WitCH 38: A Deep Hole

This one is due to commenter P.N., who raised it on another post, and the glaring issue has been discussed there. Still, for the record it should be WitCHed, and we’ve also decided to expand the WitCHiness slightly (and could have expanded it further).

The following questions appeared on 2019 Specialist Mathematics NHT, Exam 2 (CAS). The questions are followed by sample Mathematica solutions (screenshot corrected, to include final comment) provided by VCAA (presumably in the main for VCE students doing the Mathematica version of Methods). The examination report provides answers, identical to those in the Mathematica solutions, but indicates nothing further.

UPDATE (05/07/20)

The obvious problem here, of course, is that the answer for Part (b), in both the examination report and VCAA’s Mathematica solutions, is flat out wrong: the function fk will also fail to have a stationary point if k = -2 or k = 0. Nearly as bad, and plenty bad, the method in VCAA’s Mathematica solutions to Part (c) is fundamentally incomplete: for a (twice-differentiable) function f to have an inflection point at some a, it is necessary but not sufficient to have f’’(a) = 0.

That’s all pretty awful, but we believe there is worse here. The question is, how did the VCAA get it wrong? Errors can always occur, but why specifically did the error in Part (b) occur, and why, for a year and counting, wasn’t it caught? Why was a half-method suggested for Part (c), and why was this half-method presumably considered reasonable strategy for the exam? Partly, the explanation can go down to this being a question from NHT, about which, as far as we can tell, no one really gives a stuff. This VCAA screw-up, however, points to a deeper, systemic and much more important issue.

The first thing to note is that Mathematica got it wrong: the Solve function did not return the solution to the equation fk‘ = 0. What does that imply for using Mathematica and other CAS software? It implies the user should be aware that the machine is not necessarily doing what the user might reasonably think it is doing. Which is a very, very stupid property of a black box: if Solve doesn’t mean “solve”, then what the hell does it mean? Now, as it happens, Mathematica’s/VCAA’s screw-up could have been avoided by using the function Reduce instead of Solve.* That would have saved VCAA’s solutions from being wrong, but not from being garbage.

Ask yourself, what is missing from VCAA’s solutions? Yes, yes, correct answers, but what else? This is it: there are no functions. There are no equations. There is nothing, nothing at all but an unreliable black box. Here we have a question about the derivatives of a function, but nowhere are those derivatives computed, displayed or contemplated in even the smallest sense.

For the NHT problem above, the massive elephant not in the room is an expression for the derivative function:

    \[\color{red} \boldsymbol{f'_k(x) = -\frac{x^2 + 2(k+1)x +1}{(x^2-1)^2}}\]

What do you see? Yep, if your algebraic sense hasn’t been totally destroyed by CAS, you see immediately that the values k = 0 and k = -2 are special, and that special behaviour is likely to occur. You’re aware of the function, alert to its properties, and you’re led back to the simplification of fk for these special values. Then, either way or both, you are much, much less likely to screw up in the way the VCAA did.

And that always happens. A mathematician always gets a sense of solutions not just from the solution values, but also from the structure of the equations being solved. And all of this is invisible, is impossible, all of it is obliterated by VCAA’s nuclear weapon approach.

And that is insane. To expect, to effectively demand that students “solve” equations without ever seeing those equations, without an iota of concern for what the equations look like, what the equations might tell us, is mathematical and pedagogical insanity.

 

*) Thanks to our ex-student and friend and colleague Sai for explaining some of Mathematica’s subtleties. Readers will be learning more about Sai in the very near future.

WitCH 37: A Foolproof Argument

We’re amazed we didn’t know about this one, which was brought to our attention by commenter P.N.. It comes from the 2013 Specialist Mathematics Exam 2: The sole comment on this question in the Examination Report is:

“All students were awarded [the] mark for this question.”

Yep, the question is plain stuffed. We think, however, there is more here than the simple wrongness, which is why we’ve made it a WitCH rather than a PoSWW. Happy hunting.

UPDATE (11/05) Steve C’s comment below has inspired an addition:

Update (20/05/20)

The third greatest issue with the exam question is that it is wrong: none of the available answers is correct. The second greatest issue is that the wrongness is obvious: if z^3 lies in a sector then the natural guess is that z will lie in one of three equally spaced sectors of a third the width, so God knows why the alarm bells weren’t ringing. The greatest issue is that VCAA didn’t have the guts or the basic integrity to fess up: not a single word of responsibility or remorse. Assholes.

Those are the elephants stomping through the room but, as commenters as have noted, there is plenty more awfulness in this question:

  • “Letting” z = a + bi is sloppy, confusing and pointless;
  • The term “quadrant” is undefined;
  • The use of “principal” is unnecessary;
  • “argument” is better thought as the measure of an angle not the angle itself;
  • Given z is a single complex number, “the complete set of values for Arg(z)” will consist of a single number.
  • The grammar isn’t.

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 35: Overly Resolute

This WitCH (arguably a PoSWW) comes courtesy of Damien, an occasional commenter and an ex-student of ours from the nineteenth century. It is from the 2019 Specialist Mathematics Exam 2. We’ll confess, we completely overlooked the issue when going through the MAV solutions.

Update (16/02/20)

What a mess. Thanks to Damo for pointing out the problem, and thanks to the commenters for figuring out the nonsense.

In general form, the (intended) scenario of the exam question is

The vector resolute of \boldsymbol{\tilde{a}} in the direction of \boldsymbol{\tilde{b}} is \boldsymbol{\tilde{c}},

which can be pictured as follows: For the exam question, we have \boldsymbol{\tilde{a}} = \boldsymbol{\tilde{i}} + \boldsymbol{\tilde{j}} - \boldsymbol{\tilde{k}}, \boldsymbol{\tilde{b}} = m\boldsymbol{\tilde{i}} + n\boldsymbol{\tilde{j}} + p\boldsymbol{\tilde{k}} and \boldsymbol{\tilde{c}} = 2\boldsymbol{\tilde{i}} - 3\boldsymbol{\tilde{j}} + \boldsymbol{\tilde{k}}.

Of course, given \boldsymbol{\tilde{a}} and \boldsymbol{\tilde{b}} it is standard to find \boldsymbol{\tilde{c}}. After a bit of trig and unit vectors, we have (in must useful form)

\boldsymbol{\tilde{c} = \left(\dfrac{\tilde{a}\cdot \tilde{b}}{\tilde{b}\cdot \tilde{b}}\right)\tilde{b}}

The exam question, however, is different: the question is, given \boldsymbol{\tilde{a}} and \boldsymbol{\tilde{c}}, how to find \boldsymbol{\tilde{b}}.

The problem with that is, unless the vectors \boldsymbol{\tilde{a}} and \boldsymbol{\tilde{c}} are appropriately related, the scenario simply cannot occur, meaning \boldsymbol{\tilde{b}} cannot exist. Most obviously, the length of \boldsymbol{\tilde{c}} must be no greater than the length of \boldsymbol{\tilde{a}}. This requirement is clear from the triangle pictured, and can also be proved algebraically (with the dot product formula or the Cauchy-Schwarz inequality).

This implies, of course, that the exam question is ridiculous: for the vectors in the exam we have |\boldsymbol{\tilde{c}}| > |\boldsymbol{\tilde{a}}|, and that’s the end of that. In fact, the situation is more delicate; given the pictured vectors form a right-angled triangle, we require that \boldsymbol{\tilde{a}} - \boldsymbol{\tilde{c}} be perpendicular to \boldsymbol{\tilde{c}}. Which implies, once again, that the exam question is ridiculous.

Next, suppose we lucked out and began with \boldsymbol{\tilde{a}}- \boldsymbol{\tilde{c}} perpendicular to \boldsymbol{\tilde{c}}. (Of course it is very easy to check whether we’ve lucked out.) How, then, do we find \boldsymbol{\tilde{b}}? The answer is, as is made clear by the picture, “Well, duh”. The possible vectors \boldsymbol{\tilde{b}} are simply the (non-zero) scalar multiples of \boldsymbol{\tilde{c}}, and we’re done. Which shows that the mess in the intended solution, Answer A, is ridiculous.

There is a final question, however: the exam question is clearly ridiculous, but is the question also stuffed? The equations in answer A come from the equation for \boldsymbol{\tilde{c}} above and working backwards. And, these equations correctly return no solutions. Moreover, if the relationship between \boldsymbol{\tilde{a}} and \boldsymbol{\tilde{c}} had been such that there were solutions, then the A equations would have found them. So, completely ridiculous but still ok?

Nope.

The question is framed from start to end around definite, existing objects: we have THE vector resolute, resulting in THE values of m, n and p. If the VCAA had worded the question to find possible values, on the basis of a possible direction for the resolution, then, at least technically, the question would be consistent, with A a valid answer. Still an utterly ridiculous question, but consistent. But the VCAA didn’t do that and so the question isn’t that. The question is stuffed.

Further Update (26/06/20)

As commenters have noted, the Examination Report has finally appeared. And, as predicted, answer A was deemed correct, with the Report noting

Option A gives the set of equations that can be used to obtain the values of m, n and p. Explicit solution would result in a null set as it is not possible for a result of a vector to be of greater magnitude than the vector itself.

Well, it’s something. Presumably “result of a vector” was intended to be “resolute of a vector”, and the set framing is weirdly New Mathy. But, it’s something. Seriously. As John Friend notes, it is at least a small step along the way to indicating the question is not all hunky-dory.

That step, however, is way too small. We’ll close with two comments, reiterating the points made above.

1. The question is wrong

Read the question again, and read the first sentence of the Report’s comment. The question and report justification are fundamentally stuffed by the definite articles, by the language of existence. All answers should have been marked correct.

2. The question is worse than wrong

Even if the vectors \boldsymbol{\tilde{a}} and \boldsymbol{\tilde{c}} had been chosen appropriately, the question is utterly devoid of mathematical sense. It suggests a long and difficult method to solve a problem that, if indeed is solvable, is trivial. 

 

 

WitCHes in Batches

What we like about WitCHes is that they enable us to post quickly on nonsense when it occurs or when it is brought to our attention, without our needing to compose a careful and polished critique: readers can do the work in the comments. What we hate about WitCHes is that they still eventually require rounding off with a proper summation, and that’s work. We hate work.

Currently, we have a big and annoying backlog of unsummed WitCHes. That’s not great, since a timely rounding off of discussion is valuable. Our intention is to begin ticking off the unsummed WitCHes, which are listed below with brief indications of the topics. Most of these WitCHes have been properly hammered by commenters, though of course readers are always welcome to comment, including after summation. We’ll update this post as the WitCHes get ticked off. Thanks very much to all past WitCH-commenters, and we’re sorry for the delay in polishing off. We’ll attempt to keep on top of future WitCHes.

WitCH 8 (oblique asymptotes – UPDATED 05/02/20)

WitCH 10 (distance function – UPDATED 29/03/20)

WitCH 12 (trig integral)

WitCH 18 (Serena Williams)

WitCH 20 (hypothesis testing)

WitCH 21 (order of algorithms)

WitCH 22 (inflection points)

WitCH 23 (speed functions)

WitCH 24 (functional equations)

WitCH 25 (probability distributions)

WitCH 26 (function composition)

WitCH 27 (function composition – UPDATED 15/06/20)

WitCH 28 (trig graphs)

WitCH 29 (inverse derivatives – UPDATED 19/06/20)

WitCH 30 (Eddie Woo)

WitCH 31 (function composition)

WitCH 32 (PISA)

WitCH 33 (probability distributions)

WitCH 34 (numeracy guide – added 05/02/20)

WitCH 35 (vector resolutes – added 14/02/20 – UPDATED 16/02/20)

WitCH 36 (integration by substitution – added 21/04/20)

WitCH 37 (complex argument – added 11/05/20 – UPDATED 20/05/20)

WitCH 38 (stationary points – added 20/06/20)

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).}
 
and
 
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 31: Decomposing

We have a short Specialist post coming, and we’ll have more to write on the 2019 VCE exams once they’re online. But, for now, one more Mathematical Methods WitCH, from the 2019 (calculator-free) Exam 1:

Update (04/07/20)

The main crap here, of course, is part (f): as commenter John Friend puts it, what the hell is this question supposed to be testing? And, sure, the last part of the last question on an exam is allowed to be a little special, but one measly mark? Compared to the triviality of the rest of the question?

Of course, students bombed part (f). The examination report indicates that 19% of student correctly answered that there is one solution to the equation; as suggested by commenter Red Five, it’s also a pretty safe bet that the majority of students who got there did so with a Hail Mary guess. (It should be added, the students didn’t do swimmingly well on the rest of Question 9, the CAS-lobotomising having working its usual magic.)

OK, so what did examiners expect for that one measly mark? We’ll get to a reasonable solution below, but let’s first consider some unreasonable solutions.

Here is the examination report’s entire commentary on Part (f):

g(f(x) + f(g(x)) = 0 has exactly one solution.

This question was not well done. Few students attempted to draw a rough sketch of each equation and use addition of ordinates.

Gee, thanks. Drawing a “rough sketch” of either of these compositions is anything but trivial. For one measly mark. We’ll look at sketching aspects of these graphs below, but let’s get on with another unreasonable solution.

Given the weirdness of part (f), a student might hope that parts (a)-(e) provide some guidance. Let’s see.

Part (b) (for which the examination report contains an error), gets us to conclude that the composition

\boldsymbol{g(f(x)) = e^{\left(3+2x-x^2\right)}} has negative derivative when x > 1.

Part (c) leads us to the composition

\boldsymbol{f(g(x)) = \left(3 - e^x\right) \left(1 + e^x\right)}

having x-intercept when x = log(3).

Finally, Part (e) gives us that the composition f(g(x)) has the sole stationary point (0,4). How does this information help us with Part (f)? Bugger all.

So, what if we include the natural implications of our previous work? That gives us something like the following: Well, um, great. We’re left still hunting for that one measly mark.

OK, the other parts of the question are of little help, and the examiners are of no help, so what do else do we need? There are two further pieces of information we require (plus the Intermediate Value Theorem). First, note that

\boldsymbol{g(f(x)) = e^{\mbox{\bf THING}} > 0}.

Secondly, note that

\boldsymbol{f(g(x)) = \left(3 + e^x\right) \left(1 - e^x\right)} = -}\mbox{\bf HUGE} if x is huge.

Then, given we know the slopes of the compositions, we can finally complete our rough sketches: Now, let’s write S(x) for our sum function g(f(x)) + f(g(x)). We know S(x) > 0 unless one of our compositions is negative. So, the only place we could get S = 0 is if x > log(3). But S(log(3)) > 0, and eventually S is hugely negative. That means S must cross the x-axis (by IVT). But, since S is decreasing for x > 1, S can only cross the axis once, and S = 0 must have exactly one solution. 

We’ve finally earned our one measly mark. Yay?