| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Exponential model with shifted asymptote |
| Difficulty | Moderate -0.3 This is a standard exponential modelling question requiring substitution of initial conditions to find constants. Part (a) is immediate substitution at t=0. Part (b) requires differentiation and substituting the rate of change, which is routine A-level technique. The shifted asymptote ("+30") is explicitly given, removing any conceptual challenge. Slightly easier than average due to straightforward application of standard methods. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.06b Gradient of e^(kx): derivative and exponential model |
| Answer | Marks | Guidance |
|---|---|---|
| \((A =)\ 55\) | B1 | 55 only. "\(A=\)" not required. Ignore any "units" e.g. 55 °C |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left\{\frac{\mathrm{d}H}{\mathrm{d}t} =\right\} -ABe^{-Bt}\) or \(\left\{\frac{\mathrm{d}H}{\mathrm{d}t} =\right\} -\text{"55"}Be^{-Bt}\) | M1 | Differentiates to obtain form \(\pm ABe^{-Bt}\); A may already be substituted |
| \(-B \times \text{"55"} = -7.5 \Rightarrow B = \ldots \left(\frac{3}{22} = \text{awrt } 0.136\right)\) | M1 | Substitutes \(t=0\) and their \(A\) into \(\frac{\mathrm{d}H}{\mathrm{d}t}\), sets \(= \pm 7.5\), finds \(B\). Their \(\frac{\mathrm{d}H}{\mathrm{d}t}\) must not be \(H\) |
| \(H = 55e^{-0.136t} + 30\) | A1cso | Correct equation following fully correct work. Condone \(H = 55e^{-\frac{3}{22}t} + 30\) |
## Question 4:
### Part (a):
| $(A =)\ 55$ | **B1** | 55 only. "$A=$" not required. Ignore any "units" e.g. 55 °C |
### Part (b):
| $\left\{\frac{\mathrm{d}H}{\mathrm{d}t} =\right\} -ABe^{-Bt}$ or $\left\{\frac{\mathrm{d}H}{\mathrm{d}t} =\right\} -\text{"55"}Be^{-Bt}$ | **M1** | Differentiates to obtain form $\pm ABe^{-Bt}$; A may already be substituted |
| $-B \times \text{"55"} = -7.5 \Rightarrow B = \ldots \left(\frac{3}{22} = \text{awrt } 0.136\right)$ | **M1** | Substitutes $t=0$ and their $A$ into $\frac{\mathrm{d}H}{\mathrm{d}t}$, sets $= \pm 7.5$, finds $B$. Their $\frac{\mathrm{d}H}{\mathrm{d}t}$ must not be $H$ |
| $H = 55e^{-0.136t} + 30$ | **A1cso** | Correct equation following fully correct work. Condone $H = 55e^{-\frac{3}{22}t} + 30$ |
---\begin{enumerate}
\item Coffee is poured into a cup.
\end{enumerate}
The temperature of the coffee, $H ^ { \circ } \mathrm { C } , t$ minutes after being poured into the cup is modelled by the equation
$$H = A \mathrm { e } ^ { - B t } + 30$$
where $A$ and $B$ are constants.\\
Initially, the temperature of the coffee was $85 ^ { \circ } \mathrm { C }$.\\
(a) State the value of $A$.
Initially, the coffee was cooling at a rate of $7.5 ^ { \circ } \mathrm { C }$ per minute.\\
(b) Find a complete equation linking $H$ and $t$, giving the value of $B$ to 3 decimal places.
\hfill \mbox{\textit{Edexcel Paper 2 2023 Q4 [4]}}