Secret 2021 Methods Business: Exam 2 Discussion

As for yesterday, we’ve been handballed a copy of the Methods 2 exam, and we haven’t looked at it yet.

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UPDATE (12/11/21)

And, finally, Section B. It’s pretty bad, as it is every year. All in all, however, this year’s Methods exams seemed better than previous years’. Not good, but good by Methods’ poor standards.

Q1 It takes a special talent to screw up a completely standard box-volume problem. The question starts with an h x 2h sheet, which becomes a 25 x 50 sheet, which then becomes an h x 2h sheet again, and which, finally, becomes an h x h sheet. To which one can only respond, “Sheeeet!”

This re-re-repetition is an insane waste of precious of exam time. And, they screw it up. As commenters have noted, when the box morphs back to have width h we are told ”

the box’s length is still twice its width” (emphasis added).

Which would be an interesting problem but is not what was intended. Commenters have suggested this would not have caused much confusion, but we’re not so convinced. in any case, it is a bad error.

Other than that, the question is pretty much just micro-bite CAS nonsense. There is a slight issue with the domains of the volume functions; presumably open intervals were expected, but mathematicians prefer closed intervals when possible, and usually permit “degenerate” cases at the endpoints.

Q2 Three distinct questions, the first of which is standard and easy. The second question, Part (f), has the students approximating an integral of a function on the interval [-2,2], but gives a graph of the function over [-3,3]; it’s not clear whether this was intended as a potential trick, but, whatever the intention, it seems pointlessly confusing. The third question, Parts (e) and (f), starts out fine, with calculating the area trapped between y = x2 and y = √x. It then morphs into finding the area trapped between y = ax2 and y = √x and x = a, asking for which values of a this area will equal 1/3. Honestly, who the hell cares?

Q3 A weird combination of two questions. The first question involves the function \boldsymbol{q(x) =\log_e\left(x^2-1\right)-log_e\left(1-x\right)}, which is too cute. Part (b)(ii) asks

Find the equation of the line that is perpendicular to the graph of q when x = -2 and passes through the point (-2,0).

It takes real effort to write that badly.

The second question involves the function \boldsymbol{p(x) = e^{-2x}-2e^{-x} +1}. Asking students to “explain why p is not a one-to-one function” is pretty weird wording. The rest of that question seems ok but meandering, and pretty fiddly.

Q4 A standard and ok probability question. The final transformation a*f(x/b) of the pdf is ok, but is more naturally presented as a*f(bx), since the latter form results in a = b.

Q5 There are issues. See here.

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UPDATE (11/11/21)

Here are our question by question comments on the multiple choice questions. Most of the MCQ seem pretty standard, mostly CAS nonsense, and there seems to be no general consternation, so we’ve only commented if there was something particular to note.

MCQ2 Weird Methodsy phrasing, asking which “graph” is identical, rather than which function.

MCQ4 A potentially tricky max-min question, with the maximum at an endpoint, but presumably students will have just used the stupid machine.

MCQ9 A standard range of a composition question, but the function is not 1-1 and so will presumably trick a lot of students.

MCQ14 A very weird question, asking for which k the functions cos(kx – π/2) and sin(x) have the same average over [0,π]. Some good mathematics underlying the question, but asked in an incomprehensible manner, and anyway killed by the stupid machine.

MCQ16 A way over-egged trig question. Fundamentally easy, but we can’t imagine students will do well.

MCQ18 Magritte garbage. God, I hate Methods. I really, really hate Methods.

MCQ20 A somewhat strange independent probability question. The information and the possible answers are framed in terms of an unknown p, but one can obviously and easily solve for p. This would be a natural first mathematical step, and the step is entirely irrelevant to answering the question.

54 Replies to “Secret 2021 Methods Business: Exam 2 Discussion”

  1. Can anyone else confirm my suspicion that there is an error in Q1 ER? There is a reference to the box “still” being twice as long as it is wide… but I think it never was?

    1. If they were talking about the rectangle from which the box was cut I don’t see any issue (as it is 2h*h) but if it was about the folded box that is a problem (2h-2x)=/=2(h-2x).

      1. Exactly my point. “A rectangular sheet of cardboard has a width of h. Its length is twice its width.” vs “the box’s length is till twice its width”. I think the meant, the cardboard is still… but that is not what I read it as.

    2. Yes, that inconsistency had me confused for a few moments during the exam, but it seems that it is intended to be talking about the sheet rather than the folded box. Definitely an error.

  2. One thing I have been wondering for quite some time now is whether a high school mathematics exam that employs the CAS can possibly be decent, or at least salveagable.
    With proper knowledge of how to use the CAS, the mathematics used in nearly every question of the specialist and methods exams is trivial to the point where the exam is entirely about testing the skill of setting up equations (which I can admit is a valuable skill, but not one that should dominate a mathematics exam), and the skill of using the CAS.
    Perhaps this will change with the re-introduction of logic into the Specialist curriculum, though I am doubtful.

    1. Hi student,

      It is quite a skill to make questions “CAS proof”… See Sai’s comments on this matter in this blog from a couple of years ago

      At the very least the memory of the machine should be wiped prior to the exam to remove prebuilt customised mathematica code

      but it can also be done by removing the CAS altogether as is done in IB for example

      Steve R

          1. The time is better spent trying to negate the \displaystyle poisoner. Ideally they get poisoned by their own cup, along with the chef and the official food tasters (who are all accomplices). Then you can sit down and eat a decent meal.

      1. Or in NSW… yet. Here’s a quote:


        Calculator features that are not permitted

        A NESA approved scientific calculator may NOT:

        – be programmable (A calculator that can have a sequence of operations stored and then executed automatically is considered programmable)
        – graph data or store, manipulate or graph functions
        – have computer algebraic system (CAS) functionality. This functionality includes: numeric routines for differentiation and definite integration, and the solution of equations; and symbolic manipulations such as addition of algebraic expressions, binomial expansion; symbolic differentiation and integration
        – perform financial calculations: depreciation, annuities, simple and compound interest, and break-even point.”

        From: https://educationstandards.nsw.edu.au/wps/portal/nesa/11-12/hsc/rules-and-processes/approved-calculators

  3. For the moment I only have one comment to make: Question 5 part (g). This question continues the bullshit VCAA tradition over the last several years of having a 1 mark question at the end whose answer cannot possibly be mathematically justified by any student sitting the exam.

    First of all, what’s meant by the greatest possible minimum value?? Does it mean the minimum of all possible minimums, or the maximum of all possible minimums?? Because the answer to the former is -2, and the answer to the latter is -Sqrt[2].

    If we assume the former, then the answer to Q5(g) is a hopeful -2. Very anti-climactic – A hopeful guess following from part (h) gets it right. Alternatively, making a guess by looking at a couple of graphs (no need to go past a plot when a = 10) suggests the answer. That’s all VCAA wants students to do: Make a lucky guess.

    If we assume the latter, then again a guess suggested from a couple of graphs appears to be sufficient: The least negative minimum appears to occur when a = 1. In which case the minimum is -Sqrt[2]. That’s all VCAA wants students to do: Make a lucky guess. Roll up, roll up ….

    This sort of question does nothing except encourage sloppy thinking. I challenge VCAA to provide a valid mathematical proof of its answer.

    1. Just so everyone can join in, here’s the question (worth 1 mark):

      (g) Find the greatest possible minimum of \displaystyle \left[ g_a(x) = \sin{\left(\frac{x}{a}\right)} + \cos{(ax)} \right].

      1. I would interpret this as something of the form \max \{ \min \{ g_a(x) \} \}. Unless for some reason, they were discussing the magnitude of the minimum value, which would be very irritating.

      2. Oh this one is exceptionally terrible.

        Yes, “greatest possible minimum” is a very unusual phrase. Normally I would read that as the infimum, but that’s not in the curriculum I guess? The infimum is not technically the max of all the minima, because that maximum may not be defined (if there are infinitely many minima). For instance, the maximum of the set of rational numbers p/q that satisfy p^2 < 2q^2 is not a member of that set, and so doesn't exist. This important issue is what motivates the infimum (and supremum).

        "Oh but Glen" I hear you say, "That can't happen here". I would say — what is the value of a? Can it be any real number? This is the sum of two periodic functions. Do you think that the sum of two periodic functions is periodic? It isn't. "Oh but this is not any two periodic functions, these are very special periodic functions. Nice ones, and hey look at how we divide then multiply by a, surely that's fine?!?" No, it isn't fine. Take a=\pi for example. The first function has principal period 2\pi^2 and the second has principal period 2. DO YOU THINK that 2/(2\pi^2) is a rational number?

        Probably they mean the largest minimum from the set of local minima of that function, and for me, that's the main problem with it. Such a thing is not a minimum at all, a minimum for a given function should be smaller than or equal to the value of the function at any other point in its domain.

        Alternatively, they are asking for the largest value in a set that consists of either nothing (the case where the local minima cluster around a value that is larger than the rest but does not attain that value) or a single number. In which case, what is the point in asking for the "greatest"? It just IS the minimum.

        So…

        – the function in question may not attain a minimum at all for some given a, in which case the question makes no sense (and we are not told to restrict a so…)
        – they are asking for something which isn't a minimum; or
        – they are asking for the greatest number in a set with at most a single value in it.

        In any case, the students will be thoroughly confused by this and with good reason.

        More words I have for this question, appropriate for this blog they are not.

        1. In the question, a is given to be a positive integer. So wouldn’t that mean the sum would have a period of 2a\pi? Admittedly, I had also thought of something similar, that the infimum of the set of minimum values may not exist.

          1. Oh well, it was a good rant while it lasted. I was just going by what JF posted.

            If it is periodic, then the maximum of the local minima will exist, because it is picking out the largest number from a finite set of numbers. By the way, the infimum of a bounded subset of the reals will always exist (but may not be a member of that set, which is perhaps your concern).

            The question is not spectacularly awful, just regularly awful. The ambiguity with the language makes it still bad, and practically unanswerable.

              1. Hmm… I think the infimum is the incorrect thing here. a = \inf \{A\} refers to the greatest lower bound, but we’re looking at the set A = \{\min_{x \in \mathbb{R}}(g_a(x)) : a \in \mathbb{N} \}, and specifically the maximum of that set, which would be more related to the supremum (in this case maximum = supremum). The infimum of the set would likely be -2 in the case above, but would not be contained within the set, i.e the set may have no minimum. That is, under the interpretation, I suggested, in other cases, it may be very different… Hence the atrocity of this question is revealed.

              2. Sorry, Glen. I should also have included \displaystyle a was a positive integer, stated in the global preamble to Question 2. I focussed solely on part (g).

                But yes, it was a good rant! And relevant. Because that’s how you’d want students to calculate the period, rather than looking at the graph and hypothesising on the basis of what they see (although the students don’t realise that they’re hypothesising).

        1. That’s very funny, and very good. Was it one of you guys?

          I’m gathering my strength, and drugs, to look at MM2. It’ll definitely help to have the smart stackexchange guys look at the final nonsense. (All you not-quite-as-smart commenters and your links are also very helpful.)

          1. It was me that posted it haha.
            I sat the exam on Thursday, I’m personally not too fussed with the wording of it (I put -\sqrt{2}), I’m more annoyed at how they wanted us to actually solve it.
            The most practical way was to just plug in numbers and notice that the minimum value decreases (strictly) towards -2 (but never reaches -2 as the minima of each term in the function never overlap, rather they occur in a tighter neighborhood). I did this and deduced the desired value should be when a = 1 (or close to a=1). Now of course, this is really just quite poor of VCAA if that is what they intended the solution as, but when doing the question I thought to myself “knowing VCAA, they probably intend this”.

        2. It wasn’t me! But I love math.stackexchange.com and always find it a reliable source of answers. Marty, I like your suggestion that if we stop commentating here and join stackexchange, we’ll all become instantly smarter. Resistance is futile.

          It will be very interesting to see what gets posted at stackexchange. I notice that my thoughts (that are beyond the scope of the Maths Methods sillybus) haven’t as yet been assimilated:

          From the sum-to-product formulae (which follow from the compound angle formulae) we have

          \displaystyle \sin{(A)} - \sin{(B)} = 2 \cos{\left( \frac{A + B}{2} \right)} \sin{\left( \frac{A - B}{2} \right)}

          Therefore:

          \displaystyle g(x) = \sin{ \left( \frac{x}{a}\right)} + \cos{(ax)} =  \sin{ \left( \frac{x}{a}\right)} -  \sin{ \left( ax - \frac{\pi}{2}\right)}

          \displaystyle = 2 \cos{ \left( \frac{1}{2} \left[ \frac{x}{a} + ax\right] - \frac{\pi}{2} \right] \right)} \sin{ \left( \frac{1}{2} \left[ \frac{x}{a} - ax\right] + \frac{\pi}{2} \right] \right)}

          \displaystyle = -2 \cos{ \left( \frac{x}{2} \left[a + \frac{1}{a}\right] - \frac{\pi}{4} \right)} \sin{\left( \frac{x}{2} \left[a - \frac{1}{a} \right] - \frac{\pi}{4} \right).

          \displaystyle \sin{\left( \frac{x}{2} \left[a - \frac{1}{a} \right] - \frac{\pi}{4} \right) has the larger period and so acts as an ‘envelope’

          of the \displaystyle \cos{ \left( \frac{x}{2} \left[a + \frac{1}{a}\right] - \frac{\pi}{4} \right)} term.

          The phenomenon of ‘beats’ is clearly observed for \displaystyle a \geq 2.

          \displaystyle g(x) = -2 \sin{\left( \frac{x}{2} \left[a - \frac{1}{a} \right]- \frac{\pi}{4} \right) \cos{ \left( \frac{x}{2} \left[a + \frac{1}{a}\right] - \frac{\pi}{4} \right)}. … (1)

          Investigation of minimum values is now systematic and fairly simple.

          Some examples to get you started:

          Case 1: \displaystyle a = 1.

          From equation (1): \displaystyle g(x) = 2 \sin\left({\frac{\pi}{4}}\right) \cos{ \left( x - \frac{\pi}{4} \right)} = \sqrt{2} \cos{ \left( x - \frac{\pi}{4} \right)}.

          Case 2: \displaystyle a = 2 (the starting point of Question 5)

          From equation (1): \displaystyle g(x) = - 2 \sin\left({\frac{3x}{4} - \frac{\pi}{4}\right)}\cos{ \left( \frac{5x}{4} - \frac{\pi}{4} \right)}.

    2. Yes, I agree – I had a long hard look at that question, read it multiple ways but still couldn’t figure out what exactly they wanted.
      Greatest minimum is really sloppy language and can be explained so much better.

      Add to that the blaring error regarding the box’s length being twice the width rather than the sheet’s length being twice the width and I doubt that they even vet these exams.

      Apart from that – much more “doable” compared to last year’s exam.

  4. While running through the exam, the only weird thing that tripped me was 5c, which was the find the smallest positive value of h such that f(h-x) = f(x). Uh… isn’t there meant to be a for all x? There is no prior reference to x, so one must assume that it is implicitly all values of x yet it feels a bit bizarre.

  5. Q1 – when it asks for the domain of V_{box} it only constrains x>0 and V_{box}>0, so it is possible to give a non-physical domain x\in\(0, \tfrac{25}{2}) \cup (25, \infty). A bit silly though.

    Q2f – If you consider “area bounded” to mean “unsigned area bounded” then this has 3 solutions – instead of the 2 solutions you get if you use signed area. It’s not nice to try to do this by hand and the CAS has to be arm-wrestled to give the 3rd solution.

    Q3e – maybe I missed an easy approach, but this one seemed a little tricky (the two approaches I found were to: draw triangles to find possible angles for the tangent or use the tan compound angle formula that gives the angle between two lines in terms of their gradients)

    There were some very CAS heavy questions…

    1. Q2f was a CAS smashing question, since one has to notice that the point at which the two graphs intersect varies, and down the line, this mean you have to consider another case, which splits the integral into two equations (or one if you decide to use the absolute value function) and then both are solved. I wonder how time-consuming this proved, seeing as I’ve seen people report that their CAS of choice was “slow” (however this is very much word of mouth on online forums).

      For Q3e, the angle between y=x+2 (with angle \frac{\pi}{4} radians) and the tangent to the graph at x=a (p'(a) < 0) becomes \frac{\pi}{4} - \arctan(p'(a)). Set equal to \frac{\pi}{3} then smash into the calculator. At least that's what I believe it is. ( I could be wrong, I didn't think too much about the sign or which angle it was)

      1. The hand-held CAS’s can be quite slow – I always use the emulator, so I forget sometimes.
        I used to use a lot of Mathematica – but have not got a chance to teach Methods using it…

        Q3e – There are two values of a that work, did your method find both?

        1. The method above only finds one of the values of a. I believe you’d also need to set it to \frac{2\pi}{3} to obtain the other solution.

    2. For Q1, side length is defined as (h-2x) and (2h-2x). Both of these must be greater than 0 for the model to make sense. Considering this, the domain is 0<x<h/2

      1. I agree. But the question insisted on telling use the obvious physical constraints that x>0 and Vbox>0… so why not tell use the other constraints? Maybe they didn’t want use to use them? 🙂

    3. Simon, I thought it was standard in Methods that “area” means actual area, not signed area. So, I don’t take 2(f) to be ambiguous. Insane, yes, but not ambiguous. Are there indications somewhere otherwise?

  6. Thanks, everyone. I’ve had a look at the exam, and Exam 1, and I’ve been very interested in both discussions. I have thoughts, but I’ll wait until I have some proper time before updating, probably after the Specialist exams are also done.

    Good luck to all of you doing Specialist today.

  7. Each value of \displaystyle a gives a set of local minimums. There is a ‘maximum’ minimum within that set, as well as a ‘minimum’ minimum. There are many different interpretations of the wording. Could it mean the largest ‘maximum’ minimium of the all the ‘maximum’ minimums that are possible …? (Who’s on first?)

    The real issue is that a precise mathematical statement of what’s wanted can’t be given – in this case it’s not an indictment on the sillybus, it’s an indictment on the question for asking something that cannot be expressed precisely within the scope of the sillybus. This is what happens when the exams are used to push the fetish of generalisations in SACs.

    No doubt the meaning is crystal clear in the pinhead of the writer. And no doubt the Report will chastise all us idiots who didn’t understand it.

    1. Admittedly, I got caught up in this issue as well, when I was looking at the other minima as opposed to what I believe the question wants: the greatest of the global minimum. A fact like that does not seem easy in any way to prove or reach other than sketching some graphs and concluding so, if you don’t get confused by the other minima that tend towards 0. It also seems that some other people have picked up on the oddity of the question: https://www.youtube.com/watch?v=6UQgtUJbnWo (1:07:44 specifically).

        1. Yes, I’d flagged \displaystyle -2 and \displaystyle -\sqrt{2}.

          The problem with \displaystyle -2 is that it’s the \displaystyle limiting value for \displaystyle a \rightarrow \infty (good luck proving this) so it’s not a value the function can ever actually equal. Which, in my mind, makes \displaystyle -\sqrt{2} the answer VCAA \displaystyle wanted. But it’s not the answer VCAA \displaystyle unambiguously asked for.

          I don’t think \displaystyle -1 is a potential answer, it’s the result of an invalid calculation of the limit as \displaystyle a \rightarrow \infty.

          However, I agree (based on circumstantial graphical evidence) that 0 is the limiting value of one of the local minimums as \displaystyle a \rightarrow \infty. But again, it’s a limiting value, not a value that the function can ever actually equal.

  8. This was one of the most poorly worded exams that I have seen…

    Question 1
    I found the wording for the Vbox question confusing.

    Why in the f*ck did they not start with width 25 and length 50… Then in the second part of the question, use h and 2h. This honestly confused the shit out of me. The height should have been re-stated as x… and why would you use the variable “h” when h is part of the equation V=l*w*h?

    h=x, l=2h-2x, w=h-2x… Confused much? Maybe try… a, b or even m? but h?

    And the box’s length is twice it’s width??? Confused much? I was trying to get x in terms of h (in my head) and reverse the equation to get x in terms of V and h) then thought? WTF? It’s only 1 mark for the domain so they must have stuffed up…

    Question 2
    d. Is that not the most random question which relates to nothing in the rest of the question?

    Question 3
    b.ii. Find the equation of the line that is perpendicular to the graph of q when x=-2 and passes through the point (-2, 0)

    What is this? This actually confused me, and whilst I’m no genius, I thought “surely they would just say the equation to the normal at x=-2 if that’s what they wanted? Umm, no…

    Question 4
    e. I found it confusing to understand what the spin was… and maximum amount of spin? One of my smartest students got everything else right other than this question…

    Maybe “the probability of the machine obtaining the amount of spin, x, ”

    Question 5
    I thought this was one of the better questions on the paper, but the last question on the greatest possible minimum… I was confused as to the greatest possible minimum… Why didn’t they ask for the smallest possible maximum?

    This question would have been hard with a CAS, as I was using a simulated CAS on ipad, which works quickly. Luckily, due to past VCAA exams, I have taught my students how to approach these types of questions, based on previous VCAA papers… so, many of them got -root(2)…

    Here is a link to MC answers…

    And a link to Extended Response

    Remember, I am going through them trying to teach students what VCAA want, not what is mathematically correct… 😛

  9. I’ve updated the post with a few thoughts on the MCQ. We’ll update with comments on the troubled Section B tomorrow.

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