(a)     Show that the area of the shaded region is $8\sin x - 2x$ .

(b)     Find the maximum area of the shaded region.

(a)     shaded area area of triangle area of sector, i.e.     (M1)

$\left( {\frac{1}{2} \times {4^2}\sin x} \right) - \left( {\frac{1}{2}{2^2}x} \right) = 8\sin x - 2x$     A1A1AG

(b)     EITHER

any method from GDC gaining $x \approx 1.32$     (M1)(A1)

maximum value for given domain is $5.11$     A2

OR

$\frac{{{\text{d}}A}}{{{\text{d}}x}} = 8\cos x - 2$     A1

set $\frac{{{\text{d}}A}}{{{\text{d}}x}} = 0$, hence $8\cos x - 2 = 0$     M1

$\cos x = \frac{1}{4} \Rightarrow x \approx 1.32$     A1

hence ${A_{\max }} = 5.11$     A1

[7 marks]

Generally a well answered question.