OCR Further Additional Pure AS 2023 June — Question 6 9 marks

Exam BoardOCR
ModuleFurther Additional Pure AS (Further Additional Pure AS)
Year2023
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeApplied recurrence modeling
DifficultyStandard +0.8 This is a multi-part question involving a recurrence relation with real-world context. Part (a) requires straightforward substitution into the formula. Parts (b)(i)-(ii) involve interpreting tabulated data. Part (c) appears incomplete but likely asks for asymptotic analysis or finding a limiting function. The recurrence relation itself is non-standard (involving both n and P_n in complex ways), requiring careful algebraic manipulation. While computational steps are routine, the conceptual understanding of convergence behavior and the multi-step nature elevate this above average difficulty for A-level Further Maths.
Spec8.01h Modelling with recurrence: birth/death rates, INT function

6 When \(10 ^ { 6 }\) of a certain type of bacteria are detected in a blood sample of an infected animal, a course of treatment is started. The long-term aim of the treatment is to reduce the number of bacteria in such a sample to under 10000 . At this level the animal's immune system can fight the infection for itself. Once treatment has started, if the number of bacteria in a sample is 10000 or more, then treatment either continues or restarts. The model suggested to predict the progress of the course of treatment is based on the recurrence system \(P _ { n + 1 } = \frac { 2 P _ { n } } { n + 1 } + \frac { n } { P _ { n } }\) for \(n \geqslant 0\), with \(P _ { 0 } = 1000\), where \(P _ { n }\) denotes the number of bacteria (in thousands) present in the animal's body \(n\) days after the treatment was started. The table below shows the values of \(P _ { n }\), for certain chosen values of \(n\). Each value has been given correct to 2 decimal places (where appropriate).
\(n\)0123456789
\(P _ { n }\)1000200020001333.33666.67266.6725.476.642.68
\(n\)1020406080100200300400
\(P _ { n }\)3.894.676.457.849.0310.0814.2017.3620.04
  1. Find the value of \(P _ { 6 }\) correct to 2 decimal places.
  2. Using the given values for \(P _ { 0 }\) to \(P _ { 9 }\), and assuming that the model is valid,
    1. describe the effects of this course of treatment during the first 9 days,
    2. state the number of days after treatment is started when the animal's own immune system is expected to be able to fight the infection for itself.
    1. Using information from the above table, suggest a function f such that, for \(n > 10 , \mathrm { f } ( n )\) is a suitable approximation for \(P _ { n }\).
    2. Use your suggested function to estimate the number of days after treatment is started when the animal may once again require medical intervention in order to help fight off this bacterial infection.
    3. Using information from the above table and the recurrence relation, verify or correct the estimate which you found in part (c)(ii).
  3. One criticism of the system \(P _ { n + 1 } = \frac { 2 P _ { n } } { n + 1 } + \frac { n } { P _ { n } }\), with \(P _ { 0 } = 1000\), is that it gives non-integer
    values of \(P\). values of \(P _ { n }\). Suggest a modification that would correct this issue.

Question 6:
AnswerMarks Guidance
6(a) 2×𝑃5 5
P = + used with P = 266.67
6 5
6 𝑃5
so that P = 88.91 to 2d.p.
AnswerMarks
6M1
A1
AnswerMarks
[1]1.1
1.1Must consider a range of possible values of P
6
(ignoring </ at endpoints)
Shown carefully (showing that 𝑃 is monotonic in
6
the interval is not required)
SC1 88.91 (by substituting 266.67 soi)
AnswerMarks Guidance
6(b) i
[1]2.2a Must be in context (refers to the number of bacteria)
6(b) ii
[1]3.4 (∵ the number of bacteria falls below 10 000)
6(c) i
[1]3.3 Other possibilities for f may also be fine. At least
three values need to approximately fit
AnswerMarks Guidance
6(c) ii
[1]3.1b FT their f(n) even from B0 in part (c) provided >80
6(c) iii
Using 1 0 .0 8 = + to create a quadratic eqn. in x = P
99
1 0 0 x
which gives P = 10.02 …
AnswerMarks
99M1
A1
AnswerMarks
[2]3.2a
3.5bi.e. 2x2 – 1008x + 9900 = 0
It is not necessary to verify that P < 10 (working
98
back another step gives P = 9.98)
98
AnswerMarks Guidance
6(d)  2 P n 
P = I N T n + (+1)
n + 1 n + 1 P
AnswerMarks
nB1
[1]3.5c
Question 6:
6 | (a) | 2×𝑃5 5
P = + used with P = 266.67
6 5
6 𝑃5
so that P = 88.91 to 2d.p.
6 | M1
A1
[1] | 1.1
1.1 | Must consider a range of possible values of P
6
(ignoring </ at endpoints)
Shown carefully (showing that 𝑃 is monotonic in
6
the interval is not required)
SC1 88.91 (by substituting 266.67 soi)
6 | (b) | i | Number of bacteria initially increases, then declines | B1
[1] | 2.2a | Must be in context (refers to the number of bacteria)
6 | (b) | ii | On the 8th day | B1
[1] | 3.4 | (∵ the number of bacteria falls below 10 000)
6 | (c) | i | f(n) = n | B1
[1] | 3.3 | Other possibilities for f may also be fine. At least
three values need to approximately fit
6 | (c) | ii | Suggest n = 100 (accept n = 99) | B1
[1] | 3.1b | FT their f(n) even from B0 in part (c) provided >80
6 | (c) | iii | 2 x 9 9
Using 1 0 .0 8 = + to create a quadratic eqn. in x = P
99
1 0 0 x
which gives P = 10.02 …
99 | M1
A1
[2] | 3.2a
3.5b | i.e. 2x2 – 1008x + 9900 = 0
It is not necessary to verify that P < 10 (working
98
back another step gives P = 9.98)
98
6 | (d) |  2 P n 
P = I N T n + (+1)
n + 1 n + 1 P
n | B1
[1] | 3.5c
6 When $10 ^ { 6 }$ of a certain type of bacteria are detected in a blood sample of an infected animal, a course of treatment is started. The long-term aim of the treatment is to reduce the number of bacteria in such a sample to under 10000 . At this level the animal's immune system can fight the infection for itself. Once treatment has started, if the number of bacteria in a sample is 10000 or more, then treatment either continues or restarts.

The model suggested to predict the progress of the course of treatment is based on the recurrence system $P _ { n + 1 } = \frac { 2 P _ { n } } { n + 1 } + \frac { n } { P _ { n } }$ for $n \geqslant 0$, with $P _ { 0 } = 1000$, where $P _ { n }$ denotes the number of bacteria (in thousands) present in the animal's body $n$ days after the treatment was started.

The table below shows the values of $P _ { n }$, for certain chosen values of $n$. Each value has been given correct to 2 decimal places (where appropriate).

\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | c | c | c | }
\hline
$n$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline
$P _ { n }$ & 1000 & 2000 & 2000 & 1333.33 & 666.67 & 266.67 &  & 25.47 & 6.64 & 2.68 \\
\hline
\end{tabular}
\end{center}

\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | c | c | }
\hline
$n$ & 10 & 20 & 40 & 60 & 80 & 100 & 200 & 300 & 400 \\
\hline
$P _ { n }$ & 3.89 & 4.67 & 6.45 & 7.84 & 9.03 & 10.08 & 14.20 & 17.36 & 20.04 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Find the value of $P _ { 6 }$ correct to 2 decimal places.
\item Using the given values for $P _ { 0 }$ to $P _ { 9 }$, and assuming that the model is valid,
\begin{enumerate}[label=(\roman*)]
\item describe the effects of this course of treatment during the first 9 days,
\item state the number of days after treatment is started when the animal's own immune system is expected to be able to fight the infection for itself.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Using information from the above table, suggest a function f such that, for $n > 10 , \mathrm { f } ( n )$ is a suitable approximation for $P _ { n }$.
\item Use your suggested function to estimate the number of days after treatment is started when the animal may once again require medical intervention in order to help fight off this bacterial infection.
\item Using information from the above table and the recurrence relation, verify or correct the estimate which you found in part (c)(ii).
\end{enumerate}\item One criticism of the system $P _ { n + 1 } = \frac { 2 P _ { n } } { n + 1 } + \frac { n } { P _ { n } }$, with $P _ { 0 } = 1000$, is that it gives non-integer\\
values of $P$. values of $P _ { n }$.

Suggest a modification that would correct this issue.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Additional Pure AS 2023 Q6 [9]}}