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2009
P2 Q3
Topical
Functions
09 P2 Q3
2009 H2 Mathematics Paper 2 Question 3
Functions
Answers
(i)
f
−
1
(
x
)
=
a
x
b
x
−
a
{f^{-1}(x) = \frac{ax}{bx-a}}
f
−
1
(
x
)
=
b
x
−
a
a
x
f
2
(
x
)
=
x
{f^2(x)=x}
f
2
(
x
)
=
x
R
f
2
=
(
−
∞
,
a
b
)
∪
(
a
b
,
∞
)
{R_{f^2}=\left(-\infty, \frac{a}{b} \right) \cup \left(\frac{a}{b}, \infty \right)}
R
f
2
=
(
−
∞
,
b
a
)
∪
(
b
a
,
∞
)
(ii)
The composite function
f
g
{fg}
f
g
does not exist as
R
g
⊈
D
f
{R_g \not \subseteq D_f}
R
g
⊆
D
f
(iii)
x
=
0
{x=0 \;}
x
=
0
or
x
=
2
a
b
{\; x=\frac{2a}{b}}
x
=
b
2
a
Full solutions
(i)
y
=
a
x
b
x
−
a
b
x
y
−
a
y
=
a
x
b
x
y
−
a
x
=
a
y
x
(
b
y
−
a
)
=
a
y
x
=
a
y
b
y
−
a
\begin{gather*} y = \frac{ax}{bx-a} \\ bxy - ay = ax \\ bxy - ax = ay \\ x (by-a) = ay \\ x = \frac{ay}{by-a} \end{gather*}
y
=
b
x
−
a
a
x
b
x
y
−
a
y
=
a
x
b
x
y
−
a
x
=
a
y
x
(
b
y
−
a
)
=
a
y
x
=
b
y
−
a
a
y
f
−
1
(
x
)
=
a
x
b
x
−
a
■
f^{-1}(x) = \frac{ax}{bx-a} \; \blacksquare
f
−
1
(
x
)
=
b
x
−
a
a
x
■
Since
f
(
x
)
=
f
−
1
(
x
)
,
{f(x)=f^{-1}(x),}
f
(
x
)
=
f
−
1
(
x
)
,
f
2
(
x
)
=
f
f
(
x
)
=
f
f
−
1
(
x
)
=
x
■
\begin{align*} f^2 (x) &= ff(x) \\ &= ff^{-1}(x) \\ &= x \; \blacksquare \end{align*}
f
2
(
x
)
=
ff
(
x
)
=
f
f
−
1
(
x
)
=
x
■
Since
f
2
(
x
)
=
x
,
{f^2(x)=x,}
f
2
(
x
)
=
x
,
R
f
2
=
D
f
2
=
D
f
=
(
−
∞
,
a
b
)
∪
(
a
b
,
∞
)
■
\begin{align*} R_{f^2} &= D_{f^2} \\ &= D_f \\ &= \left(-\infty, \frac{a}{b} \right) \cup \left(\frac{a}{b}, \infty \right) \; \blacksquare \end{align*}
R
f
2
=
D
f
2
=
D
f
=
(
−
∞
,
b
a
)
∪
(
b
a
,
∞
)
■
(ii)
R
g
=
(
−
∞
,
0
)
∪
(
0
,
∞
)
D
f
=
(
−
∞
,
a
b
)
∪
(
a
b
,
∞
)
\begin{align*} R_g &= \left(-\infty, 0 \right) \cup \left(0, \infty \right) \\ D_f &= \left(-\infty, \frac{a}{b} \right) \cup \left(\frac{a}{b}, \infty \right) \\ \end{align*}
R
g
D
f
=
(
−
∞
,
0
)
∪
(
0
,
∞
)
=
(
−
∞
,
b
a
)
∪
(
b
a
,
∞
)
Since
a
{a}
a
is non-zero,
R
g
⊈
D
f
R_g \not \subseteq D_f
R
g
⊆
D
f
Hence the composite function
f
g
{fg}
f
g
does not exist
■
{\blacksquare}
■
(iii)
f
−
1
(
x
)
=
x
a
x
b
x
−
a
=
x
a
x
=
b
x
2
−
a
x
b
x
2
−
2
a
x
=
0
x
(
b
x
−
2
a
)
=
0
\begin{gather*} f^{-1}(x) = x \\ \frac{ax}{bx-a} = x \\ ax = bx^2 - ax \\ bx^2 - 2ax = 0 \\ x(bx-2a) = 0 \\ \end{gather*}
f
−
1
(
x
)
=
x
b
x
−
a
a
x
=
x
a
x
=
b
x
2
−
a
x
b
x
2
−
2
a
x
=
0
x
(
b
x
−
2
a
)
=
0
x
=
0
■
or
x
=
2
a
b
■
\begin{align*} x &= 0 \; \blacksquare \\ \textrm{or } \; x &= \frac{2a}{b} \; \blacksquare \end{align*}
x
or
x
=
0
■
=
b
2
a
■
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