CAIE P1 2024 June — Question 5 6 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2024
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTrig Proofs
TypeSolve equation using proven identity
DifficultyModerate -0.8 Part (a) is a straightforward algebraic proof using the Pythagorean identity sin²x = 1 - cos²x to simplify the numerator, requiring only routine manipulation. Part (b) is a direct application of the proven identity leading to a simple linear equation in cos x. Both parts involve standard techniques with no problem-solving insight required, making this easier than average.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals
  1. Prove the identity \(\frac{\sin^2 x - \cos x - 1}{1 + \cos x} \equiv -\cos x\). [3]
  2. Hence solve the equation \(\frac{\sin^2 x - \cos x - 1}{2 + 2\cos x} = \frac{1}{4}\) for \(0° \leq x \leq 360°\). [3]
Question 5:
AnswerMarks
5(a)sin2 xcosx1 1cos2 xcosx1 cos2 xcosx
 or
AnswerMarks Guidance
1cosx 1cosx 1cosxM1 For use of sin2 xcos2 x1.
Allow use of s, c, t or omission of x throughout.
cosx1cosx
=
AnswerMarks Guidance
1cosxM1 For factorising.
cosx
AnswerMarks
=A1
3
AnswerMarks
5(b)1 1  1
 cosx ⇒ x  cos1  
AnswerMarks
2 4  2M1
x120 or x240A1
A1 FTFT for 360 – their answer. A1 A0 if extra solution(s) in
2π 4π
range. SC B1 if answer in radians for both , .
3 3
3
AnswerMarks Guidance
QuestionAnswer Marks 
Question 5:
--- 5(a) ---
5(a) | sin2 xcosx1 1cos2 xcosx1 cos2 xcosx
 or
1cosx 1cosx 1cosx | M1 | For use of sin2 xcos2 x1.
Allow use of s, c, t or omission of x throughout.
cosx1cosx
=
1cosx | M1 | For factorising.
cosx
= | A1
3
--- 5(b) ---
5(b) | 1 1  1
 cosx ⇒ x  cos1  
2 4  2 | M1
x120 or x240 | A1
A1 FT | FT for 360 – their answer. A1 A0 if extra solution(s) in
2π 4π
range. SC B1 if answer in radians for both , .
3 3
3
Question | Answer | Marks | Guidance
\begin{enumerate}[label=(\alph*)]
\item Prove the identity $\frac{\sin^2 x - \cos x - 1}{1 + \cos x} \equiv -\cos x$. [3]

\item Hence solve the equation $\frac{\sin^2 x - \cos x - 1}{2 + 2\cos x} = \frac{1}{4}$ for $0° \leq x \leq 360°$. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2024 Q5 [6]}}
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Q1 5 Q2 6 Q3 5 Q4 3 Q5 6 Q6 7 Q7 8 Q8 8 Q9 6 Q10 8 Q11 13