| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Chemical reaction kinetics |
| Difficulty | Standard +0.3 This is a standard C4 differential equation question with clear scaffolding through partial fractions, separation of variables, and parameter finding. While it involves multiple steps, each part follows routine techniques (partial fractions, separating variables, substituting boundary conditions) with no novel insight required. The context is applied but the mathematics is textbook-standard, making it slightly easier than average. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((1 + 2(0.75)) / (1 - 0.75) = e^{3k}\) | M1 | Substituting \(t = 1\), \(x = 0.75\) at any stage |
| \(k = \frac{1}{3}\ln 10 (= 0.768\) (3 s.f.)) | A1 | 3sf or better |
| \(t = \ln(2.8/0.1)/3k = 1.45\) hours | A1 | 1.45 (or better) or 1 hr 27 mins |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1 + 2x = e^{3kt} - xe^{3kt}\) | ||
| \(\Rightarrow 2x + xe^{3kt} = e^{3kt} - 1\) | M1* | Multiplying out and collecting \(x\) terms (condone one error) |
| \(\Rightarrow x(2 + e^{3kt}) = e^{3kt} - 1\) | M1dep* | Factorising their \(x\) terms correctly |
| \(\Rightarrow x = (e^{3kt}-1)/(2 + e^{3kt})\) | A1 | |
| \(= (1 - e^{-3kt})/(1 + 2e^{-3kt})\) * | A1 | www (AG) — must indicate how previous line leads to result (e.g. stating or showing multiplying by \(e^{-3kt}\)) |
| When \(t \to \infty\), \(e^{-3kt} \to 0\), \(x = (1-e^{-3kt})/(1+2e^{-3kt}) \to 1/1 = 1\) | B1 | Clear indication that \(e^{-3kt} \to 0\); accept as minimum \(x \to \frac{1-0}{1+0} = 1\) or \(e^{-3kt} \to 0 \Rightarrow x \to 1\). Substitution of large values of \(t\) with no further explanation is B0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1-x}{1+2x} = e^{-3kt}\) | B1 | |
| \(1 - x = e^{-3kt} + 2xe^{-3kt}\) | M1* | Multiplying up and expanding (condone one error) |
| \(x(1 + 2e^{-3kt}) = 1 - e^{-3kt}\) | M1dep* | Factorising their \(x\) terms correctly |
| \(x = (1 - e^{-3kt})/(1 + 2e^{-3kt})\) * | A1 | www (AG) — final B mark as in scheme above |
## Question 1:
### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(1 + 2(0.75)) / (1 - 0.75) = e^{3k}$ | M1 | Substituting $t = 1$, $x = 0.75$ at any stage |
| $k = \frac{1}{3}\ln 10 (= 0.768$ (3 s.f.)) | A1 | 3sf or better |
| $t = \ln(2.8/0.1)/3k = 1.45$ hours | A1 | 1.45 (or better) or 1 hr 27 mins |
**[3 marks]**
### Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1 + 2x = e^{3kt} - xe^{3kt}$ | | |
| $\Rightarrow 2x + xe^{3kt} = e^{3kt} - 1$ | M1* | Multiplying out and collecting $x$ terms (condone one error) |
| $\Rightarrow x(2 + e^{3kt}) = e^{3kt} - 1$ | M1dep* | Factorising their $x$ terms correctly |
| $\Rightarrow x = (e^{3kt}-1)/(2 + e^{3kt})$ | A1 | |
| $= (1 - e^{-3kt})/(1 + 2e^{-3kt})$ * | A1 | www **(AG)** — must indicate how previous line leads to result (e.g. stating or showing multiplying by $e^{-3kt}$) |
| When $t \to \infty$, $e^{-3kt} \to 0$, $x = (1-e^{-3kt})/(1+2e^{-3kt}) \to 1/1 = 1$ | B1 | Clear indication that $e^{-3kt} \to 0$; accept as minimum $x \to \frac{1-0}{1+0} = 1$ or $e^{-3kt} \to 0 \Rightarrow x \to 1$. Substitution of large values of $t$ with no further explanation is B0 |
**[5 marks]**
**OR:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1-x}{1+2x} = e^{-3kt}$ | B1 | |
| $1 - x = e^{-3kt} + 2xe^{-3kt}$ | M1* | Multiplying up and expanding (condone one error) |
| $x(1 + 2e^{-3kt}) = 1 - e^{-3kt}$ | M1dep* | Factorising their $x$ terms correctly |
| $x = (1 - e^{-3kt})/(1 + 2e^{-3kt})$ * | A1 | www **(AG)** — final B mark as in scheme above |
---1 A drug is administered by an intravenous drip. The concentration, $x$, of the drug in the blood is measured as a fraction of its maximum level. The drug concentration after $t$ hours is modelled by the differential equation
$$\frac { \mathrm { d } x } { \mathrm {~d} t } = k \left( 1 + x - 2 x ^ { 2 } \right)$$
where $0 \leqslant x < 1$, and $k$ is a positive constant. Initially, $x = 0$.\\
(i) Express $\frac { 1 } { ( 1 + 2 x ) ( 1 - x ) }$ in partial fractions.\\[0pt]
[3]\\
(ii) Hence solve the differential equation to show that $\frac { 1 + 2 x } { 1 - x } = \mathrm { e } ^ { 3 k t }$.\\
(iii) After 1 hour the drug concentration reaches $75 \%$ of its maximum value and so $x = 0.75$.
Find the value of $k$, and the time taken for the drug concentration to reach $90 \%$ of its maximum value.\\
(iv) Rearrange the equation in part (ii) to show that $x = \frac { 1 - \mathrm { e } ^ { - 3 k t } } { 1 + 2 \mathrm { e } ^ { - 3 k t } }$.
Verify that in the long term the drug concentration approaches its maximum value.
\hfill \mbox{\textit{OCR MEI C4 Q1 [18]}}