| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Exponential model with shifted asymptote |
| Difficulty | Standard +0.3 This is a standard exponential modeling question requiring substitution into given equations, solving simultaneous equations with logarithms, differentiation of exponentials, and understanding translations. All techniques are routine C3 content with clear scaffolding through multiple parts. Slightly easier than average due to the structured guidance and standard methods throughout. |
| Spec | 1.06b Gradient of e^(kx): derivative and exponential model1.06i Exponential growth/decay: in modelling context1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Notes |
| \(t=10, T=18 \Rightarrow 18 = 5 + Ae^{-10k}\) and \(t=60, T=12 \Rightarrow 12 = 5 + Ae^{-60k}\) | M1 | |
| \(A = \frac{13}{e^{-10k}} = 13e^{10k}\) | M1 | |
| \(7 = 13e^{10k} \times e^{-60k}\); \(e^{-50k} = \frac{7}{13}\) | A1 | |
| \(k = -\frac{1}{50}\ln\frac{7}{13} = 0.0124\) (3sf) | M1 A1 | |
| \(A = 13e^{10 \times 0.01238} = 14.7\) (3sf) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Notes |
| \(\frac{dT}{dt} = -0.01238 \times 14.71\, e^{-0.01238t} = -0.1822e^{-0.01238t}\) | M1 A1 | |
| When \(t=20\): \(\frac{dT}{dt} = -0.1822e^{-0.01238 \times 20} = -0.142\) | M1 | |
| Temperature decreasing at rate of \(0.142\ ^\circ\)C per minute (3sf) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Notes |
| \(T = 5 + 14.71e^{-0.01238(t-60)}\) | M1 | |
| \(= 5 + 14.71e^{0.7428 - 0.01238t}\) | ||
| \(= 5 + 14.71e^{0.7428} \times e^{-0.01238t}\) | M1 | |
| \(= 5 + 30.9e^{-0.01238t},\quad B = 30.9\) (3sf) | A1 | (13) |
| Total | (72) |
# Question 9:
## Part (i)
| Working/Answer | Mark | Notes |
|---|---|---|
| $t=10, T=18 \Rightarrow 18 = 5 + Ae^{-10k}$ and $t=60, T=12 \Rightarrow 12 = 5 + Ae^{-60k}$ | M1 | |
| $A = \frac{13}{e^{-10k}} = 13e^{10k}$ | M1 | |
| $7 = 13e^{10k} \times e^{-60k}$; $e^{-50k} = \frac{7}{13}$ | A1 | |
| $k = -\frac{1}{50}\ln\frac{7}{13} = 0.0124$ (3sf) | M1 A1 | |
| $A = 13e^{10 \times 0.01238} = 14.7$ (3sf) | A1 | |
## Part (ii)
| Working/Answer | Mark | Notes |
|---|---|---|
| $\frac{dT}{dt} = -0.01238 \times 14.71\, e^{-0.01238t} = -0.1822e^{-0.01238t}$ | M1 A1 | |
| When $t=20$: $\frac{dT}{dt} = -0.1822e^{-0.01238 \times 20} = -0.142$ | M1 | |
| Temperature decreasing at rate of $0.142\ ^\circ$C per minute (3sf) | A1 | |
## Part (iii)
| Working/Answer | Mark | Notes |
|---|---|---|
| $T = 5 + 14.71e^{-0.01238(t-60)}$ | M1 | |
| $= 5 + 14.71e^{0.7428 - 0.01238t}$ | | |
| $= 5 + 14.71e^{0.7428} \times e^{-0.01238t}$ | M1 | |
| $= 5 + 30.9e^{-0.01238t},\quad B = 30.9$ (3sf) | A1 | **(13)** |
| | **Total** | **(72)** |9.\\
\includegraphics[max width=\textwidth, alt={}, center]{d1cf3850-964a-4ff1-ae25-f1bc60a6aded-3_501_1111_877_413}
The diagram shows a graph of the temperature of a room, $T ^ { \circ } \mathrm { C }$, at time $t$ minutes.\\
The temperature is controlled by a thermostat such that when the temperature falls to $12 ^ { \circ } \mathrm { C }$, a heater is turned on until the temperature reaches $18 ^ { \circ } \mathrm { C }$. The room then cools until the temperature again falls to $12 ^ { \circ } \mathrm { C }$.
For $t$ in the interval $10 \leq t \leq 60 , T$ is given by
$$T = 5 + A \mathrm { e } ^ { - k t } ,$$
where $A$ and $k$ are constants.\\
Given that $T = 18$ when $t = 10$ and that $T = 12$ when $t = 60$,\\
(i) show that $k = 0.0124$ to 3 significant figures and find the value of $A$,\\
(ii) find the rate at which the temperature of the room is decreasing when $t = 20$.
The temperature again reaches $18 ^ { \circ } \mathrm { C }$ when $t = 70$ and the graph for $70 \leq t \leq 120$ is a translation of the graph for $10 \leq t \leq 60$.\\
(iii) Find the value of the constant $B$ such that for $70 \leq t \leq 120$
$$T = 5 + B \mathrm { e } ^ { - k t }$$
\hfill \mbox{\textit{OCR C3 Q9 [13]}}