CAIE P3 2007 June — Question 5 7 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2007
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHarmonic Form
TypeIntegration using harmonic form
DifficultyStandard +0.3 This is a standard two-part harmonic form question requiring routine application of R cos(θ-α) expansion followed by straightforward integration. Part (i) involves standard coefficient matching (R=2, α=π/3), and part (ii) uses a simple substitution u=θ-π/3 with elementary integration of sec²u. While it requires multiple steps, all techniques are textbook exercises with no novel insight needed, making it slightly easier than average.
Spec1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc1.08d Evaluate definite integrals: between limits

5
  1. Express \(\cos \theta + ( \sqrt { } 3 ) \sin \theta\) in the form \(R \cos ( \theta - \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { 1 } { 2 } \pi\), giving the exact values of \(R\) and \(\alpha\).
  2. Hence show that \(\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \frac { 1 } { ( \cos \theta + ( \sqrt { } 3 ) \sin \theta ) ^ { 2 } } \mathrm {~d} \theta = \frac { 1 } { \sqrt { } 3 }\).

AnswerMarks Guidance
(i) State answer \(R = 2\)B1
Use trig formula to find \(\alpha\)M1
Obtain \(\alpha = \frac{1}{3}\pi\), or \(60°\)A1 Total: 3 marks
Guidance notes:
- [For the M1 condone a sign error in the expansion of \(\cos(\theta - \alpha)\), but the subsequent trigonometric work must be correct.]
- [SR: The answer \(\alpha = \tan^{-1}(\sqrt{3})\) earns M1 only.]
AnswerMarks Guidance
(ii) State that the integrand is of the form \(a\sec^2(\theta - \alpha)\)M1
State correct indefinite integral \(\frac{1}{a}\tan(\theta - \frac{1}{3}\pi)\)A1 \(\checkmark\)
Use limits correctly in an integral of the form \(a\tan(\theta - \alpha)\)M1
Obtain given answer correctly following full and exact workingA1 Total: 4 marks
Guidance notes:
- [The f.t. is on \(R\) and \(\alpha\).]
(i) State answer $R = 2$ | B1 |
Use trig formula to find $\alpha$ | M1 |
Obtain $\alpha = \frac{1}{3}\pi$, or $60°$ | A1 | **Total: 3 marks**

**Guidance notes:**
- [For the M1 condone a sign error in the expansion of $\cos(\theta - \alpha)$, but the subsequent trigonometric work must be correct.]
- [SR: The answer $\alpha = \tan^{-1}(\sqrt{3})$ earns M1 only.]

(ii) State that the integrand is of the form $a\sec^2(\theta - \alpha)$ | M1 |
State correct indefinite integral $\frac{1}{a}\tan(\theta - \frac{1}{3}\pi)$ | A1 $\checkmark$ |
Use limits correctly in an integral of the form $a\tan(\theta - \alpha)$ | M1 |
Obtain given answer correctly following full and exact working | A1 | **Total: 4 marks**

**Guidance notes:**
- [The f.t. is on $R$ and $\alpha$.]

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5 (i) Express $\cos \theta + ( \sqrt { } 3 ) \sin \theta$ in the form $R \cos ( \theta - \alpha )$, where $R > 0$ and $0 < \alpha < \frac { 1 } { 2 } \pi$, giving the exact values of $R$ and $\alpha$.\\
(ii) Hence show that $\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \frac { 1 } { ( \cos \theta + ( \sqrt { } 3 ) \sin \theta ) ^ { 2 } } \mathrm {~d} \theta = \frac { 1 } { \sqrt { } 3 }$.

\hfill \mbox{\textit{CAIE P3 2007 Q5 [7]}}