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2009
P1 Q1
Topical
Equations
09 P1 Q1
2009 H2 Mathematics Paper 1 Question 1
Equations and Inequalities
Answers
(i)
u
n
=
3
2
n
2
−
17
2
n
+
17
{u_n = \frac{3}{2} n^2 - \frac{17}{2} n + 17}
u
n
=
2
3
n
2
−
2
17
n
+
17
(ii)
{
n
∈
Z
:
n
≥
11
}
{\{ n \in \mathbb{Z}: n \geq 11 \}}
{
n
∈
Z
:
n
≥
11
}
Full solutions
(i)
Given
u
n
{u_n}
u
n
is a quadratic polynomial in
n
,
{n,}
n
,
u
n
=
a
n
2
+
b
n
+
c
u_n = an^2 + bn + c
u
n
=
a
n
2
+
bn
+
c
Since
u
1
=
10
,
u
2
=
6
{u_1 = 10, u_2 = 6}
u
1
=
10
,
u
2
=
6
and
u
3
=
5
,
{u_3 = 5,}
u
3
=
5
,
a
+
b
+
c
=
10
4
a
+
2
b
+
c
=
6
9
a
+
3
b
+
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}
a
+
b
+
c
4
a
+
2
b
+
c
9
a
+
3
b
+
c
=
10
=
6
=
5
Solving
(
1
)
,
(
2
)
{(1), (2)}
(
1
)
,
(
2
)
and
(
3
)
{(3)}
(
3
)
with a GC,
a
=
3
2
,
b
=
−
17
2
,
c
=
17
a=\frac{3}{2}, \; b =- \frac{17}{2}, \; c = 17
a
=
2
3
,
b
=
−
2
17
,
c
=
17
u
n
=
3
2
n
2
−
17
2
n
+
17
■
u_n = \frac{3}{2} n^2 - \frac{17}{2} n + 17 \; \blacksquare
u
n
=
2
3
n
2
−
2
17
n
+
17
■
(ii)
3
2
n
2
−
17
2
n
+
17
>
100
3
2
n
2
−
17
2
n
−
83
>
0
3
n
2
−
17
n
−
166
>
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*}
2
3
n
2
−
2
17
n
+
17
2
3
n
2
−
2
17
n
−
83
3
n
2
−
17
n
−
166
>
100
>
0
>
0
n
<
−
5.13
or
n
>
10.8
n < -5.13 \; \textrm{ or } \; n > 10.8
n
<
−
5.13
or
n
>
10.8
Since
n
{n}
n
is an integer, the set of values of
n
:
{n: }
n
:
{
n
∈
Z
:
n
≥
11
}
■
\{ n \in \mathbb{Z}: n \geq 11 \} \; \blacksquare
{
n
∈
Z
:
n
≥
11
}
■
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