Answers
Translate the curve
y=x1 by
a units in the positive
x-axis direction.
Scale the resulting curve by a factor of
a2+k parallel to the
y-axis.
Translate the resulting curve by
a units in the positive
y-axis direction.
f−1(x)=x−aax+k. f2(x)=x. f2023(1)=1−aa+k. Full solutions
(a)
By long division,
f(x)=x−aax+k=a+x−aa2+k
x1→x−a1→x−aa2+k→a+x−aa2+k
Translate the curve
y=x1 by
a units in the positive
x-axis direction.
Scale the resulting curve by a factor of a2+k parallel to the y-axis.
Translate the resulting curve by a units in the y-axis direction. ■
(b)
Let yxy−ayxy−axx(y−a)xf−1(x)=x−aax+k=ax+k=ay+k=ay+k=y−aay+k=x−aax+k■ (c)
We observe from part (b) that
f(x)=f−1(x)
f2(x)=ff(x)=ff−1(x)=x■
(d)
f2023(1)=f2021f2(1)=f2021(1)=⋯=f(1)=1−aa+k■ Question Commentary
Note that the two equations for part (a) are not structurally similar at all:
the f(x) expression has x in the numerator while
x1 does not. Attempts to apply transformations in these forms will be extremely challenging.
We thus should perform long division on the f(x) expression so the required transformation is easier to identify.
The regular rule of thumb works here in deciding the order of transformations: translation before scaling/reflection for
x transformations, and then scaling/reflection before translation for y transformations.
The rest of the question involves the idea of a "self-inverse function". This has come up numerous times in the past, but
the exact technique needed to solve part (c) and (d) was last seen in 2014 paper 1 question 1
and 2009 paper 2 question 3.