February 2020 – BAD MATHEMATICS

MAV’s Dangerous Inflection

February 24, 2020

11:02 am

15 Comments on MAV’s Dangerous Inflection

calculus, CAS, differential equations, MAV, Specialist Mathematics, VCAA, VCE

This post concerns a question on the 2019 VCE Specialist Mathematics Exam 2 and, in particular, the solution and commentary for that question available through the Mathematical Association of Victoria. As we document below, a significant part of what MAV has written on this question is confused, self-contradictory and tendentious. Thus, noting the semi-official status of MAV solutions, that these solutions play a significant role in MAV’s Meet the Assessors events, and are quite possibly written by VCE assessors, there are some troubling implications. Question 3, Section B on Exam 2 is a differential equations problem, with two independent parts. Part (a) is a routine (and pretty nice) question on exponential growth and decay.* Part (b), which is our concern, considers the differential equation

$\boldsymbol{\color{blue}\frac{{\rm d}Q}{{\rm d}t\ } = e^{t-Q}}\,,$

for t ≥ 0, along with the initial condition

$\boldsymbol{\color{blue}Q(0) =1}\,.$

The differential equation is separable, and parts (i) and (ii) of the question, worth a total of 3 marks, asks to set up the separation and use this to show the solution of the initial value problem is

$\boldsymbol{\color{blue}Q =\log_e\hspace{-1pt} \left(e^t + e -1\right)}\,.$

Part (iii), worth 2 marks, then asks to show that “the graph of Q as a function of t” has no inflection points.** Question 3(b) is contrived and bitsy and hand-holding, but not incoherent or wrong. So, pretty good by VCE standards. Unfortunately, the MAV solution and commentary to this problem is deeply problematic. The first MAV misstep, in (i), is to invert the derivative, giving

$\boldsymbol{\color{red}\frac{{\rm d}t\ }{{\rm d}Q } = e^{Q-t}}\,,$

prior to separating variables. This is a very weird extra step to include since, not only is the step not required here, it is never required or helpful in solving separable equations. Its appearance here suggests a weak understanding of this standard technique. Worse is to come in (iii). Before considering MAV’s solution, however, it is perhaps worth indicating an approach to (iii) that may be unfamiliar to many teachers and students and, possibly, the assessors. If we are interested in the inflection points of Q,*** then we are interested in the second derivative of Q. The thing to note is we can naturally obtain an expression for Q” directly from the differential equation: we differentiate the equation using the chain rule, giving

$\boldsymbol{\color{magenta}Q'' = e^{t-Q}\left(1 - Q'\right)}\,.$

Now, the exponential is never zero, and so if we can show Q’ < 1 then we’d have Q” > 0, ruling out inflection points. Such conclusions can sometimes be read off easily from the differential equation, but it does not seem to be the case here. However, an easy differentiation of the expression for Q derived in part (ii) gives

$\boldsymbol{\color{magenta}Q' =\frac{e^t}{e^t + e -1}}\,.$

The numerator is clearly smaller than the denominator, proving that Q’ < 1, and we’re done. For a similar but distinct proof, one can use the differential equation to replace the Q’ in the expression for Q”, giving

$\boldsymbol{\color{magenta}Q'' = e^{t-Q}\left(1 - e^{t-Q}\right)}\,.$

Again we want to show the second factor is positive, which amounts to showing Q > t. But that is easy to see from the expression for Q above (because the stuff in the log is greater than $\boldsymbol{e^t}$ ), and again we can conclude that Q has no inflection points. One might reasonably consider the details in the above proofs to be overly subtle for many or most VCE students. Nonetheless the approaches are natural, are typically more efficient (and are CAS-free), and any comprehensive solutions to the problem should at least mention the possibility. The MAV solutions make no mention of any such approach, simply making a CAS-driven beeline for Q” as an explicit function of t. Here are the contents of the MAV solution:

Part 1: A restatement of the equation for Q from part (ii), which is then followed by

.˙. $\boldsymbol{ \color{red}\ \frac{{\rm d}^2Q }{{\rm d}t^2\ } = \frac{e^{t+1} -e^t}{\left(e^t + e -1\right)^2} }$

Part 2: A screenshot of the CAS input-output used to obtain the conclusion of Part 1.

Part 3: The statement

Solving .˙. $\boldsymbol{\color{red} \ \frac{{\rm d}^2Q }{{\rm d}t^2\ } = 0}$ gives no solution

Part 4: A screenshot of the CAS input-output used to obtain the conclusion of Part 3.

Part 5: The half-sentence

We can see that $\boldsymbol{\color{red}\frac{{\rm d}^2Q }{{\rm d}t^2\ } > 0}$ for all t,

Part 6: A labelled screenshot of a CAS-produced graph of Q”.

Part 7: The second half of the sentence,

so Q(t) has no points of inflection

This is a mess. The ordering of the information is poor and unexplained, making the unpunctuated sentences and part-sentences extremely difficult to read. Part 3 is so clumsy it’s funny. Much more important, the MAV “solution” makes little or no mathematical sense and is utterly useless as a guide to what the VCE might consider acceptable on an exam. True, the MAV solution is followed by a commentary specifically on the acceptability question. As we shall see, however, this commentary makes things worse. But before considering that commentary, let’s itemise the obvious questions raised by the MAV solution:

Is using CAS to calculate a second derivative on a “show that” exam question acceptable for VCE purposes?
Can a stated use of CAS to “show” there are no solutions to Q” = 0 suffice for VCE purposes? If not, what is the purpose of Parts 3 and 4 of the MAV solutions?
Does copying a CAS-produced graph of Q” suffice to “show” that Q” > 0 for VCE purposes?
If the answers to the above three questions differ, why do they differ?

Yes, of course these questions are primarily for the VCAA, but first things first. The MAV solution is followed by what is intended to be a clarifying comment:

Note that any reference to CAS producing ‘no solution’ to the second derivative equalling zero would NOT qualify for a mark in this ‘show that’ question. This is not sufficient. A sketch would also be required as would stating $\boldsymbol{\color{red}e^t (e - 1) \neq 0}$ for all t.

These definitive-sounding statements are confusing and interesting, not least for their simple existence. Do these statements purport to be bankable pronouncements of VCAA assessors? If not, what is their status? In any case, given that pretty much every exam question demands that students and teachers read inscrutable VCAA tea leaves, why is it solely the solution to question 3(b) that is followed by such statements? The MAV commentary at least makes clear their answer to our second question above: quoting CAS is not sufficient to “show” that Q” = 0 has no solutions. Unfortunately, the commentary raises more questions than it answers:

Parts 3 and 4 are “not sufficient”, but are they worth anything? If so, what are they worth and, in particular, what is the import of the word “also”? If not, then why not simply declare the parts irrelevant, in which case why include those parts in the solutions at all?
If, as claimed, it is “required” to state $\boldsymbol{e^t(e-1)\neq 0}$ (which is indeed the key point of this approach and should be required), then why does the MAV solution not contain any such statement, nor even the factorisation that would naturally precede this statement?
Why is a solution “required” to include a sketch of Q”? If, in particular, a statement such as $\boldsymbol{e^t(e-1)\neq 0}$ is “required”, or in any case is included, why would the latter not in and of itself suffice?

We wouldn’t begin to suggest answers to these questions, or our four earlier questions, and they are also not the main point here. The main point is that under no circumstances should such shoddy material be the basis of VCAA assessor presentations. If the material was also written by VCAA assessors, all the worse. Of course the underlying problem is not the quality or accuracy of solutions but, rather, the fundamental idiocy of incorporating CAS into proof questions. And for that the central villain is not the MAV but the VCAA, which has permitted their glorification of technology to completely destroy the appreciation of and the teaching of proof and reason. The MAV is not primarily responsible for this nonsense. The MAV is, however, responsible for publishing it, promoting it and profiting from it, none of which should be considered acceptable. The MAV needs to put serious thought into its unhealthily close relationship with the VCAA.

*) We might ask, however, who refers to “The growth and decay” of an exponential function?

**) One might simply have referred to Q, but VCAA loves them their words.

***) Or, if preferred, the points of inflection of the graph of Q as a function of t.

Update (26/06/20)

The Examination Report is out and is basically ok; none of the nonsense and non sequiturs of the MAV solutions are included. The solution to (b)(iii) correctly focuses upon the factoring of Q”, although it needlessly worries about the sign of the denominator. There is no mention of the more natural approach to obtaining and analysing Q” but, given the question is treated by the VCAA and pretty much everyone as just another mindless exercise in pushing buttons, this is no surprise.

WitCH 35: Overly Resolute

by marty

February 14, 2020

9:53 am

55 Comments on WitCH 35: Overly Resolute

WitCH

Specialist Mathematics, VCAA, VCE, vectors, WitCH

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 most 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.

What Happened to the Evil Mathologer?

by marty

February 12, 2020

9:32 am

13 Comments on What Happened to the Evil Mathologer?

Mathologer

Mathologer, media, YouTube

I know something of the current Mathologer issue, but not much. I’ll write more soon.

ps. For those who are unaware, I always refer to the Mathologer as ‘Evil”. I support the Evil Mathologer 100%, here and always.

UPDATE (Wednesday Morning)

I’ve talked to the Evil Mathologer, and all is (sort of) OK.

Mathologer will definitely continue, with a new video under construction as we speak. As for the deleted Mathologer videos, Burkard is enquiring about that, but no promises for a quick fix. As for why the videos were deleted, I don’t want to preempt whatever Burkard may want to say or not say about that.

UPDATE (Wednesday Afternoon)

OK, the Mathologer videos are now back up! Except, there was (hopefully only) a glitch with the re-upping of the latest video. I think Burkard expects that also to be sorted out soon.

UPDATE (Thursday)

Unfortunately, the Evil Mathologer’s latest video is still not back up, and nothing really to report. I had begun to write a comment for people who know (or think they know) what happened, and how I think they should try to understand it. I thought better of it.

What I do hope to do, in the next day or so, is write some of the prehistory of the Mathologer channel. That might at least provide some perspective.

UPDATE (Friday)

Burkard’s latest video is back up!

The Troubling Cosiness of the VCAA and the MAV

by marty

February 9, 2020

10:32 am

48 Comments on The Troubling Cosiness of the VCAA and the MAV

VCAA

exams, MAV, VCAA, VCE

It seems that what amounts to VCE exam marking schemes may be available for purchase through the Mathematical Association of Victoria. This seems very strange, and we’re not really sure what is going on, but we shall give our current sense of it. (It should be noted at the outset that we are no fan of the MAV in its current form, nor of the VCAA in any form: though we are trying hard here to be straightly factual, our distaste for these organisations should be kept in mind.)

Each year, the MAV sells VCE exam solutions for the previous year’s exams. It is our understanding that it is now the MAV’s strong preference that these solutions will be written by VCAA assessors. Further, the MAV is now advertising that these solutions are “including marking allocations“. We assume that the writers are paid by the MAV for this work, and we assume that the MAV are profiting from the selling of the product, which is not cheap. Moreover, the MAV also hosts Meet the Assessors events which, again, are not cheap and are less cheap for non-members of the MAV. Again, it is reasonable to assume that the assessors and/or the MAV profit from these events.

We do not understand any of this. One would think that simple equity requires that any official information regarding VCE exams and solutions should be freely available. What we understand to be so available are very brief solutions as part of VCAA’s examiners’ reports, and that’s it. In particular, it is our understanding that VCAA marking schemes have been closely guarded secrets. If the VCAA is loosening up on that, then that’s great. If, however, VCAA assessors and/or the MAV are profiting from such otherwise unavailable information, we do not understand why anyone should regard that as acceptable. If, on the other hand, the MAV and/or the assessors are not so profiting, we do not understand the product and the access that the MAV is offering for sale.

We have written previously of the worrying relationship between the VCAA and the MAV, and there is plenty more to write. On more than one occasion the MAV has censored valid criticism of the VCAA, conduct which makes it difficult to view the MAV as a strong or objective or independent voice for Victorian maths teachers. The current, seemingly very cosy relationship over exam solutions, would only appear to make matters worse. When the VCAA stuffs up an exam question, as they do on a depressingly regular basis, why should anyone trust the MAV solutions to provide an honest summary or evaluation of that stuff up?

Again, we are not sure what is happening here. We shall do our best to find out, and commenters, who may have a better sense of MAV and VCAA workings, may comment (carefully) below.

UPDATE (13/02/20)

As John Friend has indicated in his comment, the “marking allocations” appears to be nothing but the trivial annotation of solutions with the allotted marks, not a break-down of what is required to achieve those marks. So, simply a matter of the MAV over-puffing their product. As for the appropriateness of the MAV being able to charge to “meet” VCAA assessors, and for solutions produced by assessors, those issues remain open.

We’ve also had a chance to look at the MAV 2019 Specialist solutions (not courtesy of JF, for those who like to guess such things.) More pertinent would be the Methods solutions (because of this, this, this and, especially, this.) Still, the Specialist solutions were interesting to read (quickly), and some comments are in order. In general, we thought the solutions were pretty good: well laid out with usually, though not always, the seemingly best approach indicated. There were a few important theoretical errors (see below), although not errors that affected the specific solutions. The main general and practical shortcoming is the lack of diagrams for certain questions, which would have made those solutions significantly clearer and, for the same reason, should be encouraged as standard practice.

For the benefit of those with access to the Specialist solutions (and possibly minor benefit to others), the following are brief comments on the solutions to particular questions (with section B of Exam 2 still to come); feel free to ask for elaboration in the comments. The exams are here and here.

Exam 1

Q5. There is a Magritte element to the solution and, presumably, the question.

Q6. The stated definition of linear dependence is simply wrong. The problem is much more easily done using a 3 x 3 determinant.

Q7. Part (a) is poorly set out and employs a generally invalid relationship between Arg and arctan. Parts (c) and (d) are very poorly set out, not relying upon the much clearer geometry.

Q8. A diagram, even if generic, is always helpful for volumes of revolution.

Q9. The solution to part (b) is correct, but there is an incorrect reference to the forces on the mass, rather than the ring. The expression “… the tension T is the same on both sides …” is hopelessly confused.

Q10. The question is stupid, but the solutions are probably as good as one can do.

Exam 2 (Section A)

MCQ5. The answer is clear, and much more easily obtained, from a rough diagram.

MCQ6. The formula Arg(a/b) = Arg(a) – Arg(b) is used, which is not in general true.

MCQ11. A very easy question for which two very long and poorly expressed solutions are given.

MCQ12. An (always) poor choice of formula for the vector resolute leads to a solution that is longer and significantly more prone to error. (UPDATE 14/2: For more on this question, go here.)

MCQ13. A diagram is mandatory, and the cosine rule alternative should be mentioned.

MCQ14. It is easier to first solve for the acceleration, by treating the system as a whole.

MCQ19. A slow, pointless use of CAS to check (not solve) the solution of simultaneous equations.

UPDATE (14/02/20)

For more on MCQ12, go here.

UPDATE (14/02/20)

Exam 2 (Section B)

Q1. In Part (a), the graphs are pointless, or at least a distant second choice; the choice of root is trivial, since y = tan(t) > 0. For part (b), the factorisation $\boldsymbol{x^2-2x =x(x-2)}$ should be noted. In part (c), it is preferable to begin with the chain rule in the form $\boldsymbol{dy/dt = dy/dx \times dx/dt}$ , since no inverses are then required. Part (d) is one of those annoyingly vague VCE questions, where it is impossible to know how much computation is required for full marks; the solutions include a couple of simplifications after the definite integral is established, but God knows whether these extra steps are required.

Q2. The solution to Part (c) is very poorly written. The question is (pointlessly) difficult, which means clear signposts are required in the solution; the key point is that the zeroes of the polynomial will be symmetric around (-1,0), the centre of the circle from part (b). The output of the quadratic formula is neccessarily a mess, and may be real or imaginary, but is manipulated in a clumsy manner. In particular, a factor of -1 is needlessly taken out of the root, and the expression “we expect” is used in a manner that makes no sense. The solution to the (appallingly written) Part (d) is ok, though the centre of the circle is clear just from symmetry, and we have no idea what “ve(z)” means.

Q3. There is an aspect to the solution of this question that is so bad, we’ll make it a separate post. (So, hold your fire.)

Q4. Part (a) is much easier than the notation-filled solution makes it appear.

Q5. Part (c)(i) is weird. It is a 1-point question, and so presumably just writing down the intuitive answer, as is done in the solutions, is what was expected and is perhaps reasonable. But the intuitive answer is not that intuitive, and an easy argument from considering the system as a whole (see MCQ14) seems (mathematically) preferable. For Part (c)(ii), it is more straight-forward to consider the system as a whole, making the tension redundant (see MCQ14). The first (and less preferable) solution to Part (d) is very confusing, because the two stages of computation required are not clearly separated.

Q6. It’s statistical inference: we just can’t get ourselves to care.

UPDATE (26/06/20)

The Specialist Maths examination reports are finally, finally out (here and here), so it seems worth revisiting the MAV “Assessor” solutions. In summary, the clumsiness of and errors in the MAV solutions as indicated above (and see also here and here) do not appear in the reports; in the main this is because the reports are pretty much silent on any aspect involving some subtlety. Sigh.

Some specific comments:

EXAM 1

Q5 Yes, Magritte-ish. Justifying that the critical points are extrema was not expected, meaning conscientious students wasted their time.

Q6 The error in the MAV solutions is ducked in the report.

Q7 The error in the MAV solutions is ducked in the report.

EXAM 2 (Section A)

MCQ6 The error in the MAV solutions is ducked in the report.

MCQ11 The report is silent.

MCQ12 A huge screw-up of a question, to which the report hemidemisemi confesses: see here.

MCQ14 The report suggests the better method for solving this problem.

EXAM 2 (Section B)

Q2 Jesus. This question was intrinsically confusing and very badly worded, with the students inevitably doing poorly. So, why the hell is the examination report almost completely silent? The MAV solutions were a mess, but the absence of comment in the report is disgraceful.

Q3 The solution in the report is ok, although more could have been written. But, it’s not the garbled nonsense of the MAV solution, as detailed here.

PoSWW 11: Pinpoint Inaccuracy

by marty

February 7, 2020

2:19 pm

11 Comments on PoSWW 11: Pinpoint Inaccuracy

PoSWW

calculus, exams, Germany, maxima, posww

This one comes courtesy of Christian, an occasional commenter and professional nitpicker (for which we are very grateful). It is a question from a 2016 Abitur (final year) exam for the German state of Hesse. (We know little of how the Abitur system works, and how this question may fit in. In particular, it is not clear whether the question above is a statewide exam question, or whether it is more localised.)

Christian has translated the question as follows:

A specialty store conducts an ad campaign for a particular smartphone. The daily sales numbers are approximately described by the function g with $\color{blue}\boldsymbol{g(t) = 30\cdot t\cdot e^{-0.1t}}$ , where t denotes the time in days counted from the beginning of the campaign, and g(t) is the number of sold smartphones per day. Compute the point in time when the most smartphones (per day) are sold, and determine the approximate number of sold devices on that day.

WitCH 34: Jumanji, the Numeracy Level

by marty

February 5, 2020

3:56 pm

9 Comments on WitCH 34: Jumanji, the Numeracy Level

WitCH

numeracy, WitCH

This adventure game comes courtesy of Victoria’s Department of Education and Training. Click on the graphic, or go here. (Don’t try to do it all. You will die.) If you can be bothered, you can also complete a survey here.

WitCHes in Batches

by marty

February 5, 2020

11:58 am

WitCH 33: Below Average

by marty

February 1, 2020

3:14 pm

24 Comments on WitCH 33: Below Average

WitCH

average, CAS, integration, Mathematical Methods, PDF, probability, VCA, VCAA, WitCH

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, it’s 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.