| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Model comparison/critique |
| Difficulty | Standard +0.3 This is a structured multi-part question requiring students to formulate, solve, and compare three differential equation models. While it involves multiple steps and model critique, each individual part uses standard C4 techniques (separating variables, exponential models) with clear scaffolding. The mathematical content is routine for this level, though the extended nature and real-world context add modest complexity above a typical single-technique question. |
| Spec | 1.07t Construct differential equations: in context1.08k Separable differential equations: dy/dx = f(x)g(y)1.08l Interpret differential equation solutions: in context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dV}{dt} = -k\quad (k>0)\) | B1 | D.E. |
| \(\Rightarrow V = -kt + c\) | M1 | Solving and substituting |
| When \(t=0, V=10000 \Rightarrow V = 10000 - kt\) | ||
| When \(t=5, V=5000 \Rightarrow 5k = 5000 \Rightarrow k = 1000\) | A1 | For \(c\) and \(k\) |
| \(V = 10000 - 1000t\) | ||
| When \(t=7\): \(V = 3000\) | A1 | |
| Value is £3000 after 7 years. Model breaks down since when \(t>10\) it predicts \(V\) to be negative. | B1 | |
| Total: 5 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dV}{dt} = \frac{-k}{t^{\frac{1}{2}}}\quad (k>0) = -kt^{-\frac{1}{2}}\) | B1 | D.E. |
| \(\Rightarrow V = -2kt^{\frac{1}{2}} + c\) | M1 | Solving and substituting |
| When \(t=0, V=10000 \Rightarrow c = 10000\) | ||
| When \(t=5, V=5000\): \(5000 = -2\sqrt{5}k + 10000 \Rightarrow k = 500\sqrt{5}\) | A1 | For \(c\) and \(k\) |
| \(V = 10000 - 1000\sqrt{5}\,t^{\frac{1}{2}}\) | ||
| When \(t=7\): \(V = 10000 - 1000\sqrt{35} = 4083.9\ldots\) | A1 | |
| Value is £4080 to 3 s.f. Breaks down after \(t = (2\sqrt{5})^2 = 20\) for the same reason. | B1 | |
| Total: 6 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dV}{dt} = kV \Rightarrow \int\frac{1}{V} = k\int dt\) | M1 | Solving and substituting |
| \(\Rightarrow \ln V = kt + c \Rightarrow V = Ae^{kt}\) | ||
| When \(t=0, V=10000 \Rightarrow A = 10000\) | A1 | For \(A\) |
| When \(t=5, V=5000\): \(5k = \ln\frac{1}{2} \Rightarrow k = -\frac{1}{5}\ln 2\) | A1 | For \(k\) |
| \(\Rightarrow V = 10000\left(\frac{1}{2}\right)^{\frac{t}{5}}\) | A1, A1 | |
| Value after 7 years is £3790 (to 3 s.f.). This does not break down (it just tends to zero). | B1 | |
| Total: 6 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Predicted values (approximately): model 1: £2000; model 2: £3680 (3675.4…); model 3: £3300 (3298.7…) | B1 | All 3 |
| Model 3 works best. | B1 | |
| Total: 2 marks |
## Question 8(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dV}{dt} = -k\quad (k>0)$ | B1 | D.E. |
| $\Rightarrow V = -kt + c$ | M1 | Solving and substituting |
| When $t=0, V=10000 \Rightarrow V = 10000 - kt$ | | |
| When $t=5, V=5000 \Rightarrow 5k = 5000 \Rightarrow k = 1000$ | A1 | For $c$ and $k$ |
| $V = 10000 - 1000t$ | | |
| When $t=7$: $V = 3000$ | A1 | |
| Value is £3000 after 7 years. Model breaks down since when $t>10$ it predicts $V$ to be negative. | B1 | |
| **Total: 5 marks** | | |
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## Question 8(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dV}{dt} = \frac{-k}{t^{\frac{1}{2}}}\quad (k>0) = -kt^{-\frac{1}{2}}$ | B1 | D.E. |
| $\Rightarrow V = -2kt^{\frac{1}{2}} + c$ | M1 | Solving and substituting |
| When $t=0, V=10000 \Rightarrow c = 10000$ | | |
| When $t=5, V=5000$: $5000 = -2\sqrt{5}k + 10000 \Rightarrow k = 500\sqrt{5}$ | A1 | For $c$ and $k$ |
| $V = 10000 - 1000\sqrt{5}\,t^{\frac{1}{2}}$ | | |
| When $t=7$: $V = 10000 - 1000\sqrt{35} = 4083.9\ldots$ | A1 | |
| Value is £4080 to 3 s.f. Breaks down after $t = (2\sqrt{5})^2 = 20$ for the same reason. | B1 | |
| **Total: 6 marks** | | |
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## Question 8(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dV}{dt} = kV \Rightarrow \int\frac{1}{V} = k\int dt$ | M1 | Solving and substituting |
| $\Rightarrow \ln V = kt + c \Rightarrow V = Ae^{kt}$ | | |
| When $t=0, V=10000 \Rightarrow A = 10000$ | A1 | For $A$ |
| When $t=5, V=5000$: $5k = \ln\frac{1}{2} \Rightarrow k = -\frac{1}{5}\ln 2$ | A1 | For $k$ |
| $\Rightarrow V = 10000\left(\frac{1}{2}\right)^{\frac{t}{5}}$ | A1, A1 | |
| Value after 7 years is £3790 (to 3 s.f.). This does not break down (it just tends to zero). | B1 | |
| **Total: 6 marks** | | |
---
## Question 8(iv):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Predicted values (approximately): model 1: £2000; model 2: £3680 (3675.4…); model 3: £3300 (3298.7…) | B1 | All 3 |
| Model 3 works best. | B1 | |
| **Total: 2 marks** | | |
---8 The new price of a particular make of car is $\pounds 10000$. When its age is $t$ years, the list price is $\pounds V$. When $t = 5 , V = 5000$.
Aloke, Ben and Charlie all run outlets for used cars. Each of them has a different model for the depreciation.\\
(i) Aloke claims that the rate of depreciation is constant. Write this claim as a differential equation.\\
Solve the differential equation and hence find the value of a car that is 7 years old according to this model.\\
Explain why this model breaks down for large $t$.\\
(ii) Ben believes that the rate of depreciation is inversely proportional to the square root of the age of the car. Express this claim as a differential equation and hence find the value of a car that is 7 years old according to this model.\\
Does this model ever break down?\\
(iii) Charlie believes that a better model is given by the differential equation
$$\frac { \mathrm { d } V } { \mathrm {~d} t } = k V$$
Solve this differential equation and find the value of the car after 7 years according to this model.\\
Does this model ever break down?\\
(iv) Further investigation reveals that the average value of this particular type of car when 8 years old is $\pounds 3000$.
Find the value of $V$ when $t = 8$ for the three models above. Which of the three models best predicts the value of $V$ at this time?
\hfill \mbox{\textit{OCR MEI C4 Q8 [19]}}