MAV’s Dangerous Inflection

This post concerns a question on the 2019 VCE Specialist Mathematics Exam 2 and, in particular, the solution and commentary for that question available through the Mathematical Association of Victoria. As we document below, a significant part of what MAV has written on this question is confused, self-contradictory and tendentious. Thus, noting the semi-official status of MAV solutions, that these solutions play a significant role in MAV’s Meet the Assessors events, and are quite possibly written by VCE assessors, there are some troubling implications. Question 3, Section B on Exam 2 is a differential equations problem, with two independent parts. Part (a) is a routine (and pretty nice) question on exponential growth and decay.* Part (b), which is our concern, considers the differential equation

    \[\boldsymbol{\color{blue}\frac{{\rm d}Q}{{\rm d}t\ } = e^{t-Q}}\,,\]

for t ≥ 0, along with the initial condition

    \[\boldsymbol{\color{blue}Q(0) =1}\,.\]

The differential equation is separable, and parts (i) and (ii) of the question, worth a total of 3 marks, asks to set up the separation and use this to show the solution of the initial value problem is

    \[\boldsymbol{\color{blue}Q =\log_e\hspace{-1pt} \left(e^t + e -1\right)}\,.\]

Part (iii), worth 2 marks, then asks to show that “the graph of Q as a function of t” has no inflection points.** Question 3(b) is contrived and bitsy and hand-holding, but not incoherent or wrong. So, pretty good by VCE standards. Unfortunately, the MAV solution and commentary to this problem is deeply problematic. The first MAV misstep, in (i), is to invert the derivative, giving

    \[\boldsymbol{\color{red}\frac{{\rm d}t\ }{{\rm d}Q } = e^{Q-t}}\,,\]

prior to separating variables. This is a very weird extra step to include since, not only is the step not required here, it is never required or helpful in solving separable equations. Its appearance here suggests a weak understanding of this standard technique. Worse is to come in (iii). Before considering MAV’s solution, however, it is perhaps worth indicating an approach to (iii) that may be unfamiliar to many teachers and students and, possibly, the assessors. If we are interested in the inflection points of Q,*** then we are interested in the second derivative of Q. The thing to note is we can naturally obtain an expression for Q” directly from the differential equation: we differentiate the equation using the chain rule, giving

    \[\boldsymbol{\color{magenta}Q'' = e^{t-Q}\left(1 - Q'\right)}\,.\]

Now, the exponential is never zero, and so if we can show Q’ < 1 then we’d have Q” > 0, ruling out inflection points. Such conclusions can sometimes be read off easily from the differential equation, but it does not seem to be the case here. However, an easy differentiation of the expression for Q derived in part (ii) gives

    \[\boldsymbol{\color{magenta}Q' =\frac{e^t}{e^t + e -1}}\,.\]

The numerator is clearly smaller than the denominator, proving that Q’ < 1, and we’re done. For a similar but distinct proof, one can use the differential equation to replace the Q’ in the expression for Q”, giving

    \[\boldsymbol{\color{magenta}Q'' = e^{t-Q}\left(1 - e^{t-Q}\right)}\,.\]

Again we want to show the second factor is positive, which amounts to showing Q > t. But that is easy to see from the expression for Q above (because the stuff in the log is greater than \boldsymbol{e^t}), and again we can conclude that Q has no inflection points. One might reasonably consider the details in the above proofs to be overly subtle for many or most VCE students. Nonetheless the approaches are natural, are typically more efficient (and are CAS-free), and any comprehensive solutions to the problem should at least mention the possibility. The MAV solutions make no mention of any such approach, simply making a CAS-driven beeline for Q” as an explicit function of t. Here are the contents of the MAV solution:

Part 1: A restatement of the equation for Q from part (ii), which is then followed by 

.˙.  \boldsymbol{ \color{red}\  \frac{{\rm d}^2Q }{{\rm d}t^2\ } = \frac{e^{t+1} -e^t}{\left(e^t + e -1\right)^2} } 

Part 2: A screenshot of the CAS input-output used to obtain the conclusion of Part 1.

Part 3: The statement   

Solving  .˙.  \boldsymbol{\color{red} \  \frac{{\rm d}^2Q }{{\rm d}t^2\ } = 0} gives no solution  

Part 4: A screenshot of the CAS input-output used to obtain the conclusion of Part 3.

Part 5: The half-sentence

We can see that \boldsymbol{\color{red}\frac{{\rm d}^2Q }{{\rm d}t^2\ } > 0} for all t,

Part 6: A labelled screenshot of a CAS-produced graph of Q”.

Part 7: The second half of the sentence,

so Q(t) has no points of inflection

This is a mess. The ordering of the information is poor and unexplained, making the unpunctuated sentences and part-sentences extremely difficult to read. Part 3 is so clumsy it’s funny. Much more important, the MAV “solution” makes little or no mathematical sense and is utterly useless as a guide to what the VCE might consider acceptable on an exam. True, the MAV solution is followed by a commentary specifically on the acceptability question. As we shall see, however, this commentary makes things worse. But before considering that commentary, let’s itemise the obvious questions raised by the MAV solution:
  • Is using CAS to calculate a second derivative on a “show that” exam question acceptable for VCE purposes?
  • Can a stated use of CAS to “show” there are no solutions to Q” = 0 suffice for VCE purposes? If not, what is the purpose of Parts 3 and 4 of the MAV solutions?
  • Does copying a CAS-produced graph of Q” suffice to “show” that Q” > 0 for VCE purposes?
  • If the answers to the above three questions differ, why do they differ?
Yes, of course these questions are primarily for the VCAA, but first things first. The MAV solution is followed by what is intended to be a clarifying comment:

Note that any reference to CAS producing ‘no solution’ to the second derivative equalling zero would NOT qualify for a mark in this ‘show that’ question. This is not sufficient. A sketch would also be required as would stating \boldsymbol{\color{red}e^t (e - 1) \neq 0} for all t.

These definitive-sounding statements are confusing and interesting, not least for their simple existence. Do these statements purport to be bankable pronouncements of VCAA assessors? If not, what is their status? In any case, given that pretty much every exam question demands that students and teachers read inscrutable VCAA tea leaves, why is it solely the solution to question 3(b) that is followed by such statements? The MAV commentary at least makes clear their answer to our second question above: quoting CAS is not sufficient to “show” that Q” = 0 has no solutions.  Unfortunately, the commentary raises more questions than it answers:
  • Parts 3 and 4 are “not sufficient”, but are they worth anything? If so, what are they worth and, in particular, what is the import of the word “also”? If not, then why not simply declare the parts irrelevant, in which case why include those parts in the solutions at all?
  • If, as claimed, it is “required” to state \boldsymbol{e^t(e-1)\neq 0} (which is indeed the key point of this approach and should be required), then why does the MAV solution not contain any such statement, nor even the factorisation that would naturally precede this statement?
  • Why is a solution “required” to include a sketch of Q”? If, in particular, a statement such as \boldsymbol{e^t(e-1)\neq 0} is “required”, or in any case is included, why would the latter not in and of itself suffice?
We wouldn’t begin to suggest answers to these questions, or our four earlier questions, and they are also not the main point here. The main point is that under no circumstances should such shoddy material be the basis of VCAA assessor presentations. If the material was also written by VCAA assessors, all the worse. Of course the underlying problem is not the quality or accuracy of solutions but, rather, the fundamental idiocy of incorporating CAS into proof questions. And for that the central villain is not the MAV but the VCAA, which has permitted their glorification of technology to completely destroy the appreciation of and the teaching of proof and reason. The MAV is not primarily responsible for this nonsense. The MAV is, however, responsible for publishing it, promoting it and profiting from it, none of which should be considered acceptable. The MAV needs to put serious thought into its unhealthily close relationship with the VCAA.   *) We might ask, however, who refers to “The growth and decay” of an exponential function? **) One might simply have referred to Q, but VCAA loves them their words. ***) Or, if preferred, the points of inflection of the graph of Q as a function of t.

Update (26/06/20)

The Examination Report is out and is basically ok; none of the nonsense and non sequiturs of the MAV solutions are included. The solution to (b)(iii) correctly focuses upon the factoring of Q”, although it needlessly worries about the sign of the denominator. There is no mention of the more natural approach to obtaining and analysing Q” but, given the question is treated by the VCAA and pretty much everyone as just another mindless exercise in pushing buttons, this is no surprise.

WitCH 33: Below Average

We’re not actively looking for WitCHes right now, since we have a huge backlog to update. This one, however, came up in another context and, after chatting about it with commenter Red Five, there seemed no choice. The following 1-mark multiple choice question appeared in 2019 Exam 2 (CAS) of VCE’s Mathematical Methods. The problem was to determine Pr(X > 0), the possible answers being

A. 2/3      B. 3/4      C. 4/5      D. 7/9      E. 5/6

Have fun.

Update (04/07/20)

Who writes this crap? Who writes such a problem, who proofreads such a problem, and then says “Yep, that’ll work”? Because it didn’t work, and it was never going to. The examination report indicates that 27% of students gave the correct answer, a tick or two above random guessing.
We’ll outline a solution below, but first to the crap. The main awfulness is the double-function nonsense, defining the probability distribution \boldsymbol{f} in terms of pretty make the same function \boldsymbol{p}. What’s the point of that? Well, of course \boldsymbol{f} is defined on all of \boldsymbol{R} and \boldsymbol{p} is only defined on \boldsymbol{[-a,b]}. And, what’s the point of defining \boldsymbol{f} on all of \boldsymbol{R}? There’s absolutely none. It’s completely gratuitous and, here, completely ridiculous. It is all the worse, and all the more ridiculous, since the function \boldsymbol{p} isn’t properly defined or labelled piecewise linear, or anything; it’s just Magritte crap. 
To add to the Magritte crap, commenter Oliver Oliver has pointed out the hilarious Dali crap, that the Magritte graph is impossible even on its own terms. Beginning in the first quadrant, the point \boldsymbol{(b,b)} is not quite symmetrically placed to make a 45^{\circ} angle. And, yeah, the axes can be scaled differently, but why would one do it here? But now for the Dali: consider the second quadrant and ask yourself, how are the axes scaled there? Taking a hit of acid may assist in answering that one.
Now, finally to the problem. As we indicated, the problem itself is fine, its just weird and tricky and hellishly long. And worth 1 mark. 
As commenters have pointed out, the problem doesn’t have a whole lot to do with probability. That’s just a scenario to give rise to the two equations, 
1) \boldsymbol{a^2 \ +\ \frac{b}{2}\left(2a+b\right) = 1} \qquad      \mbox{(triangle + trapezium = 1).}
2) \boldsymbol{a + b = \frac43} \qquad           \mbox{( average = 3/4).}
The problem is then to evaluate
*) \boldsymbol{\frac{b}2(2a + b)} \qquad \mbox{(trapezium).}
or, equivalently, 
**) \boldsymbol{1 - a^2 \qquad} \mbox{(1 - triangle).}
The problem is tricky, not least because it feels as if there may be an easy way to avoid the full-blown simultaneous equations. This does not appear to be the case, however. Of course, the VCAA just expects the lobotomised students to push the damn buttons which, one must admit, saves the students from being tricked.
Anyway, for the non-lobotomised among us, the simplest approach seems to be that indicated below, by commenter amca01. First multiply equation (1) by 2 and rearrange, to give
3) \boldsymbol{a^2 + (a + b)^2 = 2}.
Then, plugging in (2), we have 
4) \boldsymbol{a^2 = \frac29}.
That then plugs into **), giving the answer 7/9. 
Very nice. And a whole 90 seconds to complete, not counting the time lost making sense of all the crap. 

WitCH 19: A Powerful Solvent

The following WitCH is from VCE Mathematical Methods Exam 2, 2009. (Yeah, it’s a bit old, but the question was raised recently in a tutorial, so it’s obviously not too old.) It is a multiple choice question: The Examiners’ Report indicates that just over half of the students gave the correct answer of B. The Report also gives a brief indication of how the problem was to be approached:

    \[\mbox{\bf Solve } \boldsymbol{\frac{1}{k-0} \int\limits_0^k \left(\frac1{2x+1}\right)dx = \frac16\log_e(7) \mbox{ \bf for $\boldsymbol k$}.\ k = 3.}\]

Have fun.

Update (02/09/19)

Though undeniably weird and clunky, this question clearly annoys commenters less than me. And, it’s true that I am probably more annoyed by what the question symbolises than the question itself. In any case, the discussion below, and John’s final comment/question in particular, clarified things for me somewhat. So, as a rounding off of the post, here is an extended answer to John’s question.

Underlying my concern with the exam question is the use of “solve” to describe guessing/buttoning the solution to the (transcendental) equation \mathbf {\frac1{2k}{\boldsymbol \log} (2k+1) = \frac16{\boldsymbol \log} 7}.  John then questions whether I would similarly object to the “solving” of a quintic equation that happens to have nice roots. It is a very good question.

First of all, to strengthen John’s point, the same argument can also be made for the school “solving” of cubic and quartic equations. Yes, there are formulae for these (as the Evil Mathologer covered in his latest video), but school students never use these formulae and typically don’t know they exist. So, the existence of these formulae is irrelevant for the issue at hand.

I’m not a fan of polynomial guessing games, but I accept that such games are standard and that  “solve” is used to describe such games. Underlying these games, however, are the integer/rational root theorems (which the EM has also covered), which promise that an integer/rational coefficient polynomial has only finitely many candidate roots, and that these roots are easily enumerated. (Yes, these theorems may be a less or more explicit part of the game, but they are there and they affect the game, if only semi-consciously.) By contrast, there is typically no expectation that a transcendental equation will have somehow simple solutions, nor is there typically any method of determining candidate solutions.

I find something generally unnerving about the exam question and, in particular, the Report. It exemplifies a dilution of language which is at least confusing, and I’d suggest is actively destructive. At its weakest, “solve” means “find the solutions to”, and anything is fair game. This usage, however, loses any connotation of “solve” meaning to somehow figure out the way the equation works, to determine why the solutions are what they are. This is a huge loss.

True, the investigation of equations can continue independent of the cheapening of a particular word, but the reality is that it does not. Of course, in this manner the Solve button on CAS is the nuclear bomb that wipes out all intelligent life. The end result is a double-barrelled destruction of the way students are taught to approach an equation. First, students are taught that all that matters about an equation are the solutions.  They are trained to give the barest lip service to analysing an equation, to investigating if the equation can be attacked in a meaningful mathematical manner. Secondly, the students are taught that that there is no distinction between a precise solution and an approximation, a bunch of meaningless decimals spat out by a machine.

So, yes, the exam question above can be considered just another poorly constructed question. But the weird and “What the Hell” incorporation of a transcendental equation with an exact solution that students were supposedly meant to “solve” is emblematic of a an impoverishment of language and of mathematics that the CAS-infatuated VCAA has turned into an art form.

The Treachery of Images

Harry scowled at a picture of a French girl in a bikini. Fred nudged Harry, man-to-man. “Like that, Harry?” he asked.

“Like what?”

“The girl there.”

“That’s not a girl. That’s a piece of paper.”

“Looks like a girl to me.” Fred Rosewater leered.

“Then you’re easily fooled,” said Harry. It’s done with ink on a piece of paper. That girl isn’t lying there on the counter. She’s thousands of miles away, doesn’t even know we’re alive. If this was a real girl, all I’d have to do for a living would be to stay at home and cut out pictures of big fish.”

                       Kurt Vonnegut, God Bless you, Mr. Rosewater


It is fundamental to be able to distinguish appearance from reality. That it is very easy to confuse the two is famously illustrated by Magritte’s The Treachery of Images (La Trahison des Images):

The danger of such confusion is all the greater in mathematics. Mathematical images, graphs and the like, have intuitive appeal, but these images are mere illustrations of deep and easily muddied ideas. The danger of focussing upon the image, with the ideas relegated to the shadows, is a fundamental reason why the current emphasis on calculators and graphical software is so misguided and so insidious.

Which brings us, once again, to Mathematical Methods. Question 5 on Section Two of the second 2015 Methods exam is concerned with the function V:[0,5]\rightarrow\Bbb R, where

\phantom{\quad}  V(t) = de^{\frac{t}3} + (10-d)e^{\frac{-2t}3}\,.

Here, d \in (0,10) is a constant, with d=2 initially; students are asked to find the minimum (which occurs at t = \log_e8), and to graph V. All this is par for the course: a reasonable calculus problem thoroughly trivialised by CAS calculators. Predictably, things get worse.

In part (c)(i) of the problem students are asked to find “the set of possible values of d” for which the minimum of V occurs at t=0. (Part (c)(ii) similarly, and thus boringly and pointlessly, asks for which d the minimum occurs at t=5). Arguably, the set of possible values of d is (0,10), which of course is not what was intended; the qualification “possible” is just annoying verbiage, in which the examiners excel.

So, on to considering what the students were expected to have done for (c)(ii), a 2-mark question, equating to three minutes. The Examiners’ Report pointedly remarks that “[a]dequate working must be shown for questions worth more than one mark.” What, then, constituted “adequate working” for 5(c)(i)? The Examiners’ solution consists of first setting V'(0)=0 and solving to give d=20/3, and then … well, nothing. Without further comment, the examiners magically conclude that the answer to (c)(i) is 20/3 \leqslant d< 10.

Only in the Carrollian world of Methods could the examiners’ doodles be regarded as a summary of or a signpost to any adequate solution. In truth, the examiners have offered no more than a mathematical invocation, barely relevant to the question at hand: why should V having a stationary point at t=0 for d=20/3 have any any bearing on V for other values of d? The reader is invited to attempt a proper and substantially complete solution, and to measure how long it takes. Best of luck completing it within three minutes, and feel free to indicate how you went in the comments.

It is evident that the vast majority of students couldn’t make heads or tails of the question, which says more for them than the examiners. Apparently about half the students solved V'(0)=0 and included d = 20/3 in some form in their answer, earning them one mark. Very few students got further; 4% of students received full marks on the question (and similarly on (c)(ii)).

What did the examiners actually hope for? It is pretty clear that what students were expected to do, and the most that students could conceivably do in the allotted time, was: solve V'(0)=0 (i.e. press SOLVE on the machine); then, look at the graphs (on the machine) for two or three values of d; then, simply presume that the graphs of V for all d are sufficiently predictable to “conclude” that 20/3 is the largest value of d for which the (unique) turning point of V lies in [0,5]. If it is not immediately obvious that any such approach is mathematical nonsense, the reader is invited to answer (c)(i) for the function W:[0,5]\rightarrow\Bbb R where W(t) = (6-d)t^2 + (d-2)t.

Once upon a time, Victorian Year 12 students were taught mathematics, were taught to prove things. Now, they’re taught to push buttons and to gaze admiringly at pictures of big fish.

The Median is the Message

Our first post concerns an error in the 2016 Mathematical Methods Exam 2 (year 12 in Victoria, Australia). It is not close to the silliest mathematics we’ve come across, and not even the silliest error to occur in a Methods exam. Indeed, most Methods exams are riddled with nonsense. For several reasons, however, whacking this particular error is a good way to begin: the error occurs in a recent and important exam; the error is pretty dumb; it took a special effort to make the error; and the subsequent handling of the error demonstrates the fundamental (lack of) character of the Victorian Curriculum and Assessment Authority.

The problem, first pointed out to us by teacher and friend John Kermond, is in Section B of the exam and concerns Question 3(h)(ii). This question relates to a probability distribution with “probability density function”

    \[  f(x) =   \left\{\aligned &\frac{(210-x)e^{\frac{x-210}{20}}}{400} \qquad && 0\leqslant x \leqslant 210,\\ &0 && \text{elsewhere.} \endaligned\right.}\]

Now, anyone with a good nose for calculus is going to be thinking “uh-oh”. It is a fundamental property of a PDF that the total integral (underlying area) should equal 1. But how are all those integrated powers of e going to cancel out? Well, they don’t. What has been defined is only approximately a PDF,  with a total area of 1 - 23/2e^{21/2} \approx 0.9997. (It is easy to calculate the area exactly using integration by parts.)

Below we’ll discuss the absurdity of handing students a non-PDF, but back to the exam question. 3(h)(ii) asks the students to find the median of the “probability distribution”, correct to two decimal places. Since the question makes no sense for a non-PDF, of course the VCAA have shot themself in the foot. However, we can still attempt to make some sense of the question, which is when we discover that the VCAA has also shot themself in the other foot.

The median m of a probability distribution is the half-way point. So, in the integration context here we want the m for which

a)      \phantom{\quad}  \int\limits_0^m f(x)\,{\rm d}x = \dfrac12.

As such, this question was intended to be just another CAS exercise, and so both trivial and pointless: push the button, write down the answer and on to the next question. The problem is, the median can also be determined by the equation

b)     \phantom{\quad}  \int\limits_m^{210} f(x)\,{\rm d}x = \dfrac12,

or by the equation

c)     \phantom{\quad} \int\limits_0^m f(x)\,{\rm d}x = \int\limits_m^{210} f(x)\,{\rm d}x.

And, since our function is only approximately a PDF, these three equations necessarily give three different answers: to the demanded two decimal places the answers are respectively 176.45, 176.43 and 176.44. Doh!

What to make of this? There are two obvious questions.

1. How did the VCAA end up with a PDF which isn’t a PDF?

It would be astonishing if all of the exam’s writers and checkers failed to notice the integral was not 1. It is even more astonishing if all the writers-checkers recognised and were comfortable with a non-PDF. Especially since the VCAA can be notoriously, absurdly fussy about the form and precision of answers (see below).

2. How was the error in 3(h)(ii) not detected?

It should have been routine for this mistake to have been detected and corrected with any decent vetting. Yes, we all make mistakes. Mistakes in very important exams, however, should not be so common, and the VCAA seems to make a habit of it.

OK, so the VCAA stuffed up. It happens. What happened next? That’s where the VCAA’s arrogance and cowardice shine bright for all to see. The one and only sentence in the Examiners’ Report that remotely addresses the error is:

“As [the] function f  is a close approximation of the [???] probability density function, answers to the nearest integer were accepted”. 

The wording is clumsy, and no concession has been made that the best (and uniquely correct) answer is “The question is stuffed up”, but it seems that solutions to all of a), b) and c) above were accepted. The problem, however, isn’t with the grading of the question.

It is perhaps too much to expect an insufferably arrogant VCAA to apologise, to express anything approximating regret for yet another error. But how could the VCAA fail to understand the necessity of a clear and explicit acknowledgement of the error? Apart from demonstrating total gutlessness, it is fundamentally unprofessional. How are students and teachers, especially new teachers, supposed to read the exam question and report? How are students and teachers supposed to approach such questions in the future? Are they still expected to employ the precise definitions that they have learned? Or, are they supposed to now presume that near enough is good enough?

For a pompous finale, the Examiners’ Report follows up by snarking that, in writing the integral for the PDF, “The dx was often missing from students’ working”. One would have thought that the examiners might have dispensed with their finely honed prissiness for that one paragraph. But no. For some clowns it’s never the wrong time to whine about a missing dx.

UPDATE (16 June): In the comments below, Terry Mills has made the excellent point that the prior question on the exam is similarly problematic. 3(h)(i) asks students to calculate the mean of the probability distribution, which would normally be calculated as \int xf(x)\,{\rm d}x. For our non-PDF, however, we should should normalise by dividing by \int f(x)\,{\rm d}x. To the demanded two decimal places, that changes the answer from the Examiners’ Report’s 170.01 to 170.06.