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2021
P1 Q6
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Graphs
21 P1 Q6
2021 H2 Mathematics Paper 1 Question 6
Graphs and Transformations
Answers
(a)
(b)
Smallest
y
=
1
2
a
{y = \frac{1}{2a}}
y
=
2
a
1
(c)
Translation of
2
a
{2a}
2
a
units in the negative
x
-
{x\textrm{-}}
x
-
axis direction
Full solutions
(a)
(b)
4
a
x
−
x
2
=
−
(
x
2
−
4
a
x
)
=
−
(
(
x
−
2
a
)
2
−
4
a
2
)
=
−
(
x
−
2
a
)
2
+
4
a
2
\begin{align*} & 4ax - x^2 \\ & = - (x^2 - 4ax) \\ & = - \Big((x-2a)^2 - 4a^2 \Big) \\ &= -(x-2a)^2 + 4a^2 \end{align*}
4
a
x
−
x
2
=
−
(
x
2
−
4
a
x
)
=
−
(
(
x
−
2
a
)
2
−
4
a
2
)
=
−
(
x
−
2
a
)
2
+
4
a
2
Hence the maximum value of
4
a
x
−
x
2
{4ax - x^2}
4
a
x
−
x
2
is
4
a
2
{4a^2}
4
a
2
Smallest possible value of
y
=
1
4
a
2
=
1
2
a
■
\begin{align*} & \textrm{Smallest possible value of } y \\ & = \frac{1}{\sqrt{4a^2}} \\ & = \frac{1}{2a} \; \blacksquare \end{align*}
Smallest possible value of
y
=
4
a
2
1
=
2
a
1
■
(c)
Equation of
C
:
{C:}
C
:
y
=
1
4
a
2
−
(
x
−
2
a
)
2
y = \frac{1}{\sqrt{4a^2-(x-2a)^2}}
y
=
4
a
2
−
(
x
−
2
a
)
2
1
Hence a translation of
2
a
{2a}
2
a
units in the negative
x
-
{x\textrm{-}}
x
-
axis direction maps the graph of
C
{C}
C
onto the graph of
y
=
1
4
a
2
−
x
2
■
{y=\frac{1}{4a^2-x^2} \; \blacksquare}
y
=
4
a
2
−
x
2
1
■
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