## Question 9:

### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $u = 4 - \sqrt{x} \Rightarrow x = (4-u)^2 \Rightarrow \frac{dx}{du} = -2(4-u)$ | M1, A1 | |
| $\int \frac{dx}{4-\sqrt{x}} = \int \frac{-2(4-u)\,du}{u} = \int \left(-\frac{8}{u} + 2\right)du$ | M1, A1 | |
| $= -8\ln u + 2u \; (+c)$ | dM1 | |
| $= -8\ln\left|4-\sqrt{x}\right| + 2\left(4-\sqrt{x}\right)(+c)$ | A1 | oe |

### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Height increases when $\frac{dh}{dt} = \frac{4-\sqrt{h}}{20} > 0 \Rightarrow (0 <)\, h < 16$ | M1, A1 | |

### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dh}{dt} = \frac{4-\sqrt{h}}{20} \Rightarrow \int \frac{dh}{4-\sqrt{h}} = \int \frac{dt}{20}$ | B1 | Separates variables correctly |
| $\Rightarrow -8\ln(4-\sqrt{h}) + 2(4-\sqrt{h}) = \frac{t}{20} + c$ | M1, A1 | Uses result from part (a) |
| Substitute $t=0, h=1$: $-8\ln 3 + 6 = c$ | dM1 | |
| $\Rightarrow -8\ln(4-\sqrt{h}) + 2(4-\sqrt{h}) = \frac{t}{20} - 8\ln 3 + 6$ | A1 | oe |
| Substitute $h = 10$: $-8\ln(4-\sqrt{10}) + 2(4-\sqrt{10}) = \frac{t}{20} - 8\ln 3 + 6$ | ddM1 | Dependent on both previous M marks |
| $\Rightarrow t =$ awrt $118$ (years) | A1 | |

# Question 9:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Write $x$ in terms of $u$ AND differentiate to get $dx$ in terms of $du$ or $\frac{dx}{du}$ | M1 | Accept incorrect expressions like $x = 16-u^2 \Rightarrow \frac{dx}{du} = -2u$; minimum expectation is expression in $u$ is quadratic and derivative in $u$ is linear |
| $\frac{dx}{du} = -2(4-u)$ or $dx = -2(4-u)du$ or equivalents. Condone $dx = -8+2u$ | A1 | Accept within the integral for both marks as long as no incorrect working seen |
| Attempt to divide their $dx$ by $u$ to get integral of form $\int \frac{A}{u} + B\, du$ | M1 | |
| Fully correct integral in terms of $u$: e.g. $\int -\frac{8}{u} + 2\,du$ or $\int -8u^{-1} + 2\,du$ or $-2\left(\int \frac{4}{u}-1\right)$ | A1 | Condone omission of $du$ if intention is clear |
| Integrating $\frac{1}{u} \rightarrow \ln u$ and increasing power of any other term | dM1 | Dependent on previous M1; no need to set answer in terms of $x$; no need for $+c$ |
| $-8\ln\left\|4-\sqrt{x}\right\| + 2\left(4-\sqrt{x}\right)$ $(+c)$ or $-8\ln\left(4-\sqrt{x}\right) - 2\sqrt{x}$ $(+c)$ | A1 | No requirement for modulus signs |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Setting $\frac{dh}{dt} = 0 \Rightarrow h = \ldots$ or $\frac{dh}{dt} > 0 \Rightarrow h < \ldots$ Accept $h=16$ | M1 | |
| $h < 16$ or $0 < h < 16$ or $0 \leq h < 16$ or all values up to 16 | A1 | Correct answer can score both marks as long as no incorrect working seen |

## Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \frac{dh}{4-\sqrt{h}} = \int \frac{dt}{20}$ or equivalent | B1 | Must include $dh$ and $dt$; $\int$ could be implied |
| Attempt to integrate both sides, no need for $c$ | M1 | Follow through on their answer to part (a) for $x$ or $u$ with $h$ on lhs and $At$ on rhs |
| $-8\ln\left(4-\sqrt{h}\right) + 2\left(4-\sqrt{h}\right) = \frac{t}{20} + c$ | A1 | Fully correct answer with $+c$ |
| Substitute $t=0$, $h=1$ to find $c$ | dM1 | Minimal evidence required; accept $t=0, h=1 \Rightarrow c = \ldots$; previous M must have been awarded |
| $-8\ln\left(4-\sqrt{h}\right) + 2\left(4-\sqrt{h}\right) = \frac{t}{20} - 8\ln 3 + 6$ | A1 | Accept $\Rightarrow -8\ln\left(4-\sqrt{h}\right)+2\left(4-\sqrt{h}\right) = \frac{t}{20} + \text{awrt } 2.79$ |
| Substitute $h=10$ into equation involving $h$, $t$ and their value of $c$ to find $t$ | ddM1 | Dependent on both M marks; accept minimal evidence |
| Awrt 118 (years) | A1 | Answer without any correct working scores 0 marks; condone $x$ and $h$ interchanged |

### Part (c) Alternative:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Setting limits is equivalent to understanding there is a constant | A1 | |
| Using limits of 1 and 10 and subtracting | dM1 | |
| A fully correct expression involving just $t$ | A1 | |
| Using limits of $t$ and 0 and subtracting | ddM1 | |
| Awrt 118 (years). No working = 0 marks | A1 | |

---