| Date | November 2009 | Marks available | 7 | Reference code | 09N.2.hl.TZ0.3 |
| Level | HL only | Paper | 2 | Time zone | TZ0 |
| Command term | Find and Show that | Question number | 3 | Adapted from | N/A |
Question
The diagram below shows two concentric circles with centre O and radii 2 cm and 4 cm.
The points P and Q lie on the larger circle and \({\rm{P}}\hat {\text{O}}{\text{Q}} = x\) , where \(0 < x < \frac{\pi }{2}\) .
(a) Show that the area of the shaded region is \(8\sin x - 2x\) .
(b) Find the maximum area of the shaded region.
Markscheme
(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]
Examiners report
Generally a well answered question.
