OCR MEI C4 — Question 1 20 marks

Exam BoardOCR MEI
ModuleC4 (Core Mathematics 4)
Marks20
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPartial Fractions
TypePartial fractions for differential equations
DifficultyStandard +0.3 This is a structured multi-part question with clear signposting through each step. Part (a) involves basic range finding using sin bounds and straightforward differentiation verification. Part (b) guides students through standard partial fractions decomposition, separation of variables (with the partial fractions already set up), and finally evaluating exponential extrema. While it combines several techniques, each step is routine and heavily scaffolded, making it slightly easier than average.
Spec1.02y Partial fractions: decompose rational functions1.06f Laws of logarithms: addition, subtraction, power rules1.07b Gradient as rate of change: dy/dx notation1.07l Derivative of ln(x): and related functions1.08k Separable differential equations: dy/dx = f(x)g(y)

1 Data suggest that the number of cases of infection from a particular disease tends to oscillate between two values over a period of approximately 6 months.
  1. Suppose that the number of cases, \(P\) thousand, after time \(t\) months is modelled by the equation \(P = \frac { 2 } { 2 - \sin t }\). Thus, when \(t = 0 , P = 1\).
    1. By considering the greatest and least values of \(\sin t\), write down the greatest and least values of \(P\) predicted by this model.
    2. Verify that \(P\) satisfies the differential equation \(\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 2 } P ^ { 2 } \cos t\).
  2. An alternative model is proposed, with differential equation $$\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 2 } \left( 2 P ^ { 2 } - P \right) \cos t$$ As before, \(P = 1\) when \(t = 0\).
    1. Express \(\frac { 1 } { P ( 2 P - 1 ) }\) in partial fractions.
    2. Solve the differential equation (*) to show that $$\ln \left( \frac { 2 P } { P } \right) = \frac { 1 } { 2 } \sin t$$ This equation can be rearranged to give \(P = \frac { 1 } { 2 \mathrm { e } ^ { \frac { 1 } { 2 } \sin t } }\).
    3. Find the greatest and least values of \(P\) predicted by this model.

1 Data suggest that the number of cases of infection from a particular disease tends to oscillate between two values over a period of approximately 6 months.
\begin{enumerate}[label=(\alph*)]
\item Suppose that the number of cases, $P$ thousand, after time $t$ months is modelled by the equation $P = \frac { 2 } { 2 - \sin t }$. Thus, when $t = 0 , P = 1$.
\begin{enumerate}[label=(\roman*)]
\item By considering the greatest and least values of $\sin t$, write down the greatest and least values of $P$ predicted by this model.
\item Verify that $P$ satisfies the differential equation $\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 2 } P ^ { 2 } \cos t$.
\end{enumerate}\item An alternative model is proposed, with differential equation

$$\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 2 } \left( 2 P ^ { 2 } - P \right) \cos t$$

As before, $P = 1$ when $t = 0$.
\begin{enumerate}[label=(\roman*)]
\item Express $\frac { 1 } { P ( 2 P - 1 ) }$ in partial fractions.
\item Solve the differential equation (*) to show that

$$\ln \left( \frac { 2 P } { P } \right) = \frac { 1 } { 2 } \sin t$$

This equation can be rearranged to give $P = \frac { 1 } { 2 \mathrm { e } ^ { \frac { 1 } { 2 } \sin t } }$.
\item Find the greatest and least values of $P$ predicted by this model.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{OCR MEI C4  Q1 [20]}}