| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2011 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Harmonic Form |
| Type | Derivative involving harmonic form |
| Difficulty | Standard +0.3 This is a standard C3 harmonic form question with routine differentiation. Part (a) is textbook harmonic form conversion, part (b) applies product rule then matches to part (a), and part (c) requires setting the derivative to zero. All steps are algorithmic 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) etc1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(R^2 = 2^2 + 3^2\), \(R = \sqrt{13}\) or \(3.61...\) | M1, A1 | |
| \(\tan\alpha = \frac{3}{2}\), \(\alpha = 0.983...\) | M1, A1 | (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(f'(x) = 2e^{2x}\cos3x - 3e^{2x}\sin3x\) | M1A1A1 | |
| \(= e^{2x}(2\cos3x - 3\sin3x)\) | M1 | |
| \(= e^{2x}(R\cos(3x+\alpha))\) | ||
| \(= Re^{2x}\cos(3x+\alpha)\) | A1* | cso, (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(f'(x)=0 \Rightarrow \cos(3x+\alpha)=0\) | M1 | |
| \(3x + \alpha = \frac{\pi}{2}\) | M1 | |
| \(x = 0.196...\) | A1 | awrt 0.20, (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Notes |
| \(f'(x)=0 \Rightarrow 2\cos3x - 3\sin3x = 0\) | M1 | |
| \(\tan3x = \frac{2}{3}\) | M1 | |
| \(x = 0.196...\) | A1 | awrt 0.20, (3 marks) |
## Question 8:
### Part (a):
| Working | Mark | Notes |
|---------|------|-------|
| $R^2 = 2^2 + 3^2$, $R = \sqrt{13}$ or $3.61...$ | M1, A1 | |
| $\tan\alpha = \frac{3}{2}$, $\alpha = 0.983...$ | M1, A1 | (4 marks) |
### Part (b):
| Working | Mark | Notes |
|---------|------|-------|
| $f'(x) = 2e^{2x}\cos3x - 3e^{2x}\sin3x$ | M1A1A1 | |
| $= e^{2x}(2\cos3x - 3\sin3x)$ | M1 | |
| $= e^{2x}(R\cos(3x+\alpha))$ | | |
| $= Re^{2x}\cos(3x+\alpha)$ | A1* | cso, (5 marks) |
### Part (c):
| Working | Mark | Notes |
|---------|------|-------|
| $f'(x)=0 \Rightarrow \cos(3x+\alpha)=0$ | M1 | |
| $3x + \alpha = \frac{\pi}{2}$ | M1 | |
| $x = 0.196...$ | A1 | awrt 0.20, (3 marks) |
**Alternative for (c):**
| Working | Mark | Notes |
|---------|------|-------|
| $f'(x)=0 \Rightarrow 2\cos3x - 3\sin3x = 0$ | M1 | |
| $\tan3x = \frac{2}{3}$ | M1 | |
| $x = 0.196...$ | A1 | awrt 0.20, (3 marks) |\begin{enumerate}
\item (a) Express $2 \cos 3 x - 3 \sin 3 x$ in the form $R \cos ( 3 x + \alpha )$, where $R$ and $\alpha$ are constants, $R > 0$ and $0 < \alpha < \frac { \pi } { 2 }$. Give your answers to 3 significant figures.
\end{enumerate}
$$\mathrm { f } ( x ) = \mathrm { e } ^ { 2 x } \cos 3 x$$
(b) Show that $\mathrm { f } ^ { \prime } ( x )$ can be written in the form
$$\mathrm { f } ^ { \prime } ( x ) = R \mathrm { e } ^ { 2 x } \cos ( 3 x + \alpha )$$
where $R$ and $\alpha$ are the constants found in part (a).\\
(c) Hence, or otherwise, find the smallest positive value of $x$ for which the curve with equation $y = \mathrm { f } ( x )$ has a turning point.
\hfill \mbox{\textit{Edexcel C3 2011 Q8 [12]}}