## Question 14:

**Part (a):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $N = 90$ when $t = 0$ | B1 | AO3.4 |

**Part (b) – Way 1:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dN}{dt} = -900(3+7e^{-0.25t})^{-2}(7(-0.25)e^{-0.25t})$ $\left\{=\frac{900(0.25)(7)e^{-0.25t}}{(3+7e^{-0.25t})^2}\right\}$ | M1 | AO2.1 – attempts chain/quotient rule or implicit differentiation |
| Correct differentiation statement | A1 | AO1.1b |
| $\Rightarrow \frac{dN}{dt} = \frac{900(0.25)\left(\frac{900}{N}-3\right)}{\left(\frac{900}{N}\right)^2}$ | dM1 | AO2.1 – complete attempt eliminating $t$ to link $\frac{dN}{dt}$ and $N$ only |
| Correct algebra leading to $\frac{dN}{dt} = \frac{N(300-N)}{1200}$ * | A1* | AO1.1b |

**Part (b) – Way 2:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dN}{dt} = -900(3+7e^{-0.25t})^{-2}(7(-0.25)e^{-0.25t})$ $\left\{=\frac{900(0.25)(7)e^{-0.25t}}{(3+7e^{-0.25t})^2}\right\}$ | M1 | AO2.1 |
| Correct differentiation statement | A1 | AO1.1b |
| $\frac{N(300-N)}{1200} = \frac{\left(\frac{900}{3+7e^{-0.25t}}\right)\left(300-\frac{900}{3+7e^{-0.25t}}\right)}{1200}$ | dM1 | AO2.1 – complete substitution of $N=\frac{900}{3+7e^{-0.25t}}$ into RHS |
| LHS $= \frac{1575e^{-0.25t}}{(3+7e^{-0.25t})^2}$; RHS $= \frac{900(300(3+7e^{-0.25t})-900)}{1200(3+7e^{-0.25t})^2} = \frac{1575e^{-0.25t}}{(3+7e^{-0.25t})^2}$ and states LHS = RHS | A1* | AO1.1b |

**Part (b) – Way 3:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int\frac{1}{N(300-N)}dN = \int\frac{1}{1200}dt$ | M1 | AO2.1 – separates variables, partial fractions, integrates to give ln terms |
| $\frac{1}{300}\ln N - \frac{1}{300}\ln(300-N) = \frac{1}{1200}t\ \{+c\}$ | A1 | AO1.1b |
| Uses $t=0, N=90$ to find constant; obtains $\lambda e^{\frac{1}{4}t} = f(N)$ | dM1 | AO2.1 |
| Correct manipulation leading to $N = \frac{900}{3+7e^{-0.25t}}$ * | A1* | AO1.1b |

**Part (b) – Way 4:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $e^{-0.25t} = \frac{1}{7}\left(\frac{900}{N}-3\right) \Rightarrow e^{-0.25t} = \frac{900-3N}{7N}$; $\Rightarrow t = -4(\ln(900-3N)-\ln(7N))$ | M1 | AO2.1 – valid attempt to make $t$ subject, find two ln derivatives |
| $\frac{dt}{dN} = -4\left(\frac{-3}{900-3N}-\frac{7}{7N}\right)$ | A1 | AO1.1b |
| Forms common denominator: $\frac{dt}{dN} = 4\left(\frac{1}{300-N}+\frac{1}{N}\right) \Rightarrow \frac{dt}{dN} = 4\left(\frac{N+300-N}{N(300-N)}\right)$ | dM1 | AO2.1 |
| $\frac{dt}{dN} = \left(\frac{1200}{N(300-N)}\right) \Rightarrow \frac{dN}{dt} = \frac{N(300-N)}{1200}$ * | A1* | AO1.1b |

**Part (c):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Deduces $N=150$ (can be implied) | B1 | AO2.2a – $\frac{dN}{dt}$ maximised when $N=150$; note: $\frac{d^2N}{dt^2}=\frac{dN}{dt}\left(\frac{300}{1200}-\frac{2N}{1200}\right)=0 \Rightarrow N=150$ acceptable |
| $150 = \frac{900}{3+7e^{-0.25T}} \Rightarrow e^{-0.25T} = \frac{3}{7}$ | M1 | AO3.4 – uses model with $N=150$, proceeds to $e^{-0.25T}=k,\ k>0$ |
| $T = -4\ln\left(\frac{3}{7}\right)$ | dM1 | AO1.1b – correct use of logarithms |
| $T = -4\ln\left(\frac{3}{7}\right)$ or $T \approx$ awrt 3.4 (months) | A1 | AO1.1b |

**Notes for (c):** Ignore units for $T$. Applying $300=\frac{900}{3+7e^{-0.25t}}$ or $0=\frac{900}{3+7e^{-0.25t}}$ is M0 dM0 A0. M1 dM1 can only be gained using $N$ in range $90 < N < 300$.

**Part (d):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Either 299 or 300 | B1 | AO3.4 – accept 299 |

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