| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2021 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Show gradient condition leads to equation |
| Difficulty | Moderate -0.3 This is a standard P3 quotient rule differentiation question with straightforward follow-up parts. Part (d) requires showing a derivative equals a given form using the quotient rule, which is routine for this level. Parts (a)-(c) and (e) involve basic substitution and algebraic manipulation. While multi-part, each component is textbook-standard with no novel insight required, making it slightly easier than average. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.06i Exponential growth/decay: in modelling context1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(200\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(600\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(500 = \frac{600e^{0.3t}}{2+e^{0.3t}} \Rightarrow 100e^{0.3t} = 1000 \Rightarrow e^{0.3t} = 10\) | M1, A1 | Sets \(N=500\) and proceeds to \(Ce^{0.3t}=D\) or equivalent. A1: \(e^{0.3t}=10\); if ln's taken earlier, for \(\ln 100 + 0.3t = \ln 1000\) |
| \(t = \frac{\ln 10}{0.3} = 7 \text{ years } 8 \text{ months}\) | dM1, A1 | dM1: Full method to find \(t\) using correct log work. A1: 7 years 8 months; accept 7 years 9 months following correct \(t\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{dN}{dt} = \frac{(2+e^{0.3t})\times 180e^{0.3t} - 180e^{0.3t}\times e^{0.3t}}{(2+e^{0.3t})^2}\) | M1, A1 | M1: Uses quotient rule to obtain form \(\frac{(2+e^{0.3t})\times ae^{0.3t} - be^{0.3t}\times e^{0.3t}}{(2+e^{0.3t})^2}\). Condone missing brackets for M mark. Correct rule may be implied by \(u,v,u',v'\) then \(\frac{vu'-uv'}{v^2}\) |
| \(\frac{dN}{dt} = \frac{360e^{0.3t}}{(2+e^{0.3t})^2}\) | A1 | Withhold if \(e^{0.3t}\times e^{0.3t}\) written as \(e^{0.3t^2}\) or \(e^{0.09t}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(N = 600 - \frac{1200}{2+e^{0.3t}} \Rightarrow \frac{dN}{dt} = \frac{1200\times 0.3e^{0.3t}}{(2+e^{0.3t})^2}\) | M1, A1 | M1: splits \(\frac{600e^{0.3t}}{2+e^{0.3t}}\) into \(A \pm \frac{B}{2+e^{0.3t}}\) leading to \(\frac{dN}{dt}=\frac{ke^{0.3t}}{(2+e^{0.3t})^2}\) |
| \(\frac{dN}{dt} = \frac{360e^{0.3t}}{(2+e^{0.3t})^2}\) | A1 | Following fully correct work |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(8 = \frac{360e^{0.3t}}{(2+e^{0.3t})^2} \Rightarrow 8(e^{0.3t})^2 - 328(e^{0.3t}) + 32 = 0\) | M1 | Sets \(\frac{dN}{dt}=8\) and proceeds to quadratic in \(e^{0.3t}\) |
| \(e^{0.3t} = \frac{41+\sqrt{1665}}{2}\) | dM1 | Correct attempt to solve quadratic in \(e^{0.3t}\); condone both roots found |
| \(t = \frac{\ln\left(\frac{41+\sqrt{1665}}{2}\right)}{0.3}\) | ddM1 | Correct attempt to find \(t\) for \(e^{0.3t}=k\) where \(k>0\) using correct log work |
| \(T = \text{awrt } 12.4\) | A1 |
## Question 8:
**Part (a):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $200$ | B1 | |
**Part (b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $600$ | B1 | |
**Part (c):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $500 = \frac{600e^{0.3t}}{2+e^{0.3t}} \Rightarrow 100e^{0.3t} = 1000 \Rightarrow e^{0.3t} = 10$ | M1, A1 | Sets $N=500$ and proceeds to $Ce^{0.3t}=D$ or equivalent. A1: $e^{0.3t}=10$; if ln's taken earlier, for $\ln 100 + 0.3t = \ln 1000$ |
| $t = \frac{\ln 10}{0.3} = 7 \text{ years } 8 \text{ months}$ | dM1, A1 | dM1: Full method to find $t$ using correct log work. A1: 7 years 8 months; accept 7 years 9 months following correct $t$ |
**Part (d):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dN}{dt} = \frac{(2+e^{0.3t})\times 180e^{0.3t} - 180e^{0.3t}\times e^{0.3t}}{(2+e^{0.3t})^2}$ | M1, A1 | M1: Uses quotient rule to obtain form $\frac{(2+e^{0.3t})\times ae^{0.3t} - be^{0.3t}\times e^{0.3t}}{(2+e^{0.3t})^2}$. Condone missing brackets for M mark. Correct rule may be implied by $u,v,u',v'$ then $\frac{vu'-uv'}{v^2}$ |
| $\frac{dN}{dt} = \frac{360e^{0.3t}}{(2+e^{0.3t})^2}$ | A1 | Withhold if $e^{0.3t}\times e^{0.3t}$ written as $e^{0.3t^2}$ or $e^{0.09t}$ |
*Alt (d):*
| Answer | Mark | Guidance |
|--------|------|----------|
| $N = 600 - \frac{1200}{2+e^{0.3t}} \Rightarrow \frac{dN}{dt} = \frac{1200\times 0.3e^{0.3t}}{(2+e^{0.3t})^2}$ | M1, A1 | M1: splits $\frac{600e^{0.3t}}{2+e^{0.3t}}$ into $A \pm \frac{B}{2+e^{0.3t}}$ leading to $\frac{dN}{dt}=\frac{ke^{0.3t}}{(2+e^{0.3t})^2}$ |
| $\frac{dN}{dt} = \frac{360e^{0.3t}}{(2+e^{0.3t})^2}$ | A1 | Following fully correct work |
**Part (e):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $8 = \frac{360e^{0.3t}}{(2+e^{0.3t})^2} \Rightarrow 8(e^{0.3t})^2 - 328(e^{0.3t}) + 32 = 0$ | M1 | Sets $\frac{dN}{dt}=8$ and proceeds to quadratic in $e^{0.3t}$ |
| $e^{0.3t} = \frac{41+\sqrt{1665}}{2}$ | dM1 | Correct attempt to solve quadratic in $e^{0.3t}$; condone both roots found |
| $t = \frac{\ln\left(\frac{41+\sqrt{1665}}{2}\right)}{0.3}$ | ddM1 | Correct attempt to find $t$ for $e^{0.3t}=k$ where $k>0$ using correct log work |
| $T = \text{awrt } 12.4$ | A1 | |
---8. A scientist is studying a population of fish in a lake. The number of fish, $N$, in the population, $t$ years after the start of the study, is modelled by the equation
$$N = \frac { 600 \mathrm { e } ^ { 0.3 t } } { 2 + \mathrm { e } ^ { 0.3 t } } \quad t \geqslant 0$$
Use the equation of the model to answer parts (a), (b), (c), (d) and (e).
\begin{enumerate}[label=(\alph*)]
\item Find the number of fish in the lake at the start of the study.
\item Find the upper limit to the number of fish in the lake.
\item Find the time, after the start of the study, when there are predicted to be 500 fish in the lake. Give your answer in years and months to the nearest month.
\item Show that
$$\frac { \mathrm { d } N } { \mathrm {~d} t } = \frac { A \mathrm { e } ^ { 0.3 t } } { \left( 2 + \mathrm { e } ^ { 0.3 t } \right) ^ { 2 } }$$
where $A$ is a constant to be found.
Given that when $t = T , \frac { \mathrm {~d} N } { \mathrm {~d} t } = 8$
\item find the value of $T$ to one decimal place.\\
(Solutions relying entirely on calculator technology are not acceptable.)\\
\includegraphics[max width=\textwidth, alt={}, center]{76205772-5395-4ab2-96f9-ad9803b8388c-27_2644_1840_118_111}
\end{enumerate}
\hfill \mbox{\textit{Edexcel P3 2021 Q8 [13]}}