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.


UPDATE (23/04/22)

Our comments on the exam report are now interlaced below, in green. here. Needless to say, the examiners didn’t confess to, much less apologise for, any of their screw-ups.

UPDATE (20/04/22)

The exam report is now available, here.

UPDATE (24/12/21)

The exam is now available, here.

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.

(23/04/22) The exam report fails to even acknowledge the error. Of course. Because they’re gutless flugbongs.

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.

(23/04/22) Yes, the exam report indicates that open intervals were expected for the domains in parts (b) and (f)(i). This is not wrong, but it is close. Part (g) also included a related, off-key and anal retentive whine, about how the domain “needed to be considered”. These people are not smart. (25/04/22) Having read SRK’s comment on the report, I’ll add two further remarks here.

(1) SRK is correct, that exam report’s solution to 1(g) is incomplete. This makes the report’s whining on this question, that “students should make sure adequate working is given”, pretty ironic.

(2) Thinking about it further I still don’t think it is wrong to consider the domain to be an open interval. But it is definitely wrong to mark the closed interval incorrect. It seems clear from the exam report that this was done; it should not be done again. 

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.

(23/04/22) So, how were students supposed to “explain why” p is not one-to-one? The expected answer was apparently

‘Fails the horizontal line test’ or ‘many-to-one function’ or ‘there exist two x-values for some y-values’.

Jesus. Needless to say, unless trying to hammer some sense into a nitwit examiner, chanting such phrases is meaningless ritual, explaining nothing about anything, and certainly nothing about the specific function p. These people are so, so stupid.

There was also a notable piece of nastiness in part (f), where an answer to three decimal places was required; the correct answer was 0.750, and the report appears to suggest that 0.75 was marked incorrect. Which, technically, it is. But seriously, can’t you micro-minded nitwits arrange for your questions to not nitpick such who-gives-a-stuff nonsense?

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.

(23/04/22) For (g), the report indicates 2% of students scored the 1 mark for this question, bluntly gives the answer as -√2, without explanation, and then simply notes

An exact answer was required. −2 was a common incorrect answer.

If VCAA examiners cannot pretend to give a stuff, what possible argument is there for treating them with anything less than complete contempt? They are a disgrace.


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.

(23/04/22) Nope: with machine or otherwise, only 58% of students answered correctly. 

MCQ8 (23/04/22) God knows why only 40% of students got this trivial question correct. 

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

(23/04/22) Yep, only 56% of students answered correctly. 

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.

(23/04/22) 63% of students answered correctly, which seems high. The exam report is silent.  

MCQ15 (23/04/22) Only 48% of students got this trivial binomial question correct.

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

(23/04/22) Yep, a killing field. 31% of students answered correctly. 

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

(23/04/22) The exam report indicates the Magritte garbage that was expected. 39% of students successfully negotiated the Magritte garbage. Completely insane. 

MCQ19 (23/04/22) Dear Examiners, I don’t like you and you don’t like me. But if you actually read my blog on occasion, you might stop making idiots of yourselves so often. Your report’s explanation of this question is utter nonsense. See here. (24/04/22) As John Friend has noted in a comment, the report’s explanation is worse than we had suggested: while demonstrating that they don’t know how to prove a function is differentiable, the examiners have also gratuitously demonstrated that they do not know what “smooth” means. To quickly hammer the point, the examiners’ argument implies that the function x3/x is “smooth” at 0. And if you don’t know why that is false then Congratulations! You’ve qualified yourself to be a VCAA examiner.

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.

88 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.”


  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.


        – 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 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)}


          \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: (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.

  10. Re: Update MCQ19 (23/04/22).
    Yes, the Report’s comment \displaystyle is nonsense in the sense that the implied reason for rejecting the other options is nonsense. BUT … let it not go unsaid that we should \displaystyle commend VCAA for finally giving us its definition of ‘smooth’ 😉

    Option E is a correct option. That, together with the fact that there is (or should be, but not always where VCAA is concerned …) only one correct option means that there’s not even a need to consider and reject the other options – either for wrong reasons or for right reasons.

    You have to take the rough with the \displaystyle smooth 😉

    1. Ugh! I hadn’t even noticed that. Jesus, that is so gratuitously stupid. I’ll update the update tomorrow.

      1. Re: Update of the update (24/04/22) of MCQ 19.
        Marty, at the risk of publicly demonstrating my qualifications to be a VCAA Exam writer, surely no discussion (good or bad) about the differentiability of your example can be had until your function is properly defined. In particular, the value of the function when x = 0 …

        1. Tell it to the examiners. They seem to be claiming that if the limits of f’ from the left and right are equal then the function is “smooth” at the point in question.

          1. In fairness, the Report prefaces the limit stuff with the observation that the function f is continuous at x = 0.

            So I’d give them the reluctant benefit of the doubt that they would first want to think about the continuity of f before diving into “the limits of f’ from the left and right are equal” and their smooth business.

            1. OK, Mr. Reluctant Devil’s Advocate, let me ask some questions:

              If one is trying to prove a “hybrid” function is differentiable at the “join”, let’s say at the point x = a:

              (i) Is what the examiners are doing natural?

              That is, even throwing in the continuity, is the approach of the examiners to find the limits of the derivatives the natural approach?

              (ii) Is what the examiners are doing sufficient?

              That is, if the function is continuous and the two limits of of f’ are equal at a, does it follow that f is differentiable at a?

              (iii) Is what the examiners are doing necessary?

              That is if f is continuous and is differentiable, let’s say everywhere, does it follow that the two limits of f’ at a are equal?

              1. i) No. (But in Maths Methods, I say a contextual yes. Especially now that differentiation from first principals has been deleted from the new Stupid Design).

                ii) Yes. But more clarity is needed so that the argument is seen clearly as IF rather than IFF.

                iii) Yes, since \displaystyle \lim_{x \rightarrow a} f'(x) exists. From l’Hopital’s Rule:

                \displaystyle \lim_{x \rightarrow a} \frac{f(x) - f(a)}{x - a} = \lim_{x \rightarrow a} \frac{f'(x)}{1} = \lim_{x \rightarrow a} f'(x) = f'(a)

                1. Thanks, John.

                  Re (i), your straight “no” is correct, and your “contextual yes” is absurd.

                  Re (ii), can you provide the IF argument?

                  Re (iii), how do you know the limit of f’ exists?

                  1. i) I know. But in Methods:

                    a) We have to compromise with short cuts, like the theorem in ii).

                    b) Differentiation from first principles is deleted from the new Stupid Design, making the short cut the only cut.

                    ii) Theorem: If a function \displaystyle f(x) is continuous at \displaystyle x = a and \displaystyle \lim_{x \rightarrow a} f'(x) exists, then \displaystyle f(x) is differentiable at \displaystyle x = a and is equal to \displaystyle f'(a).

                    Yes I can prove it.
                    The Report is using this theorem (who knows whether the writer understands this). But:

                    a) the Report presents this as an ‘obvious truth’ as opposed to a theorem that can be proved (using mathematics beyond the scope of VCE). Given VCAA’s fondness for ‘proof using graphs’, maybe the writer thinks that a graph makes this obvious.

                    b) There is an incorrect implication that
                    If a function \displaystyle f(x) is continuous and differentiable at \displaystyle x = a then \displaystyle \lim_{x \rightarrow a} f'(x) exists, and is equal to \displaystyle f'(a).

                    c) The Report adds the superfluous, idiotic and meaningless statement
                    “The graph of f is smooth at x = 0”
                    instead of following up with something like “It follows from continuity and existence of the limit that …”

                    iii) Because you said f is continuous and differentiable everywhere.

                    1. (i) Nah. In any context the argument, even if correct, is garbage.

                      (ii) Do you want to convince me you can prove it?

                      (iii) Are you sure?

                    2. iii) *Sigh* Of course … No. I’m sure what I said was wrong. I should have added “… provided the limit as x -> a of f'(x) exists”. Which kind of defeats the point.

                      If f is differentiable everywhere it does NOT mean that the limit of f'(x) exists everywhere. I really sleep-walked into that one.

                      Re: Theorem in (ii). I’ll either look like an idiot or a smart-ass. Let me sleep on it.

                    3. Thanks, John. To repeat my comment in another part of the thread, I don’t expect teachers to handle these intricacies. I’m only Socraticking you because i know you like thinking about such things. However, what I do expect is that VCAA employ sufficient people who can handle these intricacies, so that teachers don’t have to stumble into them or bulldoze over them.

                    4. Continuity: \displaystyle \lim_{x \rightarrow a} f(x) = f(a).

                      Let \displaystyle \lim_{x \rightarrow a} f'(x) exist: \displaystyle \lim_{x \rightarrow a} f'(x) = L.

                      \displaystyle \lim_{x \rightarrow a} \frac{f(x) - f(a)}{x - a} = \lim_{x \rightarrow a} \frac{f'(x)}{1}

                      using continuity and l’Hopital’s rule

                      \displaystyle = \lim_{x \rightarrow a} f'(x) = L = f'(a)

                      by definition of \displaystyle \lim_{x \rightarrow a} \frac{f(x) - f(a)}{x - a}.

                    5. Thanks, John. That’s it, and obviously what the examiners had in mind. Well done.

                      One fuss-point. You wrote that continuity and l’Hopital (and the assumption that f’ has a limit) gives

                      \displaystyle \lim_{x \rightarrow a} \frac{f(x) - f(a)}{x - a} = \lim_{x \rightarrow a} \frac{f'(x)}{1}

                      Can you spell that out?

                    6. Ah, that’s interesting. Thanks, John. My proof was stupider. (I would never write your first equation, however.)

                    7. Yep, careless and lazy of me. I’m guessing you’d write something like

                      \displaystyle \lim_{x \rightarrow a} \left( f(x) - f(a) \right) \times \lim_{x \rightarrow a} \frac{1}{(x - a)} etc.

                    8. No. I wouldn’t split up the limit at all. I’d indicate the two separate limits to be 0 and then declare the ratio limit to be of “type 0/0”.

  11. I teach 11 Methods, not 12, so don’t have too many thoughts about this exam. But I am puzzled by Question 1, Section B, and the examiners report on this, since this sort of thing comes up in Year 11 also.

    For questions like this, I tell my students that closed intervals for the domain of the volume function are preferred (we allow V = 0 as a “degenerate” case, as Marty mentioned). One reason for this is it becomes easy to justify that the volume is maximised at a stationary point. But if we state that the domain of the volume function is an open interval, then I don’t think the examiners report’s argument for 1g is valid. It almost reads like the argument is “there’s a stationary point at either h/2 or h/6, but h/2 isn’t in the domain, therefore the max is at h/2”. Huh? (The remark that “some did not show adequate working” is hilarious).

    But I’m also curious how other people would handle this issue. Supposing, for whatever reason, we wanted our students to justify that a function is maximised (or minimised) at a certain point in an open interval, what method would you expect from your students, and how would you explain the validity of this method?

    1. Huh. Yes, SRK you are correct, that the solution to 1(g) in the report is incomplete. I’ll add a comment.

      On the general issue, I agree with you agreeing with me. (Well, duh.) As I wrote, I don’t think giving the domain as an open interval is wrong, but it is close. I understand the physical argument for open intervals, but there is no question that closed intervals are preferable, exactly for the reason you give.

      SRK, you wrote that in Methods 11 you permit/encourage closed intervals in such contexts. Do the texts or other sources ever back you up on this? I’d also be curious to know if any other teachers use closed intervals. Are any other teachers even aware of the argument for close intervals?

      1. Correct me if I’m wrong, but only an idiot would ask for a global max (or min) on an open interval when the global max (or min) would otherwise occur at the endpoint of that interval …
        Example: Find the global maximum of \displaystyle f(x) = x^2 on the interval \displaystyle (-1, 2). There is no global maximum. As opposed to:
        Find the global maximum of \displaystyle f(x) = x^2 on the interval \displaystyle [-1, 2].

        1. Thanks, John, but I’m not sure I understand your point.

          To be honest this stuff bores me, and I can never keep straight whether or how a student is supposed to justify that a critical point is a maximum. However, i can’t imagine that you regard the following as an acceptable argument:

          I’ve been asked to find the maximum and I’ve located the one and only critical point, and I know there’s a maximum because I’ve been asked to find it, and therefore the maximum must occur at the critical point I’ve found.

          Isn’t that, in effect, the exam report’s argument in 1(g)? Are you saying you’re ok with this? If not, what are you arguing?

          1. The global maximum of y = f(x) on a finite interval will occur either at a stationary point or an endpoint *. So you test the value of y at those points and choose the value of y that is largest. If the endpoints of the interval are not included and the global maximum would otherwise have occurred at an endpoint, then there is no global maximum.

            Example 2(a): Find the global maximum of \displaystyle y = x^3 - 3x over the interval \displaystyle (-2, 3). There is no global maximum. As opposed to

            Example 2(b): Find the global maximum of \displaystyle y = x^3 - 3x over the interval \displaystyle [-2, 3]. You test the value of y at \displaystyle x = -2, \displaystyle x = 3 and \displaystyle x = -1 and find that y is largest at \displaystyle x = 3. So the global maximum exists and is equal to 18. Or

            Example 2(c): Find the global maximum of \displaystyle y = x^3 - 3x over the interval \displaystyle (-2, 1). You test the value of y at \displaystyle x = -1, noting that it’s larger than the values of y at the endpoints \displaystyle x = -2 and \displaystyle x = 1 had they been included. So the global maximum exists and is equal to 2.

            This is how the Report should have justified the answer \displaystyle x = \frac{h}{6} for Q1(g). But remember the Report’s \displaystyle caveat:
            “This report provides sample answers or an indication of what answers may have included. Unless otherwise stated, these are not intended to be exemplary or complete responses.”
            For Q1(g) the Report’s response is neither exemplary nor complete.

            So what I’m arguing is that if its an open interval, the writer has the intelligence to ensure that the required global maximum (or minimum) exists and occurs at a stationary point lying inside the interval. In which case, I don’t understand the statement

            “but there is no question that closed intervals are preferable, exactly for the reason you give”

            What reason? Are you suggesting that closed intervals ‘idiot-proof’ the question?

            * I’m ignoring the case where the global maximum occurs at both a stationary point and an endpoint.
            * And I’m not expecting Maths Methods students to justify this.

            1. One of us is being obtuse, and I’m open-minded about which one of us it is.

              1) Closed intervals are always preferable. This is obvious, for the reason SRK gave. It has nothing to do with making a question idiot-proof, although given the VCAA examiners we have, this a non-trivial consideration.

              2) It is a mathematically poor question to tell a student to find something that, in general terms, may or may not exist, and then treat as acceptable a proof for finding this thing that critically depends upon the declaration that the thing exists. It is not wrong, but it is poor. It is teaching the wrong lesson.

              3) Let’s accept that 1(g) has an open interval in its domain. In the report’s general advice, they write “students should make sure adequate working is given” and then specifically refers to 1(g). The report’s “working” is

              V(x) = x(h-2x)^2 , V'(x) = 0,  x = h/2 or x = h/6, x = h/6 as the domain is (0, h/2).

              The report then says

              Some [students] did not show adequate working for a ‘show that’ question.

              My questions are:

              a) Do you think the report’s answer has shown adequate working?

              b) If so, can you think of a plausible way a student might have gotten to and written the correct answer with what you regard as inadequate working?

              1. a) I don’t think the Report has shown adequate working. But see its \displaystyle caveat.

                b) Many students will draw a simple graph. Some will draw it by hand. Some will substitute a value of h and draw it using a CAS. Students who draw a graph will easily justify the maximum at x = h/6 by inspection. The Report doesn’t say, but I’ll bet this is an example of the ‘adequate working’ it wanted to see. I don’t think the Report is arguing that

                “I’ve been asked to find the maximum and I’ve located the one and only critical point, and I know there’s a maximum because I’ve been asked to find it, and therefore the maximum must occur at the critical point I’ve found.”

                In fact, I don’t think such an argument would have been accepted. But we don’t know, because the marking scheme is secret. Here’s how I see it: Students are asked to do two things:
                1. To get a value of x.
                2. To show that this value corresponds to a maximum.

                1. Formula for V, dV/dx = 0 statement and solutions to dV/dx = 0: 1 mark.
                2. Justification that x = h/6 gives a maximum: 1 mark.

                However, this is all speculation. Here’s my bottom line:
                If the Report pontificates that ” Some [students] did not show adequate working for a ‘show that’ question” then, \displaystyle caveat or no \displaystyle caveat, the Report has a responsibility to bloody well make it clear what \displaystyle is adequate working.

                PS – Re: “for the reason SRK gave.” I’m still unclear what that reason is. I think my examples 2(b) and 2(c) show how the global maximum (or minimum) can be justified on an open interval.

                1. Hi, John. In quick reply to your (a) and (b). (I’ll reply to the open/closed thing separately.)

                  a) For the reasons you give, the report’s caveat is worthless.

                  b) Of course, given the examiners doesn’t fully or coherently write whatever it is they are thinking, we are both/all left guessing. But, I’m reading the tea leaves differently to you. To me, it seems that the report is suggesting they wanted students to fuss with h/6 in the domain and h/2 outside the domain; I don’t think there’s any reason to believe the examiners were expecting your 2 in anything like the manner you’ve indicated for your examples.

                  Now, as I said, I’m open-minded about what students *should* be expected to do or not do to justify that a candidate maximum really is a maximum. But I find it pretty crazy to demand fussing about h/2 and simultaneously not demand fussing about the proper work of justifying the maximum, and that’s what I read that the report is doing.

                  1. OK, now I see what you’re getting at. Yes, it’s like that optical illusion of the beautiful girl and the ugly hag:

                    And now I see the ugly hag, I can’t see the beautiful girl anymore!! So I’m inclined to agree with you. I need to find an Assessor, take them out to the backroom and get answers. Then again, maybe the new Mathematics Mangler will be forthcoming in offering the answer and maybe even getting the Report amended. (I prefer taking my chances with the former).

                2. John, your argument for 2(b) and 2(c) requires checking the not-in-the-domain endpoints, to see if these if-they-were-in-the-domain endpoints might have given larger values than the candidate interior maximum points. That is valid, but:

                  a) The validity really comes down to what SRK is saying. You are, in effect, considering the same function on the closed interval and using the fact that a maximum of the continuous function on the closed (and bounded) interval must exist, at an endpoint or stationary point (or non-differentiable point). Then, if the maximum is (only) at and endpoint, you return to the open interval, reject the not-there-anymore endpoints and declare there is no maximum.

                  b) Your aetherial endpoint framing, while valid, and common, and implicit and mandatory in VCE, is much more conceptually strained than simply using the closed interval when it can naturally be done.

  12. Thanks Marty & JF for the discussion, it has helped a lot.

    Marty, in answer to your question about textbooks / sources: I checked Cambridge Year 11 Methods; it uses closed intervals in the worked examples and in the (unproofed) worked solutions. I might have a chat with my fellow 11/12 Methods teachers and see what they think…

    In response to JF’s suggested method for justifying global maxima / minima on open intervals, while I have accepted such arguments in the past, what makes me uneasy is that – from a student’s perspective – it looks like we are playing fast and loose with the domain. We (ie. Marty and JF) might appreciate that we are considering two functions that coincide upon the open interval, but one of them is continuously extended to the endpoints. But this seems unnecessarily subtle for students at this level, and they might just get confused about what it really means for an interval to be a domain of a function if we can still substitute in values that aren’t part of the domain.

    Another approach I had in mind was to, essentially, use the first derivative test. eg. For the function V(x) = x(h-2x)^2 on (0, h/2), one could observe that V'(x) > 0 and so V(x) is strictly increasing for 0 < x ≤ h/6, whereas V'(x) < 0 and so V(x) is strictly decreasing for h/6 ≤ x < h/2, and hence V(x) is maximised at x = h/6. Although it's not clear to me that this style of argument would nicely generalise to more complicated cases (ie. those with multiple stationary points, possibly including stationary points of inflection).

    1. Thanks very much, SRK.

      If Cambridge uses closed intervals in 11 Methods then VCAA has a serious problem. One can use open intervals or closed intervals, but you cannot have the major Year 11 textbook use closed and then have VCAA insist upon open in Year 12.

      Regarding your first derivative test. Yes, that works when it works, but it won’t work generally. The only way open intervals work generally is if you take a naughty peek at the not-really-there endpoints.

    2. Hi SRK. I really liked the way Marty worded what I was trying to say:

      “You are, in effect, considering the same function on the closed interval and using the fact that a maximum of the continuous function on the closed (and bounded) interval must exist, at an endpoint or stationary point (or non-differentiable point). Then, if the maximum is (only) at [an] endpoint, you return to the open interval, reject the not-there-anymore endpoints and declare there is no maximum.”

      I wish I’d explained it like that.

      Re: Closed and open intervals. Having read what Marty wrote:

      “But, I’m reading the tea leaves differently to you. To me, it seems that the report is suggesting they wanted students to fuss with h/6 in the domain and h/2 outside the domain; I don’t think there’s any reason to believe the examiners were expecting your 2 in anything like the manner you’ve indicated for your examples.”

      I’ve converted to his opinion. Furthermore, and this is only a guess, I think VCAA deliberately used an open interval because it thought it would give students an ‘easy’ (but obliviously bogus) way of justifying the choice x = h/6. I don’t think VCAA wanted students checking endpoints (particularly for a 2 mark question, albeit a CAS-assisted “show that” question). Parts of the marking scheme get made up on the training day once a couple of hundred exams have been marked. I wonder if the ‘open interval’ won by popular vote …?

      For the record, VCAA cannot reasonably justify excluding a box of zero volume. Sure, it’s a degenerate case, but zero is still a legitimate volume. I wonder what the Report would have said if the question had been worth 3 marks …?

      1. Thanks, John. Obviously I’ve thought about (and lectured on) this stuff a lot. I don’t expect teachers to be up on the nuances here, and you and SRK have much more of a clue than most. However this is exactly the kind of nuance that textbook writers and exam writers/assessors need to be comfortable with. And they are not, which means we’re stuffed. And, they either don’t know they’re not or don’t care they’re not, which means we’re really stuffed.

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