedures with no novel problem-solving required—easier than average for C4. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.06g Equations with exponentials: solve a^x = b1.07k Differentiate trig: sin(kx), cos(kx), tan(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| \(t = 0: x = 3\) | B1 | 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| \(t = 14: x = 15 - 12e^{-1} = 10.6\) | M1, A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(-5 = -12e^{-t/14}\) | M1 | |
| \(\ln\left(\frac{5}{12}\right) = -\frac{t}{14}\) (OE) | m1 | |
| \(t = 14\ln\left(\frac{12}{5}\right)\) | A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(t = 12.256... = 12\) days | B1F | 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dx}{dt} = \frac{1}{14} \times -12e^{-t/14}\) | M1 | |
| \(= -\frac{1}{14}(x-15)\) | m1 | |
| \(= \frac{1}{14}(15-x)\) | A1 | 3 marks |
| Alt: \(t = -14\ln\left(\frac{15-x}{12}\right)\) | M1 | |
| \(\frac{dt}{dx} = \frac{-14\left(-\frac{1}{12}\right)}{\frac{15-x}{12}}\) | m1 | |
| \(\frac{dt}{dx} = \frac{14}{15-x}\) and \(\frac{dx}{dt} = \frac{1}{14}(15-x)\) | A1 | 3 marks |
| Alt: (backwards) | ||
| \(\int\frac{dx}{15-x} = \int\frac{dt}{14} = \pm 14\ln(15-x) = t + c\) | M1 | |
| \(\text{Use }(0,3): -14\ln(15-x) + 14\ln 12 = t\) | m1 | |
| Solve for \(x: x = 15 - 12e^{-t/14}\) | A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| rate of growth = 0.5 (cm per day) | B1 | 1 mark |
**4(a)(i)**
| $t = 0: x = 3$ | B1 | 1 mark | |
**4(a)(ii)**
| $t = 14: x = 15 - 12e^{-1} = 10.6$ | M1, A1 | 2 marks | or $15 - 12e^{-14/14}$ CAO |
**4(b)(i)**
| $-5 = -12e^{-t/14}$ | M1 | | substitute $x = 10$; rearrange to form $p = qe^{-t/14}$ |
| $\ln\left(\frac{5}{12}\right) = -\frac{t}{14}$ (OE) | m1 | | take lns correctly |
| $t = 14\ln\left(\frac{12}{5}\right)$ | A1 | 3 marks | must come from correct working |
**4(b)(ii)**
| $t = 12.256... = 12$ days | B1F | 1 mark | ft on a, b if $a > b$; accept $t = 12$ NMS. Accept 12 from incorrect working in b(i). Accept 13 if 12.2 or 12.3 seen |
**4(c)(i)**
| $\frac{dx}{dt} = \frac{1}{14} \times -12e^{-t/14}$ | M1 | | differentiate; allow sign error; condone $\frac{dy}{dt}$ used consistently |
| $= -\frac{1}{14}(x-15)$ | m1 | | Or $\frac{1}{14}\left(12e^{-t/14}\right)$ and $12e^{-t/14} = 15-x$ seen; OE |
| $= \frac{1}{14}(15-x)$ | A1 | 3 marks | AG – be convinced CSO |
| **Alt:** $t = -14\ln\left(\frac{15-x}{12}\right)$ | M1 | | attempt to solve given equation for $t$ |
| $\frac{dt}{dx} = \frac{-14\left(-\frac{1}{12}\right)}{\frac{15-x}{12}}$ | m1 | | differentiate wrt $x$, with $\frac{1}{15-x}$ seen; OE |
| $\frac{dt}{dx} = \frac{14}{15-x}$ and $\frac{dx}{dt} = \frac{1}{14}(15-x)$ | A1 | 3 marks | AG – be convinced |
| **Alt: (backwards)** | | | |
| $\int\frac{dx}{15-x} = \int\frac{dt}{14} = \pm 14\ln(15-x) = t + c$ | M1 | | |
| $\text{Use }(0,3): -14\ln(15-x) + 14\ln 12 = t$ | m1 | | |
| Solve for $x: x = 15 - 12e^{-t/14}$ | A1 | 3 marks | All steps shown |
**4(c)(ii)**
| rate of growth = 0.5 (cm per day) | B1 | 1 mark | Accept $\frac{7}{14}$ |4 A biologist is researching the growth of a certain species of hamster. She proposes that the length, $x \mathrm {~cm}$, of a hamster $t$ days after its birth is given by
$$x = 15 - 12 \mathrm { e } ^ { - \frac { t } { 14 } }$$
\begin{enumerate}[label=(\alph*)]
\item Use this model to find:
\begin{enumerate}[label=(\roman*)]
\item the length of a hamster when it is born;
\item the length of a hamster after 14 days, giving your answer to three significant figures.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Show that the time for a hamster to grow to 10 cm in length is given by $t = 14 \ln \left( \frac { a } { b } \right)$, where $a$ and $b$ are integers.
\item Find this time to the nearest day.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Show that
$$\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { 1 } { 14 } ( 15 - x )$$
\item Find the rate of growth of the hamster, in cm per day, when its length is 8 cm .\\
(1 mark)
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C4 2007 Q4 [11]}}