## Question 16(a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{(4-3x)(4+3x)} \equiv \frac{A}{4-3x} + \frac{B}{4+3x}$; uses suitable method to find $A$ or $B$ | M1 (1.1a) | Rearranges and substitutes values, or compares coefficients, or cover-up method, or inspection; PI by $A$ correct or $B$ correct |
| $A = \frac{1}{8}$ | A1 (1.1b) | |
| $B = \frac{1}{8}$ | A1 (1.1b) | |

**Subtotal: 3 marks**

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## Question 16(b)(i):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dV}{dt} = 0.16 - 0.36d^2$ | M1 (3.3) | Forms differential equation using $\frac{dV}{dt} = \pm 0.16 \pm 0.36d^2$ |
| $V = 1.25 \times 1.6d \Rightarrow d = \frac{V}{2}$ | B1 (3.1b) | Obtains $V = 1.25 \times 1.6d$; OE |
| $\frac{dV}{dt} = 0.16 - 0.36\left(\frac{V}{2}\right)^2 = 0.16 - 0.09V^2 = \frac{16 - 9V^2}{100}$ | M1 (3.1a) | Substitutes expression for $d$ into $\frac{dV}{dt} = \pm 0.16 \pm 0.36d^2$ to obtain differential equation in $V$ and $t$ only |
| Completes reasoned argument to show given result | R1 (2.1) | AG |

**Subtotal: 4 marks**

# Question 16(b)(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Rearranges to obtain one of: $\frac{P}{16-9V^2}dV = \frac{1}{Q}dt$ or $\frac{P}{16-9V^2}\frac{dV}{dt} = \frac{1}{Q}$ or $\frac{P}{16-9V^2} = \frac{1}{Q}\frac{dt}{dV}$ | B1 | 3.1a - where $P \times Q = 100$. If $P = 100$ no need to see $\frac{1}{Q}$ explicit with $dt$. May include integral signs: $PI\int\frac{100}{16-9V^2}dV = t$ |
| Integrates their constant integrand correctly with respect to $t$, follow through any constant | B1F | 1.1b |
| Writes $\int\frac{1}{16-9V^2}dV$ as $\int\frac{A}{4-3V}+\frac{B}{4+3V}dV$. Condone missing $dV$. PI by $-\frac{A}{3}\ln(4-3V)+\frac{B}{3}\ln(4+3V)$ | M1 | 3.1a |
| Integrates their partial fractions correctly to obtain $-\frac{A}{3}\ln(4-3V)+\frac{B}{3}\ln(4+3V)$ $(+c)$ OE. Their $A$ and $B$ may be correctly inside the natural logs, e.g. $\frac{1}{24}(-\ln(32-24V)+\ln(32+24V))$ | A1F | 1.1b |
| Typical solution: $\frac{1}{8}\int\frac{1}{4-3V}+\frac{1}{4+3V}dV = \int\frac{1}{100}dt$ giving $\frac{1}{24}(-\ln(4-3V)+\ln(4+3V)) = \frac{t}{100}+c$, then $t=0, V=0 \Rightarrow c=0$, giving $\frac{100}{24}(-\ln(4-3V)+\ln(4+3V))=t$ | R1 | 2.1 - Completes argument, including demonstrating that the constant of integration is zero |
| **Subtotal** | **5** | |

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# Question 16(b)(iii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Obtains a value for $t$ by substituting $V=1$ into their expression for $t$ from final answer in b(ii). $t = \frac{100}{24}(-\ln(4-3)+\ln(4+3)) = \frac{100}{24}\ln 7$ | M1 | 3.4 - PI by 8 minutes from a correct answer from b(ii) |
| $= 8$ minutes | A1 | 3.2a - Condone missing units |
| **Subtotal** | **2** | |
| **Question 16 Total** | **14** | |
| **Question Paper Total** | **100** | |