8. (a) Express 2 aCOS3x – 3 aSIN3x in the form R cos(3x + a), where R and α are constants, R > 0 and 0 < α < π/2 - Edexcel - A-Level Maths Pure - Question 1 - 2011 - Paper 3

Differentiating ( f(x) = e^{2x} \cos(3x) ) using the product rule:

$f' (x) = \frac{d}{dx}(e^{2x}) * \cos(3x) + e^{2x} * \frac{d}{dx}(\cos(3x))$

Calculating the derivatives:

$\frac{d}{dx}(e^{2x}) = 2e^{2x}$

and

$\frac{d}{dx}(\cos(3x)) = -3\sin(3x)$

Substituting these values in:

$f' (x) = 2e^{2x} \cos(3x) - 3e^{2x} \sin(3x)$

Factoring out ( e^{2x} ):

$f' (x) = e^{2x}(2\cos(3x) - 3\sin(3x))$

Which can be expressed in the form:

$f' (x) = R e^{2x} \cos(3x + \alpha)$

This shows that f' (x) is in the desired form.

To find the turning point, we set ( f' (x) = 0 ):

$e^{2x}(2\cos(3x) - 3\sin(3x)) = 0$

Since ( e^{2x} ) is never zero, we focus on:\n $2\cos(3x) - 3\sin(3x) = 0$

This simplifies to:

$2\cos(3x) = 3\sin(3x) \implies \tan(3x) = \frac{2}{3}$

Finding the smallest positive solution for ( 3x ):

$3x = \tan^{-1}(\frac{2}{3})$

Calculating this gives:

$x \approx \frac{1}{3} \tan^{-1}(\frac{2}{3}) \approx 0.196 \text{ (to 3 significant figures)}$