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.