Figure 1 shows a sketch of the curve C which has equation y = e^{rac{1}{3}} ext{sin } 3x, -rac{ ext{π}}{3} ext{ ≤ } x ext{ ≤ } rac{ ext{π}}{3} (a) Find the x coordinate of the turning point P on C, for which x > 0 - Edexcel - A-Level Maths Pure - Question 5 - 2012 - Paper 5

To find the x-coordinate of the turning point P, we need to set the derivative of y to zero:

1. Differentiate the equation using the product rule: $\frac{dy}{dx} = e^{\frac{1}{3}}\left(3\cos(3x) + \frac{1}{3}\sin(3x)\cdot 3\right) = e^{\frac{1}{3}}(3\cos(3x) + \sin(3x))$

2. Set the derivative equal to zero: $e^{\frac{1}{3}}(3\cos(3x) + \sin(3x)) = 0$ Since $e^{\frac{1}{3}}$ is never zero, we have: $3\cos(3x) + \sin(3x) = 0$

3. Rearranging gives: $\tan(3x) = -3$ Therefore, solving for 3x leads to: $3x = \arctan(-3) + n\pi, \text{ where } n \in \mathbb{Z}$ For x > 0, we use: $3x = \arctan(-3) + \pi \\ \Rightarrow$ $x = \frac{\arctan(-3) + \pi}{3}$ Substituting known values leads to: $x = \frac{\pi}{3}$