| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | Calculus with exponential models |
| Difficulty | Standard +0.3 This is a straightforward exponential modelling question requiring substitution into simultaneous equations, solving for constants, model critique (standard limitations), and basic differentiation. All steps are routine A-level techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.06d Natural logarithm: ln(x) function and properties1.06i Exponential growth/decay: in modelling context1.07l Derivative of ln(x): and related functions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Substitute \(t = 10\), \(h = 10\) and \(t = 85\), \(h = 200\): \(10 = a + b\ln 10\); \(200 = a + b\ln 85\); Solve simultaneously to give \(b = \left[\frac{190}{\ln\frac{85}{10}} = \frac{190}{2.14}\right] = 88.8\) | M1 | Forms two equations and attempt to solve simultaneously (BC). Allow if \(10 = a + 2.303b\) and \(200 = a + 4.443b\) used |
| E1 | AG must be 3 s.f. If by simultaneous equations solved BC, \(88.78\ldots\) or better must also be seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(a = -194\) | B1 | Accept awrt \(-194\) or \(-195\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| For small values of \(t\) the model for \(h\) predicts a negative height [which is not possible] | E1 | Argument based on negativity |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| The model predicts that the sunflower would continue to increase in height for ever, which is not possible | E1 | Argument based on contrast between ever increasing height predicted by the model and reality |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Height at \(\frac{85}{2}\) days: \(a + 88.8\ln\frac{85}{2}\) cm | M1 | 3.4 |
| [using given answers above] 139 cm which is more than 1 m | E1 | 2.2a |
| Alternative method: time to reach 1 m: \(100 = a + 88.8\ln t\) | M1 | |
| [using given answers above] 27.4 days which is less than \(\frac{85}{2}\) | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Rate of growth \(\frac{dh}{dt} = \frac{b}{t}\) | M1 | 3.1b |
| Rate 3 cm per day when \(\frac{b}{t} = 3\) | M1 | 3.4 |
| \(t = \frac{b}{3} = 29.6\) | A1 | 1.1b |
## Question 11(a)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Substitute $t = 10$, $h = 10$ and $t = 85$, $h = 200$: $10 = a + b\ln 10$; $200 = a + b\ln 85$; Solve simultaneously to give $b = \left[\frac{190}{\ln\frac{85}{10}} = \frac{190}{2.14}\right] = 88.8$ | M1 | Forms two equations and attempt to solve simultaneously (BC). Allow if $10 = a + 2.303b$ and $200 = a + 4.443b$ used |
| | E1 | AG must be 3 s.f. If by simultaneous equations solved BC, $88.78\ldots$ or better must also be seen |
---
## Question 11(a)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $a = -194$ | B1 | Accept awrt $-194$ or $-195$ |
---
## Question 11(b)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| For small values of $t$ the model for $h$ predicts a negative height [which is not possible] | E1 | Argument based on negativity |
---
## Question 11(b)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| The model predicts that the sunflower would continue to increase in height for ever, which is not possible | E1 | Argument based on contrast between ever increasing height predicted by the model and reality |
## Question 11(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Height at $\frac{85}{2}$ days: $a + 88.8\ln\frac{85}{2}$ cm | M1 | 3.4 | Also allow for 84.5 days used for 85 |
| [using given answers above] 139 cm which is more than 1 m | E1 | 2.2a | Established using a value of $h$ between 137.9 and 139 needed |
| **Alternative method:** time to reach 1 m: $100 = a + 88.8\ln t$ | M1 | | Equate to 100 and solve for $t$. Condone $h=1$ used |
| [using given answers above] 27.4 days which is less than $\frac{85}{2}$ | E1 | | Established using a value of $t$ between 27 and 28 needed |
**Total: [2]**
## Question 11(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Rate of growth $\frac{dh}{dt} = \frac{b}{t}$ | M1 | 3.1b | Attempt to differentiate to give expression of the form $\frac{k}{t}$ |
| Rate 3 cm per day when $\frac{b}{t} = 3$ | M1 | 3.4 | Equates their derivative to 3 |
| $t = \frac{b}{3} = 29.6$ | A1 | 1.1b | Allow 29 or 30 days |
**Total: [3]**
---11 The height $h \mathrm {~cm}$ of a sunflower plant $t$ days after planting the seed is modelled by $\mathrm { h } = \mathrm { a } + \mathrm { b }$ Int for $t \geqslant 9$, where $a$ and $b$ are constants. The sunflower is 10 cm tall 10 days after planting and 200 cm tall 85 days after planting.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that the value of $b$ which best models these values is 88.8 correct to $\mathbf { 3 }$ significant figures.
\item Find the corresponding value of $a$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Explain why the model is not suitable for small positive values of $t$.
\item Explain why the model is not suitable for very large positive values of $t$.
\end{enumerate}\item Show that the model indicates that the sunflower grows to 1 m in height in less than half the time it takes to grow to 2 m .
\item Find the value of $t$ for which the rate of growth is 3 cm per day.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 1 2023 Q11 [10]}}