| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2021 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Separable variables - partial fractions |
| Difficulty | Standard +0.3 This is a standard P4/FP2 separable differential equations question with routine partial fractions. Part (a) is straightforward decomposition, part (b) follows the standard method (separate, integrate, apply initial conditions), part (c) is algebraic manipulation, and part (d) requires understanding limits. While multi-step, each component uses well-practiced techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08k Separable differential equations: dy/dx = f(x)g(y) |
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| Q10 | |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{(H-5)(H+3)} = \frac{A}{H-5} + \frac{B}{H+3} \Rightarrow A = \ldots\) or \(B = \ldots\) | M1 | Implied by one correct constant |
| \(A = \frac{1}{8}\) or \(B = -\frac{1}{8}\) | A1 | |
| \(\frac{1}{(H-5)(H+3)} = \frac{1}{8(H-5)} - \frac{1}{8(H+3)}\) | A1 | Correct partial fractions; mark for correctly stated fractions in (a) or used in (b) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int\frac{40}{(H-5)(H+3)}dH = \int -1 \, dt\) | M1 | Separates variables and uses part (a) |
| \(5\ln | H-5 | - 5\ln |
| Substitutes \(t=0, H=13 \Rightarrow c = \ldots\) | M1 | Must have \(a+c\); depends on some attempt at integration |
| \(5\ln | H-5 | - 5\ln |
| \(5\ln\left(2\left | \frac{H-5}{H+3}\right | \right) = -t \Rightarrow \frac{H-5}{H+3} = \frac{1}{2}e^{-0.2t}\) |
| \(H = \dfrac{10 + 3e^{-0.2t}}{2 - e^{-0.2t}}\) | A1* | cso with sufficient working shown and no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Sets \(\frac{10+3e^{-0.2t}}{2-e^{-0.2t}} = 8 \Rightarrow e^{-0.2t} = \left(\frac{6}{11}\right)\) | M1 | Or possibly earlier version of \(H\) |
| \(t = -5\ln\left(\frac{6}{11}\right) = \) awrt 3.03 days | dM1 A1 | Correct log work leading to \(t\); depends on first M |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(k = 5\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(t = -5\ln\left(\dfrac{6}{11}\right)\) or \(t = 5\ln\left(\dfrac{11}{6}\right)\) or awrt 3.03 (days) | A1 | Depends on the first M mark |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(k = 5\) | B1 | Allow \(H = 5\) or just "5" |
# Question 10:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{(H-5)(H+3)} = \frac{A}{H-5} + \frac{B}{H+3} \Rightarrow A = \ldots$ or $B = \ldots$ | M1 | Implied by one correct constant |
| $A = \frac{1}{8}$ or $B = -\frac{1}{8}$ | A1 | |
| $\frac{1}{(H-5)(H+3)} = \frac{1}{8(H-5)} - \frac{1}{8(H+3)}$ | A1 | Correct partial fractions; mark for correctly stated fractions in (a) **or** used in (b) |
**(3 marks)**
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int\frac{40}{(H-5)(H+3)}dH = \int -1 \, dt$ | M1 | Separates variables and uses part (a) |
| $5\ln|H-5| - 5\ln|H+3| = -t(+c)$ | M1 A1ft | Integrates both sides; condone omission of $+c$ if implied by subsequent work |
| Substitutes $t=0, H=13 \Rightarrow c = \ldots$ | M1 | Must have $a+c$; depends on some attempt at integration |
| $5\ln|H-5| - 5\ln|H+3| = -t + 5\ln\left(\frac{1}{2}\right)$ | A1 | Correct equation in $H$ and $t$ |
| $5\ln\left(2\left|\frac{H-5}{H+3}\right|\right) = -t \Rightarrow \frac{H-5}{H+3} = \frac{1}{2}e^{-0.2t}$ | dddM1 | Correct attempt to make $H$ subject; **all previous M's in (b) must have been scored** |
| $H = \dfrac{10 + 3e^{-0.2t}}{2 - e^{-0.2t}}$ | A1* | cso with sufficient working shown and no errors |
**(7 marks)**
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sets $\frac{10+3e^{-0.2t}}{2-e^{-0.2t}} = 8 \Rightarrow e^{-0.2t} = \left(\frac{6}{11}\right)$ | M1 | Or possibly earlier version of $H$ |
| $t = -5\ln\left(\frac{6}{11}\right) = $ awrt 3.03 days | dM1 A1 | Correct log work leading to $t$; depends on first M |
**(3 marks)**
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $k = 5$ | B1 | |
**(1 mark)**
## Question (part c/d):
**Part (c):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $t = -5\ln\left(\dfrac{6}{11}\right)$ or $t = 5\ln\left(\dfrac{11}{6}\right)$ or awrt 3.03 (days) | A1 | Depends on the first M mark |
**Part (d):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $k = 5$ | B1 | Allow $H = 5$ or just "5" |10. (a) Write $\frac { 1 } { ( H - 5 ) ( H + 3 ) }$ in partial fraction form.
The depth of water in a storage tank is being monitored.\\
The depth of water in the tank, $H$ metres, is modelled by the differential equation
$$\frac { \mathrm { d } H } { \mathrm {~d} t } = - \frac { ( H - 5 ) ( H + 3 ) } { 40 }$$
where $t$ is the time, in days, from when monitoring began.\\
Given that the initial depth of water in the tank was 13 m ,\\
(b) solve the differential equation to show that
$$H = \frac { 10 + 3 \mathrm { e } ^ { - 0.2 t } } { 2 - \mathrm { e } ^ { - 0.2 t } }$$
(c) Hence find the time taken for the depth of water in the tank to fall to 8 m .\\
(Solutions relying entirely on calculator technology are not acceptable.)
According to the model, the depth of water in the tank will eventually fall to $k$ metres.\\
(d) State the value of the constant $k$.\\
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\hfill \mbox{\textit{Edexcel P4 2021 Q10 [14]}}