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Yearly
2016
P1 Q10
Topical
Functions
16 P1 Q10
2016 H2 Mathematics Paper 1 Question 10
Functions
Answers
(ai)
f
−
1
(
x
)
=
(
x
−
1
)
2
{f^{-1}(x)=(x-1)^2}
f
−
1
(
x
)
=
(
x
−
1
)
2
D
f
−
1
=
[
1
,
∞
)
{D_{f^{-1}}=[1,\infty)}
D
f
−
1
=
[
1
,
∞
)
(aii)
x
=
2.62
{x=2.62}
x
=
2.62
(bi)
g
(
4
)
=
6
{g(4)=6}
g
(
4
)
=
6
g
(
7
)
=
8
{g(7)=8}
g
(
7
)
=
8
g
(
12
)
=
9
{g(12)=9}
g
(
12
)
=
9
(bii)
g
{g}
g
does not have an inverse since
g
(
8
)
=
g
(
7
)
{g(8)=g(7)}
g
(
8
)
=
g
(
7
)
so
g
{g}
g
is not a one-one function
Full solutions
(ai)
y
=
1
+
x
x
=
y
−
1
x
=
(
y
−
1
)
2
\begin{gather*} y = 1 + \sqrt{x} \\ \sqrt{x} = y - 1 \\ x = (y-1)^2 \end{gather*}
y
=
1
+
x
x
=
y
−
1
x
=
(
y
−
1
)
2
f
−
1
(
x
)
=
(
x
−
1
)
2
■
f^{-1}(x) = (x-1)^2 \; \blacksquare
f
−
1
(
x
)
=
(
x
−
1
)
2
■
D
f
−
1
=
R
f
=
[
1
,
∞
)
■
\begin{align*} D_{f^{-1}} &= R_f \\ &= [1 ,\infty) \; \blacksquare \end{align*}
D
f
−
1
=
R
f
=
[
1
,
∞
)
■
(aii)
f
f
(
x
)
=
x
f
(
1
+
x
)
=
x
1
+
1
+
x
=
x
1
+
x
=
x
−
1
1
+
x
=
(
x
−
1
)
2
x
=
x
2
−
2
x
x
=
(
x
2
−
2
x
)
2
x
=
x
4
−
4
x
3
+
4
x
2
x
4
−
4
x
3
+
4
x
2
−
x
=
0
\begin{gather*} ff(x) = x \\ f(1+\sqrt{x}) = x \\ 1 + \sqrt{1+\sqrt{x}} = x \tag{*} \\ \sqrt{1 + \sqrt{x}} = x-1 \\ 1 + \sqrt{x} = (x-1)^2 \\ \sqrt{x} = x^2 - 2x \\ x = (x^2-2x)^2 \\ x = x^4 - 4x^3 + 4x^2 \\ x^4 - 4x^3 + 4x^2 - x = 0 \end{gather*}
ff
(
x
)
=
x
f
(
1
+
x
)
=
x
1
+
1
+
x
=
x
1
+
x
=
x
−
1
1
+
x
=
(
x
−
1
)
2
x
=
x
2
−
2
x
x
=
(
x
2
−
2
x
)
2
x
=
x
4
−
4
x
3
+
4
x
2
x
4
−
4
x
3
+
4
x
2
−
x
=
0
(
*
)
Since
x
≠
0
,
{x\neq 0,}
x
=
0
,
x
3
−
4
x
2
+
4
x
−
1
=
0
■
x^3 - 4 x^2 + 4 x - 1=0 \; \blacksquare
x
3
−
4
x
2
+
4
x
−
1
=
0
■
Solving with a GC,
x
=
0.382
or
x
=
1
or
x
=
2.62
\begin{align*} & x = 0.382 \\ \textrm{or} \quad & x = 1 \\ \textrm{or} \quad & x = 2.62 \\ \end{align*}
or
or
x
=
0.382
x
=
1
x
=
2.62
From
(
∗
)
,
{(*), }
(
∗
)
,
x
>
1
{x > 1}
x
>
1
x
=
2.62
■
x=2.62 \; \blacksquare
x
=
2.62
■
From
f
f
(
x
)
=
x
,
{ff(x)=x, }
ff
(
x
)
=
x
,
applying
f
−
1
{f^{-1}}
f
−
1
on both sides,
f
−
1
f
f
(
x
)
=
f
−
1
(
x
)
f
(
x
)
=
f
−
1
(
x
)
■
\begin{align*} f^{-1}ff(x) &= f^{-1}(x) \\ f(x) &= f^{-1}(x) \; \blacksquare \end{align*}
f
−
1
ff
(
x
)
f
(
x
)
=
f
−
1
(
x
)
=
f
−
1
(
x
)
■
(bi)
g
(
4
)
=
2
+
g
(
2
)
=
2
+
2
+
g
(
1
)
=
2
+
2
+
1
+
g
(
0
)
=
2
+
2
+
1
+
1
=
6
■
\begin{align*} g(4) &= 2 + g(2) \\ &= 2 + 2 + g(1) \\ &= 2 + 2 + 1 + g(0) \\ &= 2 + 2 + 1 + 1 \\ &= 6 \; \blacksquare \end{align*}
g
(
4
)
=
2
+
g
(
2
)
=
2
+
2
+
g
(
1
)
=
2
+
2
+
1
+
g
(
0
)
=
2
+
2
+
1
+
1
=
6
■
g
(
7
)
=
1
+
g
(
6
)
=
1
+
2
+
g
(
3
)
=
1
+
2
+
1
+
g
(
2
)
=
1
+
2
+
1
+
2
+
1
+
1
=
8
■
\begin{align*} g(7) &= 1 + g(6) \\ &= 1 + 2 + g(3) \\ &= 1 + 2 + 1 + g(2) \\ &= 1 + 2 + 1 + 2 + 1 + 1 &= 8 \; \blacksquare \end{align*}
g
(
7
)
=
1
+
g
(
6
)
=
1
+
2
+
g
(
3
)
=
1
+
2
+
1
+
g
(
2
)
=
1
+
2
+
1
+
2
+
1
+
1
=
8
■
g
(
12
)
=
2
+
g
(
6
)
=
2
+
2
+
1
+
2
+
1
+
1
=
9
■
\begin{align*} g(12) &= 2 + g(6) \\ &= 2 + 2 + 1 + 2 + 1 + 1 \\ &= 9 \; \blacksquare \end{align*}
g
(
12
)
=
2
+
g
(
6
)
=
2
+
2
+
1
+
2
+
1
+
1
=
9
■
(bii)
g
(
8
)
=
2
+
g
(
4
)
=
2
+
6
=
8
=
g
(
7
)
\begin{align*} g(8) &= 2 + g(4) \\ &= 2 + 6 \\ &= 8 \\ &= g(7) \end{align*}
g
(
8
)
=
2
+
g
(
4
)
=
2
+
6
=
8
=
g
(
7
)
Since
g
(
8
)
=
g
(
7
)
=
8
,
{g(8)=g(7)=8, }
g
(
8
)
=
g
(
7
)
=
8
,
g
{g}
g
is not a one-one function so
g
{g}
g
does not have an inverse
■
{\blacksquare}
■
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