It’s NHT, and I don’t care.
It’s NHT, and I don’t care.
It’s NHT, and I don’t caaaaare,
But, others might, so here you go.
General Mathematics: Exam 1, (13/06/26) Exam 2 and Exam 2 Assessment Guide (W,I).
Mathematical Methods: Exam 1, Exam 2 (13/06/26) and Assessment Guides: Exam 1 and Exam 2 (W,I).
Specialist Mathematics: Exam 1, Exam 2 (04/06/26 Assessment guides: Exam 1 and Exam 2 (W,I))
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GM Paper 1 Q11.
What happens as training time gets to 20 hours and beyond…?
There is no domain given in the question, so…
Just had a look at Specialist Exam 1. Surprised at the difficulty. Most questions were at the difficult end of the spectrum, or am I just past it?
Q6 seems to assume that a confidence interval is centred on the mean. Is this standard?
I think the skill level was on-par with previous NHT papers (they seem to be more difficult algebra-wise).
Q6, yes, that is a fairly standard assumption (in textbooks and exams).
I was influenced by “an … 90% confidence interval” rather than “the …” ? But I suppose the “an” could refer to “approximate”?
First impressions:
Specialist Exam 1:
Question 1. Constant acceleration? There are much better ways of starting the exam off with an “accessible” question.
Question 2. Why induction? Either n is even or odd, and the result follows very easily.
Question 9 reopens a can of worms: the graph of the rational function is described as having a “parabolic asymptote”.
Specialist Exam 2:
Section B, Question 3c. The study design specifies that students are only required to calculate arc length for parametrically defined curves. So this is beyond the study design. (I suppose one could easily define the curve parametrically from what is given, but I don’t think it’s reasonable to expect students to do that, given that the study design is pretty clear on this).
Oh, I missed Q9. Red Five alerted me to that question (and to the existence of the exams), but I was looking at something else.
Yeah, it’s not good. For newbies, The Cambridge can of worms is discussed here. (ps. Is that crap still in Cambridge?)
Jesus. Q1 and Q2 of SM1 are just swill. I don’t know if B3(c) of SM2 is kosher or not, but it’s trivial.
Study design Page 111: “application of integration, areas of regions bounded by curves, arc lengths for parametrically determined curves, surface area of solids of revolution, volumes of solids of revolution of a region about either coordinate axis.”
So, a reasonable person would conclude that B3(c) is not covered by the study design.
Also, the formula sheet only has the equation of an arc length when defined parametrically…
Going out on a limb here, but maybe because VCAA is staffed by nitwits?
Most textbooks I’ve looked at have examples of cartesian form as well. This does not mean the question is fair nor that the study design is not screwed.
General Exam 1, Question 36. Bit of a nitpick, but the question doesn’t specify that when the edge is added, the resulting “new” graph is connected. If the resulting planar graph isn’t connected, then I don’t think it makes sense to ask how many faces it has…
I’m not sure it’s that nitpicky. Don’t they also have to declare that the new graph is still a tree?
My reading is that only an edge is added, not also a vertex. Which means that the graph can’t be a tree any more. But I think it still needs to be connected – unless we consider disjoint collections of trees to have one face (but that’s not the intended answer, I suspect).
Yes, ok, that is probably the correct interpretation. But that would then also imply connectedness. A new edge in the middle of nowhere would also require two new vertices. Whatever, not a great question.
By definition, a tree is connected, and adding an edge (but not a vertex) cannot disconnect it. So I don’t think there’s any problem with the question.
Yes, I agree. A new edge but no new vertex tells you what is happening. But, if a good classroom exercise, I think it is too trick-up to be a good exam question.
It cannot be a tree though as no new vertices are created.
Although, I’m not sure this is made explicitly clear.
I assumed no new vertices were created which meant it could not be a tree and had to be connected (and planar).
Yes, it doesn’t say a new vertex is added, so one must assume that the number of vertices remains the same. An edge is added, so it is no longer a tree, but it is still planar, so it has 11 vertices and 11 edges, and Euler’s formula gives 2 faces. In fact, the number of vertices is irrelevant.
i do not know at what time of year these exams were given, but they strike me as the sort of exercise questions that teachers might given midterm to check out how the kids are doing (and which are taken up afterwards), rather that the sort of summative questions that used to be on state examinations that put more emphasis on integrating parts of the syllabus. A rough test if a good test of a good examination is the extent it leads to students when reviewing for the examination and looking over the syllabus for the whole year to under get the feeling of getting on top of the material and understanding it more deeply as well as an appreciation of what is most significant.
Teaching is linear — you have to do one thing after another. Understanding is nonlinear — it accretes over time as you become more familiar and as time goes on get a feeling for the context is, what is important and why it is important.
Looking at Methods Exam 2, it seems like they are trying to write anti – CAS questions. Quite a lot of the MCQ forces the use of ‘by hand’ skills and even ERQ 4 combats the CAS entirely given it doesn’t even provide a rule for the function!
Interesting. I wouldn’t ever read too much of anything into NHT, but interesting.
The specialist marking guides are released on the page.
A question for all to discuss – what does it mean by ‘turning point’?
This refers to SM2, ERQ1di.
For example:
1. does it mean stationary point (often referred to interchangably, in secondary school contexts), or
2. does it mean maximum/minimum stationary point only, or
3. does it mean a curve is changing direction at this point, or changing its curvature?
Thanks, Bodyguard. I’ve added links.
If “turning point” means something other than turning point, we have an issue.
Bodyguard I’ve looked at the question, and someone else expressed similar concerns in an email. The question looks fine to me, but you can spell out your concerns more and/or other readers may wish to weigh in.
Can you give secondary school contexts where “turning point” and “stationary point” are somehow identified?
For example – the cubic function in the ‘turning point’ form
actually has a stationary point of inflection at 
. Cambridge MM U12 textbook dodged this issue by using the phrase ‘similar to the quadratic turning point form…’ in chapter 6.
Also, the adjective ’turning’ implies the change/turn in its direction. At the SPOI one can argue that the direction of the curve has changed. Although we knew their ‘turning points’ intend to be max/min stationary points or extreme points…
Nah. That is flimsy and dodgy evidence to use against the exam question. If nothing else, I think the exam question is clear by explicitly asking for the inflection point in (d)(ii). But even standing by itself, I think (d)(i) is fine.
There are often nuances at the edges for such definitions, and there are here, as I quickly note below. But I don’t see that “turning point” in the school context can mean other that something akin to (local) maximum or minimum.
The fact that “turning” has broader meaning within mathematics and beyond, and the fact that someone (who?) may refer to the cubic thing as “turning point form”, is not really evidence of anything. I don’t think I’ve seen “turning point form” used to describe that cubic form, and it is not great, but it is really talking analogously. The expression is “turning point”, not “turning” and not “turning point form”.
The one aspect that confuses me, but is probably not worth pursuing is whether a local minimum, say, is automatically a turning point. Does the function
have a “turning point” at 
? Does the function 
? I don’t know but I don’t much care.
The main thing with *any* school semi-definition is to not overfuss it. Just stay away from the edges and everything is fine. I don’t think the exam question is near the edge.
We must not mix up mathematical meanings with common language meaning perhaps.
Enough disasters with “accurate to 2 decimal places” (which now in VCE maths means “if the answer students provide, rounded to 2 decimal places gives the correct ’rounded to 2 dp’ value, it is the correct
answer” – and as a result student copy from the calculator more digits than required and the skill of rounding is forgotten / neglected.
Any non-linear graph “changes direction at every point”, does it not?
Has there ever been a question where, say y=|x| has a T.P. ?
Thanks, Banacek. I’ll answer, but can you rephrase your first sentence? I’m not sure if you’re flagging a hypothetical, or making a suggestion, or something else.
Hi Marty. Regarding Methods Paper 1 last question, last part. This question is worrying me as it seems the correct answer is p = q – Ln ( e^q – 1) with the condition that q > 0.
Can someone confirm this and explain how to get the result in the short time allowed. Thanks.
Weird question (the exam’s, not yours).
Thanks, Rob. Unless I’m missing, something I cannot imagine more than a handful of students getting anywhere with 9(d). I did it quickly and didn’t get your answer, but either which way it seems not that long but very tricky, and fiddly, with the dickish writing of the function as a quotient, and the entirely gratuitous negative exponents. The point seems to be to notice the difference of two squares after integrating, but I cannot see who is going to notice that.
Thanks Marty.
BTW I really appreciated the now permanent post explaining the acronyms!
I don’t mean to be obscure. I’m just stupid and lazy.
My approach to this was to use difference of two squares, but I will admit it took me two starts to get anywhere:
Rearranging:
Factorising:
Now, because
: 
And the result follows.
Marty, I got the answer to this question on your best friend TI nSpire CAS.
Like you I despise this calculator except for the speed with which it can generate pseudorandom data.
I have checked the answer on my 30 year old TI 84 Silver and it seems to be correct.
So, firstly I would like to know just exactly what answer you got and how long it took you.
Time is of the essence on this exam we all know that.
Also, can anybody show me how to get this answer ?
Thanks, Rob. I think I spent two minutes, but in no way was I suggesting my answer was definitive. Will sit down to look more carefully in a little while. But, unless I missed something, I cannot see that it is a reasonable question.
Hi Rob, I looked again and your answer seems fine. I didn’t check to the last detail, but the answer I got (again) looks the same, yours is just “tidied” a little. I didn’t look closely before, and the tidying turns out to make the answer look significantly different.
The question is not long if you see to split up the fraction to integrate (also for part (a), but my guess is that many will miss that approach), and if you see the difference of two squares, after integrating, when solving for p. But I don’t see an easier way, or easier to see way, and I doubt that more than a handful of students could have gotten close with this question.
Getting the definite integral of f(x) is simple because you already know its derivative (from part (a)) – the so-called ‘integration by recognition’ trick. I’d guess 1 mark for![\displaystyle \left[ \frac{e^x - 4}{e^{2x}}\right]^q_p = 0](https://mathematicalcrap.com/wp-content/ql-cache/quicklatex.com-48cfdd05abfff4b8d473c2d869131abd_l3.svg "Rendered by QuickLaTeX.com")
which any reasonable student should get in about 10 seconds.
I’d guess 1 mark for evaluation of the definite integral and then a re-arrangement to get p-stuff on one side and q-stuff on the other side (seems obvious to do this and not very hard to do).
The third mark is the higher hanging fruit – making p the subject. It re-arranges into a quadratic in
(‘hidden quadratic’ is a standard trick that has appeared on many past exams and should be familiar to any decent student) and then get p. The question asks for “an expression” so any correct expression should get the answer mark.
In summary: 2/3 marks should be gettable for any reasonable student in a couple of minutes. The third mark is what soaks up some time.
It’s not a great question. Gratuitous in my opinion. I think it could easily and significantly be improved (without losing any raison d’etre) simply by making q a concrete number.
My *MAJOR* beef is the non-linear asymptote in SM1 Q9a. I will resist giving my opinion of the exam writing panel, vetters, other pygmies and the Maths Manager (where the buck ultimately stops). I am *certain* there has been a very clear statement in the past that ONLY linear asymptotes are included in VCE. But it seems all the VCAA FAQ Maths links have either disappeared or are broken – very convenient.
Thanks, John.
The low fruit in MM1 Q9 is not that low (and I don’t understand your “integration by recognition” here). The higher fruit is not intrinsically high but is not low: few will see the quadratic nature in (d), which is obscured by the negatives. I am pretty sure that students will have been whacked by this question.
Regarding SM1 9(a), this is NHT, so i cannot get angry as hell. Still the parabolic “asymptote”, even if not as bad as the Cambridge crap, obviously should not be there. To the extent that Monash guys were involved in vetting (no idea), I presume they simply would have naively trusted the more regular VCE-VCAA guys that VCAA wouldn’t be testing something outside the God-only-knows study design.
Regarding the disappearing FAQ stuff, that is not good, although your “very convenient” snark is silly and unhelpful. But it would indeed be worthwhile to pinpoint the “only linear” statement. Perhaps the Wayback Machine has something, and perhaps there is a link in the comments thread of my Cambridge asymptote post.
Hi Marty.
Thanks for the tips which worked out well.
I don’t quite follow John’s comments about this question. As you pointed out, finding the definite integral of f(x) is simple using partial fractions.
What is not so simple is using the idea of factorising the difference of two squares to help get the expression for p in terms of q.
When trialling this exam for time, I spent about 15 minutes on this question before giving up!
Cognitive decline?
Thanks, Rob. Definitely, seeing the difference of two squares is far from easy.
John’s suggestion is to regard the equation obtained after evaluating the integral as a quadratic equation in
. That kind of thing is seen enough in SM, although the minus signs and the q-stuff pretty well obscures it here.
Re: Monash Maths Guys. I agree. Their job, to the extent that they have one for NHT maths exams, is to check that the maths is correct. They would (should be able to) assume that the VCAA writers and vetters are ‘content matter experts’, as is typically trumpted.
I know what integration by recognition is. How does it apply here?
Whoops. It doesn’t. That was very dumb of me.
Clearly expressing
is the required approach. In which case it takes about 20 seconds to get what I’ll guess is the first mark. Not withstanding that, I stand by 2/3 easy and quick marks for any decent student (which clearly excludes the author of this comment). 3 marks = roughly 5 minutes so more than a handful of students should get close. The only reason why most might not would be having much less than 5 minutes available for part (d) (or being dumb like me).
I still think it’s gratuitous and could easily be rehabilitated with a concrete value for q (would
or similar be too tricky?). And I agree about the negatives. The last thing the last part of the last question should have is needless traps and obstacles to trip up students – it’s a bit like putting a trip wire just before the finish line of a half-marathon. Why would you do that and screw over some (many?) of those who make it to the end?
The missing GM2 exam is up, and all but one of the exam reports are up (W,I). Thank you to the reader-commenter who alerted me.
People can feel free to comment on the (possible) explanation for the GM2 delay. And, there’s an SM post coming up, as soon as I’ve checked the football scores.
Hey Marty,
Turns out NHT is “no longer offered” which is quite interesting. Perhaps there was a lack of demand and too much overhead to run.
https://www.vcaa.vic.edu.au/assessment/vce/examination-specifications-past-examinations-and-examination-reports/mathematical-methods
Thank you, Hughes. I knew it was ending but hadn’t seen an announcement.
NHT was always an absurd thing, a vanity project for Haileybury and other snot schools. I assume the thing just collapsed under the weight of its own absurdity.