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2020
P2 Q1
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20 P2 Q1
2020 H2 Mathematics Paper 2 Question 1
Equations and Inequalities
Answers
y
=
5
2
x
2
−
5
x
+
1
2
{y=\frac{5}{2} x^2 - 5 x + \frac{1}{2}}
y
=
2
5
x
2
−
5
x
+
2
1
Full solutions
y
=
a
x
2
+
b
x
+
c
y=ax^2+bx+c
y
=
a
x
2
+
b
x
+
c
Since the curve passes through
(
1
,
−
2
)
,
{(1,-2),}
(
1
,
−
2
)
,
a
+
b
+
c
=
−
2
\begin{equation}a+b+c=-2\end{equation}
a
+
b
+
c
=
−
2
d
y
d
x
=
2
a
x
+
b
\frac{\mathrm{d}y}{\mathrm{d}x}=2ax+b
d
x
d
y
=
2
a
x
+
b
Since
(
1
,
−
2
)
{(1,-2)}
(
1
,
−
2
)
is a minimum point and the gradient at
x
=
2
{x=2}
x
=
2
is
5
,
{5,}
5
,
2
a
+
b
=
0
4
a
+
b
=
5
\begin{align} && \quad 2 a + b &= 0 \\ && \quad 4 a + b &= 5 \\ \end{align}
2
a
+
b
4
a
+
b
=
0
=
5
Solving
(
1
)
,
(
2
)
{(1), (2)}
(
1
)
,
(
2
)
and
(
3
)
{(3)}
(
3
)
with a GC,
a
=
5
2
,
b
=
−
5
,
c
=
1
2
a=\frac{5}{2}, \; b=- 5, \; c=\frac{1}{2}
a
=
2
5
,
b
=
−
5
,
c
=
2
1
Equation of the curve:
y
=
5
2
x
2
−
5
x
+
1
2
■
y=\frac{5}{2} x^2 - 5 x + \frac{1}{2} \; \blacksquare
y
=
2
5
x
2
−
5
x
+
2
1
■
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