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]