CAIE P1 2023 November — Question 5 6 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2023
SessionNovember
Marks6
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Mark schemeDownload PDF ↗
TopicGeometric Sequences and Series
TypeGP with trigonometric terms
DifficultyChallenging +1.2 This question requires setting up and solving equations using the GP property (r = second/first = third/second), leading to a trigonometric equation that needs algebraic manipulation. Part (b) involves applying the sum formula with a surd common ratio and rationalizing the denominator. While it combines multiple techniques (GP properties, trigonometry, algebraic manipulation, and surd rationalization), each step follows standard procedures without requiring novel insight. The multi-step nature and combination of topics makes it moderately harder than average.
Spec1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<11.05g Exact trigonometric values: for standard angles
The first, second and third terms of a geometric progression are \(\sin\theta\), \(\cos\theta\) and \(2 - \sin\theta\) respectively, where \(\theta\) radians is an acute angle.
  1. Find the value of \(\theta\). [3]
  2. Using this value of \(\theta\), find the sum of the first 10 terms of the progression. Give the answer in the form \(\frac{b}{\sqrt{c} - 1}\), where \(b\) and \(c\) are integers to be found. [3]
Question 5:
AnswerMarks
5(a)cos 2−sin
= leading to cos2 sin=sin(2−sin) sin
AnswerMarks Guidance
sin cos*M1 OE. Forming a correct equation in  only using the terms of
the GP and an attempt to clear fractions.
1
cos2+sin2=2sin leading to sin=
 
AnswerMarks Guidance
2DM1 Correct use of cos2+sin2=1 and attempt to solve for
sin.
π
=
or 0.524
AnswerMarks Guidance
6A1 AWRT
5π
A0 for =30. Condone inclusion of and/or
6
3
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
5(b)1
a= r= 3
AnswerMarks Guidance
2B1 OE SOI
Trigonometric values need to have been evaluated but allow
decimal equivalents (0.5 and 1.73 AWRT)
 ( )10 
1− their 3
 π 
S 10 =sin their  ( ) 
 6 1− their 3 
AnswerMarks Guidance
 M1 Use of a correct formula for S with their value of .
10,
Their 3 needs to come from
cos(their) 2−sin(their)
or OE
sin(their) cos(their)
121
[S =]
AnswerMarks Guidance
10 3−1A1 −121
or 165 AWRT scores B1M1A0.
1− 3
3
AnswerMarks Guidance
QuestionAnswer Marks 
Question 5:
--- 5(a) ---
5(a) | cos 2−sin
= leading to cos2 sin=sin(2−sin) sin
sin cos | *M1 | OE. Forming a correct equation in  only using the terms of
the GP and an attempt to clear fractions.
1
cos2+sin2=2sin leading to sin=
 
2 | DM1 | Correct use of cos2+sin2=1 and attempt to solve for
sin.
π
=
or 0.524
6 | A1 | AWRT
5π
A0 for =30. Condone inclusion of and/or
6
3
Question | Answer | Marks | Guidance
--- 5(b) ---
5(b) | 1
a= r= 3
2 | B1 | OE SOI
Trigonometric values need to have been evaluated but allow
decimal equivalents (0.5 and 1.73 AWRT)
 ( )10 
1− their 3
 π 
S 10 =sin their  ( ) 
 6 1− their 3 
  | M1 | Use of a correct formula for S with their value of .
10,
Their 3 needs to come from
cos(their) 2−sin(their)
or OE
sin(their) cos(their)
121
[S =]
10 3−1 | A1 | −121
or 165 AWRT scores B1M1A0.
1− 3
3
Question | Answer | Marks | Guidance
The first, second and third terms of a geometric progression are $\sin\theta$, $\cos\theta$ and $2 - \sin\theta$ respectively, where $\theta$ radians is an acute angle.

\begin{enumerate}[label=(\alph*)]
\item Find the value of $\theta$. [3]

\item Using this value of $\theta$, find the sum of the first 10 terms of the progression. Give the answer in the form $\frac{b}{\sqrt{c} - 1}$, where $b$ and $c$ are integers to be found. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2023 Q5 [6]}}
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