CAIE Further Paper 1 2024 November — Question 5 13 marks

Exam BoardCAIE
ModuleFurther Paper 1 (Further Paper 1)
Year2024
SessionNovember
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPolar coordinates
TypeMaximum/minimum distance from pole or line
DifficultyChallenging +1.2 This is a standard Further Maths polar coordinates question with routine conversions and calculus. Part (a) is direct substitution, parts (b-c) use standard formulas, but part (d) requires optimizing y = r sin θ with the constraint, involving implicit differentiation and some algebraic manipulation—moderately challenging but well within expected FM techniques.
Spec4.09a Polar coordinates: convert to/from cartesian4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve
  1. Show that the curve with Cartesian equation \(\left(x^2 + y^2\right)^2 = 6xy\) has polar equation \(r^2 = 3\sin 2\theta\). [2]
The curve \(C\) has polar equation \(r^2 = 3\sin 2\theta\), for \(0 \leqslant \theta \leqslant \frac{1}{2}\pi\).
  1. Sketch \(C\) and state the maximum distance of a point on \(C\) from the pole. [3]
  2. Find the area of the region enclosed by \(C\). [2]
  3. Find the maximum distance of a point on \(C\) from the initial line. [6]
Question 5:
AnswerMarks Guidance
5(a)r4 =6r2sincos M1
sin2=2sincos.
AnswerMarks Guidance
r2 =3sin2A1 AG.
2
AnswerMarks Guidance
5(b)=0 B1
4
AnswerMarks Guidance
B1Single correct loop.
3B1 States maximum distance or labels sketch.
( ) ( )
Allow 3,1π but not 1π, 3 . Allow 3sf.
4 4
3
AnswerMarks
5(c)π π
32sin2d= 3 −1cos2 2
 
AnswerMarks Guidance
2 0 2 2 0M1 Forms 1r2 d. (Allow with wrong limits.)
2
3
AnswerMarks
2A1
2
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks Guidance
5(d)y=3 1 2sin 1 22sin B1
sin 1 2 2cos+sin −1 2 2cos2sin=0M1 A1 dy
Sets =0.
d
2tan
sin2cos+cos2sin=0tan2=−tan =−tan
AnswerMarks Guidance
1−tan2M1 Applies suitable trigonometric identity.
Accept sin3=0.
= 1π
AnswerMarks
3A1
5
34
=1.40
3
AnswerMarks Guidance
22A1 AEF.
6
AnswerMarks Guidance
QuestionAnswer Marks 
Question 5:
--- 5(a) ---
5(a) | r4 =6r2sincos | M1 | Substitutes x=rcos, y=rsin and applies
sin2=2sincos.
r2 =3sin2 | A1 | AG.
2
--- 5(b) ---
5(b) | =0 | B1 | Correct position and symmetrical about = 1π.
4
B1 | Single correct loop.
3 | B1 | States maximum distance or labels sketch.
( ) ( )
Allow 3,1π but not 1π, 3 . Allow 3sf.
4 4
3
--- 5(c) ---
5(c) | π π
32sin2d= 3 −1cos2 2
 
2 0 2 2 0 | M1 | Forms 1r2 d. (Allow with wrong limits.)
2
3
2 | A1
2
Question | Answer | Marks | Guidance
--- 5(d) ---
5(d) | y=3 1 2sin 1 22sin | B1
sin 1 2 2cos+sin −1 2 2cos2sin=0 | M1 A1 | dy
Sets =0.
d
2tan
sin2cos+cos2sin=0tan2=−tan =−tan
1−tan2 | M1 | Applies suitable trigonometric identity.
Accept sin3=0.
= 1π
3 | A1
5
34
=1.40
3
22 | A1 | AEF.
6
Question | Answer | Marks | Guidance
\begin{enumerate}[label=(\alph*)]
\item Show that the curve with Cartesian equation
$\left(x^2 + y^2\right)^2 = 6xy$
has polar equation $r^2 = 3\sin 2\theta$. [2]
\end{enumerate}

The curve $C$ has polar equation $r^2 = 3\sin 2\theta$, for $0 \leqslant \theta \leqslant \frac{1}{2}\pi$.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Sketch $C$ and state the maximum distance of a point on $C$ from the pole. [3]

\item Find the area of the region enclosed by $C$. [2]

\item Find the maximum distance of a point on $C$ from the initial line. [6]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 1 2024 Q5 [13]}}
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