| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Exponential model with shifted asymptote |
| Difficulty | Standard +0.3 This is a standard exponential modelling question with straightforward parts: (i) requires substituting t=0 to find a=75; (ii) asks for interpretation of the asymptote (room temperature); (iii) uses differentiation to find k from the given rate; (iv) solves a logarithmic equation; (v) requires basic understanding that a lower initial temperature means smaller coefficient. All techniques are routine for C3/C4 level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.06i Exponential growth/decay: in modelling context1.07j Differentiate exponentials: e^(kx) and a^(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| \((a =)75\) | B1 | AO3.3 |
| Answer | Marks | Guidance |
|---|---|---|
| 25 is the value that \(T\) approaches after a long time; so therefore it is the ambient temperature | B1 | AO2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| \(-ake^{-kt}\) | B1 | AO3.1a |
| \(-ak = -15\) | M1 | AO3.4 |
| \(k = \frac{1}{5}\) | A1ft | AO1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(45 = 25 + 75e^{-\frac{1}{5}t} \Rightarrow 75e^{-\frac{1}{5}t} = 20\) | M1 | AO1.1 |
| \((eg)\ -\frac{1}{5}t = \ln\left(\frac{4}{15}\right) \Rightarrow t = ...\) | M1 | AO1.1 |
| After 6.6 mins | A1 | AO3.2a |
| Answer | Marks | Guidance |
|---|---|---|
| Decrease the value of \(a\) | B1 | AO3.5c |
## Question 6:
### Part (i):
$(a =)75$ | **B1** | AO3.3 |
### Part (ii):
25 is the value that $T$ approaches after a long time; so therefore it is the ambient temperature | **B1** | AO2.2a | oe e.g. room temperature, minimum, lowest, etc. Not e.g. initial, etc.
### Part (iii):
$-ake^{-kt}$ | **B1** | AO3.1a | Correct rate of change of $T$
$-ak = -15$ | **M1** | AO3.4 | Substitute $t = 0$ into their rate of change and equate with $+/-15$
$k = \frac{1}{5}$ | **A1ft** | AO1.1 | oe FT their $\frac{15}{a}$
### Part (iv):
$45 = 25 + 75e^{-\frac{1}{5}t} \Rightarrow 75e^{-\frac{1}{5}t} = 20$ | **M1** | AO1.1 | Substitute $T = 45$ and subtract 25 from both sides; their $a$ and $k$
$(eg)\ -\frac{1}{5}t = \ln\left(\frac{4}{15}\right) \Rightarrow t = ...$ | **M1** | AO1.1 | Take logs correctly and attempt to solve for $t$
After 6.6 mins | **A1** | AO3.2a | Cao (no FT on this mark) with units; $6.6087792...$
### Part (v):
Decrease the value of $a$ | **B1** | AO3.5c | Ignore mention of changes to $k$ and/or 25
---6 A pan of water is heated until it reaches $100 ^ { \circ } \mathrm { C }$. Once the water reaches $100 ^ { \circ } \mathrm { C }$, the heat is switched off and the temperature $T ^ { \circ } \mathrm { C }$ of the water decreases. The temperature of the water is modelled by the equation
$$T = 25 + a \mathrm { e } ^ { - k t }$$
where $t$ denotes the time, in minutes, after the heat is switched off and $a$ and $k$ are positive constants.\\
(i) Write down the value of $a$.\\
(ii) Explain what the value of 25 represents in the equation $T = 25 + a \mathrm { e } ^ { - k t }$.
When the heat is switched off, the initial rate of decrease of the temperature of the water is $15 ^ { \circ } \mathrm { C }$ per minute.\\
(iii) Calculate the value of $k$.\\
(iv) Find the time taken for the temperature of the water to drop from $100 ^ { \circ } \mathrm { C }$ to $45 ^ { \circ } \mathrm { C }$.\\
(v) A second pan of water is heated, but the heat is turned off when the water is at a temperature of less than $100 ^ { \circ } \mathrm { C }$. Suggest how the equation for the temperature as the water cools would be modified by this.
\hfill \mbox{\textit{OCR PURE Q6 [9]}}