Find \(f'(x)\) .

Hence

(i)     show that \(q = - 2\) ;

(ii)    verify that A is a minimum point.

Find the maximum value of \(f(x)\) .

The function \(f(x)\) can be written in the form \(r\cos (x - a)\) .

Write down the value of r and of a .

\(f'(x) = - \sin x + \sqrt 3 \cos x\)     A1A1     N2

[2 marks]

(i) at A, \(f'(x) = 0\)     R1

correct working     A1

e.g. \(\sin x = \sqrt 3 \cos x\)

\(\tan x = \sqrt 3 \)     A1

\(x = \frac{\pi }{3}\) , \(\frac{{4\pi }}{3}\)     A1

attempt to substitute their x into \(f(x)\)     M1

e.g. \(\cos \left( {\frac{{4\pi }}{3}} \right) + \sqrt 3 \sin \left( {\frac{{4\pi }}{3}} \right)\)

correct substitution     A1

e.g. \( - \frac{1}{2} + \sqrt 3 \left( { - \frac{{\sqrt 3 }}{2}} \right)\)

correct working that clearly leads to \( - 2\)     A1

e.g. \( - \frac{1}{2} - \frac{3}{2}\)

 \(q = - 2\)     AG     N0

(ii) correct calculations to find \(f'(x)\) either side of \(x = \frac{{4\pi }}{3}\)     A1A1

e.g. \(f'(\pi ) = 0 - \sqrt 3 \) ,  \(f'(2\pi ) = 0 + \sqrt 3 \)
  

\(f'(x)\) changes sign from negative to positive     R1

so A is a minimum     AG     N0

[10 marks]

max when \(x = \frac{\pi }{3}\)     R1

correctly substituting \(x = \frac{\pi }{3}\) into \(f(x)\)     A1

e.g. \(\frac{1}{2} + \sqrt 3 \left( {\frac{{\sqrt 3 }}{2}} \right)\)

max value is 2     A1     N1

[3 marks]

\(r = 2\) , \(a = \frac{\pi }{3}\)     A1A1     N2

[2 marks]