## Question 14:

### Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\{u = 4 - \sqrt{h}\} \Rightarrow \frac{du}{dh} = -\frac{1}{2}h^{-\frac{1}{2}}$ or $\frac{dh}{du} = -2(4-u)$ or $\frac{dh}{du} = -2\sqrt{h}$ | B1 | Allow $du = -\frac{1}{2}h^{-\frac{1}{2}}dh$, $dh = -2(4-u)du$, $dh = -2\sqrt{h}\,du$ |
| $\left\{\int \frac{dh}{4-\sqrt{h}} =\right\} \int \frac{-2(4-u)}{u}\,du$ | M1 | Complete method for applying $u = 4-\sqrt{h}$; must give form $\int \frac{k(4-u)}{u}\,du,\ k\neq 0$ |
| $= \int\left(-\frac{8}{u} + 2\right)du$ | M1 | Proceeds to obtain integral of form $\int\left(\frac{A}{u}+B\right)\{du\};\ A,B \neq 0$ |
| $= -8\ln u + 2u\ \{+c\}$ | M1, A1 | $\int\left(\frac{A}{u}+B\right)\{du\} \to D\ln u + Eu;\ A,B,D,E\neq 0$ |
| $= -8\ln\|4-\sqrt{h}\| + 2(4-\sqrt{h}) + c = -8\ln\|4-\sqrt{h}\| - 2\sqrt{h} + k$ | A1* | Dependent on all previous marks; substitutes $u=4-\sqrt{h}$; must combine $2(4)$ and $+c$ to give $+k$; condone brackets instead of modulus |

### Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left\{\frac{dh}{dt} = \frac{t^{0.25}(4-\sqrt{h})}{20} = 0\right\} \Rightarrow 4 - \sqrt{h} = 0$ | M1 | Uses model context; tree grows until $\frac{dh}{dt}=0$; accept $\frac{dh}{dt}>0 \Rightarrow 4-\sqrt{h}>0$ |
| Deduces any of $0<h<16$, $0\leq h<16$, $0<h\leq 16$, $0\leq h\leq 16$, $h<16$, $h\leq 16$ or all values up to 16 | A1 | Accept $h=16$ or $16$ used in inequality statement |

### Part (c) — Way 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \frac{1}{(4-\sqrt{h})}\,dh = \int \frac{1}{20}t^{0.25}\,dt$ | B1 | Separates variables correctly; condone absence of integral signs |
| $-8\ln\|4-\sqrt{h}\| - 2\sqrt{h} = \frac{1}{25}t^{1.25}\ \{+c\}$ | M1, A1 | Integrates $t^{0.25}$ to give $\lambda t^{1.25},\ \lambda\neq 0$; correct integration |
| $\{t=0, h=1\} \Rightarrow -8\ln(4-1) - 2\sqrt{1} = \frac{1}{25}(0)^{1.25} + c$ | M1 | Some evidence of applying $t=0$ and $h=1$ to model containing constant of integration |
| $\Rightarrow c = -8\ln(3)-2 \Rightarrow -8\ln\|4-\sqrt{h}\| - 2\sqrt{h} = \frac{1}{25}t^{1.25} - 8\ln(3) - 2$ | dM1 | Dependent on previous M; complete process finding constant then applying $h=12$ |
| $\{h=12\} \Rightarrow -8\ln\|4-\sqrt{12}\| - 2\sqrt{12} = \frac{1}{25}t^{1.25} - 8\ln(3) - 2$ | | |
| $t^{1.25} = 221.2795202... \Rightarrow t = \sqrt[1.25]{221.2795...}$ or $t = (221.2795...)^{0.8}$ | M1 | Rearranges to make $t^{\text{their }1.25} = ...$; correct method to give $t=...;\ t>0$ |
| $t = 75.154... \Rightarrow t = 75.2\ \text{(years)}\ (3\text{ sf})$ or awrt $75.2\ \text{(years)}$ | A1 | |

### Part (c) — Way 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_1^{12} \frac{20}{(4-\sqrt{h})}\,dh = \int_0^T t^{0.25}\,dt$ | B1 | Separates variables correctly with limits |
| $\left[20(-8\ln\|4-\sqrt{h}\| - 2\sqrt{h})\right]_1^{12} = \left[\frac{4}{5}t^{1.25}\right]_0^T$ | M1, A1 | Same as Way 1 (ignore limits) |
| $20(-8\ln(4-\sqrt{12})-2\sqrt{12}) - 20(-8\ln(4-1)-2\sqrt{1}) = \frac{4}{5}T^{1.25} - 0$ | M1, dM1 | Applies limits 1 and 12 to $h$-expression and 0 and $T$ appropriately |
| $T^{1.25} = 221.2795202... \Rightarrow T = \sqrt[1.25]{221.2795...}$ or $T=(221.2795...)^{0.8}$ | M1 | Same as Way 1 |
| $T = 75.154... \Rightarrow T = 75.2\ \text{(years)}\ (3\text{ sf})$ or awrt $75.2\ \text{(years)}$ | A1 | |