2009 H2 Mathematics Paper 1 Question 1

Equations and Inequalities

Answers

un=32n2172n+17{u_n = \frac{3}{2} n^2 - \frac{17}{2} n + 17}
{nZ:n11}{\{ n \in \mathbb{Z}: n \geq 11 \}}

Full solutions

(i)

Given un{u_n} is a quadratic polynomial in n,{n,}
un=an2+bn+cu_n = an^2 + bn + c
Since u1=10,u2=6{u_1 = 10, u_2 = 6} and u3=5,{u_3 = 5,}
a+b+c=104a+2b+c=69a+3b+c=5\begin{align} && \quad a + b + c &= 10 \\ && \quad 4 a + 2 b + c &= 6 \\ && \quad 9 a + 3 b + c &= 5 \\ \end{align}
Solving (1),(2){(1), (2)} and (3){(3)} with a GC,
a=32,  b=172,  c=17a=\frac{3}{2}, \; b =- \frac{17}{2}, \; c = 17
un=32n2172n+17  u_n = \frac{3}{2} n^2 - \frac{17}{2} n + 17 \; \blacksquare

(ii)

32n2172n+17>10032n2172n83>03n217n166>0\begin{align*} {\textstyle \frac{3}{2} n^2 - \frac{17}{2} n + 17} &> 100 \\ {\textstyle \frac{3}{2} n^2 - \frac{17}{2} n - 83} &> 0 \\ 3 n^2 - 17 n - 166 &> 0 \\ \end{align*}
n<5.13   or   n>10.8n < -5.13 \; \textrm{ or } \; n > 10.8
Since n{n} is an integer, the set of values of n:{n: }
{nZ:n11}  \{ n \in \mathbb{Z}: n \geq 11 \} \; \blacksquare