CAIE P3 2023 June — Question 4 5 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2023
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAddition & Double Angle Formulae
TypeEquation with half angles
DifficultyStandard +0.8 This question requires using the double angle formula to express cos x in terms of cos(x/2), leading to a quadratic equation in cos(x/2). Students must then solve the quadratic, apply inverse cosine carefully to the half-angle, and find all solutions in the given range. The half-angle manipulation and ensuring all solutions are found makes this moderately harder than standard trigonometric equations.
Spec1.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals

4 Solve the equation \(2 \cos x - \cos \frac { 1 } { 2 } x = 1\) for \(0 \leqslant x \leqslant 2 \pi\).

Question 4:
AnswerMarks Guidance
AnswerMark Guidance
Use correct double angle formula to obtain an equation in \(\cos\left(\frac{x}{2}\right)\) only*M1 e.g. \(2\left(2\cos^2\left(\frac{x}{2}\right)-1\right) - \cos\left(\frac{x}{2}\right) = 1\)
Obtain a 3 term quadratic in \(\cos\left(\frac{x}{2}\right)\)A1 e.g. \(4\cos^2\left(\frac{x}{2}\right) - \cos\left(\frac{x}{2}\right) - 3 = 0\). Allow \(4\cos^2 u - \cos u - 3 = 0\), condone \(\frac{x}{2} = x\)
Obtain \(\cos\left(\frac{x}{2}\right) = -\frac{3}{4}\) and \(\cos\left(\frac{x}{2}\right) = 1\)A1 Allow answer in \(u\) e.g. \((4\cos u + 3)(\cos u - 1)\) and condone \(\frac{x}{2} = x\)
Solve for the original \(x\)DM1 Must see evidence of doubling, not halving
Obtain \(x = 0\) and \(4.84\) and no others in the intervalA1 Ignore any answers outside interval. Accept AWRT 4.84. Accept \(1.54\pi\). Must be in radians. 277.2 indicates M1 but is A0
Alternative Method:
AnswerMarks Guidance
AnswerMark Guidance
Use correct double angle formula to obtain an equation in \(\cos x\) only*M1 e.g. \(2\cos x - 1 = \sqrt{\frac{\cos x + 1}{2}}\)
Obtain a 3 term quadratic in \(\cos x\)A1 e.g. \(8\cos^2 x - 9\cos x + 1 = 0\)
Obtain \(\cos x = \frac{1}{8}\) and \(\cos x = 1\)A1
Solve for \(x\)DM1
Obtain answers \(x = 0\) and \(4.84\) and no others in the intervalA1 Ignore any answers outside interval. Accept AWRT 4.84. Must be in radians. 277.2 is A0
Total: 5
## Question 4:

| Answer | Mark | Guidance |
|--------|------|----------|
| Use correct double angle formula to obtain an equation in $\cos\left(\frac{x}{2}\right)$ only | *M1 | e.g. $2\left(2\cos^2\left(\frac{x}{2}\right)-1\right) - \cos\left(\frac{x}{2}\right) = 1$ |
| Obtain a 3 term quadratic in $\cos\left(\frac{x}{2}\right)$ | A1 | e.g. $4\cos^2\left(\frac{x}{2}\right) - \cos\left(\frac{x}{2}\right) - 3 = 0$. Allow $4\cos^2 u - \cos u - 3 = 0$, condone $\frac{x}{2} = x$ |
| Obtain $\cos\left(\frac{x}{2}\right) = -\frac{3}{4}$ and $\cos\left(\frac{x}{2}\right) = 1$ | A1 | Allow answer in $u$ e.g. $(4\cos u + 3)(\cos u - 1)$ and condone $\frac{x}{2} = x$ |
| Solve for the **original** $x$ | DM1 | Must see evidence of doubling, not halving |
| Obtain $x = 0$ and $4.84$ and no others in the interval | A1 | Ignore any answers outside interval. Accept AWRT 4.84. Accept $1.54\pi$. Must be in radians. 277.2 indicates M1 but is A0 |

**Alternative Method:**

| Answer | Mark | Guidance |
|--------|------|----------|
| Use correct double angle formula to obtain an equation in $\cos x$ only | *M1 | e.g. $2\cos x - 1 = \sqrt{\frac{\cos x + 1}{2}}$ |
| Obtain a 3 term quadratic in $\cos x$ | A1 | e.g. $8\cos^2 x - 9\cos x + 1 = 0$ |
| Obtain $\cos x = \frac{1}{8}$ and $\cos x = 1$ | A1 | |
| Solve for $x$ | DM1 | |
| Obtain answers $x = 0$ and $4.84$ and no others in the interval | A1 | Ignore any answers outside interval. Accept AWRT 4.84. Must be in radians. 277.2 is A0 |
| **Total: 5** | | |

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4 Solve the equation $2 \cos x - \cos \frac { 1 } { 2 } x = 1$ for $0 \leqslant x \leqslant 2 \pi$.\\

\hfill \mbox{\textit{CAIE P3 2023 Q4 [5]}}