C3 June 2012 Q3
3.
Figure 1 shows a sketch of the curve \(C\) which has equation
\[y = \mathrm{e}^{x\sqrt{3}}\sin 3x, \quad -\frac{\pi}{3} \leqslant x \leqslant \frac{\pi}{3}\]
Give your answer as a multiple of \(\pi\). (6)
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \sqrt{3}e^{x\sqrt{3}}\sin 3x + 3e^{x\sqrt{3}}\cos 3x\) | M1A1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0 \qquad e^{x\sqrt{3}}(\sqrt{3}\sin 3x + 3\cos 3x) = 0\) | M1 |
| \(\tan 3x = -\sqrt{3}\) | A1 |
| \(3x = \dfrac{2\pi}{3} \Rightarrow x = \dfrac{2\pi}{9}\) | M1A1 |
| (6) |
Notes
M1 Applies the product rule vu’+uv’ to \(e^{x\sqrt{3}}\sin 3x\). If the rule is quoted it must be correct and there must have been some attempt to differentiate both terms. If the rule is not quoted (nor implied by their working, ie. terms are written out u=…,u’=….,v=….,v’=….followed by their vu’+uv’ ) only accept answers of the form \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = Ae^{x\sqrt{3}}\sin 3x + e^{x\sqrt{3}} \times \pm B\cos 3x\)
A1 Correct expression for \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \sqrt{3}e^{x\sqrt{3}}\sin 3x + 3e^{x\sqrt{3}}\cos 3x\)
M1 Sets their \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\), factorises out or divides by \(e^{x\sqrt{3}}\) producing an equation in sin3x and cos3x
A1 Achieves either \(\tan 3x = -\sqrt{3}\) or \(\tan 3x = -\dfrac{3}{\sqrt{3}}\)
M1 Correct order of arctan, followed by \(\div 3\).
Accept \(3x = \dfrac{5\pi}{3} \Rightarrow x = \dfrac{5\pi}{9}\) or \(3x = \dfrac{-\pi}{3} \Rightarrow x = \dfrac{-\pi}{9}\) but not \(x = \arctan\left(\dfrac{-\sqrt{3}}{3}\right)\)
A1 CS0 \(x = \dfrac{2\pi}{9}\) Ignore extra solutions outside the range. Withhold mark for extra inside the range.
Alternative in part (a) using the form \(R\sin(3x + \alpha)\) JUST LAST 3 MARKS
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \sqrt{3}e^{x\sqrt{3}}\sin 3x + 3e^{x\sqrt{3}}\cos 3x\) | M1A1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0 \qquad e^{x\sqrt{3}}(\sqrt{3}\sin 3x + 3\cos 3x) = 0\) | M1 |
| \((\sqrt{12})\sin\left(3x + \dfrac{\pi}{3}\right) = 0\) | A1 |
| \(3x = \dfrac{2\pi}{3} \Rightarrow x = \dfrac{2\pi}{9}\) | M1A1 |
| (6) |
A1 Achieves either \((\sqrt{12})\sin\left(3x + \dfrac{\pi}{3}\right) = 0\) or \((\sqrt{12})\cos\left(3x - \dfrac{\pi}{6}\right) = 0\)
M1 Correct order of arcsin or arcos, etc to produce a value of \(x\)
Eg accept \(3x + \dfrac{\pi}{3} = 0\) or \(\pi\) or \(2\pi \Rightarrow x = \ldots\)
A1 Cao \(x = \dfrac{2\pi}{9}\) Ignore extra solutions outside the range. Withhold mark for extra inside the range.
Alternative to part (a) squaring both sides JUST LAST 3 MARKS
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \sqrt{3}e^{x\sqrt{3}}\sin 3x + 3e^{x\sqrt{3}}\cos 3x\) | M1A1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0 \qquad e^{x\sqrt{3}}(\sqrt{3}\sin 3x + 3\cos 3x) = 0\) | M1 |
| \(\sqrt{3}\sin 3x = -3\cos 3x \Rightarrow \cos^2(3x) = \dfrac{1}{4}\) or \(\sin^2(3x) = \dfrac{3}{4}\) | A1 |
| \(x = \dfrac{1}{3}\arccos\left(\pm\sqrt{\dfrac{1}{4}}\right)\) oe | M1 |
| \(x = \dfrac{2\pi}{9}\) | A1 |
| Scheme | Marks |
|---|---|
| At \(x = 0\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3\) | B1 |
| Equation of normal is \(-\dfrac{1}{3} = \dfrac{y - 0}{x - 0}\) or any equivalent \(y = -\dfrac{1}{3}x\) | M1A1 |
| (3) | |
| (9 marks) |
Notes
B1 Sight of 3 for the gradient
M1 A full method for finding an equation of the normal.
Their tangent gradient \(m\) must be modified to \(-\dfrac{1}{m}\) and used together with (0, 0).
Eg \(-\dfrac{1}{\text{their } 'm'} = \dfrac{y - 0}{x - 0}\) or equivalent is acceptable
A1 \(y = -\dfrac{1}{3}x\) or any correct equivalent including \(-\dfrac{1}{3} = \dfrac{y - 0}{x - 0}\).