OCR MEI Further Pure Core 2021 November — Question 14 14 marks

Exam BoardOCR MEI
ModuleFurther Pure Core (Further Pure Core)
Year2021
SessionNovember
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPolar coordinates
TypeMaximum/minimum distance from pole or line
DifficultyStandard +0.8 This is a Further Maths polar coordinates question requiring optimization via calculus (finding maximum r), proving a curve is a circle by converting to Cartesian form or completing the square, and identifying geometric properties. While the techniques are standard for FM students, the multi-part nature and need to work fluently between polar and Cartesian representations makes it moderately challenging.
Spec4.09a Polar coordinates: convert to/from cartesian4.09b Sketch polar curves: r = f(theta)
14 A curve has polar equation \(\mathrm { r } = \mathrm { a } ( \cos \theta + 2 \sin \theta )\), where \(a\) is a positive constant and \(0 \leqslant \theta \leqslant \pi\).
  1. Determine the polar coordinates of the point on the curve which is furthest from the pole.
    1. Show that the curve is a circle whose radius should be specified.
    2. Write down the polar coordinates of the centre of the circle.
Question 14:
AnswerMarks Guidance
14(a) Let c o s 2 s i n R c o s ( )     + = −
=Rcoscos+Rsinsin
Rcos=1, Rsin=2
R= 5
t a n 2  =
1 .1 0 7  =
so r 5 a c o s ( 1 .1 0 7 )  = −
r is maximum when c o s ( 1 .1 0 7 ) 1  − =
1 .1 0 7 r a d   =
AnswerMarks
polar coordinates are  5 a , 1 .1 0 7 M1A1
M1
A1
M1
A1
AnswerMarks
A13.1a
1.1
1.1
1.1
3.1a
1.1
AnswerMarks
1.1or arctan(2)
or arctan(2)
AnswerMarks
Alternative solution2.24 or better
d r
a ( s i n 2 c o s )   = − +
d 
M1A1
M1
d r
r is maximum when 0 =
d 
AnswerMarks
2 c o s s i n 0    − + =M1
t a n 2 1 .1 0 7    =  =
A1B1
when 1 .1 0 7 , r 5 a o r 2 .2 3 6 a  = =
B1
polar coordinates are [2.236a, 1.107]
[7]
AnswerMarks Guidance
14(b) (i)
 x 2 + y 2 = a x + 2 a y
 ( x − 12 a ) 2 + ( y − a ) 2 = 54 a 2
This is the cartesian equation of a circle
AnswerMarks
radius 12 5 aM1
M1
A1
M1
A1
A1
AnswerMarks
[6]3.1a
3.1a
1.1
2.1
2.2a
AnswerMarks
3.2aattempt to find cartesian eqn
multiplying by r
completing the square
AnswerMarks Guidance
14(b) (ii)
[1]1.1 
Question 14:
14 | (a) | Let c o s 2 s i n R c o s ( )     + = −
=Rcoscos+Rsinsin
Rcos=1, Rsin=2
R= 5
t a n 2  =
1 .1 0 7  =
so r 5 a c o s ( 1 .1 0 7 )  = −
r is maximum when c o s ( 1 .1 0 7 ) 1  − =
1 .1 0 7 r a d   =
polar coordinates are  5 a , 1 .1 0 7  | M1A1
M1
A1
M1
A1
A1 | 3.1a
1.1
1.1
1.1
3.1a
1.1
1.1 | or arctan(2)
or arctan(2)
Alternative solution | 2.24 or better
d r
a ( s i n 2 c o s )   = − +
d 
M1A1
M1
d r
r is maximum when 0 =
d 
2 c o s s i n 0    − + = | M1
t a n 2 1 .1 0 7    =  =
A1B1
when 1 .1 0 7 , r 5 a o r 2 .2 3 6 a  = =
B1
polar coordinates are [2.236a, 1.107]
[7]
14 | (b) | (i) | r 2 a r c o s 2 a r s i n   = +
 x 2 + y 2 = a x + 2 a y
 ( x − 12 a ) 2 + ( y − a ) 2 = 54 a 2
This is the cartesian equation of a circle
radius 12 5 a | M1
M1
A1
M1
A1
A1
[6] | 3.1a
3.1a
1.1
2.1
2.2a
3.2a | attempt to find cartesian eqn
multiplying by r
completing the square
14 | (b) | (ii) | centre  12 5 a , 1 .1 0 7  | B1
[1] | 1.1
14 A curve has polar equation $\mathrm { r } = \mathrm { a } ( \cos \theta + 2 \sin \theta )$, where $a$ is a positive constant and $0 \leqslant \theta \leqslant \pi$.
\begin{enumerate}[label=(\alph*)]
\item Determine the polar coordinates of the point on the curve which is furthest from the pole.
\item \begin{enumerate}[label=(\roman*)]
\item Show that the curve is a circle whose radius should be specified.
\item Write down the polar coordinates of the centre of the circle.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure Core 2021 Q14 [14]}}
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