| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2021 |
| Session | March |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Harmonic Form |
| Type | Integration using harmonic form |
| Difficulty | Standard +0.8 This is a multi-part harmonic form question requiring (a) standard R-cos(x-α) conversion, (b) finding minimum value and corresponding x, and (c) integration of 1/(harmonic form)² with a triple angle. Part (c) requires recognizing the substitution u=3θ and applying the sec²u integration result, which is non-routine for A-level. The combination of techniques and the squared harmonic form in the integral makes this moderately challenging. |
| Spec | 1.05l Double angle formulae: and compound angle formulae1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State \(R = 10\) | B1 | |
| Use appropriate trigonometry to find \(\alpha\) | M1 | |
| Obtain \(\alpha = \frac{1}{6}\pi\) | A1 | |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State \(-6\) | B1 FT | Following *their* value of \(R\) |
| Attempt to find \(x\) from their \(\cos(x - \alpha) = -1\) | M1 | |
| Obtain \(x - \frac{1}{6}\pi = -\pi\) and hence \(-\frac{5}{6}\pi\) | A1 | |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State integrand of form \(k_1\sec^2\!\left(3\theta - \frac{1}{6}\pi\right)\) | \*M1 | |
| Integrate to obtain form \(k_2\tan\!\left(3\theta - \frac{1}{6}\pi\right)\) | DM1 | |
| Obtain \(\dfrac{1}{300}\tan\!\left(3\theta - \frac{1}{6}\pi\right) + c\) | A1 | |
| Total | 3 |
## Question 7:
**Part (a):**
| Answer | Mark | Guidance |
|--------|------|----------|
| State $R = 10$ | B1 | |
| Use appropriate trigonometry to find $\alpha$ | M1 | |
| Obtain $\alpha = \frac{1}{6}\pi$ | A1 | |
| **Total** | **3** | |
**Part (b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| State $-6$ | B1 FT | Following *their* value of $R$ |
| Attempt to find $x$ from their $\cos(x - \alpha) = -1$ | M1 | |
| Obtain $x - \frac{1}{6}\pi = -\pi$ and hence $-\frac{5}{6}\pi$ | A1 | |
| **Total** | **3** | |
**Part (c):**
| Answer | Mark | Guidance |
|--------|------|----------|
| State integrand of form $k_1\sec^2\!\left(3\theta - \frac{1}{6}\pi\right)$ | \*M1 | |
| Integrate to obtain form $k_2\tan\!\left(3\theta - \frac{1}{6}\pi\right)$ | DM1 | |
| Obtain $\dfrac{1}{300}\tan\!\left(3\theta - \frac{1}{6}\pi\right) + c$ | A1 | |
| **Total** | **3** | |7
\begin{enumerate}[label=(\alph*)]
\item Express $5 \sqrt { 3 } \cos x + 5 \sin x$ in the form $R \cos ( x - \alpha )$, where $R > 0$ and $0 < \alpha < \frac { 1 } { 2 } \pi$.
\item As $x$ varies, find the least possible value of
$$4 + 5 \sqrt { 3 } \cos x + 5 \sin x$$
and determine the corresponding value of $x$ where $- \pi < x < \pi$.
\item Find $\int \frac { 1 } { ( 5 \sqrt { 3 } \cos 3 \theta + 5 \sin 3 \theta ) ^ { 2 } } d \theta$.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2021 Q7 [9]}}