| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Rate of change in exponential model |
| Difficulty | Standard +0.3 This is a straightforward application of differentiation to exponential functions. Part (a) requires simple substitution at t=0, while part (b) involves differentiating two exponential expressions and equating them—standard AS-level calculus with no conceptual challenges beyond recognizing that 'rate of change' means differentiation. |
| Spec | 1.06i Exponential growth/decay: in modelling context1.07b Gradient as rate of change: dy/dx notation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| At \(t=0\), \(V_A=100+20=120 \Rightarrow p=2\times"120"\) | M1 | Attempts to find price per gram of metal \(A\) at \(t=0\), then doubles. Expression must be evaluated |
| \((p=)\ 240\) | A1 | 240 only. Withhold if \(V_B=240\) but \(p\) stated as different value |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\ldots e^{0.04T}=\ldots e^{-0.02T}\) | M1 | Attempts to set \(\pm\frac{dV_A}{dt}\) equal to \(\pm\frac{dV_B}{dt}\). Look for equation of form \(pe^{0.04T}=qe^{-0.02T}\). Do not allow \(\ldots Te^{0.04T}=\ldots Te^{-0.02T}\) |
| \(0.8e^{0.04T}="4.8"e^{-0.02T}\) | A1ft | Follow through on positive value of \(p\). May be implied by further work |
| \(\ldots e^{0.04T}=\ldots e^{-0.02T}\Rightarrow e^{\pm 0.06T}=\ldots\) | dM1 | Rearranges to \(re^{\pm 0.06T}=s\) where \(r\times s>0\). Dependent on first M1. Condone slips |
| \((T=)\) awrt \(29.9\) (months) | A1cso |
## Question 11:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| At $t=0$, $V_A=100+20=120 \Rightarrow p=2\times"120"$ | M1 | Attempts to find price per gram of metal $A$ at $t=0$, then doubles. Expression must be evaluated |
| $(p=)\ 240$ | A1 | 240 only. Withhold if $V_B=240$ but $p$ stated as different value |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\ldots e^{0.04T}=\ldots e^{-0.02T}$ | M1 | Attempts to set $\pm\frac{dV_A}{dt}$ equal to $\pm\frac{dV_B}{dt}$. Look for equation of form $pe^{0.04T}=qe^{-0.02T}$. Do not allow $\ldots Te^{0.04T}=\ldots Te^{-0.02T}$ |
| $0.8e^{0.04T}="4.8"e^{-0.02T}$ | A1ft | Follow through on positive value of $p$. May be implied by further work |
| $\ldots e^{0.04T}=\ldots e^{-0.02T}\Rightarrow e^{\pm 0.06T}=\ldots$ | dM1 | Rearranges to $re^{\pm 0.06T}=s$ where $r\times s>0$. Dependent on first M1. Condone slips |
| $(T=)$ awrt $29.9$ (months) | A1cso | |\begin{enumerate}
\item The prices of two precious metals are being monitored.
\end{enumerate}
The price per gram of metal $A , \pounds V _ { A }$, is modelled by the equation
$$V _ { A } = 100 + 20 \mathrm { e } ^ { 0.04 t }$$
where $t$ is the number of months after monitoring began.\\
The price per gram of metal $B , \pounds V _ { B }$, is modelled by the equation
$$V _ { B } = p \mathrm { e } ^ { - 0.02 t }$$
where $p$ is a positive constant and $t$ is the number of months after monitoring began.\\
Given that $V _ { B } = 2 V _ { A }$ when $t = 0$\\
(a) find the value of $p$
When $t = T$, the rate of increase in the price per gram of metal $A$ was equal to the rate of decrease in the price per gram of metal $B$\\
(b) Find the value of $T$, giving your answer to one decimal place.\\
(Solutions based entirely on calculator technology are not acceptable.)
\hfill \mbox{\textit{Edexcel AS Paper 1 2024 Q11 [6]}}