| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Exponential model with shifted asymptote |
| Difficulty | Moderate -0.3 This is a standard AS-level exponential modelling question with three routine parts: finding a constant from initial conditions, solving an exponential equation using logarithms, and finding a rate of change using differentiation. All techniques are textbook applications with no novel insight required, making it slightly easier than average. |
| Spec | 1.02z Models in context: use functions in modelling1.06a Exponential function: a^x and e^x graphs and properties1.06g Equations with exponentials: solve a^x = b1.07j Differentiate exponentials: e^(kx) and a^(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((k=)\ 0.8\) | B1 | Completes the equation for the model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(1=0.8+1.4e^{-0.5t}\Rightarrow1.4e^{-0.5t}=0.2\) | M1 | Use model with \(P=1\) and \(k\) from (a), proceed to \(Ae^{\pm0.5t}=B\); condone negative \(A\) or \(B\) |
| \(-0.5t=\ln\left(\frac{0.2}{1.4}\right)\Rightarrow t=...\) | M1 | Correct log work to solve \(Ae^{\pm0.5t}=B\); eg \(t=2\ln7\) or \(-2\ln\left(\frac{1}{7}\right)\) acceptable; cannot proceed directly to awrt 3.9 without intermediate working |
| awrt \(3.9\) minutes | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{\mathrm{d}P}{\mathrm{d}t}=-0.7e^{-0.5t}\), then \(\left(\frac{\mathrm{d}P}{\mathrm{d}t}\right)_{t=2}=-0.7e^{-0.5\times2}\) | M1 | Links rate of change to gradient; differentiates to form \(Ae^{-0.5t}\); substitutes \(t=2\); do not accept \(Ate^{-0.5t}\) as derivative; do not accept substituting \(t=2\) then proceeding from \(e^{-1}\) to \(e^{-2}\) |
| \(=\) awrt \(0.258\) (kg/cm² per minute) | A1 | Condone awrt \(-0.258\); units not required; awrt \(\pm0.258\) with no working is M0A0 |
## Question 8:
**Part (a):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $(k=)\ 0.8$ | B1 | Completes the equation for the model |
**Part (b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $1=0.8+1.4e^{-0.5t}\Rightarrow1.4e^{-0.5t}=0.2$ | M1 | Use model with $P=1$ and $k$ from (a), proceed to $Ae^{\pm0.5t}=B$; condone negative $A$ or $B$ |
| $-0.5t=\ln\left(\frac{0.2}{1.4}\right)\Rightarrow t=...$ | M1 | Correct log work to solve $Ae^{\pm0.5t}=B$; eg $t=2\ln7$ or $-2\ln\left(\frac{1}{7}\right)$ acceptable; cannot proceed directly to awrt 3.9 without intermediate working |
| awrt $3.9$ minutes | A1 | |
**Part (c):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{\mathrm{d}P}{\mathrm{d}t}=-0.7e^{-0.5t}$, then $\left(\frac{\mathrm{d}P}{\mathrm{d}t}\right)_{t=2}=-0.7e^{-0.5\times2}$ | M1 | Links rate of change to gradient; differentiates to form $Ae^{-0.5t}$; substitutes $t=2$; do not accept $Ate^{-0.5t}$ as derivative; do not accept substituting $t=2$ then proceeding from $e^{-1}$ to $e^{-2}$ |
| $=$ awrt $0.258$ (kg/cm² per minute) | A1 | Condone awrt $-0.258$; units not required; awrt $\pm0.258$ with no working is M0A0 |\begin{enumerate}
\item In this question you must show all stages of your working.
\end{enumerate}
\section*{Solutions relying entirely on calculator technology are not acceptable.}
The air pressure, $P \mathrm {~kg} / \mathrm { cm } ^ { 2 }$, inside a car tyre, $t$ minutes from the instant when the tyre developed a puncture is given by the equation
$$P = k + 1.4 \mathrm { e } ^ { - 0.5 t } \quad t \in \mathbb { R } \quad t \geqslant 0$$
where $k$ is a constant.\\
Given that the initial air pressure inside the tyre was $2.2 \mathrm {~kg} / \mathrm { cm } ^ { 2 }$\\
(a) state the value of $k$.
From the instant when the tyre developed the puncture,\\
(b) find the time taken for the air pressure to fall to $1 \mathrm {~kg} / \mathrm { cm } ^ { 2 }$
Give your answer in minutes to one decimal place.\\
(c) Find the rate at which the air pressure in the tyre is decreasing exactly 2 minutes from the instant when the tyre developed the puncture.\\
Give your answer in $\mathrm { kg } / \mathrm { cm } ^ { 2 }$ per minute to 3 significant figures.
\hfill \mbox{\textit{Edexcel AS Paper 1 2022 Q8 [6]}}