| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2015 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Exponential model with shifted asymptote |
| Difficulty | Moderate -0.3 This is a straightforward exponential modelling question requiring substitution (part a), solving for λ using logarithms (part b), and another substitution with logarithms (part c). All steps are standard C3 techniques with clear signposting, making it slightly easier than average despite being multi-part. |
| 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 = b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((\theta =)\ 20\) | B1 | Sight of \((\theta=)20\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Sub \(t=40, \theta=70\): \(70=120-100e^{-40\lambda}\), proceeding to \(e^{\pm 40\lambda}=A\) | M1 | Allow sign slips and copying errors |
| \(e^{-40\lambda} = 0.5\) (or \(e^{40\lambda}=2\)) or exact equivalent | A1 | |
| Taking ln's and proceeding to \(\lambda=\ldots\) | M1 | May be implied by correct decimal awrt 0.017 or \(\lambda=\frac{\ln 0.5}{-40}\) |
| \(\lambda = \frac{\ln 2}{40}\) | A1 | Accept equivalents \(\frac{\ln a}{b}\), \(a,b \in \mathbb{Z}\) e.g. \(\frac{\ln 4}{80}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\theta=100\): \(T = \frac{\ln 0.2}{-\text{their }\lambda}\) or \(T=\frac{\ln 5}{\text{their }\lambda}\) | M1 | Allow inequalities; substitutes \(\theta=100\) into \(\theta=120-100e^{-\lambda t}\) and proceeds to \(T=\ldots\) |
| \(T =\) awrt \(93\) | A1 | Watch: candidates losing minus sign in (b) using \(\lambda=\frac{\ln\frac{1}{2}}{40}\) then reaching \(T=-93\) scores M1 A0 |
# Question 4 (part a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(\theta =)\ 20$ | B1 | Sight of $(\theta=)20$ |
# Question 4 (part b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sub $t=40, \theta=70$: $70=120-100e^{-40\lambda}$, proceeding to $e^{\pm 40\lambda}=A$ | M1 | Allow sign slips and copying errors |
| $e^{-40\lambda} = 0.5$ (or $e^{40\lambda}=2$) or exact equivalent | A1 | |
| Taking ln's and proceeding to $\lambda=\ldots$ | M1 | May be implied by correct decimal awrt 0.017 or $\lambda=\frac{\ln 0.5}{-40}$ |
| $\lambda = \frac{\ln 2}{40}$ | A1 | Accept equivalents $\frac{\ln a}{b}$, $a,b \in \mathbb{Z}$ e.g. $\frac{\ln 4}{80}$ |
# Question 4 (part c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\theta=100$: $T = \frac{\ln 0.2}{-\text{their }\lambda}$ or $T=\frac{\ln 5}{\text{their }\lambda}$ | M1 | Allow inequalities; substitutes $\theta=100$ into $\theta=120-100e^{-\lambda t}$ and proceeds to $T=\ldots$ |
| $T =$ awrt $93$ | A1 | Watch: candidates losing minus sign in (b) using $\lambda=\frac{\ln\frac{1}{2}}{40}$ then reaching $T=-93$ scores M1 A0 |\begin{enumerate}
\item Water is being heated in an electric kettle. The temperature, $\theta ^ { \circ } \mathrm { C }$, of the water $t$ seconds after the kettle is switched on, is modelled by the equation
\end{enumerate}
$$\theta = 120 - 100 \mathrm { e } ^ { - \lambda t } , \quad 0 \leqslant t \leqslant T$$
(a) State the value of $\theta$ when $t = 0$
Given that the temperature of the water in the kettle is $70 ^ { \circ } \mathrm { C }$ when $t = 40$,\\
(b) find the exact value of $\lambda$, giving your answer in the form $\frac { \ln a } { b }$, where $a$ and $b$ are integers.
When $t = T$, the temperature of the water reaches $100 ^ { \circ } \mathrm { C }$ and the kettle switches off.\\
(c) Calculate the value of $T$ to the nearest whole number.\\
\hfill \mbox{\textit{Edexcel C3 2015 Q4 [7]}}