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?

WitCH 29: Bad Roots

This one is double-barrelled. A strange multiple choice question appeared in the 2019 NHT Mathematical Methods Exam 2 (CAS). We had thought to let it pass, but a similar question appeared in last week’s Methods exam (no link yet, but the Study Design is here). So, here we go. First, the NHT question: The examination report indicates the correct answer, C, and provides a suggested solution:

\Large\color{blue} \boldsymbol{ g(x)=f^{-1}(x)=\frac{x^{\frac15}-b}{a},\ g'(x) = \frac{x^{-\frac45}}{5a},\ g'(1) = \frac1{5a}}

And, here’s last week’s question (with no examination report yet available):

Update (19/06/20)

As commenters have noted, it is very difficult to understand any purpose to these questions. They obviously suggest the inverse function theorem, testing the knowledge of and application of the formula g'(d) = 1/f'(c), where f(c) =d. The trouble is, the inverse function theorem is not part of the curriculum, appearing only implicitly as a dodgy version of the chain rule, and is typically only applied in Leibniz form.

As indicated by the solution in the first examination report, the intent seems to have been for students to have explicitly computed the inverses, although probably with their idiot machines. (The second examination report has now appeared, but is silent on the intended method.) Moreover, as JF noted below, the algebra in the first question makes the IFT approach somewhat fiddly. But, what is the point of pushing a method that is generally cumbersome, and often impossible, to apply?

To add to the nonsense, below is a sample solution for the first question, provided by VCAA to students undertaking the Mathematica version of Methods. So, the VCAA has suggested two approaches, one which is generally ridiculous and another which is outside the curriculum. That makes it all as clear as dumb mud.

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 27: Uncomposed

Ah, so much crap …

Tons of nonsense to post on, and the Evil Mathologer is breathing down our neck. We’ll have (at least) three posts on last week’s Mathematical Methods exams. This one is by no means the worst to come, but it fits in with our previous WitCH, so let’s quickly get it going. It is from Exam 1. (No link yet, but the Study Design is here.)

Update (15/06/20)

The examination report (and exam) is out, so it’s time to wade into this swamp. Before doing so, we’ll note the number of students who sank; according to the examination report, the average score on this question was 0.14 + 0.09 + 0.14 ≈ 0.4 marks out of 4. Justified or not, students had absolutely no clue what to do. Now, into the swamp.

The main wrongness is in Part (b), but we’ll begin at the beginning: the very first sentence of Part (a) is a mess. Who on Earth writes

“The function f: R \to R, f(x)  is a polynomial function …”?

It’s like writing

“The Prime Minister Scott Morrison of Australia, Scott Morrison is a crap Prime Minister”.

Yes, you may properly want to emphasise that Scott Morrison is the Prime Minister of Australia, and he is crap, but that’s not the way to do it. This is nitpicking, of course, but there are two reasons to do so. The first reason is there is no reason not to: why forgive the gratuitously muddled wording of the very first sentence of an exam question? From these guys? Forget it. The second reason is that the only possible excuse for this ridiculous wording is to emphasise that the domain of f is all of R, which turns out to be entirely pointless.

Now, to Part (a) proper. This may come as a surprise to the VCAA overlords, but functions do not have “rules”, at least not unique ones.  The functions f(x) = -4x^2\left(x^2 - 1\right) and h(x) = 4x^2-4x^4, for example, are the exact same function. Yes, this is annoying, but we’re sorry, that’s the, um, rule. Again this is nitpicking and, again, we have no sympathy for the overlords. If they insist that a function should be regarded as a suitable set of ordered pairs then they have to live with that choice. Yes, eventually ordered pairs are the precise and useful way to define functions, but in school it’s pretty much just a pedantic pain in the ass.

To be fair, we’re not convinced that the clumsiness in the wording of Part (a) contributed significantly to students doing poorly. That is presumably much more do to with the corruption of students’ arithmetic and algebraic skills, the inevitable consequence of VCAA and ACARA calculatoring the curriculum to death.

On to Part (b), where, having found f(x) = -4x^2\left(x^2 - 1\right) or whatever, we’re told that g is “a function with the same rule as f”. This is ridiculous and meaningless. It is ridiculous because we never did anything with f in the first place, and so it would have been a hell of lot clearer to have simply begun the damn question with g on some unknown domain E. It is meaningless because we cannot determine anything about the domain E from the information provided. The point is, in VCE the composition \log(g(x)) is either defined (if the range g(E) is wholly contained in the positive reals), or it isn’t (otherwise). End of story.  Which means that in VCE the concept of “maximal domain” makes no sense for a composition. Which means Part (b) makes no sense whatsoever. Yes, this is annoying, but we’re sorry, that’s the, um, rule.

Finally, to Part (c). Taking (b) as intended rather than written, Part (c) is ok, just some who-really-cares domain trickery.

In summary, the question is attempting and failing to test little more than a pedantic attention to boring detail, a test that the examiners themselves are demonstrably incapable of passing.

WitCH 26: Imminent Domain

The following WitCH is pretty old, but it came up in a tutorial yesterday, so what the Hell. (It’s also a good warm-up for another WitCH, to appear in the next day or so.) It comes from the 2011 Mathematical Methods Exam 1:

For part (a), the Examination Report indicates that f(g)(x) =([x+2][x+8]), leading to c = 2 and d = 8, or vice versa. The Report indicates that three quarters of students scored 2/2, “However, many [students] did not state a value for c and d”.

For Part (b), the Report indicates that 84% of students scored 0/2. After indicating the intended answer, (-∞,-8) U (-2,∞) (-∞,-8] U [-2,∞) or R\backslash(-8,-2), the Report goes on to comment:

“This question was very poorly done. Common incorrect responses included [-3,3] (the domain of  f(x); x ≥ -2 (as the ‘intersection’ of  x ≥ -8 with x ≥ -2); or x ≥ -8 (as the ‘union’ of x ≥ -8 with x ≥ -2). Those who attempted to use the properties of composite functions tended to get confused. Students needed to look for a domain that would make the square root function work.”

The Report does not indicate how students got “confused”, although the composition of functions is briefly discussed in the Study Design (page 72).

WitCH 24: The Fix is In

We’ve finally found some time to take a look at VCAA’s 2019 NHT exams. They’re generally bad in the predictable ways, and they include some specific and seemingly now standard weirdness that we’ll try to address soon in a more systematic manner. WitCHwise, we were tempted by a number of questions, but we’ve decided to keep it to two or three.

Our first NHT WitCH is from the final question on Exam 2 (CAS) of Mathematical Methods:

As usual, the NHT “Report” indicates nothing of how students went, and little of what was expected. In regard to part f, the Report writes,

p(x) = q(x) = x, p'(x) = q'(x) = 1, k = 1/e

For part g, all that the Report provides is the answer, k = 1.

The VCAA also provides sample Mathematica solutions to schools trialling Methods CBE. For the questions above, these solutions are as follows:

Make of it what you will.

WitCH 23: Speed Bump

Our second WitCH of the day also comes from the 2017 VCE Specialist Mathematics Exam 2. (Clearly an impressive exam, and we haven’t even gotten to the bit about using inverse trig functions to design a brooch.) It is courtesy of the mysterious SRK, who raised it in the discussion of an earlier WitCH.

Question 5 of Section B of the (CAS) exam concerns a boat and a jet ski. Though SRK was concerned with one particular aspect, the entire question is worth pondering:

The  Examiner’s Report indicates an average student score of 1.4 on part a, and comments,

Students plotted the initial positions correctly but significant numbers of students did not label the direction of motion or clearly identify the jet ski and the boat. Both requirements were explicitly stated in the question.

For part i, the Report indicates an average score of 1.3, and comments,

Most students found correct expressions for velocity vectors. The most common error was to equate these velocity vectors rather than equating speeds. 

For part ii, the Report gives the intended answer as (3,3). The Report indicates that slightly under half of students were awarded the mark, and comments,

Some answers were not given in coordinate form.

For part i, the Report suggests the answer {\sqrt{(\sin t - 2\cos t)^2 + (1 + \sin t + \cos t)^2}} (with the displayed answer adorned by a weird, extra root sign). The report indicates that a little over half of the students were awarded the mark, and comments,

A variety of correct forms was given by students; many of these were likely produced by CAS technology, including expressions involving double angles. Students should take care when transcribing expressions from technology output as errors frequently occur, particularly regarding the number and placement of brackets. Some incorrect answers retained vectors in the expression.

For Part ii, the Report indicates the intended answer of 0.33, and that 15% of students were awarded the mark for this question. The Report comments,

Many students found this question difficult. Incorrect answers involving other locally minimum values were frequent.

The Report indicates an average score of 1.3 on part d, and comments;

Most students correctly equated the vector components and solved for t . Many went on to give decimal approximations rather than supplying the exact forms. Students are reminded of the instruction saying that an exact answer is required unless otherwise specified.

Lots there. Get hunting.