SPS SPS SM Pure 2025 February — Question 11 10 marks

Exam BoardSPS
ModuleSPS SM Pure (SPS SM Pure)
Year2025
SessionFebruary
Marks10
TopicExponential Functions
TypeExponential model with shifted asymptote
DifficultyStandard +0.3 This is a straightforward applied exponential question with standard techniques: evaluating at t=0, finding limits as t→∞, differentiation to find maximum (solving e^(-0.2t) = (9/8)e^(-0.9t)), algebraic manipulation to show an equation, and simple iteration. All parts are routine A-level procedures with no novel insight required, making it slightly easier than average.
Spec1.06a Exponential function: a^x and e^x graphs and properties1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07n Stationary points: find maxima, minima using derivatives1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams

11. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{bc7fb499-9462-40ae-88f4-87fc60f6a005-26_462_586_148_593} \captionsetup{labelformat=empty} \caption{Figure 4}
\end{figure} The heart rate of a horse is being monitored.
The heart rate \(H\), measured in beats per minute (bpm), is modelled by the equation $$H = 32 + 40 \mathrm { e } ^ { - 0.2 t } - 20 \mathrm { e } ^ { - 0.9 t }$$ where \(t\) minutes is the time after monitoring began.
Figure 4 is a sketch of \(H\) against \(t\). \section*{Use the equation of the model to answer parts (a) to (e).}
  1. State the initial heart rate of the horse. In the long term, the heart rate of the horse approaches \(L \mathrm { bpm }\).
  2. State the value of \(L\). The heart rate of the horse reaches its maximum value after \(T\) minutes.
  3. Find the value of \(T\), giving your answer to 3 decimal places.
    (Solutions based entirely on calculator technology are not acceptable.) The heart rate of the horse is 37 bpm after \(M\) minutes.
  4. Show that \(M\) is a solution of the equation $$t = 5 \ln \left( \frac { 8 } { 1 + 4 \mathrm { e } ^ { - 0.9 t } } \right)$$ Using the iteration formula $$t _ { n + 1 } = 5 \ln \left( \frac { 8 } { 1 + 4 \mathrm { e } ^ { - 0.9 t _ { n } } } \right) \quad \text { with } \quad t _ { 1 } = 10$$
    1. find, to 4 decimal places, the value of \(t _ { 2 }\)
    2. find, to 4 decimal places, the value of \(M\)

11.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{bc7fb499-9462-40ae-88f4-87fc60f6a005-26_462_586_148_593}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}

The heart rate of a horse is being monitored.\\
The heart rate $H$, measured in beats per minute (bpm), is modelled by the equation

$$H = 32 + 40 \mathrm { e } ^ { - 0.2 t } - 20 \mathrm { e } ^ { - 0.9 t }$$

where $t$ minutes is the time after monitoring began.\\
Figure 4 is a sketch of $H$ against $t$.

\section*{Use the equation of the model to answer parts (a) to (e).}
\begin{enumerate}[label=(\alph*)]
\item State the initial heart rate of the horse.

In the long term, the heart rate of the horse approaches $L \mathrm { bpm }$.
\item State the value of $L$.

The heart rate of the horse reaches its maximum value after $T$ minutes.
\item Find the value of $T$, giving your answer to 3 decimal places.\\
(Solutions based entirely on calculator technology are not acceptable.)

The heart rate of the horse is 37 bpm after $M$ minutes.
\item Show that $M$ is a solution of the equation

$$t = 5 \ln \left( \frac { 8 } { 1 + 4 \mathrm { e } ^ { - 0.9 t } } \right)$$

Using the iteration formula

$$t _ { n + 1 } = 5 \ln \left( \frac { 8 } { 1 + 4 \mathrm { e } ^ { - 0.9 t _ { n } } } \right) \quad \text { with } \quad t _ { 1 } = 10$$
\item \begin{enumerate}[label=(\roman*)]
\item find, to 4 decimal places, the value of $t _ { 2 }$
\item find, to 4 decimal places, the value of $M$
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{SPS SPS SM Pure 2025 Q11 [10]}}