| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2011 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Show constant equals specific form |
| Difficulty | Moderate -0.3 This is a straightforward exponential decay question requiring substitution to find p, solving a logarithmic equation to show k equals a given form, and differentiation followed by equation solving. All techniques are standard C3 material with clear signposting and no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.06i Exponential growth/decay: in modelling context1.07j Differentiate exponentials: e^(kx) and a^(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(p = 7.5\) | B1 | |
| (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2.5 = 7.5e^{-4k}\) | M1 | |
| \(e^{-4k} = \frac{1}{3}\) | M1 | |
| \(-4k = \ln\!\left(\frac{1}{3}\right)\), \(-4k = -\ln(3)\) | dM1 | |
| \(k = \frac{1}{4}\ln(3)\) | A1* | |
| (4) | See notes for additional correct solutions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dm}{dt} = -kpe^{-kt}\) | M1A1ft | ft on their \(p\) and \(k\) |
| \(-\frac{1}{4}\ln 3 \times 7.5e^{-\frac{1}{4}(\ln 3)t} = -0.6\ln 3\) | ||
| \(e^{-\frac{1}{4}(\ln 3)t} = \frac{2.4}{7.5} = (0.32)\) | M1A1 | |
| \(-\frac{1}{4}(\ln 3)t = \ln(0.32)\) | dM1 | |
| \(t = 4.1486\ldots\) — 4.15 or awrt 4.1 | A1 | |
| (6) | 11 Marks |
# Question 5:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $p = 7.5$ | B1 | |
| | (1) | |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2.5 = 7.5e^{-4k}$ | M1 | |
| $e^{-4k} = \frac{1}{3}$ | M1 | |
| $-4k = \ln\!\left(\frac{1}{3}\right)$, $-4k = -\ln(3)$ | dM1 | |
| $k = \frac{1}{4}\ln(3)$ | A1* | |
| | (4) | See notes for additional correct solutions |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dm}{dt} = -kpe^{-kt}$ | M1A1ft | ft on their $p$ and $k$ |
| $-\frac{1}{4}\ln 3 \times 7.5e^{-\frac{1}{4}(\ln 3)t} = -0.6\ln 3$ | | |
| $e^{-\frac{1}{4}(\ln 3)t} = \frac{2.4}{7.5} = (0.32)$ | M1A1 | |
| $-\frac{1}{4}(\ln 3)t = \ln(0.32)$ | dM1 | |
| $t = 4.1486\ldots$ — 4.15 or awrt 4.1 | A1 | |
| | (6) | **11 Marks** |5. The mass, $m$ grams, of a leaf $t$ days after it has been picked from a tree is given by
$$m = p \mathrm { e } ^ { - k t }$$
where $k$ and $p$ are positive constants.\\
When the leaf is picked from the tree, its mass is 7.5 grams and 4 days later its mass is 2.5 grams.
\begin{enumerate}[label=(\alph*)]
\item Write down the value of $p$.
\item Show that $k = \frac { 1 } { 4 } \ln 3$.
\item Find the value of $t$ when $\frac { \mathrm { d } m } { \mathrm {~d} t } = - 0.6 \ln 3$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2011 Q5 [11]}}