| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Time to reach target in exponential model |
| Difficulty | Moderate -0.3 This is a straightforward exponential modelling question requiring basic substitution (part a), routine differentiation and evaluation (part b), and solving an exponential equation using logarithms (part c). All techniques are standard A-level procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.06b Gradient of e^(kx): derivative and exponential model1.06g Equations with exponentials: solve a^x = b1.06i Exponential growth/decay: in modelling context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| 265 thousand | B1 | Accept 265 thousand or 265 000 or 265 k but not just 265 |
| Total: (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Attempts \(\frac{dN_b}{dt}=11e^{0.05t}\) | M1 | Differentiates to form \(ke^{0.05t}\), \(k>0, k\neq 220\) |
| Substitutes \(t=10\) into their \(\frac{dN_b}{dt}\) | M1 | lhs must have implied differentiation |
| \(\frac{dN_b}{dt}=\) awrt 18.1, approximately 18 thousand per year * | A1* | Requires correct lhs, intermediate line of \(11e^{0.05\times10}\) or awrt 18.1, and conclusion referencing 18 000 or 18 thousand |
| Total: (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Sets \(45+220e^{0.05t}=10+800e^{-0.05t}\Rightarrow 220e^{0.05t}+35-800e^{-0.05t}=0\) | M1 | |
| Correct quadratic \(\Rightarrow 220(e^{0.05t})^2+35e^{0.05t}-800=0\) | A1 | |
| \(e^{0.05t}=1.829,(-1.988)\Rightarrow 0.05t=\ln(1.829)\) | M1 | |
| \(T=12.08\) | A1 | |
| Total: (4) |
## Question 10:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| 265 thousand | B1 | Accept 265 thousand or 265 000 or 265 k but not just 265 |
| **Total: (1)** | | |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Attempts $\frac{dN_b}{dt}=11e^{0.05t}$ | M1 | Differentiates to form $ke^{0.05t}$, $k>0, k\neq 220$ |
| Substitutes $t=10$ into their $\frac{dN_b}{dt}$ | M1 | lhs must have implied differentiation |
| $\frac{dN_b}{dt}=$ awrt 18.1, approximately 18 thousand per year * | A1* | Requires correct lhs, intermediate line of $11e^{0.05\times10}$ or awrt 18.1, and conclusion referencing 18 000 or 18 thousand |
| **Total: (3)** | | |
### Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Sets $45+220e^{0.05t}=10+800e^{-0.05t}\Rightarrow 220e^{0.05t}+35-800e^{-0.05t}=0$ | M1 | |
| Correct quadratic $\Rightarrow 220(e^{0.05t})^2+35e^{0.05t}-800=0$ | A1 | |
| $e^{0.05t}=1.829,(-1.988)\Rightarrow 0.05t=\ln(1.829)$ | M1 | |
| $T=12.08$ | A1 | |
| **Total: (4)** | | |\begin{enumerate}
\item A scientist is studying the number of bees and the number of wasps on an island.
\end{enumerate}
The number of bees, measured in thousands, $N _ { b }$, is modelled by the equation
$$N _ { b } = 45 + 220 \mathrm { e } ^ { 0.05 t }$$
where $t$ is the number of years from the start of the study.\\
According to the model,\\
(a) find the number of bees at the start of the study,\\
(b) show that, exactly 10 years after the start of the study, the number of bees was increasing at a rate of approximately 18 thousand per year.
The number of wasps, measured in thousands, $N _ { w }$, is modelled by the equation
$$N _ { w } = 10 + 800 \mathrm { e } ^ { - 0.05 t }$$
where $t$ is the number of years from the start of the study.\\
When $t = T$, according to the models, there are an equal number of bees and wasps.\\
(c) Find the value of $T$ to 2 decimal places.
\hfill \mbox{\textit{Edexcel Paper 1 2022 Q10 [8]}}