WitCH 36: Sub Standard

This WitCH is a companion to our previous, MitPY post, and is a little different from most of our WitCHes. Typically in a WitCH the sin is unarguable, and it is only the egregiousness of the sin that is up for debate. In this case, however, there is room for disagreement, along with some blatant sinning. It comes, predictably, from Cambridge’s Specialist Mathematics 3 & 4 (2020).

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 28: Tone Deaf

We haven’t yet had a chance to go through the 2019 VCE exams, but this question was flagged to me independently by two colleagues: let’s call them Dr. Death and Simon the Likeable. It’s from Mathematical Methods Exam 2 (CAS). (No link yet.)

UPDATE (05/07/20)

Even ignoring the stuff-ups, this question is ugly and pointless; the pseudo-applied framing is ugly and pointless; the CASification is ugly and pointless; the back-to-front integral is ugly and pointless; the matrix equation is ugly and pointless; the transformation is really ugly and really pointless. Part (f) is the pinnacle of ugliness and pointlessness, but the entire question is swill, from beginning to end.

And then there’s Part (e). “This question was not answered well” the examiners solemnly intone. Gee, really? Do you think your question being completely stuffed might have had something to do with it? Do you think maybe having a transformation of x when there’s not an x in sight may have been just a tad confusing? Do you think that the transformation then resulting in a function of t was maybe not the smartest move? Do you think writing an integral backwards was perhaps just a little too cute? Do you think possibly referring to the area of, rather than to the value of, an integral was slightly clunky? And, most importantly, do you think perhaps asking a question for which there is an infinite and impenetrable jungle of answers may have been an exercise in canyon-sized incompetence?

But, sure, those troublesome students didn’t answer your question well.

Part (e) was intended to have students find a transformation of the function f that effectively switches the behaviour on the intervals [0,4] and [4,6] to the intervals [2,6] and [0,2].  Ignoring the fact that the intended question was asked in an absurdly opaque manner, and ignoring the fact that no motivation for the intended question was either provided or is imaginable, the question asked was entirely different, and was ridiculous.

Writing the transformation out,

    \[\boldsymbol{\left\{\aligned &X = ax + c \\ &Y = by + d, \right.\endaligned}\]

we then have

    \[\boldsymbol{\left\{\aligned &x = \frac{X - c}{a} \\ & y = \frac{Y - d}{b}. \right.\endaligned}\]

So, the function y = f(t) y = f(x) can be written

    \[\boldsymbol{\dfrac{Y - d}{b} = f\!\left(\dfrac{X - c}{a}\right).}\]

Solving for Y, that means our transformed function Y = g(X) can be written

    \[\boldsymbol{g(X) = b\, f\!\left(\dfrac{X - c}{a}\right) + d.}\]

Well, this is our function g unless a = 0, in which case g doesn’t exist. Whatever. Back to the swill.

Using the result from Part (d), we have Part (e) asking for a, b, c and d such that

    \[\boldsymbol{\int\limits_2^0 + \int\limits_2^6 \ \left[ b\, f\! \left(\dfrac{X - c}{a}\right) + d\right]  \, {\rm d}X \ = \ \dfrac{15}{\pi}.}\]

What then are the solutions to this equation? The examination report lists a couple of families and then blithely remarks “There are other solutions”. Really? Then why didn’t you list them, you clowns? 

We’ll tell you why. Because the complete solution to this monster is a God Almighty multi-infinite mess. As a starting idea, pick any three of the variables, say a and b and c, to be whatever you want, and then try to adjust the fourth variable, d, to solve the equation. We’ll offer a prize for anyone who can give a complete solution. 

This question is as good an example as there can be of the pointlessness, the ugliness and the monumental klutziness of VCAA’s swamp mathematics.

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 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 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.

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.