| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2013 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Harmonic Form |
| Type | Find value where max/min occurs |
| Difficulty | Standard +0.3 This is a standard C3 harmonic form question with routine application of R cos(θ - α) conversion followed by straightforward max/min analysis. The multi-part structure and requirement to find where the maximum occurs adds slight complexity beyond pure recall, but follows a well-practiced algorithm with no novel insight required. |
| Spec | 1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| \(R^2 = 6^2 + 8^2 \Rightarrow R = 10\) | M1A1 | Using Pythagoras' Theorem with 6 and 8 to find \(R\). Accept \(R^2 = 6^2 + 8^2\). If \(a\) has been found first accept \(R = \pm \frac{8}{\sin \alpha}\) or \(R = \pm \frac{6}{\cos \alpha}\). |
| \(\tan \alpha = \frac{8}{6} \Rightarrow \alpha = \text{awrt } 0.927\) | M1A1 | For \(\tan \alpha = \pm \frac{8}{6}\) or \(\tan \alpha = \pm \frac{6}{8}\). If \(R\) is used then only accept \(\sin \alpha = \pm \frac{8}{R}\) or \(\cos \alpha = \pm \frac{6}{R}\). \(\alpha = \text{awrt } 0.927\). Note that \(53.1°\) is \(A0\). |
| Answer | Marks | Guidance |
|---|---|---|
| \(p(x) = \frac{4}{12 - 'R'} \cdot\) | M1 | Award for \(p(x) = \frac{4}{12 - 'R'}\). |
| Cao \(p(x)_{\max} = 2\). | A1 | The answer is acceptable for both marks as long as no incorrect working is seen. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\theta - \text{'their } \alpha' = \pi\) | M1 | For setting \(\theta -\) 'their \(\alpha\) ' \(= \pi\) and proceeding to \(\theta=...\) If working exclusively in degrees accept \(\theta -\) 'their \(\alpha\) ' \(= 180\). Do not accept mixed units. |
| \(\theta = \text{awrt } 4.07\) | A1 | \(\theta = \text{awrt } 4.07\). If the final A mark in part (a) is lost for 53.1, then accept awrt 233.1. |
| (8 marks) |
**(a)**
$R^2 = 6^2 + 8^2 \Rightarrow R = 10$ | M1A1 | Using Pythagoras' Theorem with 6 and 8 to find $R$. Accept $R^2 = 6^2 + 8^2$. If $a$ has been found first accept $R = \pm \frac{8}{\sin \alpha}$ or $R = \pm \frac{6}{\cos \alpha}$. | (4 marks)
$\tan \alpha = \frac{8}{6} \Rightarrow \alpha = \text{awrt } 0.927$ | M1A1 | For $\tan \alpha = \pm \frac{8}{6}$ or $\tan \alpha = \pm \frac{6}{8}$. If $R$ is used then only accept $\sin \alpha = \pm \frac{8}{R}$ or $\cos \alpha = \pm \frac{6}{R}$. $\alpha = \text{awrt } 0.927$. Note that $53.1°$ is $A0$. | | |
**(b) Note that (b)(i) and (b)(ii) can be marked together**
**(i)**
$p(x) = \frac{4}{12 - 'R'} \cdot$ | M1 | Award for $p(x) = \frac{4}{12 - 'R'}$. |
Cao $p(x)_{\max} = 2$. | A1 | The answer is acceptable for both marks as long as no incorrect working is seen. | (2 marks)
**(ii)**
$\theta - \text{'their } \alpha' = \pi$ | M1 | For setting $\theta -$ 'their $\alpha$ ' $= \pi$ and proceeding to $\theta=...$ If working exclusively in degrees accept $\theta -$ 'their $\alpha$ ' $= 180$. Do not accept mixed units. |
$\theta = \text{awrt } 4.07$ | A1 | $\theta = \text{awrt } 4.07$. If the final A mark in part (a) is lost for 53.1, then accept awrt 233.1. | (2 marks)
| (8 marks) |
---\begin{enumerate}
\item (a) Express $6 \cos \theta + 8 \sin \theta$ in the form $R \cos ( \theta - \alpha )$, where $R > 0$ and $0 < \alpha < \frac { \pi } { 2 }$.
\end{enumerate}
Give the value of $\alpha$ to 3 decimal places.\\
(b)
$$\mathrm { p } ( \theta ) = \frac { 4 } { 12 + 6 \cos \theta + 8 \sin \theta } , \quad 0 \leqslant \theta \leqslant 2 \pi$$
Calculate\\
(i) the maximum value of $\mathrm { p } ( \theta )$,\\
(ii) the value of $\theta$ at which the maximum occurs.\\
\hfill \mbox{\textit{Edexcel C3 2013 Q4 [8]}}