Update (04/07/20)
Of course, students bombed part (f). The examination report indicates that 19% of student correctly answered that there is one solution to the equation; as suggested by commenter Red Five, it’s also a pretty safe bet that the majority of students who got there did so with a Hail Mary guess. (It should be added, the students didn’t do swimmingly well on the rest of Question 9, the CAS-lobotomising having working its usual magic.)
OK, so what did examiners expect for that one measly mark? We’ll get to a reasonable solution below, but let’s first consider some unreasonable solutions.
Here is the examination report’s entire commentary on Part (f):
g(f(x) + f(g(x)) = 0 has exactly one solution.
This question was not well done. Few students attempted to draw a rough sketch of each equation and use addition of ordinates.
Gee, thanks. Drawing a “rough sketch” of either of these compositions is anything but trivial. For one measly mark. We’ll look at sketching aspects of these graphs below, but let’s get on with another unreasonable solution.
Given the weirdness of part (f), a student might hope that parts (a)-(e) provide some guidance. Let’s see.
Part (b) (for which the examination report contains an error), gets us to conclude that the composition
has negative derivative when x > 1.
Part (c) leads us to the composition
having x-intercept when x = log(3).
Finally, Part (e) gives us that the composition f(g(x)) has the sole stationary point (0,4). How does this information help us with Part (f)? Bugger all.
So, what if we include the natural implications of our previous work? That gives us something like the following: 
Well, um, great. We’re left still hunting for that one measly mark.
OK, the other parts of the question are of little help, and the examiners are of no help, so what do else do we need? There are two further pieces of information we require (plus the Intermediate Value Theorem). First, note that
Secondly, note that
if x is huge.
Then, given we know the slopes of the compositions, we can finally complete our rough sketches: 
Now, let’s write S(x) for our sum function g(f(x)) + f(g(x)). We know S(x) > 0 unless one of our compositions is negative. So, the only place we could get S = 0 is if x > log(3). But S(log(3)) > 0, and eventually S is hugely negative. That means S must cross the x-axis (by IVT). But, since S is decreasing for x > 1, S can only cross the axis once, and S = 0 must have exactly one solution.
We’ve finally earned our one measly mark. Yay?
I know there will be lots of people with lots to say, so I’m going to comment on the one thing that irks me above all else in this question (and indeed, out of the whole exam):
What the hell is part(f) meant to be testing?? There is 1 mark and it’s given solely for a correct answer. A student can write down ‘one’ as a total guess, with absolutely no justification, and get that mark. Making part(f) worth 1 mark (so that no justification is required) is beyond stupid. It makes me want to scream.
Part(f) reeks of having been appended at the last minute simply to make the total exam marks add up to 40.
By the way, as a general observation I have never seen an exam with so many marks given to the lame and the infirm. Look at Q7(a) ….
I will repeat the same question I have asked in multiple places (it irks me a lot…):
By VCAA’s definition, does “the rule” include the domain or not???
If yes, why is part (a) one mark?
If no, why call it “the rule” and not “the equation”?
Hi RF. Good question. Once upon a time there was a lot of ambiguity with what VCAA wanted with this sort of wording.
But I think the following (quoted from https://www.vcaa.vic.edu.au/curriculum/vce/vce-study-designs/mathematicalmethods/advice-for-teachers/Pages/Units3and4PerformanceCriteria.aspx ) clears things up (or at least holds VCAA to account):
“Use of correct terminology, including set notation, to specify relations and functions, such as domain, co-domain, range and rule.”
So VCAA clearly considers the domain as separate from the ‘rule’ – so rule means only the equation. Whether this should be considered correct or not is an argument for another day.
Thanks, RF and JF. Of course it’s not the main issue with this exam question, but I agree that use of the expression “the rule” is, and always is, irritating. Th expression is also, by the way, meaningless.
I have no idea what the last part is trying to assess. I wouldn’t mind betting a significant number of students guessed 1 (correctly) with little thought as to why.
Exactly, RF. And that’s what I heard when I put a cup up to the wall ….
Either VCAA was:
A. too gutless to commit more than 1 mark to a challenging (but fair*) question,
B. too stupid to realise how pointless making it worth only 1 mark is,
C. too apathetic to care if 1 mark made sense or not because it was added at the last minute simply to make the total exam marks add up to 40, or
D. too lazy, just wanting a question that was really easy to mark.
(Gutless, stupid, apathetic and lazy works for me).
As I said earlier, beyond stupid. I doubt it’s something that ‘Meet the Assessors’ would want to dwell on ….
By the way, you’re wrong on one count – you credit too much thought to the significant number of students ….
I would like to explain (mainly for the benefit of the school students who read Marty’s blogs and will otherwise be thwarted by relying either on the MAV solutions written by the ‘Meet the Assessors’, the Examiners Report, or the free solutions) why I said in Option A above that the question is challenging but fair:
y = g(f(x)) = e^(-quadratic) and so the graph should be recognised as having a bell-shape (that is, the same shape as the graph of a normal pdf). Furthermore, it then follows from part(b) that the peak occurs at x = 1 (so no need to even complete the square). So a rough graph is simple to sketch.
y = f(g(x)) has an x-intercept and turning point at (0, 4) (from part (d) and part (e)) and has an obvious horizontal asymptote y = 3. Furthermore y –> 3 as x –> -oo and y –> -oo as x –> +oo. So sketching a rough graph is a reasonable expectation.
Now there are two options: Either use addition of ordinates and note by inspection that the result has only one x-intercept, or draw y = g(f(x)) and y = -f(g(x)) and note by inspection that the graphs only intersect once.
Correction of typo: y = f(g(x)) has an x-intercept at (log(3), 0) and turning point at (0, 4).
As an aside, I would like to point out that, yet again, NESA (NSW Education Standards) have released their 2019 exams within days of them being sat by students: https://educationstandards.nsw.edu.au/wps/portal/nesa/11-12/resources/hsc-exam-papers
And they always release their marking guidelines, in a timely manner.
And, yet again, it will likely be many months before VCAA releases its exams (using the excuse that there are ‘copyright issues’ that have to be resolved by their ‘legal team’) …. So for those people who think lazy and apathetic are too harsh labels, think again. (And don’t even think that the Examiners Reports will be available before the end of Term 2 next year. Lazy, apathetic, gutless).