Diagram 1. Whole Question 
This is a question taken from Anderson Junior College ,
one of Singapore ’s
aboveaverage junior colleges (in terms of the calibre of student intake). By now
this college is reputed to set the most difficult examination questions in Singapore . It seems that they are trying to give the top
junior colleges a run for their money, so to speak. The question is difficult because it really
tests students’ understanding of the concepts.
If you do not understand what is happening, you would be totally lost –
even your graphing calculator (GC), the student’s favorite psychological
crutch, would not be of much help.
Diagram 2. Part (i) of Question 
Part (i) of the question tests students’
understanding of 1to1 functions (a.k.a. oneone
functions or injective functions). A function is invertible if and only if the
function is oneone. [ For the current A
level H2 syllabus, it is assumed that the codomain is always the same as the range,
so there is no need to worry about survjectivity.
] Using the GC to graph the
function f and to
obtain the local maximum point, one sees that the required domain is 2 < x
< 0.
Ans: k =
2. [ I am using black for explanations and blue
for written answers. ]
The domain is highlighted in yellow in the
diagram below. If the yellow region were
to extend to the left beyond this point, it would be possible for a horizontal
line to cross two points in the yellow region.
That would make the function f not 1to1 and hence not invertible.
Diagram 3. Graph of f 
Diagram 4. Part (ii) of Question 
For part (ii), we recall that
a composite function exists if and
only if the range of the first function
(read from right to left) is a subset of the domain of the second function.
We require
We require
Range of g Í Domain
of f^{1}

The domain of f^{1} (the second function) is actually the range
of f.
From the above diagram, the relevant part of f(x) goes from ¥ to f(2) = 2 + ln 4 = 2 ln 2 – 2 » 0.614. The
range of g is everything from 1
downwards (see diagram below: imagine taking every possible point of R,
the domain of g on the xaxis and
shooting them over to the yaxis). We write
Range of g = (¥, 1]
Domain of f^{1} = (¥, 2 ln 2 – 2]
Domain of f^{1} = (¥, 2 ln 2 – 2]
Since Range of g Í Domain of f^{1},
therefore f^{1}g exists.
Diagram 5. Range of g 
Finding the range of
composite functions is something that many students have difficulty with. There are two methods: the direct method and the twostep method. The direct method is usually difficult or
infeasible. In this case, finding range
from the graph of y = f^{1}g(x) is practically impossible, because there is no
simple formula for f^{1}.
The
twostep
method:
Step
1. Find the range of the first
function.
Step
2. Transfer this range to the xaxis of the graph of the second
function and map every point therein over to the yaxis. 
Step 1 has been
done already. We have found that Range of g = (¥, 1] .
Diagram 6. The Twostep Method 
For step 2,
although the formula for f^{1} is impossible to find (it’s a pretty nasty
question, isn’t it?), we know that this graph is a reflection of the graph
of y
= f(x) (shown in blue on the diagram on the right) in
the line y = x, and we can sketch
this (shown in red on the diagram on the right). Now transfer the range of g from the yaxis of your first diagram over to the
xaxis of this diagram on the right
(shown in green). Now imagine taking
every possible point of this set and mapping it over to the yaxis of the second graph. The problem is: how to find f^{1}(1) when we don’t even know the formula for f^{1}?
(really evil problem, isn’t it?). One way to deal with this is to make an
educated guess for f(what?) = 1.
Notice that f(1) = 1.
Therefore f^{1}(1) =
1.
What if your intuition really sucks and
you cannot make a guess? The GC can come
to your rescue. Set up the graph of y =
f(x)
with the restricted domain and then intersect that with the graph
of y
= 1. The intesection is at x =
1, which means f(1) = 1, or f^{1}(1)
= 1, as above.
Diagram 7. Using Intersection on the GC 
Going back to diagram 6: From the range of
g on the xaxis of the graph on the
right, the points will land on every point from 0 down to f^{1}(1) = 1,
including 1 but excluding 0 (because xaxis is an asymptote for y =
f^{1}(x)). We answer thus:
Range of f^{1}g = [1,
0)
Diagram 8. Part (iii) of Question 
Part (iii) tests students
understanding of increasing and decreasing functions.
f is an increasing function means
whenever a
> b, f(a)
> f(b)
f is an decreasing function means
whenever a
> b, f(a)
< f(b)

These are in fact the definitions of increasing and decreasing
functions. One can recognise an
increasing function from its graph by the up slope (positive gradients) as you
move from left to right. For a
decreasing function, the slope will be down as you move from left to right
(negative gradients). [An interesting
note: if
f is a decreasing function, then f^{1}
will also be a decreasing function.]
In our case,
the graph of f is downsloping, so it is a decreasing
function.
Since f is a
decreasing function, whenever a
> b, f(a)
< f(b).
Once again we seem
to have the pernicious problem of not knowing formula for f^{1}. How to solve the inequality then? Well, we can apply f to
both sides of the inequality and the inequality reverses (because f is a
decreasing function). Note that f
and f^{1} “cancel”
as functions i.e. ff^{1}(w) = w for whatever the w is as long as it is welldefined. Hence we proceed as follows
f^{1}g(x) > 1
ff^{1}g(x) < f(1)
g(x) < 1
1  x^{2} < 1
 x^{2} < 0
This latter inequality is the last trick on the questionsetter
sleeve desgined to unsettle the student.
How do you solve this inequality?
Do you need to equate or intersect with anything? Anyway, what is the meaning of solve?
To
solve
an inequality means to find all the possible values of x such that when you substitute each value
into the inequality, the inequality becomes a true statement.

Note here that if we
substituted x = 0, we would get 0 < 0, which is not true. However, if we substituted
any other real number, x^{2} would always be a positive number, the LHS
would always be a negative number, which is less than zero. Conclusion:
x can be any real number except 0.
Ans: x Î R \ {0}
Note that the written
solution (the parts typed in blue) is actually very short, although the
explanation is rather long, because a lot of deep thinking is involved.
Reflection
Let us think back on the
lessons learnt while solving this particularly difficult problem.
* the reason why this problem seems difficult
is because it tests students’ understanding of
concepts (which most are weak in). From experience with many cohorts of students, the
JC teachers know what concepts students are weak in and they like to set questions that
exploit the chinks in students armours.
concepts (which most are weak in). From experience with many cohorts of students, the
JC teachers know what concepts students are weak in and they like to set questions that
exploit the chinks in students armours.
* remember the horizontal line test and domain
restriction to get a 1to1 function, so that the
function is invertible.
function is invertible.
* remember the condition for the existence of
composite functions
* inverse functions swap the domain and range
with the original functions
* the twostep method is recommended for
finding range of composite functions
* increasing functions preserve inequalities,
while decreasing functions reverse
inequalities. You can recognise a decreasing function from the downward
inequalities. You can recognise a decreasing function from the downward
slope of its graph as you go from left to
right.
Although you do not have the
formula for f^{1},
* the value of
f^{1}(1) can be found by intersecting the graph
of y
= f(x) with the horizontal line
y = 1.
* f and f^{1}
“cancel” each other.y = 1.
Finally,
* what is the meaning of “solve
an inequality”?
* How to solve inequalities like x^{2}
< 0? What about x^{2}
> 0? x^{2} > 0?
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