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

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


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.