CAIE P1 2024 November — Question 6 6 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2024
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRadians, Arc Length and Sector Area
TypeSimultaneous equations with arc/area
DifficultyStandard +0.8 This question requires setting up two simultaneous equations from perimeter and area conditions (involving both linear and quadratic terms in r and θ), then solving a non-linear system with constraints. While the individual formulas for arc length and sector area are standard, the algebraic manipulation to solve the system and verify constraints requires careful multi-step reasoning beyond routine application.
Spec1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta
\includegraphics{figure_6} The diagram shows a metal plate \(OABCDEF\) consisting of sectors of two circles, each with centre \(O\). The radii of sectors \(AOB\) and \(EOF\) are \(r\) cm and the radius of sector \(COD\) is \(2r\) cm. Angle \(AOB =\) angle \(EOF = \theta\) radians and angle \(COD = 2\theta\) radians. It is given that the perimeter of the plate is 14 cm and the area of the plate is 10 cm\(^2\). Given that \(r \geqslant \frac{3}{2}\) and \(\theta < \frac{3}{4}\), find the values of \(r\) and \(\theta\). [6]
Question 6:
AnswerMarks Guidance
6[Perimeter =] r+r+r+2r2+r+r+r =4r+6r B1
1 1 1
[Area =] r2+ (2r)22+ r2 = 5r2
 
AnswerMarks Guidance
2 2 2B1
4r+6r=14 and 5r2=10M1* ar+br=14 and cr2=10where a, b and c are constants0.
Terms may be uncollected.
EITHER
14 −4r  10 
5r2 =10 or 4r+6r =14
AnswerMarks Guidance
6r 5r2 DM1 Eliminateto get an equation in r.
 2r2 −7r+6=0  (r−2)(2r−3)=0
AnswerMarks Guidance
 DM1 Factorise or other accepted method for solving their 3-term
quadratic.
OR
 14  2  10   10 
5  =10 or 4    +6    =14
4+6 5 5
AnswerMarks Guidance
   DM1 Eliminate r to get an equation in.
[1 82−25+8=0] (9−8)(2−1)=0DM1 Factorise or other accepted method for solving their 3-term
quadratic.
Then
AnswerMarks Guidance
r=2 and=0.5B1 3 8
Condone extra answers r = and= .
2 9
6
AnswerMarks Guidance
QuestionAnswer Marks 
Question 6:
6 | [Perimeter =] r+r+r+2r2+r+r+r =4r+6r | B1
1 1 1
[Area =] r2+ (2r)22+ r2 = 5r2
 
2 2 2 | B1
4r+6r=14 and 5r2=10 | M1* | ar+br=14 and cr2=10where a, b and c are constants0.
Terms may be uncollected.
EITHER
14 −4r  10 
5r2 =10 or 4r+6r =14
6r 5r2  | DM1 | Eliminateto get an equation in r.
 2r2 −7r+6=0  (r−2)(2r−3)=0
  | DM1 | Factorise or other accepted method for solving their 3-term
quadratic.
OR
 14  2  10   10 
5  =10 or 4    +6    =14
4+6 5 5
    | DM1 | Eliminate r to get an equation in.
[1 82−25+8=0] (9−8)(2−1)=0 | DM1 | Factorise or other accepted method for solving their 3-term
quadratic.
Then
r=2 and=0.5 | B1 | 3 8
Condone extra answers r = and= .
2 9
6
Question | Answer | Marks | Guidance
\includegraphics{figure_6}

The diagram shows a metal plate $OABCDEF$ consisting of sectors of two circles, each with centre $O$. The radii of sectors $AOB$ and $EOF$ are $r$ cm and the radius of sector $COD$ is $2r$ cm. Angle $AOB =$ angle $EOF = \theta$ radians and angle $COD = 2\theta$ radians.

It is given that the perimeter of the plate is 14 cm and the area of the plate is 10 cm$^2$.

Given that $r \geqslant \frac{3}{2}$ and $\theta < \frac{3}{4}$, find the values of $r$ and $\theta$. [6]

\hfill \mbox{\textit{CAIE P1 2024 Q6 [6]}}
This paper (10 questions)
View full paper
Q1 5 Q2 5 Q3 5 Q4 6 Q5 10 Q6 6 Q7 8 Q8 10 Q9 10 Q10 10