| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Integration with substitution given |
| Difficulty | Standard +0.3 This is a structured multi-part question with clear guidance at each step. Parts (i)-(iii) involve routine differentiation and algebraic manipulation. Parts (iv)-(vi) use standard partial fractions and integration techniques with the method essentially prescribed. While it requires multiple techniques (differentiation, partial fractions, integration, limits), each step is straightforward and well-signposted, making it slightly easier than average for C4 level. |
| Spec | 1.02y Partial fractions: decompose rational functions1.07b Gradient as rate of change: dy/dx notation1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = a(1+kt)^{-1} \Rightarrow \frac{dx}{dt} = -ka(1+kt)^{-2}\) | M1, A1 | Chain rule (or quotient rule) |
| \(= -ka(x/a)^2 = -kx^2/a\) | E1 | Substitution for \(x\) |
| [3] | ||
| OR: \(kt = a/x - 1\), \(t = a/kx - 1/k\), \(dt/dx = -a/kx^2\) | M1, A1 | |
| \(\Rightarrow dx/dt = -kx^2/a\) | E1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| When \(t=0\), \(x = a \Rightarrow a = 2.5\) | B1 | \(a = 2.5\) |
| When \(t=1\), \(x = 1.6 \Rightarrow 1.6 = 2.5/(1+k)\) | M1 | |
| \(\Rightarrow 1 + k = 1.5625\) | A1 | |
| \(\Rightarrow k = 0.5625\) | ||
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| In the long term, \(x \to 0\) | B1 | or, for example, they die out |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2y - y^2} = \frac{1}{y(1-y)} \cdot \frac{1}{?} = \frac{A}{y} + \frac{B}{2-y}\) | M1 | partial fractions |
| \(1 = A(2-y) + By\) | M1 | evaluating constants by substituting values, equating coefficients or cover-up |
| \(y=0 \Rightarrow 2A = 1 \Rightarrow A = \frac{1}{2}\) | A1 | |
| \(y=2 \Rightarrow 2B = 1 \Rightarrow B = \frac{1}{2}\) | A1 | |
| \(\Rightarrow \frac{1}{2y-y^2} = \frac{1}{2y} + \frac{1}{2(2-y)}\) | ||
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int \frac{1}{2y-y^2}dy = \int dt\) | M1 | Separating variables |
| \(\int\left[\frac{1}{2y} + \frac{1}{2(2-y)}\right]dy = \int dt\) | B1 ft | \(\frac{1}{2}\ln y - \frac{1}{2}\ln(2-y)\), ft their \(A,B\) |
| \(\Rightarrow \frac{1}{2}\ln y - \frac{1}{2}\ln(2-y) = t + c\) | ||
| When \(t=0\), \(y=1 \Rightarrow 0 - 0 = 0 + c \Rightarrow c = 0\) | A1 | evaluating the constant |
| \(\Rightarrow \ln\frac{y}{2-y} = 2t\) | E1 | |
| \(\frac{y}{2-y} = e^{2t}\) | M1 | Anti-logging |
| \(\Rightarrow y = 2e^{2t} - ye^{2t}\) | ||
| \(\Rightarrow y + ye^{2t} = 2e^{2t}\) | DM1 | Isolating \(y\) |
| \(\Rightarrow y(1 + e^{2t}) = 2e^{2t}\) | ||
| \(\Rightarrow y = \frac{2e^{2t}}{1+e^{2t}} = \frac{2}{1+e^{-2t}}\) | E1 | |
| [7] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| As \(t \to \infty\), \(e^{-2t} \to 0 \Rightarrow y \to 2\), so long term population is 2000 | B1 | or \(y = 2\) |
| [1] |
# Question 3:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = a(1+kt)^{-1} \Rightarrow \frac{dx}{dt} = -ka(1+kt)^{-2}$ | M1, A1 | Chain rule (or quotient rule) |
| $= -ka(x/a)^2 = -kx^2/a$ | E1 | Substitution for $x$ |
| **[3]** | | |
| OR: $kt = a/x - 1$, $t = a/kx - 1/k$, $dt/dx = -a/kx^2$ | M1, A1 | |
| $\Rightarrow dx/dt = -kx^2/a$ | E1 | |
| **[3]** | | |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| When $t=0$, $x = a \Rightarrow a = 2.5$ | B1 | $a = 2.5$ |
| When $t=1$, $x = 1.6 \Rightarrow 1.6 = 2.5/(1+k)$ | M1 | |
| $\Rightarrow 1 + k = 1.5625$ | A1 | |
| $\Rightarrow k = 0.5625$ | | |
| **[3]** | | |
## Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| In the long term, $x \to 0$ | B1 | or, for example, they die out |
| **[1]** | | |
## Part (iv)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2y - y^2} = \frac{1}{y(1-y)} \cdot \frac{1}{?} = \frac{A}{y} + \frac{B}{2-y}$ | M1 | partial fractions |
| $1 = A(2-y) + By$ | M1 | evaluating constants by substituting values, equating coefficients or cover-up |
| $y=0 \Rightarrow 2A = 1 \Rightarrow A = \frac{1}{2}$ | A1 | |
| $y=2 \Rightarrow 2B = 1 \Rightarrow B = \frac{1}{2}$ | A1 | |
| $\Rightarrow \frac{1}{2y-y^2} = \frac{1}{2y} + \frac{1}{2(2-y)}$ | | |
| **[4]** | | |
## Part (v)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \frac{1}{2y-y^2}dy = \int dt$ | M1 | Separating variables |
| $\int\left[\frac{1}{2y} + \frac{1}{2(2-y)}\right]dy = \int dt$ | B1 ft | $\frac{1}{2}\ln y - \frac{1}{2}\ln(2-y)$, ft their $A,B$ |
| $\Rightarrow \frac{1}{2}\ln y - \frac{1}{2}\ln(2-y) = t + c$ | | |
| When $t=0$, $y=1 \Rightarrow 0 - 0 = 0 + c \Rightarrow c = 0$ | A1 | evaluating the constant |
| $\Rightarrow \ln\frac{y}{2-y} = 2t$ | E1 | |
| $\frac{y}{2-y} = e^{2t}$ | M1 | Anti-logging |
| $\Rightarrow y = 2e^{2t} - ye^{2t}$ | | |
| $\Rightarrow y + ye^{2t} = 2e^{2t}$ | DM1 | Isolating $y$ |
| $\Rightarrow y(1 + e^{2t}) = 2e^{2t}$ | | |
| $\Rightarrow y = \frac{2e^{2t}}{1+e^{2t}} = \frac{2}{1+e^{-2t}}$ | E1 | |
| **[7]** | | |
## Part (vi)
| Answer/Working | Mark | Guidance |
|---|---|---|
| As $t \to \infty$, $e^{-2t} \to 0 \Rightarrow y \to 2$, so long term population is 2000 | B1 | or $y = 2$ |
| **[1]** | | |3 Some years ago an island was populated by red squirrels and there were no grey squirrels. Then grey squirrels were introduced.
The population $x$, in thousands, of red squirrels is modelled by the equation
$$x = \frac { a } { 1 + k t }$$
where $t$ is the time in years, and $a$ and $k$ are constants. When $t = 0 , x = 2.5$.\\
(i) Show that $\frac { \mathrm { d } x } { \mathrm {~d} t } = - \frac { k x ^ { 2 } } { a }$.\\
(ii) Given that the initial population of 2.5 thousand red squirrels reduces to 1.6 thousand after one year, calculate $a$ and $k$.\\
(iii) What is the long-term population of red squirrels predicted by this model?
The population $y$, in thousands, of grey squirrels is modelled by the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} t } = 2 y - y ^ { 2 }$$
When $t = 0 , y = 1$.\\
(iv) Express $\frac { 1 } { 2 y - y ^ { 2 } }$ in partial fractions.\\
(v) Hence show by integration that $\ln \left( \frac { y } { 2 y } \right) = 2 t$.
Show that $y = \frac { 2 } { 1 + \mathrm { e } ^ { - 2 t } }$.\\
(vi) What is the long-term population of grey squirrels predicted by this model?
\hfill \mbox{\textit{OCR MEI C4 Q3 [19]}}