| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2008 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Harmonic Form |
| Type | Express and solve equation |
| Difficulty | Moderate -0.3 This is a standard harmonic form question requiring routine application of R-cos(x-α) conversion using R²=a²+b² and tan α=b/a, followed by straightforward equation solving and identifying maximum values. While it involves multiple parts, each step follows a well-practiced procedure with no novel insight required, making it slightly easier than average. |
| Spec | 1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(R^2 = 5^2 + 12^2\) | M1 | |
| \(R = 13\) | A1 | |
| \(\tan \alpha = \frac{12}{5}\) | M1 | |
| \(\alpha \approx 1.176\) | cao A1 | (4) |
| (b) \(\cos(x - \alpha) = \frac{6}{13}\) | M1 | |
| \(x - \alpha = \arccos\frac{6}{13} = 1.091\ldots\) | A1 | |
| \(x = 1.091\ldots + 1.176\ldots \approx 2.267\ldots\) awrt 2.3 | A1 | |
| \(x - \alpha = -1.091\ldots\) accept \(\ldots 5.19\ldots\) for M | M1 | |
| \(x = -1.091\ldots + 1.176\ldots \approx 0.0849\ldots\) awrt 0.084 or 0.085 | A1 | (5) |
| (c)(i) \(R_{\max} = 13\) fit their \(R\) | B1 ft | |
| (ii) At the maximum, \(\cos(x - \alpha) = 1\) or \(x - \alpha = 0\) | M1 | |
| \(x = \alpha = 1.176\ldots\) awrt 1.2, fit their \(\alpha\) | A1 ft | (3) [12] |
**(a)** $R^2 = 5^2 + 12^2$ | M1 |
$R = 13$ | A1 |
$\tan \alpha = \frac{12}{5}$ | M1 |
$\alpha \approx 1.176$ | cao A1 | (4)
**(b)** $\cos(x - \alpha) = \frac{6}{13}$ | M1 |
$x - \alpha = \arccos\frac{6}{13} = 1.091\ldots$ | A1 |
$x = 1.091\ldots + 1.176\ldots \approx 2.267\ldots$ awrt 2.3 | A1 |
$x - \alpha = -1.091\ldots$ accept $\ldots 5.19\ldots$ for M | M1 |
$x = -1.091\ldots + 1.176\ldots \approx 0.0849\ldots$ awrt 0.084 or 0.085 | A1 | (5)
**(c)(i)** $R_{\max} = 13$ fit their $R$ | B1 ft |
**(ii)** At the maximum, $\cos(x - \alpha) = 1$ or $x - \alpha = 0$ | M1 |
$x = \alpha = 1.176\ldots$ awrt 1.2, fit their $\alpha$ | A1 ft | (3) [12]
---2.
$$f ( x ) = 5 \cos x + 12 \sin x$$
Given that $\mathrm { f } ( x ) = R \cos ( x - \alpha )$, where $R > 0$ and $0 < \alpha < \frac { \pi } { 2 }$,
\begin{enumerate}[label=(\alph*)]
\item find the value of $R$ and the value of $\alpha$ to 3 decimal places.
\item Hence solve the equation
$$5 \cos x + 12 \sin x = 6$$
for $0 \leqslant x < 2 \pi$.
\item \begin{enumerate}[label=(\roman*)]
\item Write down the maximum value of $5 \cos x + 12 \sin x$.
\item Find the smallest positive value of $x$ for which this maximum value occurs.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2008 Q2 [12]}}