# Question 12(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{V(25-V)} = \frac{P}{V} + \frac{Q}{25-V}$, e.g. $1 = P(25-V) + QV$ | M1 | Sets up partial fractions and uses correct method to find at least one constant; do **not** condone $1 = PV + Q(25-V)$ |
| $\frac{1}{V(25-V)} = \frac{1}{25V} + \frac{1}{25(25-V)}$ | A1 | Correct partial fractions in any form; mark not just for constants but correctly written fractions, seen in (a) **or used in** (b) |
| **(2 marks)** | | |

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# Question 12(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \frac{1}{V(25-V)}\,dV = \frac{1}{25}\ln V - \frac{1}{25}\ln(25-V)$ | M1 | Reaches form $p\ln\alpha V \pm q\ln\beta(25-V)$; must attempt partial fractions first |
| $\frac{1}{25}\ln V - \frac{1}{25}\ln(25-V) = \frac{1}{10}t\,(+c)$ | A1ft | Fully correct equation following their $P$ and $Q$; "$+c$" not required here |
| $t=0, V=20 \Rightarrow \frac{1}{25}\ln 20 - \frac{1}{25}\ln 5 = c \Rightarrow c = \frac{1}{25}\ln 4$ | M1 | Substitutes $t=0$, $V=20$ to find constant |
| $V=24 \Rightarrow t = \frac{2}{5}\ln 24 - \frac{2}{5}\ln 1 - \frac{2}{5}\ln 4$ | dM1 | Substitutes $V=24$ into their equation with found constant |
| $= 43$ (or exact $24\ln 6$) | A1 | |
| **(5 marks)** | | |

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# Question 12(c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\ln V - \ln(25-V) = 2.5t + \ln 4$ | M1 | Multiplies through by 25 and combines log terms correctly |
| $\ln\frac{V}{4(25-V)} = 2.5t \Rightarrow \frac{V}{4(25-V)} = e^{2.5t}$ | | |
| $V = 4e^{2.5t}(25-V) \Rightarrow V + 4Ve^{2.5t} = 100e^{2.5t}$ | M1 | Rearranges to make $V$ the subject |
| $V = \frac{100e^{2.5t}}{1+4e^{2.5t}} = \frac{100}{e^{-2.5t}+4}$ | A1 | Correct expression for $V$ |
| **(3 marks)** | | |

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# Question 12(d):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $25$ (microlitres) | B1 | |
| Since as $t \to \infty$, $e^{-2.5t} \to 0$ | B1 | Correct reasoning |
| **(2 marks)** | | |

# Mark Scheme Extraction

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## Question 12 (Differential Equation - continued):

### Part (b) - Integration:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{25}\ln V - \frac{1}{25}\ln(625-25V) = \frac{1}{10}t(+c)$ | M1A1 | Allow $( )$ or $\| \|$ around ln arguments; condone "log" for ln |
| $\ln V - \ln(25-V) = 2.5t(+c)$ | | Equivalent forms accepted |
| $\frac{2}{5}\ln 25V - \frac{2}{5}\ln(25-V) = t(+c)$ | | Multiple forms acceptable |
| General form: $P\ln\alpha V - Q\ln\beta(25-V) = \frac{1}{10}t(+c)$ | | Do **not** condone missing brackets unless implied by later work |
| Uses $t=0$, $V=20$ to find constant of integration | M1 | Not formally dependent but needs some attempt to integrate both sides |
| Uses $V=24$ to find value for $t$ | dM1 | Depends on previous method mark; depends on attempt to integrate both sides |
| $t = 43$ or awrt $43.0$ or exact value $24\ln 6$ | A1 | Units not required; if given must be minutes; note in hours $t = 0.7167...$ scores A0 |

**Alternative for final 3 marks:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left[\frac{1}{25}\ln V - \frac{1}{25}\ln(25-V)\right]_{20}^{24} = \left[\frac{1}{10}t\right]_0^T \Rightarrow \frac{1}{25}\ln 24 - \frac{1}{25}\ln 4 = \frac{1}{10}T$ | M1 | Applies limits 20 and 24 to lhs and 0 to "$T$" on rhs |
| $T = \frac{10}{25}\ln 24 - \frac{10}{25}\ln 4 = ...$ Solves to find $T$ | dM1 | Depends on previous method mark |
| $t = 43$ or awrt $43.0$ or exact value $24\ln 6$ | A1 | Units not required; if given must be minutes |

**Note:**
$$\int \frac{1}{25V} + \frac{1}{25(25-V)}\,dV = \int \frac{1}{25V} - \frac{1}{25V-625}\,dV = \frac{1}{25}\ln V - \frac{1}{25}\ln(25V-625) = \frac{1}{10}t(+c)$$
is also correct integration and scores M1A1.

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### Part (c) - Eliminating logarithms:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses fully correct log work to eliminate all ln's including from e.g. $e^{\ln 4}$ | M1 | Condone sign or coefficient slips only |
| Proceeds from equation of form $\frac{...V}{...(25-V)} = ...e^{-t}$ using correct algebra: $V = ...$ e.g. $\frac{...V}{...(25-V)} = ...e^{-t} \Rightarrow ...V = ...(25-V)...e^{-t} \Rightarrow (...\pm...)V = ...e^{-t} \Rightarrow V = ...$ | M1 | Condone sign/coefficient slips only |
| Correct expression (not just values for constants) | A1 | |

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### Part (d) - Long-term value:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct value of 25 seen | B1 | Allow e.g. $<25$ or $\leq 25$; condone $>25$ but following mark not available |
| Depends on correct final equation in (c) e.g. $V = \frac{100e^{2.5t}}{1+4e^{2.5t}}$; considers behaviour as $t\to\infty$, e.g. as $t\to\infty$, $e^{-2.5t}\to 0$ | B1 | **and one of:** |
| OR: $V < 25$ as $\ln(25-V)$ not possible when $V \geq 25$ | | |
| OR: Verifies the 25 using a value of $t$ | | |

**Using the differential equation:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct value of 25 seen | B1 | Allow $<25$ or $\leq 25$; condone $>25$ but following mark unavailable |
| When $V=25$, $\frac{dV}{dt}=0$ **or** $\frac{dV}{dt}<0$ if $V>25$ | B1 | |

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