OCR Further Pure Core 2 2019 June — Question 9 11 marks

Exam BoardOCR
ModuleFurther Pure Core 2 (Further Pure Core 2)
Year2019
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPolar coordinates
TypeMaximum/minimum distance from pole or line
DifficultyChallenging +1.2 This is a Further Maths polar coordinates question requiring (a) integration for area and (b) optimization using calculus. While it involves multiple techniques (polar area formula, differentiation, solving transcendental equations), these are standard procedures for FM students. The optimization is straightforward once dr/dθ=0 is set up, and the exact form suggests the algebra works out cleanly. Moderately above average difficulty due to FM content and multi-step nature, but follows predictable patterns.
Spec4.09a Polar coordinates: convert to/from cartesian4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve
  1. Find the exact area enclosed by the curve.
  2. Show that the greatest value of \(r\) on the curve is \(\sqrt { \frac { \sqrt { 3 } } { 2 } } \mathrm { e } ^ { \frac { 1 } { 6 } }\).
Question 9:
AnswerMarks Guidance
9(a) DR
1 2
1 π 23 𝑐𝑐 𝑐𝑐 𝑐𝑐𝜃𝜃
�1�√ 𝑐𝑐𝑠𝑠𝑛𝑛𝑐𝑐𝑒𝑒 c o sθ � d𝑐𝑐
2A= ∫sinθe3 dθ
2
0
π
1 3 2 cosθ 
= ×− e3 
2 2 
0
3 2 2
−
e3 −e 3
AnswerMarks
4 M1
*A1
dep*M
1
A1
AnswerMarks
[4]3.1a
2.1
1.1a
AnswerMarks
1.1Correct form, in terms of θ,
Integrand has been squared out.
Must include limits (can be seen
later)
Might be as result of substitution
Allow coefficient error for M1
AnswerMarks
iswM1 can be implied by 1.0757...
BC
1
3 2 3 2 u 
eg 4   eu  − 3 2 or 4   e3   oe
3 −1
1 1 1 cosθ
(sinθ)2(− sinθ)e3
3
dr 1 − 1 1 cosθ
= (sinθ) 2e3 (3cosθ−2sin2θ)
dθ 6
dr
=0⇒3cosθ−2sin2θ=0
dθ
2cos2θ+3cosθ−2=0
cosθ = ½, –2
cosθ≠−2
3 3 1 × 1 3 1
⇒sinθ= ⇒r = e3 2 = e6
AnswerMarks
2 2 2dep*M
1
M1
*A1
dep*A1
A1
AnswerMarks
[7]2.2a
2.1
1.1
2.3
AnswerMarks
2.2aSetting r to zero and
factorising/cancelling to produce
a quadrati’c equation in cos and/or
sin
Use of cos2 + sin2 = 1 to find 3
term quadratic equation in cosθ.
Solving quadratic correctly
Explicitly rejecting root
AG. At least one intermediate
AnswerMarks
step must be seen.Or could be in sin2θ;
4sin4θ+9sin2θ−9=0
sin2θ = ¾, –3
Rejects sin2θ = –3 and
𝑐𝑐𝑠𝑠𝑛𝑛 𝑐𝑐 =
Ca√n 3 be awarded even if rejection
− 2
of root(s) was implicit.
AnswerMarks
(b)1  π π
ln +cosx=0⇒x= (or − )
2  3 3
π 2
 
3 3
∴ln − ≈0
2 3
3 π2 3 3
ln − ≈0⇒π≈ 27ln  =3 3ln 
AnswerMarks
2 27 2 2B1
M1
A1
AnswerMarks
[3]1.1
3.1a
AnswerMarks
3.2aFinding either ±π/3 as a root.
Allow 60° for B1. Ignore other
roots
Substituting their root, in radians,
into their Maclaurin series and
equating (approximately) to 0.
Could see ± but must be removed
by final conclusion.
Must use approximately equals
AnswerMarks
symbol (not just equals symbol)Or equating their expression
(approximately) to 0 and
rearranging for x:
3 x2 3
ln − ≈0⇒x≈ 3ln 
2 3 2
π 3 3
≈ 3ln ⇒π≈3 3ln 
3 2 2
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Cambridge
CB2 8EA
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Question 9:
9 | (a) | DR
1 2
1 π 23 𝑐𝑐 𝑐𝑐 𝑐𝑐𝜃𝜃
�1�√ 𝑐𝑐𝑠𝑠𝑛𝑛𝑐𝑐𝑒𝑒 c o sθ � d𝑐𝑐
2A= ∫sinθe3 dθ
2
0
π
1 3 2 cosθ 
= ×− e3 
2 2 
0
3 2 2
−
e3 −e 3
4  | M1
*A1
dep*M
1
A1
[4] | 3.1a
2.1
1.1a
1.1 | Correct form, in terms of θ,
Integrand has been squared out.
Must include limits (can be seen
later)
Might be as result of substitution
Allow coefficient error for M1
isw | M1 can be implied by 1.0757...
BC
1
3 2 3 2 u 
eg 4   eu  − 3 2 or 4   e3   oe
3 −1
1 1 1 cosθ
(sinθ)2(− sinθ)e3
3
dr 1 − 1 1 cosθ
= (sinθ) 2e3 (3cosθ−2sin2θ)
dθ 6
dr
=0⇒3cosθ−2sin2θ=0
dθ
2cos2θ+3cosθ−2=0
cosθ = ½, –2
cosθ≠−2
3 3 1 × 1 3 1
⇒sinθ= ⇒r = e3 2 = e6
2 2 2 | dep*M
1
M1
*A1
dep*A1
A1
[7] | 2.2a
2.1
1.1
2.3
2.2a | Setting r to zero and
factorising/cancelling to produce
a quadrati’c equation in cos and/or
sin
Use of cos2 + sin2 = 1 to find 3
term quadratic equation in cosθ.
Solving quadratic correctly
Explicitly rejecting root
AG. At least one intermediate
step must be seen. | Or could be in sin2θ;
4sin4θ+9sin2θ−9=0
sin2θ = ¾, –3
Rejects sin2θ = –3 and
𝑐𝑐𝑠𝑠𝑛𝑛 𝑐𝑐 =
Ca√n 3 be awarded even if rejection
− 2
of root(s) was implicit.
(b) | 1  π π
ln +cosx=0⇒x= (or − )
2  3 3
π 2
 
3 3
∴ln − ≈0
2 3
3 π2 3 3
ln − ≈0⇒π≈ 27ln  =3 3ln 
2 27 2 2 | B1
M1
A1
[3] | 1.1
3.1a
3.2a | Finding either ±π/3 as a root.
Allow 60° for B1. Ignore other
roots
Substituting their root, in radians,
into their Maclaurin series and
equating (approximately) to 0.
Could see ± but must be removed
by final conclusion.
Must use approximately equals
symbol (not just equals symbol) | Or equating their expression
(approximately) to 0 and
rearranging for x:
3 x2 3
ln − ≈0⇒x≈ 3ln 
2 3 2
π 3 3
≈ 3ln ⇒π≈3 3ln 
3 2 2
PPMMTT
OCR (Oxford Cambridge and RSA Examinations)
The Triangle Building
Shaftesbury Road
Cambridge
CB2 8EA
OCR Customer Contact Centre
Education and Learning
Telephone: 01223 553998
Facsimile: 01223 552627
Email: general.qualifications@ocr.org.uk
www.ocr.org.uk
For staff training purposes and as part of our quality assurance
programme your call may be recorded or monitored
Oxford Cambridge and RSA Examinations
is a Company Limited by Guarantee
Registered in England
Registered Office; The Triangle Building, Shaftesbury Road, Cambridge, CB2 8EA
Registered Company Number: 3484466
OCR is an exempt Charity
OCR (Oxford Cambridge and RSA Examinations)
Head office
Telephone: 01223 552552
Facsimile: 01223 552553
© OCR 2019
\begin{enumerate}[label=(\alph*)]
\item Find the exact area enclosed by the curve.
\item Show that the greatest value of $r$ on the curve is $\sqrt { \frac { \sqrt { 3 } } { 2 } } \mathrm { e } ^ { \frac { 1 } { 6 } }$.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Pure Core 2 2019 Q9 [11]}}
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Q1 7 Q2 8 Q3 5 Q4 5 Q5 11 Q6 6 Q7 7 Q8 8 Q9 11 Q10 7