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 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 10: Malfunction

It’s a long, long time since we’ve had a WitCH. They have been not-so-slowly accumulating, however. And now, since we’re temporarily free of the Evil Mathologer, it is the WitCHing hour. Due mostly to the hard work of Damo, all of the outstanding WitCHes have been resolved, with the exception of WitCH 8. That one will take time: it’s a jungle of half-maths. Our new WitCHes are not so tricky, although there is perhaps more to be said than indicated at first glance. The first of our new batch of WitCHes is from the VCE 2018 Specialist Exam 1:

The Examiners’ Report gives the answer as \int_0^{\frac34}\left(2-t^2\right)dt. The Report also indicates that the average score on this question was 1.3/5, with 98% of students scoring 3 or lower, and over a third of students scoring 0. Happy WitCHing.

Update (16/02/20)

This problem is ridiculous and, more importantly, it is wrong. First, the wrongness.

As indicated by the examination report, the examiners imagined that they were, in essence, asking for students to determine the speed function \boldsymbol{v(t)} of the particle. The distance is given by \boldsymbol{d=\int\limits_0^{3/4} v(t)\,{\rm d}t}, and a non-trivial calculation gives \boldsymbol{v(t) = 2 +0t - t^2}. Then, the coefficients \boldsymbol{a = -1, b = 0, c = 2} can be read off. 

That is not, however, the question the examiners asked. What did the examiners really ask? They asked for integers \boldsymbol{a,b,c} for which \boldsymbol{d=\int\limits_0^{3/4} at^2+bt+c\,{\rm d}t}. But

    \[\boldsymbol{d=\int\limits_0^{3/4} 2  - t^2\,{\rm d}t = \dfrac{87}{64}\,.}\]

and

    \[\boldsymbol{\int\limits_0^{3/4} at^2+bt+c\,{\rm d}t = \frac{9}{64}a+\frac{9}{32}b+\frac{3}{4}c \,.}\]

So, multiplying out the fractions and cancelling out a 3, what the examiners really asked for were integer solutions to the equation

    \[\boldsymbol{3a+ 6b+ 16c  = 29\,.}\]

This equation has infinitely many integer solutions, meaning the examination report is missing infinity minus one valid solutions.

This is a flat out, undeniable error (which the Trumpian VCAA will never concede), but is it a problem? As commenters here have noted, there is little chance of a VCE student being actively misled to chase the infinitely many solutions. In, particular, the method to find all solutions requires first finding the particular solution the examiners had in mind. We are not convinced such direct concerns should be so quickly dismissed, and we discuss this further below. Still, the extra solutions require thought to even contemplate, and significant work to compute, which is an important point.

Whatever the immediate practical concerns, however, mathematicians are aghast at this error. They are aghast because the exam question is simply not testing mathematics. Yes, the students went through the ritual and attempted to compute what was intended and were graded accordingly. And, yes, teachers can now coach current and future students on the required ritual. But none of that is mathematics and, indeed, it is worse: it is antimathematics. It is teaching students to ignore mathematical meaning, to see no value in mathematical precision, to respect only ritual.

OK, that is the awful wrongness of the exam question. Now, the sundry ridiculousnesses:

  • The question is badly and needlessly opaque. There is no a priori reason to imagine the distance as being given by the integral of a quadratic. Asking for (more accurately, attempting to ask for) the speed function in this overly cute manner adds no value, only confusion. The confusion is enhanced by the arbitrariness of the 3/4 limit and, especially, by the pointless specification that the coefficients of the quadratic be integers.
  • Independent of the opacity, the wording of the question is lazy and clumsy. The distanced travelled “in three-quarters of a second” is not the same as the distance travelled in the first three-quarters of a second and, indeed, is not anything. The phrases “moving along a curve” and “travels along a curve” are just verbiage. The units are pointless.
  • The question would be much more natural as an arc-length question, rather than a distance question.
  • The answer in the examination report is incorrect, even in the intended terms. The question asked for the values of the coefficients, not the integral. Yes, this is a nitpick, but it is exactly the kind of nitpick that the examiners routinely employ in their sanctimonious whacking of VCE students. So screw ’em. Sauce for the gander.
  • Last, and far from least, there is something very strange about the score distribution for the question. The average score was 1.3/5, which is depressing, although not surprising: computing the speed (without CAS) requires a level of care and facility beyond most CAS-drunk students, and the question contains a hidden absolute value to negotiate. What is strange is that, whereas 2% of students received the full 5/5 for the question, apparently 0% of students received 4/5. It is difficult to see how that could occur with any sensible grading scheme.

Inverted Logic

The 2018 Northern Hemisphere Mathematical Methods exams (1 and 2) are out. We didn’t spot any Magritte-esque lunacy, which was a pleasant surprise. In general, the exam questions were merely trivial, clumsy, contrived, calculator-infested and loathsomely ugly. So, all in all not bad by VCAA standards.

There, was, however, one notable question. The final multiple choice question on Exam 2 reads as follows:

Let f be a one-to-one differentiable function such that f (3) = 7, f (7) = 8, f′(3) = 2 and f′(7) = 3. The function g is differentiable and g(x) = f –1(x) for all x. g′(7) is equal to …

The wording is hilarious, at least it is if you’re not a frazzled Methods student in the midst of an exam, trying to make sense of such nonsense. Indeed, as we’ll see below, the question turned out to be too convoluted even for the examiners.

Of course –1 is a perfectly fine and familiar name for the inverse of f. It takes a special cluelessness to imagine that renaming –1 as g is somehow required or remotely helpful. The obfuscating wording, however, is the least of our concerns.

The exam question is intended to be a straight-forward application of the inverse function theorem. So, in Leibniz form dx/dy = 1/(dy/dx), though the exam question effectively requires the more explicit but less intuitive function form, 

    \[\boldsymbol {\left(f^{-1}\right)'(b) = \frac1{f'\left(f^{-1}(b)\right)}}.}\]

IVT is typically stated, and in particular the differentiability of –1 can be concluded, with suitable hypotheses. In this regard, the exam question needlessly hypothesising that the function g is differentiable is somewhat artificial. However it is not so simple in the school context to discuss natural hypotheses for IVT. So, underlying the ridiculous phrasing is a reasonable enough question.

What, then, is the problem? The problem is that IVT is not explicitly in the VCE curriculum. Really? Really.

Even ignoring the obvious issue this raises for the above exam question, the subliminal treatment of IVT in VCE is absurd. One requires plenty of inverse derivatives, even in a first calculus course. Yet, there is never any explicit mention of IVT in either Specialist or Methods, not even a hint that there is a common question with a universal answer.

All that appears to be explicit in VCE, and more in Specialist than Methods, is application of the chain rule, case by isolated case. So, one assumes the differentiability of –1 and and then differentiates –1(f(x)) in Leibniz form. For example, in the most respected Methods text the derivative of y = log(x) is somewhat dodgily obtained using the chain rule from the (very dodgily obtained) derivative of x = ey.

It is all very implicit, very case-by-case, and very Leibniz. Which makes the above exam question effectively impossible.

How many students actually obtained the correct answer? We don’t know since the Examiners’ Report doesn’t actually report anything. Being a multiple choice question, though, students had a 1 in 5 chance of obtaining the correct answer by dumb luck. Or, sticking to the more plausible answers, maybe even a 1 in 3 or 1 in 2 chance. That seems to be how the examiners stumbled upon the correct answer.

The Report’s solution to the exam question reads as follows (as of September 20, 2018):

f(3) = 7, f'(3) = 8, g(x) = f –1(x) , g‘(x) = 1/2 since

f'(x) x f'(y) = 1, g(x) = f'(x) = 1/f'(y).

The awfulness displayed above is a wonder to behold. Even if it were correct, the suggested solution would still bear no resemblance to the Methods curriculum, and it would still be unreadable. And the answer is not close to correct.

To be fair, The Report warns that its sample answers are “not intended to be exemplary or complete”. So perhaps they just forgot the further warning, that their answers are also not intended to be correct or comprehensible.

It is abundantly clear that the VCAA is incapable of putting together a coherent curriculum, let alone one that is even minimally engaging. Apparently it is even too much to expect the examiners to be familiar with their own crappy curriculum, and to be able to examine it, and to report on it, fairly and accurately.

Fixations and Madness

Our sixth and final post on the 2017 VCE exam madness is on some recurring nonsense in Mathematical Methods. The post will be relatively brief, since a proper critique of every instance of the nonsense would be painfully long, and since we’ve said it all before.

The mathematical problem concerns, for a given function f, finding the solutions to the equation

    \[\boldsymbol{(1)\qquad\qquad f(x) \ = \ f^{-1}(x)\,.}\]

This problem appeared, in various contexts, on last month’s Exam 2 in 2017 (Section B, Questions 4(c) and 4(i)), on the Northern Hemisphere Exam 1 in 2017 (Questions 8(b) and 8(c)), on Exam 2 in 2011 (Section 2, Question 3(c)(ii)), and on Exam 2 in 2010 (Section 2, Question 1(a)(iii)).

Unfortunately, the technique presented in the three Examiners’ Reports for solving equation (1) is fundamentally wrong. (The Reports are here, here and here.) In synch with this wrongness, the standard textbook considers four misleading examples, and its treatment of the examples is infused with wrongness (Chapter 1F). It’s a safe bet that the forthcoming Report on the 2017 Methods Exam 2 will be plenty wrong.

What is the promoted technique? It is to ignore the difficult equation above, and to solve instead the presumably simpler equation

    \[ \boldsymbol{(2) \qquad\qquad  f(x) \ = \  x\,,}\]

or perhaps the equation

    \[\boldsymbol{(2)' \qquad\qquad f^{-1}(x)\ = \ x \,.}\]

Which is wrong.

It is simply not valid to assume that either equation (2) or (2)’ is equivalent to (1). Yes, as long as the inverse of f exists then equation (2)’ is equivalent to equation (2): a solution x to (2)’ will also be a solution to (2), and vice versa. And, yes, then any solution to (2) and (2)’ will also be a solution to (1). The converse, however, is in general false: a solution to (1) need not be a solution to (2) or (2)’.

It is easy to come up with functions illustrating this, or think about the graph above, or look here.

OK, the VCAA might argue that the exams (and, except for a couple of up-in-the-attic exercises, the textbook) are always concerned with functions for which solving (2) or (2)’ happens to suffice, so what’s the problem? The problem is that this argument would be idiotic.

Suppose that we taught students that roots of polynomials are always integers, instructed the students to only check for integer solutions, and then carefully arranged for the students to only encounter polynomials with integer solutions. Clearly, that would be mathematical and pedagogical crap. The treatment of equation (1) in Methods exams, and the close to universal treatment in Methods more generally, is identical.

OK, the VCAA might continue to argue that the students have their (stupifying) CAS machines at hand, and that the graphs of the particular functions under consideration make clear that solving (2) or (2)’ suffices. There would then be three responses:

(i) No one tests whether Methods students do anything like a graphical check, or anything whatsoever.

(ii) Hardly any Methods students do do anything. The overwhelming majority of students treat equations (1), (2) and (2)’ as automatically equivalent, and they have been given explicit license by the Examiners’ Reports to do so. Teachers know this and the VCAA knows this, and any claim otherwise is a blatant lie. And, for any reader still in doubt about what Methods students actually do, here’s a thought experiment: imagine the 2018 Methods exam requires students to solve equation (1) for the function f(x) = (x-2)/(x-1), and then imagine the consequences.

(iii) Even if students were implicitly or explicitly arguing from CAS graphics, “Look at the picture” is an absurdly impoverished way to think about or to teach mathematics, or pretty much anything. The power of mathematics is to be able take the intuition and to either demonstrate what appears to be true, or demonstrate that the intuition is misleading. Wise people are wary of the treachery of images; the VCAA, alas, promotes it.

The real irony and idiocy of this situation is that, with natural conditions on the function f, equation (1) is equivalent to equations (2) and (2)’, and that it is well within reach of Methods students to prove this. If, for example, f is a strictly increasing function then it can readily be proved that the three equations are equivalent. Working through and applying such results would make for excellent lessons and excellent exam questions.

Instead, what we have is crap. Every year, year after year, thousands of Methods students are being taught and are being tested on mathematical crap.