Math Repository
about
topic
al
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ly
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2011
P1 Q2
Topical
Equations
11 P1 Q2
2011 H2 Mathematics Paper 1 Question 2
Equations and Inequalities
Answers
(i)
a
=
0.215
,
{a=0.215,\;}
a
=
0.215
,
b
=
−
0.490
,
{b=-0.490,\;}
b
=
−
0.490
,
c
=
3.281
{c=3.281}
c
=
3.281
(ii)
{
x
∈
R
:
x
>
1.14
}
{\{ x \in \mathbb{R}: x > 1.14 \}}
{
x
∈
R
:
x
>
1.14
}
Full solutions
(i)
(
−
1.5
)
2
a
−
1.5
b
+
c
=
4.5
2.
1
2
a
+
2.1
b
+
c
=
3.2
3.
4
2
a
+
3.4
b
+
c
=
4.1
\begin{align} && \quad (-1.5)^2 a -1.5 b + c &= 4.5 \\ && \quad 2.1^2 a + 2.1 b + c &= 3.2 \\ && \quad 3.4^2 a + 3.4 b + c &= 4.1 \\ \end{align}
(
−
1.5
)
2
a
−
1.5
b
+
c
2.
1
2
a
+
2.1
b
+
c
3.
4
2
a
+
3.4
b
+
c
=
4.5
=
3.2
=
4.1
Solving
(
1
)
,
(
2
)
{(1), (2)}
(
1
)
,
(
2
)
and
(
3
)
{(3)}
(
3
)
with a GC,
a
=
0.215
(3 dp)
■
b
=
−
0.490
(3 dp)
■
c
=
3.281
(3 dp)
■
\begin{align*} a&=0.215 \textrm{ (3 dp)} \; \blacksquare \\ b&=-0.490 \textrm{ (3 dp)} \; \blacksquare \\ c&=3.281 \textrm{ (3 dp)} \; \blacksquare \end{align*}
a
b
c
=
0.215
(3 dp)
■
=
−
0.490
(3 dp)
■
=
3.281
(3 dp)
■
(ii)
f
(
x
)
=
0.215
x
2
−
0.490
x
+
3.281
f
′
(
x
)
=
0.430
x
−
0.490
\begin{gather*} f(x) = 0.215x^2 -0.490x + 3.281 \\ f'(x) = 0.430x -0.490 \\ \end{gather*}
f
(
x
)
=
0.215
x
2
−
0.490
x
+
3.281
f
′
(
x
)
=
0.430
x
−
0.490
For
f
(
x
)
{f(x)}
f
(
x
)
to be increasing,
f
′
(
x
)
>
0
0.430
x
−
0.490
>
0
x
>
1.14
(3 sf)
\begin{gather*} f'(x) > 0 \\ 0.430x -0.490 > 0 \\ x > 1.14 \textrm{ (3 sf)} \end{gather*}
f
′
(
x
)
>
0
0.430
x
−
0.490
>
0
x
>
1.14
(3 sf)
Set of values of
x
{x}
x
for which
f
(
x
)
{f(x)}
f
(
x
)
is an increasing function:
{
x
∈
R
:
x
>
1.14
}
■
\{ x \in \mathbb{R}: x > 1.14 \} \; \blacksquare
{
x
∈
R
:
x
>
1.14
}
■
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