OCR Further Pure Core 1 2022 June — Question 5 11 marks

Exam BoardOCR
ModuleFurther Pure Core 1 (Further Pure Core 1)
Year2022
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPolar coordinates
TypeShow polar curve has Cartesian form
DifficultyChallenging +1.2 This is a standard Further Maths polar coordinates question requiring conversion to Cartesian form using standard substitutions (r²=x²+y², sin2θ=2sinθcosθ=(2xy/r²)), verifying symmetry, and computing loop areas via polar integration. While it involves multiple parts and algebraic manipulation, the techniques are routine for FM students and follow predictable patterns. The 'show that' format provides the target, reducing problem-solving demand.
Spec4.09a Polar coordinates: convert to/from cartesian4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve
5 The diagram below shows the curve \(C\) with polar equation \(r = 3 ( 1 - \sin 2 \theta )\) for \(0 \leqslant \theta \leqslant 2 \pi\). \includegraphics[max width=\textwidth, alt={}, center]{23e58e5e-bbaa-4932-aad0-89b3de6647b2-5_728_963_303_239}
  1. Show that a cartesian equation of \(C\) is \(\left( x ^ { 2 } + y ^ { 2 } \right) ^ { 3 } = 9 ( x - y ) ^ { 4 }\).
  2. Show that the line with equation \(\mathrm { y } = \mathrm { x }\) is a line of symmetry of \(C\).
  3. In this question you must show detailed reasoning. Find the exact area of each of the loops of \(C\).
Question 5:
AnswerMarks Guidance
5(a) AG
A1
AnswerMarks
A11.1
1.1Fully shown (AG). Use of double angle formula and
oe
𝑦𝑦 =
Any correct form of the equation with eliminated.
𝑟𝑟sin𝜃𝜃, 𝑥𝑥 = 𝑟𝑟cos𝜃𝜃
𝜃𝜃
Fully shown
 2xy
r =31− 
 r2 
⇒r3 =3 ( r2 −2xy ) =3 ( x− y )2
⇒r6 =9 ( x− y )4
⇒ ( x2 + y2 )3 =9 ( x− y )4
[3]
AnswerMarks Guidance
(b)(b) ⇒ ( y2 +x2 )3 =9 ( y−x )4
⇒ ( x2 + y2 )3 =9 ( −( x− y ))4
=9 ( −14 )( x− y )4
AnswerMarks Guidance
⇒ ( x2 + y2 )3 =9 ( x− y )4M1
A1M1 3.1a
2.43.1a Interchange of and
𝑥𝑥 𝑦𝑦
Showing/explaining clearly that interchange of and leaves
the equation unchanged.
AnswerMarks Guidance
𝑥𝑥 𝑦𝑦Interchange of and
A12.4 𝑥𝑥 𝑦𝑦
Showing/explaining clearly that interchange of and leaves
the equation unchanged.
Alternative method
AnswerMarks Guidance
M1M1 Showing that is symmetrical about (or
1
). May be done graphically or using trig identities.
sin2𝜃𝜃 𝜃𝜃 = 4𝜋𝜋 𝜃𝜃 =
AnswerMarks
5Showing that is symmetrical about (or
1
). May be done graphically or using trig identities.
sin2𝜃𝜃 𝜃𝜃 = 4𝜋𝜋 𝜃𝜃 =
1 1
soirn 2� 𝜋𝜋+𝛼𝛼� = sin2� 𝜋𝜋−𝛼𝛼�
4 4
5 5
So sin2�4𝜋𝜋 (a+nd𝛼𝛼 � = sin 2ar�e 4 li𝜋𝜋ne−s 𝛼𝛼of�
1 5
symmetry). So is a line of symmetry.
𝜃𝜃 = 4𝜋𝜋 𝜃𝜃 = 4𝜋𝜋
AnswerMarks Guidance
𝑦𝑦 = 𝑥𝑥5 5
So sin2�4𝜋𝜋 (a+nd𝛼𝛼 � = sin 2ar�e 4 li𝜋𝜋ne−s 𝛼𝛼of�A1 4𝜋𝜋
so has lines of symmetry at and
1 5
symmetry). So is a line of symmetry.
AnswerMarks
𝜃𝜃 = 4𝜋𝜋 𝜃𝜃 = 4𝜋𝜋1
𝑟𝑟 = 3 a(n 1 d − th s e i n lin 2 e 𝜃𝜃 ) has polar equation 𝜃𝜃 = 4𝜋𝜋 and
5 1
.
𝜃𝜃 = 4𝜋𝜋 𝑦𝑦 = 𝑥𝑥 𝜃𝜃 = 4𝜋𝜋
5
Any mention of is unnecessary since it is implied by
𝜃𝜃 = 4𝜋𝜋
5
the case.
𝜃𝜃 = 4𝜋𝜋
AnswerMarks
[2]1
4
M1
A1
A1
1.1
1.1
Fully shown (AG). Use of double angle formula and
oe
𝑦𝑦 =
Any correct form of the equation with eliminated.
𝑟𝑟sin𝜃𝜃, 𝑥𝑥 = 𝑟𝑟cos𝜃𝜃
𝜃𝜃
Fully shown
AnswerMarks
(c)DR
5π
1 4
A= ∫ 9 ( 1−sin2θ)2 dθ
2
π
4
5π
1 4 ( )
= ∫ 9 1−2sin2θ+sin22θ dθ
2
π
4
5π
1 4  1 
= ∫ 91−2sin2θ+ ( 1−cos4θ) dθ
2  2 
π
4
5π
93 1  4
= θ+cos2θ− sin4θ
 
22 8 
π
4
935π π  27π
= − +0+0 =
  
22 4 4   4
AnswerMarks
Other loop is the sameB1
M1
M1*
M1dep
M1
AnswerMarks
A12.2a
1.1
1.1
1.1
3.1a
AnswerMarks
2.2aFor identifying appropriate limits correctly
1
Use of A= ∫r2dθ
2
1
Expand correctly and use of sin22θ= ( 1−cos4θ) Ignore
2
limits
Integrate
Substituting correct limits and subtracting
Answer for each loop which must be stated
Alternative method for last 2 marks
Using limits 0 to 2π
2π
93 1 
= θ+cos2θ− sin4θ
 
22 8 
0
9 27π
= ( 3π+0+0 ) = for both loops
2 2
27π
⇒ for one loop by symmetry in y = x
AnswerMarks
4Substituting correct limits and subtracting
Answer for each loop which must be stated
M1
A1
[6]
Alternative method for last 2 marks
Using limits 0 to 2π
2π
93 1 
= θ+cos2θ− sin4θ
 
22 8 
0
9 27π
= ( 3π+0+0 ) = for both loops
2 2
27π
⇒ for one loop by symmetry in y = x
4
Substituting correct limits and subtracting
Answer for each loop which must be stated
Question 5:
5 | (a) | AG | M1
A1
A1 | 1.1
1.1 | Fully shown (AG). Use of double angle formula and
oe
𝑦𝑦 =
Any correct form of the equation with eliminated.
𝑟𝑟sin𝜃𝜃, 𝑥𝑥 = 𝑟𝑟cos𝜃𝜃
𝜃𝜃
Fully shown
 2xy
r =31− 
 r2 
⇒r3 =3 ( r2 −2xy ) =3 ( x− y )2
⇒r6 =9 ( x− y )4
⇒ ( x2 + y2 )3 =9 ( x− y )4
[3]
(b) | (b) | ⇒ ( y2 +x2 )3 =9 ( y−x )4
⇒ ( x2 + y2 )3 =9 ( −( x− y ))4
=9 ( −14 )( x− y )4
⇒ ( x2 + y2 )3 =9 ( x− y )4 | M1
A1 | M1 | 3.1a
2.4 | 3.1a | Interchange of and
𝑥𝑥 𝑦𝑦
Showing/explaining clearly that interchange of and leaves
the equation unchanged.
𝑥𝑥 𝑦𝑦 | Interchange of and
A1 | 2.4 | 𝑥𝑥 𝑦𝑦
Showing/explaining clearly that interchange of and leaves
the equation unchanged.
Alternative method
M1 | M1 | Showing that is symmetrical about (or
1
). May be done graphically or using trig identities.
sin2𝜃𝜃 𝜃𝜃 = 4𝜋𝜋 𝜃𝜃 =
5 | Showing that is symmetrical about (or
1
). May be done graphically or using trig identities.
sin2𝜃𝜃 𝜃𝜃 = 4𝜋𝜋 𝜃𝜃 =
1 1
soirn 2� 𝜋𝜋+𝛼𝛼� = sin2� 𝜋𝜋−𝛼𝛼�
4 4
5 5
So sin2�4𝜋𝜋 (a+nd𝛼𝛼 � = sin 2ar�e 4 li𝜋𝜋ne−s 𝛼𝛼of�
1 5
symmetry). So is a line of symmetry.
𝜃𝜃 = 4𝜋𝜋 𝜃𝜃 = 4𝜋𝜋
𝑦𝑦 = 𝑥𝑥 | 5 5
So sin2�4𝜋𝜋 (a+nd𝛼𝛼 � = sin 2ar�e 4 li𝜋𝜋ne−s 𝛼𝛼of� | A1 | 4𝜋𝜋
so has lines of symmetry at and
1 5
symmetry). So is a line of symmetry.
𝜃𝜃 = 4𝜋𝜋 𝜃𝜃 = 4𝜋𝜋 | 1
𝑟𝑟 = 3 a(n 1 d − th s e i n lin 2 e 𝜃𝜃 ) has polar equation 𝜃𝜃 = 4𝜋𝜋 and
5 1
.
𝜃𝜃 = 4𝜋𝜋 𝑦𝑦 = 𝑥𝑥 𝜃𝜃 = 4𝜋𝜋
5
Any mention of is unnecessary since it is implied by
𝜃𝜃 = 4𝜋𝜋
5
the case.
𝜃𝜃 = 4𝜋𝜋
[2] | 1
4
M1
A1
A1
1.1
1.1
Fully shown (AG). Use of double angle formula and
oe
𝑦𝑦 =
Any correct form of the equation with eliminated.
𝑟𝑟sin𝜃𝜃, 𝑥𝑥 = 𝑟𝑟cos𝜃𝜃
𝜃𝜃
Fully shown
(c) | DR
5π
1 4
A= ∫ 9 ( 1−sin2θ)2 dθ
2
π
4
5π
1 4 ( )
= ∫ 9 1−2sin2θ+sin22θ dθ
2
π
4
5π
1 4  1 
= ∫ 91−2sin2θ+ ( 1−cos4θ) dθ
2  2 
π
4
5π
93 1  4
= θ+cos2θ− sin4θ
 
22 8 
π
4
935π π  27π
= − +0+0 =
  
22 4 4   4
Other loop is the same | B1
M1
M1*
M1dep
M1
A1 | 2.2a
1.1
1.1
1.1
3.1a
2.2a | For identifying appropriate limits correctly
1
Use of A= ∫r2dθ
2
1
Expand correctly and use of sin22θ= ( 1−cos4θ) Ignore
2
limits
Integrate
Substituting correct limits and subtracting
Answer for each loop which must be stated
Alternative method for last 2 marks
Using limits 0 to 2π
2π
93 1 
= θ+cos2θ− sin4θ
 
22 8 
0
9 27π
= ( 3π+0+0 ) = for both loops
2 2
27π
⇒ for one loop by symmetry in y = x
4 | Substituting correct limits and subtracting
Answer for each loop which must be stated
M1
A1
[6]
Alternative method for last 2 marks
Using limits 0 to 2π
2π
93 1 
= θ+cos2θ− sin4θ
 
22 8 
0
9 27π
= ( 3π+0+0 ) = for both loops
2 2
27π
⇒ for one loop by symmetry in y = x
4
Substituting correct limits and subtracting
Answer for each loop which must be stated
5 The diagram below shows the curve $C$ with polar equation $r = 3 ( 1 - \sin 2 \theta )$ for $0 \leqslant \theta \leqslant 2 \pi$.\\
\includegraphics[max width=\textwidth, alt={}, center]{23e58e5e-bbaa-4932-aad0-89b3de6647b2-5_728_963_303_239}
\begin{enumerate}[label=(\alph*)]
\item Show that a cartesian equation of $C$ is $\left( x ^ { 2 } + y ^ { 2 } \right) ^ { 3 } = 9 ( x - y ) ^ { 4 }$.
\item Show that the line with equation $\mathrm { y } = \mathrm { x }$ is a line of symmetry of $C$.
\item In this question you must show detailed reasoning.

Find the exact area of each of the loops of $C$.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Pure Core 1 2022 Q5 [11]}}
This paper (9 questions)
View full paper
Q1 6 Q2 9 Q3 11 Q4 4 Q5 11 Q6 6 Q7 10 Q8 13 Q9 5