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21 P1 Q1
2021 H2 Mathematics Paper 1 Question 1
Equations and Inequalities
Answers
a
=
4
,
{a=4,\;}
a
=
4
,
b
=
−
6
,
{b=- 6,\;}
b
=
−
6
,
c
=
0
,
{c=0,\;}
c
=
0
,
d
=
7
,
{d=7,\;}
d
=
7
,
Full solutions
y
=
a
x
3
+
b
x
2
+
c
x
+
d
y=ax^3+bx^2+cx+d
y
=
a
x
3
+
b
x
2
+
c
x
+
d
Since the curve passes through
(
1
,
5
)
{(1,5)}
(
1
,
5
)
and
(
−
1
,
−
3
)
{(-1,-3)}
(
−
1
,
−
3
)
,
a
+
b
+
c
+
d
=
5
−
a
+
b
−
c
+
d
=
−
3
\begin{align} && \quad a + b + c + d &= 5 \\ && \quad -a + b - c + d &= -3 \\ \end{align}
a
+
b
+
c
+
d
−
a
+
b
−
c
+
d
=
5
=
−
3
d
y
d
x
=
3
a
x
2
+
2
b
x
+
c
\frac{\mathrm{d}y}{\mathrm{d}x}=3ax^2+2bx+c
d
x
d
y
=
3
a
x
2
+
2
b
x
+
c
Since the graph has a turning point at
x
=
1
,
{x=1,}
x
=
1
,
3
a
+
2
b
+
c
=
0
\begin{equation}3a+2b+c=0\end{equation}
3
a
+
2
b
+
c
=
0
∫
0
1
f
(
x
)
d
x
=
6
[
a
x
4
4
+
b
x
3
3
+
c
x
2
2
+
d
x
]
0
1
=
6
\begin{gather*} \int_0^1 f(x) \; \mathrm{d}x = 6 \\ \left[ \frac{ax^4}{4} + \frac{bx^3}{3} + \frac{cx^2}{2} + dx \right]_0^1 = 6 \end{gather*}
∫
0
1
f
(
x
)
d
x
=
6
[
4
a
x
4
+
3
b
x
3
+
2
c
x
2
+
d
x
]
0
1
=
6
1
4
a
+
1
3
b
+
1
2
c
+
d
=
6
\begin{equation}\qquad \frac{1}{4}a+\frac{1}{3}b+\frac{1}{2}c+d=6\end{equation}
4
1
a
+
3
1
b
+
2
1
c
+
d
=
6
Solving
(
1
)
,
(
2
)
,
(
3
)
{(1), (2), (3)}
(
1
)
,
(
2
)
,
(
3
)
and
(
4
)
{(4)}
(
4
)
with a GC,
a
=
4
■
b
=
−
6
■
c
=
0
■
d
=
7
■
\begin{align*} &a = 4 \; \blacksquare \\ &b = - 6 \; \blacksquare \\ &c = 0 \; \blacksquare \\ &d = 7 \; \blacksquare \\ \end{align*}
a
=
4
■
b
=
−
6
■
c
=
0
■
d
=
7
■
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