CAIE P3 2013 June — Question 4 7 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2013
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 techniques: (i) uses the R cos(x-α) expansion formula with straightforward coefficient matching, and (ii) applies a standard substitution u = tan(x-α) to integrate sec²(x-α), which is a well-practiced method in P3. The exact answer verification adds minimal difficulty. Slightly easier than average due to its predictable structure.
Spec1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc1.08h Integration by substitution

4
  1. Express \(( \sqrt { } 3 ) \cos x + \sin x\) in the form \(R \cos ( x - \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { 1 } { 2 } \pi\), giving the exact values of \(R\) and \(\alpha\).
  2. Hence show that $$\int _ { \frac { 1 } { 6 } \pi } ^ { \frac { 1 } { 2 } \pi } \frac { 1 } { ( ( \sqrt { } 3 ) \cos x + \sin x ) ^ { 2 } } \mathrm {~d} x = \frac { 1 } { 4 } \sqrt { } 3$$

AnswerMarks Guidance
(i) State \(R = 2\)B1
Use trig formula to find \(\alpha\)M1
Obtain \(\alpha = \frac{1}{6}\pi\) with no errors seenA1 [3 marks]
(ii) Substitute denominator of integrand and state integral \(k \tan(x - \alpha)\)M1*
State correct indefinite integral \(\frac{1}{4}\tan\left(x - \frac{1}{6}\pi\right)\)A1
Substitute limitsM1 (dep*)
Obtain the given answer correctlyA1 [4 marks total]
**(i)** State $R = 2$ | B1 |
Use trig formula to find $\alpha$ | M1 |
Obtain $\alpha = \frac{1}{6}\pi$ with no errors seen | A1 | [3 marks]

**(ii)** Substitute denominator of integrand and state integral $k \tan(x - \alpha)$ | M1* |
State correct indefinite integral $\frac{1}{4}\tan\left(x - \frac{1}{6}\pi\right)$ | A1 |
Substitute limits | M1 (dep*) |
Obtain the given answer correctly | A1 | [4 marks total]

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4 (i) Express $( \sqrt { } 3 ) \cos x + \sin x$ in the form $R \cos ( x - \alpha )$, where $R > 0$ and $0 < \alpha < \frac { 1 } { 2 } \pi$, giving the exact values of $R$ and $\alpha$.\\
(ii) Hence show that

$$\int _ { \frac { 1 } { 6 } \pi } ^ { \frac { 1 } { 2 } \pi } \frac { 1 } { ( ( \sqrt { } 3 ) \cos x + \sin x ) ^ { 2 } } \mathrm {~d} x = \frac { 1 } { 4 } \sqrt { } 3$$

\hfill \mbox{\textit{CAIE P3 2013 Q4 [7]}}