Some Special Madness

Our second post on the 2017 VCE exam madness concerns a question on the first Specialist Mathematics exam. Typically Specialist exams, particularly the first exams, don’t go too far off the rails; it’s usually more “meh” than madness. (Not that “meh” is an overwhelming endorsement of what is nominally a special mathematics subject.) This year, however, the Specialist exams have some notably Methodsy bits. The following nonsense was pointed out to us by John, a friend and colleague.

The final question, Question 10, on the first Specialist exam concerns the function \boldsymbol{f(x) = \sqrt{\arccos(x/2)}}, on its maximal domain [-2,2]. In part (c), students are asked to determine the volume of the solid of revolution formed when the region under the graph of f is rotated around the x-axis. This leads to the integral

    \[V \ = \ \pi \int\limits_{-2}^{2}  \arccos(x/2)\, {\rm d}x\,.\]

Students don’t have their stupifying CAS machines in this first exam, so how to do the integral? It is natural to consider integration by parts, but unfortunately this standard and powerful technique is no longer part of the VCE curriculum. (Why not? You’ll have to ask the clowns at ACARA and the VCAA.)

No matter. The VCAA examiners love to have the students to go through a faux-parts computation. So, in part (a) of the question, students are asked to check the derivative of \boldsymbol{x\arccos(x/a)}. Setting a = 2 in the resulting equation, this gives

    \[ \frac{{\rm d}\phantom{x}}{{\rm d}{x}}\left(x\arccos(x/2)\right)= \arccos(x/2) - \dfrac{x}{\sqrt{4-x^2}}\,.\]

We can now integrate and rearrange, giving

    \[ \aligned V \ &= \ \pi\left\Big[\!x\arccos(x/2)\!\!\right\Big]\limits_{-2}^{2} \quad +\quad \pi \int\limits_{-2}^{2} \dfrac{x}{\sqrt{4-x^2}}\, {\rm d}x\\[2\jot] \ &= \ 2\pi^2\quad +\quad \pi \int\limits_{-2}^{2} \dfrac{x}{\sqrt{4-x^2}}\, {\rm d}x\,.\endaligned\]

So, all that remains is to do that last integral, and … uh oh.

It is easy to integrate \boldsymbol{x/\sqrt{4-x^2}} indefinitely by substitution, but the problem is that our definite(ish) integral is improper at both endpoints. And, unfortunately, improper integrals are no longer part of the VCE curriculum. (Why not? You’ll have to ask the clowns at ACARA and the VCAA.) Moreover, even if improper integrals were available, the double improperness is fiddly: we are not permitted to simply integrate from some –b to b and then let b tend to 2.

So, what is a Specialist student to do? One can hope to argue that the integral is zero by odd symmetry, but the improperness is again an issue. As an example indicating the difficulty, the integral \boldsymbol{\int\limits_{-2}^2 x/(4-x^2)\,{\rm d}x} is not equal to 0. (The TI Inspire falsely computes the integral to be 0, which is less than inspiring.) Any argument which arrives at the answer 0 for integrating \boldsymbol{x/(4-x^2)} is invalid, and is thus prima facie invalid for integrating \boldsymbol{x/\sqrt{4-x^2}} as well.

Now, in fact \boldsymbol{\int\limits_{-2}^2 x/\sqrt{4-x^2}\,{\rm d}x} is equal to zero, and so \boldsymbol{V = 2\pi^2}. In particular, it is possible to argue that the fatal problem with \boldsymbol{x/(4-x^2)} does not occur for our integral, and so both the substitution and symmetry approaches can be made to work. The argument, however, is subtle, well beyond what is expected in a Specialist course.

Note also that this improperness could have been avoided, with no harm to the question, simply by taking the original domain to be, for example, [-1,1]. Which was exactly the approach taken on Question 5 of the 2017 Northern Hemisphere Specialist Exam 1. God knows why it wasn’t done here, but it wasn’t and the consequently the examiners have trouble ahead.

The blunt fact is, Specialist students cannot validly compute \boldsymbol{\int\limits_{-2}^2 x/\sqrt{4-x^2}\,{\rm d}x} with any technique they would have seen in a standard Specialist class. They must either argue incompletely by symmetry or ride roughshod over the improperness. The Examiners’ Report will be a while coming out, though presumably the examiners will accept either argument. But here is a safe prediction: the Report will either contain mealy-mouthed nonsense or blatant mathematical falsehoods. The only alternative is for the examiners to make a clear admission that they stuffed up. Which won’t happen.

Finally, the irony. Look again at the original integral for V. Though this integral arose in the calculation of a volume, it can still be interpreted as the area under the graph of the function y = arccos(x/2):

But now we can consider the corresponding area under the inverse function y = 2cos(x):

It follows that

    \[V \ = \  \pi \int\limits_{-2}^{2}  \arccos(x/2)\, {\rm d}x \ = \ \pi \int\limits_{0}^{\pi} \left[  2\cos(x) - (-2)\right]  \, {\rm d}x \ = \ 2\pi^2\,.\]


This inverse function trick is standard for Specialist (and Methods) students, and so the students can readily calculate the volume V in this manner. True, reinterpreting the integral for V as an area is a sharp conceptual shift, but with appropriate wording it could have made for a very good Specialist question.

In summary, the Specialist Examiners guided the students to calculate V with a jerry-built technique, leading to an integral that the students cannot validly compute, all the while avoiding a simpler approach well within the students’ grasp. Well played, Examiners, well played.


There’s Madness in the Methods

Yes, we’ve used that title before, but it’s a damn good title. And there is so much madness in Mathematical Methods to cover. And not only Methods. Victoria’s VCE exams are coming to an end, the maths exams are done, and there is all manner of new and astonishing nonsense to consider. This year, the Victorian Curriculum and Assessment Authority have outdone themselves.

Over the next week we’ll put up a series of posts on significant errors in the 2017 Methods, Specialist Maths and Further Maths exams, including in the mid-year Northern Hemisphere examsBy “significant error” we mean more than just a pointless exercise in button-pushing, or tone-deaf wording, or idiotic pseudomodelling, or aimless pedantry, all of which is endemic in VCE maths exams. A “significant error” in an exam question refers to a fundamental mathematical flaw with the phrasing, or with the intended answer, or with the (presumed or stated) method that students were supposed to use. Not all the errors that we shall discuss are large, but they are all definite errors, they are errors that would have (or at least should have) misled some students, and none of these errors should have occurred. (It is courtesy of diligent (and very annoyed) maths teachers that I learned of most of these questions.) Once we’ve documented the errors, we’ll post on the reasons that the errors are so prevalent, on the pedagogical and administrative climate that permits and encourages them.

Our first post concerns Exam 1 of Mathematical Methods. In the final question, Question 9, students consider the function \boldsymbol{ f(x) =\sqrt{x}(1-x)} on the closed interval [0,1], pictured below. In part (b), students are required to show that, on the open interval (0,1), “the gradient of the tangent to the graph of f” is (1-3x)/(2\sqrt{x}). A clumsy combination of calculation and interpretation, but ok. The problem comes when students then have to consider tangents to the graph.

In part (c), students take the angle θ in the picture to be 45 degrees. The pictured tangents then have slopes 1 and -1, and the students are required to find the equations of these two tangents. And therein lies the problem: it turns out that the “derivative”  of f is equal to -1 at the endpoint x = 1. However, though the natural domain of the function \sqrt{x}(1-x)} is [0,∞), the students are explicitly told that the domain of f is [0,1].

This is obvious and unmitigated madness.

Before we hammer the madness, however, let’s clarify the underlying mathematics.

Does the derivative/tangent of a suitably nice function exist at an endpoint? It depends upon who you ask. If the “derivative” is to exist then the standard “first principles” definition must be modified to be a one-sided limit. So, for our function f above, we would define

    \[f'(1) = \lim_{h\to0^-}\frac{f(1+h) - f(1)}{h}\,.\]

This is clearly not too difficult to do, and with this definition we find that f'(1) = -1, as implied by the Exam question. (Note that since f naturally extends to the right of =1, the actual limit computation can be circumvented.) However, and this is the fundamental point, not everyone does this.

At the university level it is common, though far from universal, to permit differentiability at the endpoints. (The corresponding definition of continuity on a closed interval is essentially universal, at least after first year.) At the school level, however, the waters are much muddier. The VCE curriculum and the most popular and most respected Methods textbook appear to be completely silent on the issue. (This textbook also totally garbles the related issue of derivatives of piecewise defined (“hybrid”) functions.) We suspect that the vast majority of Methods teachers are similarly silent, and that the minority of teachers who do raise the issue would not in general permit differentiability at an endpoint.

In summary, it is perfectly acceptable to permit derivatives/tangents to graphs at their endpoints, and it is perfectly acceptable to proscribe them. It is also perfectly acceptable, at least at the school level, to avoid the issue entirely, as is done in the VCE curriculum, by most teachers and, in particular, in part (b) of the Exam question above.

What is blatantly unacceptable is for the VCAA examiners to spring a completely gratuitous endpoint derivative on students when the issue has never been raised. And what is pure and unadulterated madness is to spring an endpoint derivative after carefully and explicitly avoiding it on the immediately previous part of the question.

The Victorian Curriculum and Assessment Authority has a long tradition of scoring own goals. The question above, however, is spectacular. Here, the VCAA is like a goalkeeper grasping the ball firmly in both hands, taking careful aim, and flinging the ball into his own net.

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.

Factoring in the Stupidity

It is very brave to claim that one has found the stupidest maths exam question of all time. And the claim is probably never going to be true: there will always be some poor education system, in rural Peru or wherever, doing something dumber than anything ever done before. For mainstream exams in wealthy Western countries, however, New Zealand has come up with something truly exceptional.

Last year, New Zealand students at Year 11 sat one of two algebra exams administered by the New Zealand Qualifications Authority. The very first question on the second exam reads:

A rectangle has an area of  \bf x^2+5x-36. What are the lengths of the sides of the rectangle in terms of  \bf x.

The real problem here is to choose the best answer, which we can probably all agree is sides of length \pi and (x^2+5x-36)/\pi.

OK, clearly what was intended was for students to factorise the quadratic and to declare the factors as the sidelengths of the rectangle. Which is mathematical lunacy. It is simply wrong.

Indeed, the question would arguably still have been wrong, and would definitely still have been awful, even if it had been declared that x has a unit of length: who wants students to be thinking that the area of a rectangle uniquely determines its sidelengths? But, even that tiny sliver of sense was missing.

So, what did students do with this question? (An equivalent question, 3(a)(i), appeared on the first exam.) We’re guessing that, seeing no alternative, the majority did exactly what was intended and factorised the quadratic. So, no harm done? Hah! It is incredible that such a question could make it onto a national exam, but it gets worse.

The two algebra exams were widely and strongly criticised, by students and teachers and the media. People complained that the exams were too difficult and too different in style from what students and teachers had been led to expect. Both types of criticism may well have been valid. For all of the public criticism of the exams, however, we could find no evidence of the above question or its Exam 1 companion being flagged. Plenty of complaining about hard questions, plenty of complaining about unexpected questions, but not a word about straight out mathematical crap.

So, not only do questions devoid of mathematical sense appear on a nationwide exam. It then appears that the entire nation of students is being left to accept that this is what mathematics is: meaningless autopilot calculation. Well done, New Zealand. You’ve made the education authorities in rural Peru feel very much better about themselves.

Accentuate the Negative

Each year about a million Australian school students are required to sit the Government’s NAPLAN tests. Produced by ACARA, the same outfit responsible for the stunning Australian Curriculum, these tests are expensive, annoying and pointless. In particular it is ridiculous for students to sit a numeracy test, rather than a test on arithmetic or more broadly on mathematics. It guarantees that the general focus will be wrong and that specific weirdnesses will abound. The 2017 NAPLAN tests, conducted last week, have not disappointed. Today, however, we have other concerns.

Wading into NAPLAN’s numeracy quagmire, one can often find a nugget or two of glowing wrongness. Here is a question from the 2017 Year 9 test:

In this inequality is a whole number.

\color{blue} \dfrac7{n} \boldsymbol{<} \dfrac57

What is the smallest possible value for n to make this inequality true?

The wording is appalling, classic NAPLAN. They could have simply asked:

What is the smallest whole number n for which \color{red} \dfrac7{n} \boldsymbol{<} \dfrac57\, ?

But of course the convoluted wording is the least of our concerns. The fundamental problem is that the use of the expression “whole number” is disastrous.

Mathematicians would avoid the expression “whole number”, but if pressed would most likely consider it a synonym for “integer”, as is done in the Australian Curriculum (scroll down) and some dictionaries. With this interpretation, where the negative integers are included, the above NAPLAN question obviously has no solution. Sometimes, including in, um, the Australian Curriculum (scroll down), “whole number” is used to refer to only the nonnegative integers or, rarely, to only the positive integers. With either of these interpretations the NAPLAN question is pretty nice, with a solution n = 10. But it remains the case that, at best, the expression “whole number” is irretrievably ambiguous and the NAPLAN question is fatally flawed.

Pointing out an error in a NAPLAN test is like pointing out one of Donald Trump’s lies: you feel you must, but doing so inevitably distracts from the overall climate of nonsense and nastiness. Still, one can hope that ACARA will be called on this, will publicly admit that they stuffed up, and will consider employing a competent mathematician to vet future questions. Unfortunately, ACARA is just about as inviting of criticism and as open to admitting error as Donald Trump.