AQA Further Paper 2 2020 June — Question 14 11 marks

Exam BoardAQA
ModuleFurther Paper 2 (Further Paper 2)
Year2020
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPolar coordinates
TypeArea between two polar curves
DifficultyHard +2.3 This Further Maths question requires sketching a rose curve, finding intersection points by solving a transcendental equation involving sin θ and cos 2θ, then computing a complex area using polar integration with non-standard limits. The 9-mark part (b) demands sophisticated manipulation to express the result in the given form with α = sin⁻¹((√5-1)/2), requiring multiple integration techniques and trigonometric identities. This goes well beyond standard polar area questions.
Spec4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve
The diagram shows the polar curve \(C_1\) with equation \(r = 2 \sin \theta\) The diagram also shows part of the polar curve \(C_2\) with equation \(r = 1 + \cos 2\theta\) \includegraphics{figure_14}
  1. On the diagram above, complete the sketch of \(C_2\) [2 marks]
  2. Show that the area of the region shaded in the diagram is equal to $$k\pi + m\alpha - \sin 2\alpha + q \sin 4\alpha$$ where \(\alpha = \sin^{-1} \left( \frac{\sqrt{5} - 1}{2} \right)\), and \(k\), \(m\) and \(q\) are rational numbers. [9 marks]
Question 14:
AnswerMarks
14(a)Draws another quarter
of the curve in the
AnswerMarks Guidance
correct location.1.1a M1
Draws the complete
AnswerMarks Guidance
curve.1.1b A1
AnswerMarks
14(b)Selects a method to find the
area of the shaded region
by splitting it into two
regions with a line from the
pole to the point of
intersection of and
AnswerMarks Guidance
PI by correct lim 𝐶𝐶1its 𝐶𝐶23.1a B1
1+co s2𝜃𝜃 =2sin𝜃𝜃
2
2cos 𝜃𝜃 =2sin𝜃𝜃
2
2−2sin 𝜃𝜃 =2sin𝜃𝜃
2
sin 𝜃𝜃+sin𝜃𝜃−1 = 0
−1±√5
is acute so sin𝜃𝜃=
2
as defined in the questio−n1. √5−1
𝜃𝜃 𝜃𝜃 =sin � 2 �=𝛼𝛼
𝛼𝛼
1 2
𝐴𝐴1 = 2 � 0 (2sin𝜃𝜃) d𝜃𝜃
𝛼𝛼
𝐴𝐴1 = � 0 (1−cos2𝜃𝜃)d𝜃𝜃
𝛼𝛼
1
𝐴𝐴1 = �𝜃𝜃−2 sin2𝜃𝜃 �
0
1
𝐴𝐴1 = 𝛼𝛼−2 sin2𝛼𝛼
𝜋𝜋
2
1 2
𝐴𝐴2 = � (1+cos2𝜃𝜃) d𝜃𝜃
2 𝛼𝛼
𝜋𝜋
2
1 2
𝐴𝐴2 = � (1+2cos2𝜃𝜃+cos 2𝜃𝜃)d𝜃𝜃
2 𝛼𝛼
𝜋𝜋
2
1 1
𝐴𝐴2 = � �1+2cos2𝜃𝜃+ (1+cos4𝜃𝜃)�d𝜃𝜃
2 𝛼𝛼 2
𝜋𝜋
2
3 1 1
𝐴𝐴2 = � 𝜃𝜃+ sin2𝜃𝜃+ sin4𝜃𝜃�
4 2 16 𝛼𝛼
3𝜋𝜋 3 1 1
Area e2nc losed
𝐴𝐴 = − � 𝛼𝛼+ sin2𝛼𝛼+ sin4𝛼𝛼�
8 4 2 16
=𝐴𝐴1 +𝐴𝐴2
3𝜋𝜋 1 1
= + 𝛼𝛼−sin2𝛼𝛼− sin4𝛼𝛼
8 4 16
Forms an equation for the
AnswerMarks Guidance
intersection of and1.1a M1
Solves the equ𝐶𝐶a 1 tion to𝐶𝐶 f 2 ind
the value of at
−1+√5
the point of intersection.
AnswerMarks Guidance
sin𝜃𝜃 = 22.2a A1
Uses an integral of the form
to find an area
1 2
enclosed by a polar curve.
AnswerMarks Guidance
2∫𝑟𝑟 𝑑𝑑𝜃𝜃3.1a B1
Selects a method to
integrate or
by using a do2uble ang2le
AnswerMarks Guidance
formula sin 𝜃𝜃 cos 2𝜃𝜃3.1a M1
Correctly integrates
AnswerMarks Guidance
21.1b A1
∫C(o2rrseinct𝜃𝜃ly) in𝑑𝑑te𝜃𝜃grates1.1b A1
2
S∫u(b1s+titcuotess2 𝜃𝜃lim) it𝑑𝑑s𝜃𝜃 into one of
their integrals to determine
AnswerMarks Guidance
an area.1.1a M1
Completes a rigorous
argument to show that the
area of the shaded region is
3 1
𝜋𝜋+ 𝛼𝛼−sin2𝛼𝛼
8 4
AnswerMarks Guidance
12.1 R1
− sinT4o𝛼𝛼tal11
QMarking Instructions AO 
Question 14:
--- 14(a) ---
14(a) | Draws another quarter
of the curve in the
correct location. | 1.1a | M1
Draws the complete
curve. | 1.1b | A1
--- 14(b) ---
14(b) | Selects a method to find the
area of the shaded region
by splitting it into two
regions with a line from the
pole to the point of
intersection of and
PI by correct lim 𝐶𝐶1its 𝐶𝐶2 | 3.1a | B1 | Point of intersection:
1+co s2𝜃𝜃 =2sin𝜃𝜃
2
2cos 𝜃𝜃 =2sin𝜃𝜃
2
2−2sin 𝜃𝜃 =2sin𝜃𝜃
2
sin 𝜃𝜃+sin𝜃𝜃−1 = 0
−1±√5
is acute so sin𝜃𝜃=
2
as defined in the questio−n1. √5−1
𝜃𝜃 𝜃𝜃 =sin � 2 �=𝛼𝛼
𝛼𝛼
1 2
𝐴𝐴1 = 2 � 0 (2sin𝜃𝜃) d𝜃𝜃
𝛼𝛼
𝐴𝐴1 = � 0 (1−cos2𝜃𝜃)d𝜃𝜃
𝛼𝛼
1
𝐴𝐴1 = �𝜃𝜃−2 sin2𝜃𝜃 �
0
1
𝐴𝐴1 = 𝛼𝛼−2 sin2𝛼𝛼
𝜋𝜋
2
1 2
𝐴𝐴2 = � (1+cos2𝜃𝜃) d𝜃𝜃
2 𝛼𝛼
𝜋𝜋
2
1 2
𝐴𝐴2 = � (1+2cos2𝜃𝜃+cos 2𝜃𝜃)d𝜃𝜃
2 𝛼𝛼
𝜋𝜋
2
1 1
𝐴𝐴2 = � �1+2cos2𝜃𝜃+ (1+cos4𝜃𝜃)�d𝜃𝜃
2 𝛼𝛼 2
𝜋𝜋
2
3 1 1
𝐴𝐴2 = � 𝜃𝜃+ sin2𝜃𝜃+ sin4𝜃𝜃�
4 2 16 𝛼𝛼
3𝜋𝜋 3 1 1
Area e2nc losed
𝐴𝐴 = − � 𝛼𝛼+ sin2𝛼𝛼+ sin4𝛼𝛼�
8 4 2 16
=𝐴𝐴1 +𝐴𝐴2
3𝜋𝜋 1 1
= + 𝛼𝛼−sin2𝛼𝛼− sin4𝛼𝛼
8 4 16
Forms an equation for the
intersection of and | 1.1a | M1
Solves the equ𝐶𝐶a 1 tion to𝐶𝐶 f 2 ind
the value of at
−1+√5
the point of intersection.
sin𝜃𝜃 = 2 | 2.2a | A1
Uses an integral of the form
to find an area
1 2
enclosed by a polar curve.
2∫𝑟𝑟 𝑑𝑑𝜃𝜃 | 3.1a | B1
Selects a method to
integrate or
by using a do2uble ang2le
formula sin 𝜃𝜃 cos 2𝜃𝜃 | 3.1a | M1
Correctly integrates
2 | 1.1b | A1
∫C(o2rrseinct𝜃𝜃ly) in𝑑𝑑te𝜃𝜃grates | 1.1b | A1
2
S∫u(b1s+titcuotess2 𝜃𝜃lim) it𝑑𝑑s𝜃𝜃 into one of
their integrals to determine
an area. | 1.1a | M1
Completes a rigorous
argument to show that the
area of the shaded region is
3 1
𝜋𝜋+ 𝛼𝛼−sin2𝛼𝛼
8 4
1 | 2.1 | R1
− sinT4o𝛼𝛼tal | 11
Q | Marking Instructions | AO | Marks | Typical Solution
The diagram shows the polar curve $C_1$ with equation $r = 2 \sin \theta$

The diagram also shows part of the polar curve $C_2$ with equation $r = 1 + \cos 2\theta$

\includegraphics{figure_14}

\begin{enumerate}[label=(\alph*)]
\item On the diagram above, complete the sketch of $C_2$
[2 marks]

\item Show that the area of the region shaded in the diagram is equal to
$$k\pi + m\alpha - \sin 2\alpha + q \sin 4\alpha$$
where $\alpha = \sin^{-1} \left( \frac{\sqrt{5} - 1}{2} \right)$, and $k$, $m$ and $q$ are rational numbers.
[9 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further Paper 2 2020 Q14 [11]}}
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