| Date | November 2015 | Marks available | 4 | Reference code | 15N.1.hl.TZ0.10 |
| Level | HL only | Paper | 1 | Time zone | TZ0 |
| Command term | Find | Question number | 10 | Adapted from | N/A |
Question
A given polynomial function is defined as \(f(x) = {a_0} + {a_1}x + {a_2}{x^2} + \ldots + {a_n}{x^n}\). The roots of the polynomial equation \(f(x) = 0\) are consecutive terms of a geometric sequence with a common ratio of \(\frac{1}{2}\) and first term 2.
Given that \({a_{n - 1}} = - 63\) and \({a_n} = 16\) find
the degree of the polynomial;
the value of \({a_0}\).
Markscheme
the sum of the roots of the polynomial \( = \frac{{63}}{{16}}\) (A1)
\(2\left( {\frac{{1 - {{\left( {\frac{1}{2}} \right)}^n}}}{{1 - \frac{1}{2}}}} \right) = \frac{{63}}{{16}}\) M1A1
Note: The formula for the sum of a geometric sequence must be equated to a value for the M1 to be awarded.
\(1 - {\left( {\frac{1}{2}} \right)^n} = \frac{{63}}{{64}} \Rightarrow {\left( {\frac{1}{2}} \right)^n} = \frac{1}{{64}}\)
\(n = 6\) A1
[4 marks]
\(\frac{{{a_0}}}{{{a_n}}} = 2 \times 1 \times \frac{1}{2} \times \frac{1}{4} \times \frac{1}{8} \times \frac{1}{{16}},{\text{ (}}{{\text{a}}_n} = 16)\) M1
\({a_0} = 16 \times 2 \times 1 \times \frac{1}{2} \times \frac{1}{4} \times \frac{1}{8} \times \frac{1}{{16}}\)
\({a_0} = {2^{ - 5}}\;\;\;\left( { = \frac{1}{{32}}} \right)\) A1
[2 marks]
Total [6 marks]
