Edexcel PURE 2024 October — Question 8

Exam BoardEdexcel
ModulePURE
Year2024
SessionOctober
PaperDownload PDF ↗
TopicFixed Point Iteration
TypeDerive stationary point equation
DifficultyStandard +0.3 This is a structured multi-part question on exponential functions and fixed-point iteration. Parts (a)-(b) are trivial substitution/limits, part (c) requires standard differentiation and solving, part (d) is algebraic rearrangement (shown), and part (e) is straightforward calculator iteration. All techniques are routine for Further Maths students, making this slightly easier than average.
Spec1.06a Exponential function: a^x and e^x graphs and properties1.06b Gradient of e^(kx): derivative and exponential model1.07j Differentiate exponentials: e^(kx) and a^(kx)1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams1.09d Newton-Raphson method
8. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{b9472037-c143-4b68-86e2-801f71029773-24_472_595_246_735} \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 \(M\)
Question 8:
AnswerMarks Guidance
8 (a)52 b.p.m. B1
(1)
AnswerMarks Guidance
(b)32 b.p.m. B1
(1)
AnswerMarks
(c)d H − 0 .2 t − 0 .9 t
= − 8 e + 1 8 e
AnswerMarks
d tM1, A1
− 0 .2 t − 0 .9 t 0 .7 t
AnswerMarks
Sets − 8 e + 1 8 e = 0  4 e = 9dM1
9
 0 . 7 t = l n  t = . . .
AnswerMarks
4M1
T = 1.158 (minutes)A1
(5)
AnswerMarks
(d)− 0 .9 t
− 0 .2 t − 0 .9 t − 0 .2 t 1 + 4 e
3 7 = 3 2 + 4 0 e − 2 0 e  e =
AnswerMarks
8M1
 
0 .2 t 8 8
 e =  t = 5 l n
− 0 .9 t − 0 .9 t
AnswerMarks
1 + 4 e 1 + 4 eA1*
(2)
AnswerMarks
(e) 8 
t =5ln =....
2 −0.910
AnswerMarks
1+4e M1
( )
t = a w r t 1 0 . 3 9 4 7
AnswerMarks
2A1
(M )= 1 0 . 3 9 5 5A 1
(3)

Total 12

…………………………………………………………………………………………………………
There may be answers produced from a calculator. In cases such as this we are going to apply the rule
‘answers without full working may not gain credit.’
dH
−0.2t −0.9t
= −8 e +18 e = 0 t =1.158scores 1,1,0,0,0
dt
………………………………………………………………………………………………….
(d)
M1: Sets H = 3 7 and proceeds to make e
 0 .2 t
the subject
A1*: Proceeds to the given answer showing all necessary steps.
Note that this is a given answer so the steps must be logical without huge jumps
Acceptable proofs:
Example 1
3 7 = 3 2 + 4 0 e
− 0 .2 t
− 2 0 e
− 0 .9 t
 e
− 0 .2 t
=
1 + 4
8
e
− 0 .9 t
− 0
t
t
.
=
=
2 t
−
5
=
5
l n
l
l n
n


1
1

1
+
+
+
4
4
4
8
8
e
− 0 .9
e
8
t − 0 .9
e
t − 0 .9
t



Example 2
3 7 = 3 2 + 4 0 e t −0.2 − 2 0 e t −0.9  e t −0.2 = 1 + 4
8
e t −0.9  e t 0.2 =
1 + 4
8
e t −0.9
 0 .2 t = ln 
1 + 4
8
e t −0.9
  t = 5 l n 
1 + 4
8
e − 0 .9 t

Unacceptable proof:
−0.9t  −0.9t   −0.9t 
−0.2t −0.9t −0.2t
1+4e 1+4e 1+4e
37 =32+40e −20e e =  −0.2t =ln  t = −5ln 
8  8   8 
   
 
8
t =5ln 
 −0.9t 
1+4e 
They need to deal with 0.2 and −
PMT
sign in separate steps
…………………………………………………………………………………………………………
Alt (d) lns can be taken earlier. The M1 is scored for making 0.2 t the subject. See below
−0.2t −0.9t
37 =32+40e −20e
( ) ( )
−0.2t −0.9t −0.9t −0.9t
40e =5+20e ln40−0.2t =ln 5+20e 0.2t =ln40−ln 5+20e
We would need to see three further steps. E.g.
     
40 40 8
0.2t =ln t =5ln t =5ln 
 −0.9t   −0.9t   −0.9t 
5+20e  5+20e  1+4e 
……………………………………………………………………………………………………………………………
…………..
(e)
M1: Uses the iteration formula once. Allow for the embedded 10 leading to a value or sight of t =awrt10.4
2
( )
A1: t =awrt10.3947
2
Question 8:
--- 8 (a) ---
8 (a) | 52 b.p.m. | B1
(1)
(b) | 32 b.p.m. | B1
(1)
(c) | d H − 0 .2 t − 0 .9 t
= − 8 e + 1 8 e
d t | M1, A1
− 0 .2 t − 0 .9 t 0 .7 t
Sets − 8 e + 1 8 e = 0  4 e = 9 | dM1
9
 0 . 7 t = l n  t = . . .
4 | M1
T = 1.158 (minutes) | A1
(5)
(d) | − 0 .9 t
− 0 .2 t − 0 .9 t − 0 .2 t 1 + 4 e
3 7 = 3 2 + 4 0 e − 2 0 e  e =
8 | M1
 
0 .2 t 8 8
 e =  t = 5 l n
− 0 .9 t − 0 .9 t
1 + 4 e 1 + 4 e | A1*
(2)
(e) |  8 
t =5ln =....
2 −0.910
1+4e  | M1
( )
t = a w r t 1 0 . 3 9 4 7
2 | A1
(M )= 1 0 . 3 9 5 5 | A 1
(3)
Total 12
…………………………………………………………………………………………………………
There may be answers produced from a calculator. In cases such as this we are going to apply the rule
‘answers without full working may not gain credit.’
dH
−0.2t −0.9t
= −8 e +18 e = 0 t =1.158scores 1,1,0,0,0
dt
………………………………………………………………………………………………….
(d)
M1: Sets H = 3 7 and proceeds to make e
 0 .2 t
the subject
A1*: Proceeds to the given answer showing all necessary steps.
Note that this is a given answer so the steps must be logical without huge jumps
Acceptable proofs:
Example 1
3 7 = 3 2 + 4 0 e
− 0 .2 t
− 2 0 e
− 0 .9 t
 e
− 0 .2 t
=
1 + 4
8
e
− 0 .9 t



− 0
t
t
.
=
=
2 t
−
5
=
5
l n
l
l n
n


1

1

1
+
+
+
4
4
4
8
8
e
− 0 .9
e
8
t − 0 .9
e
t − 0 .9
t



Example 2
3 7 = 3 2 + 4 0 e t −0.2 − 2 0 e t −0.9  e t −0.2 = 1 + 4
8
e t −0.9  e t 0.2 =
1 + 4
8
e t −0.9
 0 .2 t = ln 
1 + 4
8
e t −0.9
  t = 5 l n 
1 + 4
8
e − 0 .9 t

Unacceptable proof:
−0.9t  −0.9t   −0.9t 
−0.2t −0.9t −0.2t
1+4e 1+4e 1+4e
37 =32+40e −20e e =  −0.2t =ln  t = −5ln 
8  8   8 
   
 
8
t =5ln 
 −0.9t 
1+4e 
They need to deal with 0.2 and −
PMT
sign in separate steps
…………………………………………………………………………………………………………
Alt (d) lns can be taken earlier. The M1 is scored for making 0.2 t the subject. See below
−0.2t −0.9t
37 =32+40e −20e
( ) ( )
−0.2t −0.9t −0.9t −0.9t
40e =5+20e ln40−0.2t =ln 5+20e 0.2t =ln40−ln 5+20e
We would need to see three further steps. E.g.
     
40 40 8
0.2t =ln t =5ln t =5ln 
 −0.9t   −0.9t   −0.9t 
5+20e  5+20e  1+4e 
……………………………………………………………………………………………………………………………
…………..
(e)
M1: Uses the iteration formula once. Allow for the embedded 10 leading to a value or sight of t =awrt10.4
2
( )
A1: t =awrt10.3947
2
8.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{b9472037-c143-4b68-86e2-801f71029773-24_472_595_246_735}
\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{Edexcel PURE 2024 Q8}}
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