| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Session | Specimen |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Exponential model with shifted asymptote |
| Difficulty | Moderate -0.3 This is a standard C3 exponential modelling question with straightforward parts: substitution (a,d), interpretation of horizontal asymptote (b), sketching (c), and chain rule differentiation (e,f). All techniques are routine applications with no novel problem-solving required, making it slightly easier than average but not trivial due to the multi-part structure and need to interpret the model. |
| Spec | 1.01d Proof by contradiction |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(T = 80\) | B1 | (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(e^{-0.1t} \geq 0\) or equivalent | B1 | (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| Negative exponential shape | M1 | |
| \(t \geq 0\), starts at "80", clearly not \(\to x\)-axis | A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(60 = 20 + 60e^{-0.1t} \Rightarrow 60e^{-0.1t} = 40\) | M1 | |
| \(\Rightarrow -0.1t = \ln\left(\frac{2}{3}\right)\) | M1A1 | |
| \(t = 4.1\) | A1 | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(\frac{dT}{dt} = -6e^{-0.1t}\) | M1A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| Using \(\frac{dT}{dt} = -1.8\) | B1 | |
| Solving for \(t\), or using value of \(e^{-0.1t}\) (0.3) | M1 | |
| \(T = 38\) | A1 | (3) |
## Question 6:
### Part (a):
| Answer/Working | Marks | Notes |
|---|---|---|
| $T = 80$ | B1 | **(1)** |
### Part (b):
| Answer/Working | Marks | Notes |
|---|---|---|
| $e^{-0.1t} \geq 0$ or equivalent | B1 | **(1)** |
### Part (c):
| Answer/Working | Marks | Notes |
|---|---|---|
| Negative exponential shape | M1 | |
| $t \geq 0$, starts at "80", clearly not $\to x$-axis | A1 | **(2)** |
### Part (d):
| Answer/Working | Marks | Notes |
|---|---|---|
| $60 = 20 + 60e^{-0.1t} \Rightarrow 60e^{-0.1t} = 40$ | M1 | |
| $\Rightarrow -0.1t = \ln\left(\frac{2}{3}\right)$ | M1A1 | |
| $t = 4.1$ | A1 | **(4)** |
### Part (e):
| Answer/Working | Marks | Notes |
|---|---|---|
| $\frac{dT}{dt} = -6e^{-0.1t}$ | M1A1 | **(2)** |
### Part (f):
| Answer/Working | Marks | Notes |
|---|---|---|
| Using $\frac{dT}{dt} = -1.8$ | B1 | |
| Solving for $t$, or using value of $e^{-0.1t}$ (0.3) | M1 | |
| $T = 38$ | A1 | **(3)** |
---6. As a substance cools its temperature, $T ^ { \circ } \mathrm { C }$, is related to the time ( $t$ minutes) for which it has been cooling. The relationship is given by the equation
$$T = 20 + 60 \mathrm { e } ^ { - 0.1 t } , t \geq 0$$
\begin{enumerate}[label=(\alph*)]
\item Find the value of $T$ when the substance started to cool.
\item Explain why the temperature of the substance is always above $20 ^ { \circ } \mathrm { C }$.
\item Sketch the graph of $T$ against $t$.
\item Find the value, to 2 significant figures, of $t$ at the instant $T = 60$.
\item Find $\frac { \mathrm { d } T } { \mathrm {~d} t }$.
\item Hence find the value of $T$ at which the temperature is decreasing at a rate of $1.8 ^ { \circ } \mathrm { C }$ per minute.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 Q6 [13]}}