| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2014 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Show constant equals specific form |
| Difficulty | Moderate -0.3 This is a standard exponential modeling question requiring substitution of given values to find constants, then solving an exponential equation. The algebra is straightforward (substituting t=0 and t=15, solving simultaneous equations, then using logarithms), making it slightly easier than average but still requiring multiple steps and understanding of exponential decay models. |
| Spec | 1.06f Laws of logarithms: addition, subtraction, power rules1.06i Exponential growth/decay: in modelling context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| When \(t=0\), \(\theta=85\) so \(85=A+60\), \(A=25\) | B1 (1) | Gives answer \(A=25\); any work seen should be correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(58="25"+60e^{-k15}\), so \("33"=60e^{-k15}\), \(e^{-k15}=\frac{"33"}{60}\) | M1 | Uses values 58 and 15 with their \(A\) to form equation in \(k\) and isolates \(e^{-k15}\) |
| \(-15k=\ln\left(\frac{"11"}{20}\right)\) or \(15k=\ln\left(\frac{20}{"11"}\right)\) | M1 | Uses logs correctly following correct log rules, only applying log to positive quantities |
| \(k=-\frac{1}{15}\ln\left(\frac{11}{20}\right)=\frac{1}{15}\ln\left(\frac{20}{11}\right)\) | A1cso* (3) | Needs step between \(-15k=\ln\left(\frac{"11"}{20}\right)\) and printed answer; use of decimals 0.03985 as part of proof gives A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(50="25"+60e^{-kt}\), \(e^{-kt}=\frac{25}{60}\) (or awrt 0.42) | M1, A1 | M1: Uses 50 with their \(A\) and makes \(e^{-kt}\) subject; A1: correct numerical fraction |
| \(t=\frac{\ln\left(\frac{"25"}{60}\right)}{-k}\) | M1 | Uses logs correctly then rearranges correctly |
| \(\frac{\ln\left(\frac{25}{60}\right)}{-\frac{1}{15}\ln\left(\frac{20}{11}\right)}=(21.96)=\) 22 mins (approx) or \(t=22\) | A1 (4) | awrt 22 minutes; accept 11.22 (24hr clock) or \(t=22\); not \(t=22\) degrees C |
# Question 4:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| When $t=0$, $\theta=85$ so $85=A+60$, $A=25$ | B1 (1) | Gives answer $A=25$; any work seen should be correct |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $58="25"+60e^{-k15}$, so $"33"=60e^{-k15}$, $e^{-k15}=\frac{"33"}{60}$ | M1 | Uses values 58 and 15 with their $A$ to form equation in $k$ and isolates $e^{-k15}$ |
| $-15k=\ln\left(\frac{"11"}{20}\right)$ or $15k=\ln\left(\frac{20}{"11"}\right)$ | M1 | Uses logs correctly following correct log rules, only applying log to positive quantities |
| $k=-\frac{1}{15}\ln\left(\frac{11}{20}\right)=\frac{1}{15}\ln\left(\frac{20}{11}\right)$ | A1cso* (3) | Needs step between $-15k=\ln\left(\frac{"11"}{20}\right)$ and printed answer; use of decimals 0.03985 as part of proof gives A0 |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $50="25"+60e^{-kt}$, $e^{-kt}=\frac{25}{60}$ (or awrt 0.42) | M1, A1 | M1: Uses 50 with their $A$ and makes $e^{-kt}$ subject; A1: correct numerical fraction |
| $t=\frac{\ln\left(\frac{"25"}{60}\right)}{-k}$ | M1 | Uses logs correctly then rearranges correctly |
| $\frac{\ln\left(\frac{25}{60}\right)}{-\frac{1}{15}\ln\left(\frac{20}{11}\right)}=(21.96)=$ 22 mins (approx) or $t=22$ | A1 (4) | awrt 22 minutes; accept 11.22 (24hr clock) or $t=22$; not $t=22$ degrees C |
---\begin{enumerate}
\item A pot of coffee is delivered to a meeting room at 11 am . At a time $t$ minutes after 11 am the temperature, $\theta ^ { \circ } \mathrm { C }$, of the coffee in the pot is given by the equation
\end{enumerate}
$$\theta = A + 60 \mathrm { e } ^ { - k t }$$
where $A$ and $k$ are positive constants.
Given also that the temperature of the coffee at 11 am is $85 ^ { \circ } \mathrm { C }$ and that 15 minutes later it is $58 ^ { \circ } \mathrm { C }$,\\
(a) find the value of $A$.\\
(b) Show that $k = \frac { 1 } { 15 } \ln \left( \frac { 20 } { 11 } \right)$\\
(c) Find, to the nearest minute, the time at which the temperature of the coffee reaches $50 ^ { \circ } \mathrm { C }$.\\
\hfill \mbox{\textit{Edexcel C3 2014 Q4 [8]}}