Date | May 2009 | Marks available | 8 | Reference code | 09M.1.hl.TZ2.9 |
Level | HL only | Paper | 1 | Time zone | TZ2 |
Command term | Explain and Show that | Question number | 9 | Adapted from | N/A |
Question
A triangle has sides of length \(({n^2} + n + 1)\), \((2n + 1)\) and \(({n^2} - 1)\) where \(n > 1\).
(a) Explain why the side \(({n^2} + n + 1)\) must be the longest side of the triangle.
(b) Show that the largest angle, \(\theta \), of the triangle is \(120^\circ \).
Markscheme
(a) a reasonable attempt to show either that \({n^2} + n + 1 > 2n + 1\) or
\({n^2} + n + 1 > {n^2} - 1\) M1
complete solution to each inequality A1A1
(b) \(\cos \theta = \frac{{{{(2n + 1)}^2} + {{({n^2} - 1)}^2} - {{({n^2} + n + 1)}^2}}}{{2(2n + 1)({n^2} - 1)}}\) M1A1
\( = \frac{{ - 2{n^3} - {n^2} + 2n + 1}}{{2(2n + 1)({n^2} - 1)}}\) M1
\( = - \frac{{(n - 1)(n + 1)(2n + 1)}}{{2(2n + 1)({n^2} - 1)}}\) A1
\( = - \frac{1}{2}\) A1
\(\theta = 120^\circ \) AG
[8 marks]
Examiners report
There were very few complete and accurate answers to part a). The most common incorrect response was to state the triangle inequality and feel that this was sufficient.
Many substituted a particular value for n and illustrated the result. Most students recognised the need for the Cosine rule and applied it correctly. Many then expanded and simplified to the correct answer. There was significant fudging in the middle on some papers. There were many good responses to this question.