OCR MEI Further Pure Core 2024 June — Question 9 8 marks

Exam BoardOCR MEI
ModuleFurther Pure Core (Further Pure Core)
Year2024
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPolar coordinates
TypeArea enclosed by polar curve
DifficultyChallenging +1.2 This is a standard Further Maths polar curves question requiring a sketch of r = a sin 3θ and finding the area of one loop using the polar area formula. While it requires understanding of when r is negative and setting up the correct integral limits (0 to π/3 for one loop), the integration itself is routine using the double angle formula. The 'show detailed reasoning' requirement and multi-step nature elevate it slightly above average difficulty, but it follows a well-practiced template for polar area questions.
Spec4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve

9 A curve has polar equation \(r = \operatorname { asin } 3 \theta\), for \(0 \leqslant \theta \leqslant \pi\), where \(a\) is a positive constant.
  1. Sketch the curve. Indicate the parts of the curve where \(r\) is negative by using a broken line.
  2. In this question you must show detailed reasoning. Determine the area of one of the loops of the curve.

Question 9:
AnswerMarks Guidance
9(a) B1*
B1*
B1dep
AnswerMarks
[3]1.1
1.1
AnswerMarks
1.1One loop in correct position (requires initial line drawn)
Exactly three loops in correct position
Lower loop only shown with a broken line. Any
coordinates on the curve must be correct (either polar or
cartesian)
AnswerMarks Guidance
9(b) DR
13 
A 12 a 2 s i n 2 3 d   = 
0
1
A=3 1a 2 (1−cos6)d
0 4
13 
14 a 2 16 s i n 6   =  − 
0
11 a 2  =
AnswerMarks
2M1
A1
M1
A1
A1
AnswerMarks
[5]1.1
1.1
3.1a
1.1
AnswerMarks
1.1A 12 a 2 s i n 2 3 d   =  . Condone missing d𝜃.
Limits correct. Accept alternative limits between 0 and 𝜋
provided correct multiplication or division of integral is
seen or implied at some stage.
Double angle formula used correctly in their integral.
Condone missing d𝜃.
1
𝑘[𝜃− sin6𝜃]. Condone incorrect or missing limits
6
SC B4 for an otherwise fully correct answer using limits
outside the range 0 to 𝜋.
Question 9:
9 | (a) | B1*
B1*
B1dep
[3] | 1.1
1.1
1.1 | One loop in correct position (requires initial line drawn)
Exactly three loops in correct position
Lower loop only shown with a broken line. Any
coordinates on the curve must be correct (either polar or
cartesian)
9 | (b) | DR
13 
A 12 a 2 s i n 2 3 d   = 
0
1
A=3 1a 2 (1−cos6)d
0 4
13 
14 a 2 16 s i n 6   =  − 
0
11 a 2  =
2 | M1
A1
M1
A1
A1
[5] | 1.1
1.1
3.1a
1.1
1.1 | A 12 a 2 s i n 2 3 d   =  . Condone missing d𝜃.
Limits correct. Accept alternative limits between 0 and 𝜋
provided correct multiplication or division of integral is
seen or implied at some stage.
Double angle formula used correctly in their integral.
Condone missing d𝜃.
1
𝑘[𝜃− sin6𝜃]. Condone incorrect or missing limits
6
SC B4 for an otherwise fully correct answer using limits
outside the range 0 to 𝜋.
9 A curve has polar equation $r = \operatorname { asin } 3 \theta$, for $0 \leqslant \theta \leqslant \pi$, where $a$ is a positive constant.
\begin{enumerate}[label=(\alph*)]
\item Sketch the curve. Indicate the parts of the curve where $r$ is negative by using a broken line.
\item In this question you must show detailed reasoning.

Determine the area of one of the loops of the curve.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure Core 2024 Q9 [8]}}