CAIE P3 2018 June — Question 6 6 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2018
SessionJune
Marks6
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Mark schemeDownload PDF ↗
TopicRadians, Arc Length and Sector Area
TypeTangent and sector - two tangents from external point
DifficultyStandard +0.3 This is a straightforward application of standard circle geometry formulas (arc length, sector area, triangle area) with tangent properties. Part (i) requires setting up an equation by equating areas, leading to a simple algebraic result. Part (ii) involves direct substitution into formulas. Both parts are routine calculations with no novel insight required, making this slightly easier than average.
Spec1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta3.04a Calculate moments: about a point
\includegraphics{figure_6} The diagram shows points \(A\) and \(B\) on a circle with centre \(O\) and radius \(r\). The tangents to the circle at \(A\) and \(B\) meet at \(T\). The shaded region is bounded by the minor arc \(AB\) and the lines \(AT\) and \(BT\). Angle \(AOB\) is \(2\theta\) radians.
  1. In the case where the area of the sector \(AOB\) is the same as the area of the shaded region, show that \(\tan \theta = 2\theta\). [3]
  2. In the case where \(r = 8\) cm and the length of the minor arc \(AB\) is 19.2 cm, find the area of the shaded region. [3]
Question 6:
AnswerMarks Guidance
6(i)Use correct method for finding the area of a segment and area of
semicircle and form an equation in θM1 πa2 1 1
e.g. = a2θ− a2sinθ
4 2 2
AnswerMarks Guidance
State a correct equation in any formA1 Given answer so check working carefully
Obtain the given answer correctlyA1
3
AnswerMarks Guidance
6(ii)Calculate values of a relevant expression or pair of expressions
at θ=2.2 and θ = 2.4M1 e.g. f ( θ )= π 2 +sinθ    f f ( ( 2 2 . . 2 4 ) ) = = 2 2 . . 3 2 7 4 . . . . . .< > 2 2 . . 4 2
( )=θ− π  f ( 2 . 2 ) = − 0 . 1 7 . . .< 0
or f θ 2 −sinθ   f ( 2 . 4 ) = + 0 . 1 5 . . . > 0
AnswerMarks
Complete the argument correctly with correct calculated valuesA1
2
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
6(iii)1
Use θ = π+ sinθ correctly at least once
AnswerMarks Guidance
n+1 2 nM1 e.g.
2.2 2.3 2.4
2.3793 2.3165 2.2463
2.2614 2.3054 2.3512
2.3417 2.3129 2.2814
2.2881 2.3079 2.3288
2.3244 2.2970
2.3000 2.3185
2.3165 2.3041
2.3054 2.3138
2.3129 2.3072
AnswerMarks
Obtain final answer 2.31A1
Show sufficient iterations to 4 d.p. to justify 2.31 to 2 d.p. or
AnswerMarks
show there is a sign change in the interval (2.305, 2.315)A1
3
AnswerMarks Guidance
2.22.3 2.4
2.37932.3165 2.2463
2.26142.3054 2.3512
2.34172.3129 2.2814
2.28812.3079 2.3288
2.32442.2970
2.30002.3185
2.31652.3041
2.30542.3138
2.31292.3072
QuestionAnswer Marks 
Question 6:
--- 6(i) ---
6(i) | Use correct method for finding the area of a segment and area of
semicircle and form an equation in θ | M1 | πa2 1 1
e.g. = a2θ− a2sinθ
4 2 2
State a correct equation in any form | A1 | Given answer so check working carefully
Obtain the given answer correctly | A1
3
--- 6(ii) ---
6(ii) | Calculate values of a relevant expression or pair of expressions
at θ=2.2 and θ = 2.4 | M1 | e.g. f ( θ )= π 2 +sinθ    f f ( ( 2 2 . . 2 4 ) ) = = 2 2 . . 3 2 7 4 . . . . . .< > 2 2 . . 4 2
( )=θ− π  f ( 2 . 2 ) = − 0 . 1 7 . . .< 0
or f θ 2 −sinθ   f ( 2 . 4 ) = + 0 . 1 5 . . . > 0
Complete the argument correctly with correct calculated values | A1
2
Question | Answer | Marks | Guidance
--- 6(iii) ---
6(iii) | 1
Use θ = π+ sinθ correctly at least once
n+1 2 n | M1 | e.g.
2.2 2.3 2.4
2.3793 2.3165 2.2463
2.2614 2.3054 2.3512
2.3417 2.3129 2.2814
2.2881 2.3079 2.3288
2.3244 2.2970
2.3000 2.3185
2.3165 2.3041
2.3054 2.3138
2.3129 2.3072
Obtain final answer 2.31 | A1
Show sufficient iterations to 4 d.p. to justify 2.31 to 2 d.p. or
show there is a sign change in the interval (2.305, 2.315) | A1
3
2.2 | 2.3 | 2.4
2.3793 | 2.3165 | 2.2463
2.2614 | 2.3054 | 2.3512
2.3417 | 2.3129 | 2.2814
2.2881 | 2.3079 | 2.3288
2.3244 | 2.2970
2.3000 | 2.3185
2.3165 | 2.3041
2.3054 | 2.3138
2.3129 | 2.3072
Question | Answer | Marks | Guidance
\includegraphics{figure_6}

The diagram shows points $A$ and $B$ on a circle with centre $O$ and radius $r$. The tangents to the circle at $A$ and $B$ meet at $T$. The shaded region is bounded by the minor arc $AB$ and the lines $AT$ and $BT$. Angle $AOB$ is $2\theta$ radians.

\begin{enumerate}[label=(\roman*)]
\item In the case where the area of the sector $AOB$ is the same as the area of the shaded region, show that $\tan \theta = 2\theta$. [3]

\item In the case where $r = 8$ cm and the length of the minor arc $AB$ is 19.2 cm, find the area of the shaded region. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE P3 2018 Q6 [6]}}
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