The Arc Enemy

Our previous post was on good guys making a silly, funny and inconsequential mistake. This post is not.

Question B1 of Exam 2 for 2018 Northern Hemisphere Specialist Mathematics begins innocently enough. In part (a), students are required to graph the function \boldsymbol{f(x) = 10\arccos(2-2x)} over its maximal domain. Then, things begin to get stupid.

In part (b), the graph of f is rotated around the y-axis, to model a vase. Students are required to find the volume of this stupid vase, by setting up the integral and then pushing the stupid buttons on their stupid calculators. So, a reasonable integration question lost in ridiculous pseudomodelling and brainless button-pushing. Whatever. Just standard VCE crap. Then, things stay stupid.

Part (c) is a related rates question. In principle a good problem, though it’s hard to imagine anyone ever requiring dh/dt when the water depth is exactly \boldsymbol{5\pi} cm. Whatever. Standard VCE crap. Then, things get really, really stupid.

Part (d) of the problem has a bee climbing from the bottom of the vase to the top. Students are required to find the minimum distance the bee needs to travel.

Where to begin with this idiotic, 1-mark question. Let’s begin with the bee.

Why is it a bee? Why frame a shortest walk question in terms of a bug with wings? Sure, the question states that the bug is climbing, and the slight chance of confusion is overshadowed by other, much greater issues with the question. But still, why would one choose a flying bug to crawl up a vase? It’s not importantly stupid, but it is gratuitously, hilariously stupid.

Anyway, we’re stuck with our stupid bee climbing up our stupid vase. What distance does our stupid bee travel? Well, obviously our stupid, non-flying bee should should climb as “up” as possible, without veering left or right, correct?

No and yes.

It is true that a bottom-to-top shortest path (geodesic) on a surface of revolution is a meridian. The proof of this, however, is very far from obvious; good luck explaining it to your students. But of course this is only Specialist Mathematics, so it’s not like we should expect the students to be inquisitive or critical or questioning assumptions or anything like that.

Anyway, our stupid non-flying bee climbs “up” our stupid vase. The distance our stupid bee travels is then the arc length of the graph of the original function f, and the required distance is given by the integral

    \[\boldsymbol{{\Huge \int\limits_{\frac12}^{\frac32}}\sqrt{1+\left[\tfrac{20}{1 - (2-2x)^2}\right]^2}}\ {\bf d}\boldsymbol{x}\]

The integral is ugly. More importantly, the integral is (doubly) improper and thus has no required meaning for Specialist students. Pretty damn stupid, and a stupidity we’ve seen not too long ago. It gets stupider.

Recall that this is a 1-mark question, and it is clearly expected to have the stupid calculator do the work. Great, sort of. The calculator computes integrals that the students are not required to understand but, apart from being utterly meaningless crap, everything is fine. Except, the calculators are really stupid.

Two brands of CAS calculators appear to be standard in VCE. Brand A will readily compute the integral above. Unfortunately, Brand A calculators will also compute improper integrals that don’t exist. Which is stupid. Brand B calculators, on the other hand, will not directly compute improper integrals such as the one above; instead, one first has to de-improper the integral by changing the limits to something like 0.50001 and 1.49999. Which is ugly and stupid. It also requires students to recognise the improperness in the integral, which they are supposedly not required to understand. Which is really stupid. (The lesser known Brand C appears to be less stupid with improper integrals.)

There is a stupid way around this stupidity. The arc length can also be calculated in terms of the inverse function of f, which avoid the improperness and then all is good. All is good, that is, except for the thousands of students who happen to have a Brand B calculator and who naively failed to consider that a crappy, 1-mark button-pushing question might require them to hunt for a Specialist-valid and B-compatible approach.

The idiocy of VCE exams is truly unlimited.

Inverted Logic

The 2018 Northern Hemisphere Mathematical Methods exams (1 and 2) are out. We didn’t spot any Magritte-esque lunacy, which was a pleasant surprise. In general, the exam questions were merely trivial, clumsy, contrived, calculator-infested and loathsomely ugly. So, all in all not bad by VCAA standards.

There, was, however, one notable question. The final multiple choice question on Exam 2 reads as follows:

Let f be a one-to-one differentiable function such that f (3) = 7, f (7) = 8, f′(3) = 2 and f′(7) = 3. The function g is differentiable and g(x) = f –1(x) for all x. g′(7) is equal to …

The wording is hilarious, at least it is if you’re not a frazzled Methods student in the midst of an exam, trying to make sense of such nonsense. Indeed, as we’ll see below, the question turned out to be too convoluted even for the examiners.

Of course –1 is a perfectly fine and familiar name for the inverse of f. It takes a special cluelessness to imagine that renaming –1 as g is somehow required or remotely helpful. The obfuscating wording, however, is the least of our concerns.

The exam question is intended to be a straight-forward application of the inverse function theorem. So, in Leibniz form dx/dy = 1/(dy/dx), though the exam question effectively requires the more explicit but less intuitive function form, 

    \[\boldsymbol {\left(f^{-1}\right)'(b) = \frac1{f'\left(f^{-1}(b)\right)}}.}\]

IVT is typically stated, and in particular the differentiability of –1 can be concluded, with suitable hypotheses. In this regard, the exam question needlessly hypothesising that the function g is differentiable is somewhat artificial. However it is not so simple in the school context to discuss natural hypotheses for IVT. So, underlying the ridiculous phrasing is a reasonable enough question.

What, then, is the problem? The problem is that IVT is not explicitly in the VCE curriculum. Really? Really.

Even ignoring the obvious issue this raises for the above exam question, the subliminal treatment of IVT in VCE is absurd. One requires plenty of inverse derivatives, even in a first calculus course. Yet, there is never any explicit mention of IVT in either Specialist or Methods, not even a hint that there is a common question with a universal answer.

All that appears to be explicit in VCE, and more in Specialist than Methods, is application of the chain rule, case by isolated case. So, one assumes the differentiability of –1 and and then differentiates –1(f(x)) in Leibniz form. For example, in the most respected Methods text the derivative of y = log(x) is somewhat dodgily obtained using the chain rule from the (very dodgily obtained) derivative of x = ey.

It is all very implicit, very case-by-case, and very Leibniz. Which makes the above exam question effectively impossible.

How many students actually obtained the correct answer? We don’t know since the Examiners’ Report doesn’t actually report anything. Being a multiple choice question, though, students had a 1 in 5 chance of obtaining the correct answer by dumb luck. Or, sticking to the more plausible answers, maybe even a 1 in 3 or 1 in 2 chance. That seems to be how the examiners stumbled upon the correct answer.

The Report’s solution to the exam question reads as follows (as of September 20, 2018):

f(3) = 7, f'(3) = 8, g(x) = f –1(x) , g‘(x) = 1/2 since

f'(x) x f'(y) = 1, g(x) = f'(x) = 1/f'(y).

The awfulness displayed above is a wonder to behold. Even if it were correct, the suggested solution would still bear no resemblance to the Methods curriculum, and it would still be unreadable. And the answer is not close to correct.

To be fair, The Report warns that its sample answers are “not intended to be exemplary or complete”. So perhaps they just forgot the further warning, that their answers are also not intended to be correct or comprehensible.

It is abundantly clear that the VCAA is incapable of putting together a coherent curriculum, let alone one that is even minimally engaging. Apparently it is even too much to expect the examiners to be familiar with their own crappy curriculum, and to be able to examine it, and to report on it, fairly and accurately.

VCAA Plays Dumb and Dumber

Late last year we posted on Madness in the 2017 VCE mathematics exams, on blatant errors above and beyond the exams’ predictably general clunkiness. For one (Northern Hemisphere) exam, the subsequent VCAA Report had already appeared; this Report was pretty useless in general, and specifically it was silent on the error and the surrounding mathematical crap. None of the other reports had yet appeared.

Now, finally, all the exam reports are out. God only knows why it took half a year, but at least they’re out. We have already posted on one particularly nasty piece of nitpicking nonsense, and now we can review the VCAA‘s own assessment of their five errors:

 

So, the VCAA responds to five blatant errors with five Trumpian silences. How should one describe such conduct? Unprofessional? Arrogant? Cowardly? VCAA-ish? All of the above?

 

Little Steps for Little Minds

Here’s a quick but telling nugget of awfulness from Victoria’s 2017 VCE maths exams. Q9 of the first (non-calculator) Methods Exam is concerned with the function

    \[\boldsymbol {f(x) = \sqrt{x}(1-x)\,.}\]

In Part (b) of the question students are asked to show that the gradient of the tangent to the graph of f” equals \boldsymbol{ \frac{1-3x}{2\sqrt{x}} } .

A normal human being would simply have asked for the derivative of f, but not much can go wrong, right? Expanding and differentiating, we have

    \[\boldsymbol {f'(x) = \frac{1}{2\sqrt{x}} - \frac32\sqrt{x}=\frac{1-3x}{2\sqrt{x}}\,.}\]

Easy, and done.

So, how is it that 65% of Methods students scored 0 on this contrived but routine 1-point question? Did they choke on “the gradient of the tangent to the graph of f” and go on to hunt for a question written in English?

The Examiners’ Report pinpoints the issue, noting that the exam question required a step-by-step demonstration …. And, [w]hen answering ‘show that’ questions, students should include all steps to demonstrate exactly what was done (emphasis added). So the Report implies, for example, that our calculation above would have scored 0 because we didn’t explicitly include the step of obtaining a common denominator.

Jesus H. Christ.

Any suggestion that our calculation is an insufficient answer for a student in a senior maths class is pedagogical and mathematical lunacy. This is obvious, even ignoring the fact that Methods questions way too often are flawed and/or require the most fantastic of logical leaps. And, of course, the instruction that “all steps” be included is both meaningless and utterly mad, and the solution in the Examiners’ Report does nothing of the sort. (Exercise: Try to include all steps in the computation and simplification of f’.)

This is just one 1-point question, but such infantilising nonsense is endemic in Methods. The subject is saturated with pointlessly prissy language and infuriating, nano-step nitpicking, none of which bears the remotest resemblance to real mathematical thought or expression.

What is the message of such garbage? For the vast majority of students, who naively presume that an educational authority would have some expertise in education, the message is that mathematics is nothing but soulless bookkeeping, which should be avoided at all costs. For anyone who knows mathematics, however, the message is that Victorian maths education is in the clutches of a heartless and entirely clueless antimathematical institution.

A Lack of Moral Authority

The Victorian Minister for Education has announced that the state’s senior school curriculum will undergo a review. The stated focus of the review is to consider whether “there should be a more explicit requirement for students to meet minimum standards of literacy and numeracy …“. The review appears to be strongly supported by industry, with a representative of the Australian Industry Group noting that “many companies complained school leavers made mistakes in spelling and grammar, and could not do basic maths“.

Dumb and dumber.

First, let’s note that Victorian schools have 12 years (plus prep) to teach the 3 Rs. That works out to 4 years (plus prep/3) per R, yet somehow it’s not working. Somehow the standards are sufficiently low that senior students can scale an exhausting mountain of assignments and exams, and still too many students come out lacking basic skills.

Secondly, the Minister has determined that the review will be conducted by the VCAA, the body already responsible for Victorian education.

If the definition of insanity is doing the same thing over and over and expecting different results, then the definition of insane governance is expecting the arrogant clown factory responsible for years of educational idiocy to have any willingness or ability to fix it.

Fixations and Madness

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

    \[\boldsymbol{(1)\qquad\qquad f(x) \ = \ f^{-1}(x)\,.}\]

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

    \[ \boldsymbol{(2) \qquad\qquad  f(x) \ = \  x\,,}\]

or perhaps the equation

    \[\boldsymbol{(2)' \qquad\qquad f^{-1}(x)\ = \ x \,.}\]

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.

The Madness of Crowd Models

Our fifth and penultimate post on the 2017 VCE exam madness concerns Question 3 of Section B on the Northern Hemisphere Specialist Mathematics Exam 2. The question begins with the logistic equation for the proportion P of a petrie dish covered by bacteria:

    \[\boldsymbol{\frac{{\rm d} P}{{\rm d} t\ }= \frac{P}{2}\left(1 - P\right)\,\qquad 0 < P < 1\,.}\]

This is not a great start, since it’s a little peculiar using the logistic equation to model an area proportion, rather than a population or a population density. It’s also worth noting that the strict inequalities on P are unnecessary and rule out of consideration the equilibrium (constant) solutions P = 0 and P = 1.

Clunky framing aside, part (a) of Question 3 is pretty standard, requiring the solving of the above (separable) differential equation with initial condition P(0) = 1/2. So, a decent integration problem trivialised by the presence of the stupifying CAS machine. After which things go seriously off the rails.

The setting for part (b) of the question has a toxin added to the petri dish at time t = 1, with the bacterial growth then modelled by the equation

    \[\boldsymbol{\frac{{\rm d} P}{{\rm d} t\ }= \frac{P}{2}\left(1 - P\right) - \frac{\sqrt{P}}{20}\,.}\]

Well, probably not. The effect of toxins is most simply modelled as depending linearly on P, and there seems to be no argument for the square root. Still, this kind of fantasy modelling is par for the VCAA‘s crazy course. Then, however, comes Question 3(b):

Find the limiting value of P, which is the maximum possible proportion of the Petri dish that can now be covered by the bacteria.

The question is a mess. And it’s wrong.

The Examiners’ “Report” (which is not a report at all, but merely a list of short answers) fails to indicate what students did or how well they did on this short, 2-mark question. Presumably the intent was for students to find the limit of P by finding the maximal equilibrium solution of the differential equation. So, setting dP/dt = 0 implies that the right hand side of the differential equation is also 0. The resulting equation is not particularly nice, a quartic equation for Q = √P. Just more silly CAS stuff, then, giving the largest solution P = 0.894 to the requested three decimal places.

In principle, applying that approach here is fine. There are, however, two major problems.

The first problem is with the wording of the question: “maximum possible proportion” simply does not mean maximal equilibrium solution, nor much of anything. The maximum possible proportion covered by the bacteria is P = 1. Alternatively, if we follow the examiners and needlessly exclude = 1 from consideration, then there is no maximum possible proportion, and P can just be arbitrarily close to 1. Either way, a large initial P will decay down to the maximal equilibrium solution.

One might argue that the examiners had in mind a continuation of part (a), so that the proportion begins below the equilibrium value and then rises towards it. That wouldn’t rescue the wording, however. The equilibrium solution is still not a maximum, since the equilibrium value is never actually attained. The expression the examiners are missing, and possibly may even have heard of, is least upper bound. That expression is too sophisticated to be used on a school exam, but whose problem is that? It’s the examiners who painted themselves into a corner.

The second issue is that it is not at all obvious – indeed it can easily fail to be true – that the maximal equilibrium solution for P will also be the limiting value of P. The garbled information within question (b) is instructing students to simply assume this. Well, ok, it’s their question. But why go to such lengths to impose a dubious and impossible-to-word assumption, rather than simply asking directly for an equilibrium solution?

To clarify the issues here, and to show why the examiners were pretty much doomed to make a mess of things, consider the following differential equation:

    \[\boldsymbol{\frac{{\rm d} P}{{\rm d} t\ }= 3P - 4P^2 - \sqrt{P}\,.}\]

By setting Q = √P, for example, it is easy to show that the equilibrium solutions are P = 0 and P = 1/4. Moreover, by considering the sign of dP/dt for P above and below the equilibrium P = 1/4, it is easy to obtain a qualitative sense of the general solutions to the differential equation:

In particular, it is easy to see that the constant solution P = 1/4 is a semi-stable equilibrium: if P(0) is slightly below 1/4 then P(t) will decay to the stable equilibrium P = 0.

This type of analysis, which can readily be performed on the toxin equation above, is simple, natural and powerful. And, it seems, non-existent in Specialist Mathematics. The curriculum  contains nothing that suggests or promotes any such analysis, nor even a mention of equilibrium solutions. The same holds for the standard textbook, in which for, for example, the equation for Newton’s law of cooling is solved (clumsily), but there’s not a word of insight into the solutions.

And this explains why the examiners were doomed to fail. Yes, they almost stumbled into writing a good, mathematically rich exam question. The paper thin curriculum, however, wouldn’t permit it.

 

A Madness for all Seasons

Our fourth post on the  2017 VCE exam madness will be similar to our previous post: a quick whack of a straight-out error. This error was flagged by a teacher friend, David. (No, not that David.)

The 11th multiple choice question on the first Further Mathematics Exam reads as follows:

Which one of the following statistics can never be negative? 

A. the maximum value in a data set

B. the value of a Pearson correlation coefficient

C. the value of a moving mean in a smoothed time series

D. the value of a seasonal index

E. the value of a slope of a least squares line fitted to a scatterplot

Before we get started, a quick word on the question’s repeated use of the redundant “the value of”.

Bleah!

Now, on with answering the question.

It is pretty obvious that the statistics in A, B, C and E can all be negative, so presumably the intended answer is D. However, D is also wrong: a seasonal index can also be negative. Unfortunately the explanation of “seasonal index” in the standard textbook is lost in a jungle of non-explanation, so to illustrate we’ll work through a very simple example.

Suppose a company’s profits and losses over the four quarters of a year are as follows:

    \[ \begin{tabular} {| c | c | c | c |}\hline {\bf\phantom{S}Summer \phantom{I}} &{\bf\phantom{S}Autumn \phantom{I}} &{\bf\phantom{S}Winter \phantom{I}} &{\bf\phantom{S}Spring \phantom{I}} \\  \hline {\bf \$6000} & {\bf -\$1000} & {\bf -\$2000} & {\bf \$5000}\\ \hline \end{tabular}\]

So, the total profit over the year is $8,000, and then the average quarterly profit is $2000. The seasonal index (SI) for each quarter is then that quarter’s profit (or loss) divided by the average quarterly profit:

    \[ \begin{tabular} {| c | c | c | c |}\hline {\bf Summer SI} &{\bf Autumn SI} &{\bf Winter SI} &{\bf Spring SI} \\  \hline {\bf 3} & {\bf -0.5} & {\bf -1.0} & {\bf 2.5}\\ \hline \end{tabular}\]

Clearly this example is general, in the sense that in any scenario where the seasonal data are both positive and negative, some of the seasonal indices will be negative. So, the exam question is not merely technically wrong, with a contrived example raising issues: the question is wrong wrong.

Now, to be fair, this time the VCAA has a defense. It appears to be more common to apply seasonal indices in contexts where all the data are one sign, or to use absolute values to then consider magnitudes of deviations. It also appears that most or all examples Further students would have studied included only positive data.

So, yes, the VCAA (and the Australian Curriculum) don’t bother to clarify the definition or permitted contexts for seasonal indices. And yes, the definition in the standard textbook implicitly permits negative seasonal indices. And yes, by this definition the exam question is plain wrong. But, hopefully most students weren’t paying sufficient attention to realise that the VCAA weren’t paying sufficient attention, and so all is ok.

Well, the defense is something like that. The VCAA can work on the wording.

 

Further Madness

Our third post on the 2017 VCE exam madness will be brief, on a question containing a flagrant error.

The first question in the matrix module of Further Mathematics’ Exam 2 is concerned with a school canteen selling pies, rolls and sandwiches over three separate weeks. The number of items sold is set up as a 3 x 3 matrix, one row for each week and one column for each food choice. The last part, (c)(ii), of the question then reads:

The matrix equation below shows that the total value of all rolls and sandwiches sold in these three weeks is $915.60 

    \[   \boldsymbol{L \times\begin{bmatrix} 491.55 \\ 428.00\\ 487.60 \end{bmatrix} \ = \ [915.60]}\]

Matrix L in this equation is of order 1 x 3.

Write down matrix L.

This 1-mark question is presumably meant to be a gimme, with answer L = [0 1 1]. Unfortunately the question is both weird and wrong. (And lacking in punctuation. Guys, it’s not that hard.) The wrongness comes from the examiners having confused their rows and columns. As is made clear in the the previous part, (c)(i), of the question, the  3 x 1 matrix of numbers indicates the total earnings from each of the three weeks, not from each of the three food choices. So, the equation indicates the total value of all products sold in weeks 2 and 3.

There’s not much to say about such an obvious error. It is very easy to confuse rows and columns, and we’ve all done it on occasion, but if VCAA’s vetting cannot catch this kind of mistake then it cannot be relied upon to catch anything. The only question is how the Examiners’ Report will eventually address the error. The VCAA is well-practised in cowardly silence and weasel-wording, but it would be exceptionally Trumplike to attempt such tactics here.

Error aside, the question is artificial, and it is not clear that the matrix equation “shows” much of anything. Yes, 0-1 and on-or-off matrices are important and useful, but the use of such a matrix in this context is contrived and confusing. Not a hanging offence, and benign by VCAA’s standards, but the question is pretty silly. And, not forgetting, wrong.

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 not 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

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

Done.

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.