Our sixth and final post on the 2017 VCE exam madness is on some recurring nonsense in Mathematical Methods. The post will be relatively brief, since a proper critique of every instance of the nonsense would be painfully* *long, and since we’ve said it all before.

The mathematical problem concerns, for a given function f, finding the solutions to the equation

This problem appeared, in various contexts, on last month’s Exam 2 in 2017 (Section B, Questions 4(c) and 4(i)), on the Northern Hemisphere Exam 1 in 2017 (Questions 8(b) and 8(c)), on Exam 2 in 2011 (Section 2, Question 3(c)(ii)), and on Exam 2 in 2010 (Section 2, Question 1(a)(iii)).

Unfortunately, the technique presented in the three Examiners’ Reports for solving equation (1) is fundamentally wrong. (The Reports are here, here and here.) In synch with this wrongness, the standard textbook considers four misleading examples, and its treatment of the examples is infused with wrongness (Chapter 1F). It’s a safe bet that the forthcoming Report on the 2017 Methods Exam 2 will be plenty wrong.

What is the promoted technique? It is to ignore the difficult equation above, and to solve instead the presumably simpler equation

or perhaps the equation

Which is wrong.

It is simply not valid to assume that either equation (2) or (2)’ is equivalent to (1). Yes, as long as the inverse of *f* exists then equation (2)’ is equivalent to equation (2): a solution *x* to (2)’ will also be a solution to (2), and vice versa. And, yes, then any solution to (2) and (2)’ will also be a solution to (1). The converse, however, is in general false: **a**** solution to (1) need not be a solution to (2) or (2)’**.

It is easy to come up with functions illustrating this, or think about the graph above, or look here.

OK, the VCAA might argue that the exams (and, except for a couple of up-in-the-attic exercises, the textbook) are always concerned with functions for which solving (2) or (2)’ happens to suffice, so what’s the problem? The problem is that this argument would be idiotic.

Suppose that we taught students that roots of polynomials are always integers, instructed the students to *only* check for integer solutions, and then carefully arranged for the students to only encounter polynomials with integer solutions. Clearly, that would be mathematical and pedagogical crap. The treatment of equation (1) in Methods exams, and the close to universal treatment in Methods more generally, is identical.

OK, the VCAA might continue to argue that the students have their (stupifying) CAS machines at hand, and that the graphs of the particular functions under consideration make clear that solving (2) or (2)’ suffices. There would then be three responses:

(i) No one *tests* whether Methods students do anything like a graphical check, or anything whatsoever.

(ii) Hardly any Methods students *do *do anything. The overwhelming majority of students treat equations (1), (2) and (2)’ as automatically equivalent, and they have been given explicit license by the Examiners’ Reports to do so. Teachers know this and the VCAA knows this, and any claim otherwise is a blatant lie. And, for any reader still in doubt about what Methods students actually do, here’s a thought experiment: imagine the 2018 Methods exam requires students to solve equation (1) for the function f(x) = (x-2)/(x-1), and then imagine the consequences.

(iii) Even if students were implicitly or explicitly arguing from CAS graphics, “Look at the picture” is an absurdly impoverished way to think about or to teach mathematics, or pretty much anything. The power of mathematics is to be able take the intuition and to either *demonstrate *what appears to be true, *or *demonstrate that the intuition is misleading. Wise people are wary of the treachery of images; the VCAA, alas, promotes it.

The real irony and idiocy of this situation is that, with natural conditions on the function f, equation (1) *is* equivalent to equations (2) and (2)’, and that it is well within reach of Methods students to *prove* this. If, for example, *f* is a strictly increasing function then it can readily be proved that the three equations are equivalent. Working through and applying such results would make for excellent lessons and excellent exam questions.

Instead, what we have is crap. Every year, year after year, thousands of Methods students are being taught and are being tested on mathematical crap.

Nice analysis, as always, Marty.

I think we can depend on the Examiner’s Report to NOT mention any of this, or any of what I’m about to add. I’d like to add two things (I could add a whole lot more but life is short and opportunities should be left for others):

1. There’s a very simple proof (totally within the scope of the Maths Methods Study Design) that if f is a strictly increasing function then the points of intersection of y = f(x) and y = f^-1(x) lie on the line y = x:

Assume a point (a, b) that is an intersection point of y = f(x) and y = f^-1(x).

Then f(a) = b and f^-1(a) = b => a = f(b).

Case 1: Let a < b. Then f(b) b. Then f(b) > f(a) therefore f is not increasing on [b, a] which is a contradiction since f is a strictly increasing function.

Therefore a = b and so the intersection point lies on y = x.

Note:

(a) Cases 1 and 2 are symmetric and so either one is sufficient to conclude a = b.

(b) *Strictly* increasing rather than increasing is required to ensure existence of the inverse function.

This proof would make a nice context to and add relevance for teaching strictly increasing functions – at the moment the idea of strictly increasing functions is in the Study Design for no obvious reason. Just another bit of piecemeal added material.

2. Part (i) (ii) provides a nice opportunity for applying the above theorem. By starting with f(x) = x the conjecture that the coordinates of the second intersection point approach (-2, -2) can be *proved*. From which it can be *proved* that the limiting bounded area is 4, rather than looking at some pictures and then giving a *conjectured* answer. A lost opportunity for VCAA. The proof is left as a simple exercise.

The question is only worth 1 mark, so finding an answer by fair means or foul is encouraged here. This is an total indictment on VCAA , which consistently gets on its high horse and says that correct answers obtained by incorrect mathematics receive zero. While I agree with the sentiment, it’s totally sanctimonious and hypocritical when VCAA is more or less forcing students to answer exam questions simply by looking at some pictures and then making a conjecture. That is, forcing students to *incorrectly* answer the question. It’s incorrect because conjecture without *proof* is NOT correct mathematics.

“…correct answers obtained by incorrect Mathematics receive zero.”

Is this statement (presumably a copy-and-paste from an official VCAA publication somewhere) not itself a contradiction of the method mark/answer mark system by which 99% of marks are awarded in Methods papers? The extra 1% is a margin of error because I have seen a reasoning mark in a marking scheme recently but it may have been a HSC or even IB paper (the question was quite nice, so balance of probability dictates it probably wasn’t VCE)

Thanks Number 8. I’m not sure if John was quoting or being rhetorical. Either way, what constitutes acceptable reasoning in Methods is anyone’s guess.

It’s not a direct quote but it wouldn’t be hard to go to the reports and find this sort of statement. And who really knows how the marking scheme operates (accept the assessors who are sworn to confidentiality), it’s all top hush ….

BY the way, my proof keeps publishing wrong, I give up. I’m sure readers can figure out what’s meant.

I should clarify: What’s not correct mathematics is conjecture that is not clearly stated as conjecture. The Goldbach conjecture is good (great) mathematics, but I don’t think any competent mathematician would consider that being tested up to 400,000,000,000,000 constitutes a proof. But this is the sort of ‘proof’ that VCAA is encouraging with these sort of questions ….

Wonderful! Love it!

A bit like the “3 is a prime, 5 is a prime, 7 is a prime… all odd numbers are prime.” type of proof.

That could be a whole new blog post Marty: “When is a proof not a proof?” Answer: when it is on a VCAA marking scheme.

Yes. John, I think this a very important point, more generally on Methods. Whatever students are doing in Methods, and whatever its value, it’s not mathematical reasoning. It’s not establishing the truth of anything.

And yet, if f is invertible (i.e. if f^(-1) exists), the simple solution of “solve f^(2)(x) = x” is a perfectly valid way to find all solutions to “f(x) = f^(-1)(x)”. Why *that* isn’t taught, I have no idea.

Thanks, Eddie. Yes, for invertible functions the equation f

^{2}(x) = x is equivalent to (1), and is potentially much easier to solve. And, of course the iteration of a function is much more natural.Why isn’t it taught? I suspect the equations have no intrinsic interest for the VCAA examiners. It appears they’re more concerned with the pretty picture made by graphing f

^{-1}and f together, than they are with actual function behaviour.It’s also sad to see the number of copycat questions appearing in school assessments and commercial trial exams that propagate and further entrench this method of solution. As someone who is often vetting this stuff, I am consistently having to alert the writer that such a method needs to be justified with the prior observation that either f or f^-1 is a strictly increasing function.