| Exam Board | AQA |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2023 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Harmonic Form |
| Type | Find value where max/min occurs |
| Difficulty | Standard +0.8 This is a multi-part optimization problem requiring geometric decomposition, algebraic manipulation to derive an area formula, calculus to find maxima with justification using second derivative test, and proportional reasoning. While the individual techniques are A-level standard, the combination of geometry, trigonometry, and calculus across multiple connected parts with 14 total marks makes this moderately challenging, requiring sustained problem-solving over several steps. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta1.05q Trig in context: vectors, kinematics, forces1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks |
|---|---|
| 7(a) | Forms an expression for the |
| Answer | Marks | Guidance |
|---|---|---|
| Condone missing brackets | 3.1b | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Condone r = 5 | 3.1b | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Condone missing brackets | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| brackets | 2.1 | R1 |
| Subtotal | 4 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 7(b)(i) | Differentiates wrt θ |
| Answer | Marks | Guidance |
|---|---|---|
| 8 | 3.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| O E | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| dθ | 2.4 | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| 8 | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| AG | 2.2a | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| the minimum | 2.4 | R1 |
| Subtotal | 6 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 7(b)(ii) | π |
| Answer | Marks | Guidance |
|---|---|---|
| or AWFW [11.9, 12] | 3.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| ISW | 1.1b | A1 |
| Subtotal | 2 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 7(c) | States the angle would be the |
| Answer | Marks | Guidance |
|---|---|---|
| same | 3.5c | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| factor of 4 | 3.5c | E1 |
| Subtotal | 2 | |
| Question 7 Total | 14 | |
| Q | Marking instructions | AO |
Question 7:
--- 7(a) ---
7(a) | Forms an expression for the
area of one sector or both
sectors
1 π−θ π−θ
2 2
e.g r or r
2 2 2
1
r2(π−θ)
or
2
OE
Allow substitution of r = 2.5
Condone r = 5
Condone missing brackets | 3.1b | M1 | Area of sectors =
1 π−θ
2× (2.5) 2
2 2
Area of rhombus =
1
2× (2.5)2sinθ
2
π−θ
A =(2.5) 2 +(2.5)2sinθ
2
25 25
A = (π−θ)+ sinθ
8 4
25
A = (π−θ+2sinθ)
8
Forms an expression for the
area of half rhombus or full
rhombus
e.g 1 r2sinθ or r 2 sinθ
2
Allow substitution of r = 2.5
Condone r = 5 | 3.1b | M1
Substitutes r = 2.5 to get a
correct expression for area of
both sectors or full rhombus
Condone missing brackets | 1.1b | A1
Completes reasoned argument
by calculating correct total area
with at least one correct
intermediate step and no error
seen to show the given result.
AG
Allow recovery of missing
brackets | 2.1 | R1
Subtotal | 4
Q | Marking instructions | AO | Marks | Typical solution
--- 7(b)(i) ---
7(b)(i) | Differentiates wrt θ
Condone sign errors and
25
omission of
8 | 3.1a | M1 | dA 25 25
=− + cosθ
dθ 8 4
dA
Max area occurs when =0
dθ
25 25
− + cosθ=0
8 4
25 25
cosθ=
4 8
1 π
cosθ= θ=
2 3
∴
π
When θ=
3
2
d A
=−5.41 < 0 so maximum.
dθ2
25
(−1+2cosθ)
Obtains
8
O E | 1.1b | A1
Explains maximum or stationary
dA
or turning occurs when =0
dθ
dA
Label must be seen
dθ | 2.4 | E1
25
(−1+2cosθ)
Equates their
8
to 0 and rearranges to obtain a
value forcosθ or θwhen cos is
not seen
25
Condone omission of
8 | 1.1a | M1
1 −1 1
Obtains cosθ= or cos
2 2
π
OE and shows that θ=
3
AG | 2.2a | A1
Uses second derivative to obtain
25 3
− or AWRT 5 and
8
completes argumen−t to show
π
maximum occurs when θ=
3
Allow gradient test
To be awarded R1, marks
M1A1M1A1 must be scored as
the minimum | 2.4 | R1
Subtotal | 6
Q | Marking instructions | AO | Marks | Typical solution
--- 7(b)(ii) ---
7(b)(ii) | π
Substitutes θ= into
3
25
A = (π−θ+2sinθ) fully
8
or AWFW [11.9, 12] | 3.4 | M1 | 25 π π
A= π− +2sin
8 3 3
252π
= + 3
8 3
Obtains the correct exact area
π
ACF with sin evaluated
3
ISW | 1.1b | A1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 7(c) ---
7(c) | States the angle would be the
π
same or the angle will still be
3
or (b)(i) stays the same
Condone the answer will be the
same | 3.5c | E1 | The angle would be the same.
The area would be quadrupled.
States the area would be
quadrupled or area is
252π
+ 3 or their answer
2 3
in (b)(ii) multiplied by 4
OE
Allow (b)(ii) increased by scale
factor of 4 | 3.5c | E1
Subtotal | 2
Question 7 Total | 14
Q | Marking instructions | AO | Marks | Typical solutionA new design for a company logo is to be made from two sectors of a circle, $ORP$ and $OQS$, and a rhombus $OSTR$, as shown in the diagram below.
\includegraphics{figure_7}
The points $P$, $O$ and $Q$ lie on a straight line and the angle $ROS$ is $\theta$ radians.
A large copy of the logo, with $PQ = 5$ metres, is to be put on a wall.
\begin{enumerate}[label=(\alph*)]
\item Show that the area of the logo, $A$ square metres, is given by
$$A = \frac{25}{8}(\pi - \theta + 2\sin\theta)$$
[4 marks]
\item \begin{enumerate}[label=(\roman*)]
\item Show that the maximum value of $A$ occurs when $\theta = \frac{\pi}{3}$
Fully justify your answer.
[6 marks]
\item Find the exact maximum value of $A$
[2 marks]
\end{enumerate}
\item Without further calculation, state how your answers to parts (b)(i) and (b)(ii) would change if $PQ$ were increased to 10 metres.
[2 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 3 2023 Q7 [14]}}