AQA AS Paper 2 2024 June — Question 10 11 marks

Exam BoardAQA
ModuleAS Paper 2 (AS Paper 2)
Year2024
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicExponential Equations & Modelling
Typeln(y) vs x: find constants from two points
DifficultyModerate -0.3 This is a standard AS-level exponential modelling question with routine logarithmic manipulation and differentiation. Part (a)(i) is direct log manipulation, (a)(ii) reads values from a graph, (b)(i) is straightforward chain rule differentiation, (b)(ii) substitutes into the derivative, and (c) requires a simple comment on model validity. All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average.
Spec1.06b Gradient of e^(kx): derivative and exponential model1.06i Exponential growth/decay: in modelling context1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates
A singer has a social media account with a number of followers. The singer releases a new song and the number of followers grows exponentially. The number of followers, \(F\), may be modelled by the formula $$F = ae^{kt}$$ where \(t\) is the number of days since the song was released and \(a\) and \(k\) are constants. • Two days after the song is released the account has 2050 followers. • Five days after the song is released the account has 9200 followers. On the graph below ln \(F\) has been plotted against \(t\) for these two pieces of data. A line has been drawn passing through these two data points. \includegraphics{figure_2}
    1. Show that \(\ln F = \ln a + kt\) [2 marks]
    2. Using the graph, estimate the value of the constant \(a\) and the value of the constant \(k\) [4 marks]
    1. Show that \(\frac{dF}{dt} = kF\) [2 marks]
    2. Using the model, estimate the rate at which the number of followers is increasing 5 days after the song is released. [2 marks]
  3. The singer claims that 30 days after the song is released, the account will have more than a billion followers. Comment on the singer's claim. [1 mark]
Question 10:
AnswerMarks Guidance
10(a)(i)Takes logarithms of both sides
PI1.1a M1
ln F = ln aekt
ln F = ln a + ln ekt
ln F = ln a + kt
Derives the required equation
Must see either:
ln F = ln a + ln ekt
Or
AnswerMarks Guidance
ln F = ln a + ktlne2.1 R1
Subtotal2
QMarking instructions AO
AnswerMarks
10(a)(ii)Equates ln a to intercept value
6.55 ≤ lna ≤ 6.65
Or
Uses one/two points on graph to
AnswerMarks Guidance
form 1/2 equations in ln a and k3.1b M1
a = 735
gradient = 0.5 = k
Equates k to a gradient value
Or
Solves equation(s) or uses F to
find a value of a and/or a value
AnswerMarks Guidance
of k1.1a M1
Obtains a value for a
AnswerMarks Guidance
AWFW 700 to 7731.1b A1
Obtains a value for k
AnswerMarks Guidance
AWFW 0.45 to 0.551.1b A1
Subtotal4
QMarking instructions AO
AnswerMarks
10(b)(i)ekt
Recalls rate of change of is
kekt
Condone done numerically
AnswerMarks Guidance
using their a and k1.2 M1
= kaekt = kF
d t
Obtains given result in algebraic
terms only
AnswerMarks Guidance
AG2.1 R1
Subtotal2
QMarking instructions AO
AnswerMarks
10(b)(ii)Substitutes their value for k
provided 0.45 ≤ k ≤ 0.55 and
AnswerMarks Guidance
F = 9200 into the differential3.4 M1
At t = 5 = 0.5 × 9200
d t
= 4600 per day
dF
Obtains value for
dt
AnswerMarks Guidance
AWFW 4140 to 50603.2a A1
Subtotal2
QMarking instructions AO
AnswerMarks
10(c)Makes a suitable comment in
context to explain why the
number of followers will not
be/unlikely to be more than 1
billion.
E.g.
Realistically the number of
followers is too large.
In reality, the number of
followers will not continue to
grow exponentially
Singer may do something which
causes followers to unfollow or
lose followers
Song may become less popular
Song may not be to that number
of people’s taste
New songs may be released
No individual has a social media
following of over 1 billion people
Model only based upon days 2
and 5 so too much extrapolation
People are fickle/lose interest
If any incorrect statement is
included with a correct
AnswerMarks Guidance
statement(s) then E03.5a E1
over one billion followers as that is too
high a number.
So, the exponential model will not to
be valid after 30 days.
AnswerMarks Guidance
Subtotal1
Question 10 Total11
QMarking instructions AO 
Question 10:
--- 10(a)(i) ---
10(a)(i) | Takes logarithms of both sides
PI | 1.1a | M1 | F = aekt
ln F = ln aekt
ln F = ln a + ln ekt
ln F = ln a + kt
Derives the required equation
Must see either:
ln F = ln a + ln ekt
Or
ln F = ln a + ktlne | 2.1 | R1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 10(a)(ii) ---
10(a)(ii) | Equates ln a to intercept value
6.55 ≤ lna ≤ 6.65
Or
Uses one/two points on graph to
form 1/2 equations in ln a and k | 3.1b | M1 | ln a = 6.6
a = 735
gradient = 0.5 = k
Equates k to a gradient value
Or
Solves equation(s) or uses F to
find a value of a and/or a value
of k | 1.1a | M1
Obtains a value for a
AWFW 700 to 773 | 1.1b | A1
Obtains a value for k
AWFW 0.45 to 0.55 | 1.1b | A1
Subtotal | 4
Q | Marking instructions | AO | Marks | Typical solution
--- 10(b)(i) ---
10(b)(i) | ekt
Recalls rate of change of is
kekt
Condone done numerically
using their a and k | 1.2 | M1 | d F
= kaekt = kF
d t
Obtains given result in algebraic
terms only
AG | 2.1 | R1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 10(b)(ii) ---
10(b)(ii) | Substitutes their value for k
provided 0.45 ≤ k ≤ 0.55 and
F = 9200 into the differential | 3.4 | M1 | d F
At t = 5 = 0.5 × 9200
d t
= 4600 per day
dF
Obtains value for
dt
AWFW 4140 to 5060 | 3.2a | A1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 10(c) ---
10(c) | Makes a suitable comment in
context to explain why the
number of followers will not
be/unlikely to be more than 1
billion.
E.g.
Realistically the number of
followers is too large.
In reality, the number of
followers will not continue to
grow exponentially
Singer may do something which
causes followers to unfollow or
lose followers
Song may become less popular
Song may not be to that number
of people’s taste
New songs may be released
No individual has a social media
following of over 1 billion people
Model only based upon days 2
and 5 so too much extrapolation
People are fickle/lose interest
If any incorrect statement is
included with a correct
statement(s) then E0 | 3.5a | E1 | It is unlikely that the singer will have
over one billion followers as that is too
high a number.
So, the exponential model will not to
be valid after 30 days.
Subtotal | 1
Question 10 Total | 11
Q | Marking instructions | AO | Marks | Typical solution
A singer has a social media account with a number of followers. The singer releases a new song and the number of followers grows exponentially.

The number of followers, $F$, may be modelled by the formula
$$F = ae^{kt}$$
where $t$ is the number of days since the song was released and $a$ and $k$ are constants.

• Two days after the song is released the account has 2050 followers.
• Five days after the song is released the account has 9200 followers.

On the graph below ln $F$ has been plotted against $t$ for these two pieces of data.

A line has been drawn passing through these two data points.

\includegraphics{figure_2}

\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that $\ln F = \ln a + kt$
[2 marks]

\item Using the graph, estimate the value of the constant $a$ and the value of the constant $k$
[4 marks]
\end{enumerate}

\item \begin{enumerate}[label=(\roman*)]
\item Show that $\frac{dF}{dt} = kF$
[2 marks]

\item Using the model, estimate the rate at which the number of followers is increasing 5 days after the song is released.
[2 marks]
\end{enumerate}

\item The singer claims that 30 days after the song is released, the account will have more than a billion followers.

Comment on the singer's claim.
[1 mark]
\end{enumerate}

\hfill \mbox{\textit{AQA AS Paper 2 2024 Q10 [11]}}
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