| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Long-term behaviour and limits |
| Difficulty | Moderate -0.8 This is a straightforward application of exponential functions requiring only routine substitution (part a), differentiation and evaluation (part b), and recognizing that e^(-0.2t) → 0 as t → ∞ (part c). All techniques are standard AS-level procedures with no problem-solving insight needed. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.07k Differentiate trig: sin(kx), cos(kx), tan(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(h = 2.3 - 1.7e^0\) | M1 | Substitutes \(t=0\) into \(h = 2.3 - 1.7e^{-0.2t}\) |
| \(0.6\) {m} or \(60\) cm | A1 | Allow \(0.6\), \(0.6\) m, \(60\) cm or \(\frac{3}{5}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dh}{dt} = 0.34e^{-0.2t}\) | M1 | Links rate of change to gradient; differentiates to \(ke^{-0.2t}\), \(k \neq -1.7\) |
| At \(t=4 \Rightarrow 0.34e^{-0.8} = 0.15277...\) {m/year} | dM1 | Substitutes \(t=4\) into \(ke^{-0.2t}\), \(k \neq -1.7\) |
| \(0.153\) {m per year} \(= 15.3\) cm {per year} | A1* | Requires sight of \(\frac{dh}{dt} = 0.34e^{-0.2t}\); answer awrt \(0.153\) m/yr and converted to awrt \(15.3\) cm/yr |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2.3\) (m) | B1 | Allow \(2.3\), \(2.3\) m, \(230\) cm; \(2.2\dot{9}\) and \(2.2999...\) acceptable; \(2.29999999\) is not |
# Question 11:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $h = 2.3 - 1.7e^0$ | M1 | Substitutes $t=0$ into $h = 2.3 - 1.7e^{-0.2t}$ |
| $0.6$ {m} or $60$ cm | A1 | Allow $0.6$, $0.6$ m, $60$ cm or $\frac{3}{5}$ |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dh}{dt} = 0.34e^{-0.2t}$ | M1 | Links rate of change to gradient; differentiates to $ke^{-0.2t}$, $k \neq -1.7$ |
| At $t=4 \Rightarrow 0.34e^{-0.8} = 0.15277...$ {m/year} | dM1 | Substitutes $t=4$ into $ke^{-0.2t}$, $k \neq -1.7$ |
| $0.153$ {m per year} $= 15.3$ cm {per year} | A1* | Requires sight of $\frac{dh}{dt} = 0.34e^{-0.2t}$; answer awrt $0.153$ m/yr and converted to awrt $15.3$ cm/yr |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2.3$ (m) | B1 | Allow $2.3$, $2.3$ m, $230$ cm; $2.2\dot{9}$ and $2.2999...$ acceptable; $2.29999999$ is not |
---\begin{enumerate}
\item The height, $h$ metres, of a plant, $t$ years after it was first measured, is modelled by the equation
\end{enumerate}
$$h = 2.3 - 1.7 \mathrm { e } ^ { - 0.2 t } \quad t \in \mathbb { R } \quad t \geqslant 0$$
Using the model,\\
(a) find the height of the plant when it was first measured,\\
(b) show that, exactly 4 years after it was first measured, the plant was growing at approximately 15.3 cm per year.
According to the model, there is a limit to the height to which this plant can grow.\\
(c) Deduce the value of this limit.
\hfill \mbox{\textit{Edexcel AS Paper 1 2023 Q11 [6]}}