| Date | May 2014 | Marks available | 7 | Reference code | 14M.1.hl.TZ2.9 |
| Level | HL only | Paper | 1 | Time zone | TZ2 |
| Command term | Find and Show that | Question number | 9 | Adapted from | N/A |
Question
The first three terms of a geometric sequence are \(\sin x,{\text{ }}\sin 2x\) and \(4\sin x{\cos ^2}x,{\text{ }} - \frac{\pi }{2} < x < \frac{\pi }{2}\).
(a) Find the common ratio r.
(b) Find the set of values of x for which the geometric series \(\sin x + \sin 2x + 4\sin x{\cos ^2}x + \ldots \) converges.
Consider \(x = \arccos \left( {\frac{1}{4}} \right),{\text{ }}x > 0\).
(c) Show that the sum to infinity of this series is \(\frac{{\sqrt {15} }}{2}\).
Markscheme
(a) \(\sin x,{\text{ }}\sin 2x{\text{ and }}4\sin x{\cos ^2}x\)
\(r = \frac{{2\sin x\cos x}}{{\sin x}} = 2\cos x\) A1
Note: Accept \(\frac{{\sin 2x}}{{\sin x}}\).
[1 mark]
(b) EITHER
\(\left| r \right| < 1 \Rightarrow \left| {2\cos x} \right| < 1\) M1
OR
\( - 1 < r < 1 \Rightarrow - 1 < 2\cos x < 1\) M1
THEN
\(0 < \cos x < \frac{1}{2}{\text{ for }} - \frac{\pi }{2} < x < \frac{\pi }{2}\)
\( - \frac{\pi }{2} < x < - \frac{\pi }{3}{\text{ or }}\frac{\pi }{3} < x < \frac{\pi }{2}\) A1A1
[3 marks]
(c) \({S_\infty } = \frac{{\sin x}}{{1 - 2\cos x}}\) M1
\({S_\infty } = \frac{{\sin \left( {\arccos \left( {\frac{1}{4}} \right)} \right)}}{{1 - 2\cos \left( {\arccos \left( {\frac{1}{4}} \right)} \right)}}\)
\( = \frac{{\frac{{\sqrt {15} }}{4}}}{{\frac{1}{2}}}\) A1A1
Note: Award A1 for correct numerator and A1 for correct denominator.
\( = \frac{{\sqrt {15} }}{2}\) AG
[3 marks]
Total [7 marks]
