Edexcel C4 — Question 1 8 marks

Exam BoardEdexcel
ModuleC4 (Core Mathematics 4)
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDifferential equations
TypeFirst-order integration
DifficultyModerate -0.3 This is a straightforward C4 differential equations question requiring standard techniques: finding when dn/dt = 0 for part (a), integrating with an initial condition for part (b), and a simple interpretation for part (c). The integration is routine (exponential and constant), and no novel problem-solving is needed. Slightly easier than average due to the mechanical nature of all parts.
Spec1.07n Stationary points: find maxima, minima using derivatives1.08k Separable differential equations: dy/dx = f(x)g(y)1.08l Interpret differential equation solutions: in context

The number of people, \(n\), in a queue at a Post Office \(t\) minutes after it opens is modelled by the differential equation $$\frac{dn}{dt} = e^{0.5t} - 5, \quad t \geq 0.$$
  1. Find, to the nearest second, the time when the model predicts that there will be the least number of people in the queue. [3]
  2. Given that there are 20 people in the queue when the Post Office opens, solve the differential equation. [4]
  3. Explain why this model would not be appropriate for large values of \(t\). [1]

AnswerMarks Guidance
(a) \(\frac{dn}{dt} = 0 \Rightarrow e^{0.5t} = 5\)M1
\(t = 2\ln 5 = 3.219\) mins \(= 3\) mins \(13\) secsM1 A1
(b) \(\int dn = \int (e^{0.5t} - 5) \, dt\)M1 A1
\(n = 2e^{0.5t} - 5t + c\)M1 A1
\(t = 0, n = 20 \Rightarrow 20 = 2 + c, \quad c = 18\)M1
\(n = 2e^{0.5t} - 5t + 18\)A1
(c) as \(t\) increases, \(n\) rapidly becomes very large \(\therefore\) not realisticB1 (8)
**(a)** $\frac{dn}{dt} = 0 \Rightarrow e^{0.5t} = 5$ | M1 |
$t = 2\ln 5 = 3.219$ mins $= 3$ mins $13$ secs | M1 A1 |

**(b)** $\int dn = \int (e^{0.5t} - 5) \, dt$ | M1 A1 |
$n = 2e^{0.5t} - 5t + c$ | M1 A1 |
$t = 0, n = 20 \Rightarrow 20 = 2 + c, \quad c = 18$ | M1 |
$n = 2e^{0.5t} - 5t + 18$ | A1 |

**(c)** as $t$ increases, $n$ rapidly becomes very large $\therefore$ not realistic | B1 | (8) |

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The number of people, $n$, in a queue at a Post Office $t$ minutes after it opens is modelled by the differential equation
$$\frac{dn}{dt} = e^{0.5t} - 5, \quad t \geq 0.$$

\begin{enumerate}[label=(\alph*)]
\item Find, to the nearest second, the time when the model predicts that there will be the least number of people in the queue. [3]

\item Given that there are 20 people in the queue when the Post Office opens, solve the differential equation. [4]

\item Explain why this model would not be appropriate for large values of $t$. [1]
\end{enumerate}

\hfill \mbox{\textit{Edexcel C4  Q1 [8]}}