WitCH 51: Logging the Possibilities

December 5, 2020

6:42 pm

37 Comments on WitCH 51: Logging the Possibilities

algebra, logarithms, Mathematical Methods, periodic functions, trigonometry, VCAA, VCE, WitCH

The question below is from the second 2020 Methods exam (not online), discussed here. You may wish to brush up on your modal logic before attempting the question.

37 Replies to “WitCH 51: Logging the Possibilities”

Glen says:

December 5, 2020 at 7:57 pm

Gross question.

As a first reaction, since we require the function $f$ to be $1$ -periodic, we must have $a = 2k\pi$ for some integer $k$ .

Reading further, I’d have guessed only $D$ and $E$ stand a chance at being correct. This is because if the range of $g$ is $[-1,0]$ then the range of $f$ must be $[1/2,1]$ . The values of $\theta$ such that $\cos \theta$ covers $[1/2,1]$ are any set containing $[-\pi/3,0]$ or $[0,\pi/3]$ . Note that we only need the set to cover these intervals, so something like $[-\pi/9,\pi/3]$ is also admissible (this is important for option $E$ ).

For $D$ , $a = \pi$ , and for $E$ , $a = 4\pi/3$ . Neither of these are integer multiples of $2k\pi$ , and so I’m stuck. None of the remaining options contain zero, and so I guess that there is no correct answer to this question.

I did this a little quick, apologies for any errors.

Reply
1. John Friend says:
  
  December 5, 2020 at 9:34 pm
  
  The way I see it, the period could be be 1/1 or 1/2 or 1/3 or … (in fact, anything of the form 1/n where n is a positive integer). But 1/1 = 1 is simplest so try that:
  
  Period = 1 therefore $a = 2 \pi$ and so $f(x) = \cos{(2 \pi x)}$ .
  
  Require $\displaystyle -1 \leq \log_{2} (\cos{(2 \pi x)}) \leq 0 \implies \frac{1}{2} \leq \cos{(2 \pi x)} \leq 1$ and only Option B fits the bill.
  
  Of course, for other functions with the required property (that is, functions with a period 1/2 or 1/3 or …) there is no correct option. So the question only works for the special case that VCAA had in mind. It’s likely that VCAA did not realise this was a special case and that no option was true in general, otherwise we would have seen some additional convoluted wording trying to specify that $a$ had the smallest possible value or such-like.
  
  Modal logic is not required. You have to think like the enemy and use crazy VCAA logic. https://www.youtube.com/watch?v=aGPTfxdw_lw
  
  Reply
  1. marty says:
    
    December 5, 2020 at 9:49 pm
    
    We can’t know that the period is 1. We are told, in effect, that the period is 1/n.
    
    This isn’t my main issue with the question. My main issue is that I think the question is insane. But I also think it can reasonably be argued that the question is screwed.
    
    Reply
  2. Glen says:
    
    December 5, 2020 at 10:18 pm
    
    Oh, I see, for anyone following my reasoning, just note that not only sets that cover $[0,\pi/3]$ or $[-\pi/3,0]$ apply, but any set that covers an interval of the form $[2k\pi,2k\pi+\pi/3]$ or $[2k\pi-\pi/3,2k\pi]$ .
    
    The reasoning I gave rules out $D$ and $E$ but as you say $B$ fits, because it satisfies the $k=1$ case.
    
    As a remark, the $1/n$ periodicity talk confuses me. I think you’re just saying the same thing that I am, I only used a $k$ instead of an $n$ . We are told the function $f$ is $1$ -periodic, but that can be achieved of course by imposing $1/2$ or $1/3$ or whatever periodicity of the form $1/n$ on $f$ . It is still $1$ -periodic (also $2$ -periodic…). For our particular function here that means $a = 2k\pi$ (or $2n\pi$ if you prefer).
    
    Marty — why do you say the periodicity may not be 1? From what I can see, it is definitely 1, but it may also be something smaller. We don’t know the principal period — is that what you mean?
    
    The question is definitely fucked regardless from my point of view. It took some time to check those stupid intervals, and even then, I fucked it up. It’s just mean.
    
    Reply
    1. marty says:
      
      December 5, 2020 at 10:26 pm
      
      Sorry, Glen. I was vague. Yes, I was using “the period”, with the definite article, to mean the minimal/principal period.
      
      Reply
Banacek Spaces says:

December 5, 2020 at 11:04 pm

Let’s delete “for all” and replace it with “iff”.
Works better?

Reply
1. marty says:
  
  December 5, 2020 at 11:57 pm
  
  You mean in the exam question? No, that doesn’t work, since there’s also an implicit “for all x”. For most x there are other h that will work.
  
  Reply
  1. Banacek Spaces says:
    
    December 6, 2020 at 7:27 pm
    
    So not much better… will not be posting after midnight again.
    One may expect then iff and tan x next year
    
    So, the last line of defence (” your posts leave me no choice but to defend this … “).
    It is supposed to be enigma-tic.
    I have a recollection of code breaking documentary and a member of one team remembering how there were so many permutations that they gave up. Another team started with an assupmtion the one of the discs ( i think there were 3 discs) had letters in the order: A, B, C,… They broke the code. And it is rather known story
    KISS? Not strictly mathematical, but a winner.
    
    Back to the exam. Q20, the last in the section. They are to discriminate. In the pockets of some smart students, i have no doubts.
    
    Reply
    1. marty says:
      
      December 6, 2020 at 7:47 pm
      
      Nice pun, although I’d suggest the question is more tic than enigma.
      
      Sure, as the last MCQ, it’s intended to be testing, and one gives the examiners some license. But not license to smash into a cliff. The question above has so many layers of deciphering, it’s a parody. And, as I’ve been arguing here, I think it has one more layer than was intended, making the question technically wrong.
      
      Reply
student-teacher says:

December 6, 2020 at 8:17 am

I think that, as Glen said, we know that we must have a = $2k\pi$ for some integer k (and can well assume k is positive because cosine is even). In order for the range of the function to be $[-1, 0]$ on an interval, we need that the interval be less than $\frac{1}{3 k}$ wide. The intervals widths given are A: 1/6, B: 1/6, C: 1/3, D: 1/3, E: 1/3. So that means we need $k=1$ or $k=2$ for any of these to be possible. Then we have two possible functions and can check each interval. How long are students supposed to spend on each multiple choice question? This might take me too long.

Reply
1. marty says:
  
  December 6, 2020 at 8:32 am
  
  Thanks, S-T. Yes, one can work that none of the other intervals works for any k. And, yes, the whole thing is a long nightmare for 1 Mark.
  
  But more than that, I think, like JF, you’re reading the question too generously. We don’t know k = 1, so we don’t know *any* of the intervals are “possible”.
  
  Reply
  1. John Friend says:
    
    December 6, 2020 at 9:12 am
    
    Marty, I’m not being generous. I’m being pragmatic. We don’t know that the period *isn’t* 1 … If we assume that the period is 1, then there’s a correct option. None of the options are correct otherwise.
    
    So where the question is fucked is that the ‘minimal-period’ assumption is expected and required. So you’re forced to work backwards: *If* there’s a correct answer (and of course we expect that there is) then the period must be 1. And if the period is 1, then the correct answer is B … The ‘minimal-period’ should have been given data, not an assumption.
    
    It’s another case of a 1 mark MCQ trying to be too cute by using too many ideas. The arms race. What’s being tested here? Why not just define $\displaystyle f(x) = \cos{(2 \pi x)}$ ? Then you can properly test one idea. Or why not discard the composite function stuff and just ask what the possible periods of $\displaystyle f(x) = \cos{(ax)}$ are?
    
    What do we learn from the spread of students that chose each option for this question? What do we learn if only 17% of students chose option B. Jackshit.
    
    Reply
    1. marty says:
      
      December 6, 2020 at 9:23 am
      
      So, you’re arguing that it is “possible” that $\boldsymbol{k=1}$ ?
      
      Reply
      1. John Friend says:
        
        December 6, 2020 at 10:04 am
        
        Yes.
        
        *If* there’s a correct answer (and of course we expect that there is) then the period must be 1. And if the period is 1, then the correct answer is B.
        
        (I get the feeling I’m walking into a trap here. I’ve predicated my argument on the assumption that there is a correct option …)
        
        Reply
        
        marty says:
        
        December 6, 2020 at 10:09 am
        
        So, it is “possible” that $\boldsymbol{k=1}$ , in which case it is “possible” that the domain is as in B.
        
        Then, if it is possible that something is possible, is that thing possible?
        
        Reply
        
        John Friend says:
        
        December 6, 2020 at 11:01 am
        
        It is “certain” that there is a correct option in which case it is “certain” that the domain is as in B.
        
        Re: “if it is possible that something is possible, is that thing possible?”
        
        Yes, there is a non-zero probability for that thing … (Feel free to spring the trap now).
        
        Reply
        
        marty says:
        
        December 6, 2020 at 11:22 am
        
        Well, “trap” is too strong, but I’ll admit to having fun. This question is exactly what I was pondering with my obscure reference to modal logic in the post.
        
        It seems, as you agree, that a reasonable axiom of “possibility” is that if something is possibly possible then it is possible. But, it is not difficult to query that axiom and to conclude that it maybe shouldn’t be accepted so automatically.
        
        I guess this a byproduct of my weird semi-training as a logician, but as soon I read MCQ20, my reaction was
        
        a) What the fuck is this?
        
        and
        
        b) “The answer’s gonna depend upon your modal axiom system”.
        
        Obviously, a question isn’t great if it inspires (a), but of course it happens all the damn time in VCE. But I’m astonished and bemused to have come across a question that inspired (b).
        
        Reply
2. John (No) Friend (of VCAA) says:
  
  December 6, 2020 at 8:52 am
  
  There are many different ways of doing Q20 I suppose. But the pragmatic bottom line in a VCAA MCQ is getting ‘the’ (sometimes ‘a’ …!) correct option.
  
  Sometimes this means rejecting all the wrong options. If one option remains, that’s the answer (or more precisely, the answer VCAA intended). You might not even be able to prove why it’s the answer. It’s the answer simply because all others are wrong and therefore can’t be the answer.
  
  So if you try and find the correct option through rigorous mathematical analysis and thought, you’re often going to waste a lot of time on an MCQ and you won’t finish the exam.
  
  That is the sad reality of the examination landscape we’re currently forced to train our students to traverse.
  
  So for Q20, your excellent mathematical analysis is going to take too long. You have to use a method that gets the answer in 90 seconds. I have offered a method that achieves this.
  
  Yes, it’s crazy. insanity. There’s an arms race (particularly among commercial trial exams) – who can come up with the trickiest stupidest wordiest MCQ. Many MCQ’s on the VCAA exam are clearly worth more than 1 mark. The Section A on Exam 2 is fucked. Some questions are too long, others are trivialised by the CAS-poison. More generally, Exam 2 in its current form is fucked. Marty is an idealist and wants CAS and Exam 2 gone. I admire and applaud his uncompromising principals. I’m a pragmatist – I think CAS is here to stay so I want an Exam 2 that’s not fucked. Unfortunately, until the VCAA decks are swept clean of its idiot infestation living in the Magic Faraway Tree (and Mathematics is not the only subject that needs a hard broom taken to it …) we are all stucked and fucked.
  
  Reply
  1. marty says:
    
    December 6, 2020 at 9:03 am
    
    John, I’ll answer further in a little while, but this is the same as the slippery slope problem. The necessity of quickly clawing out some plausible answer to an exam question doesn’t preclude criticising the question for being insane and/or wrong.
    
    Reply
    1. marty says:
      
      December 6, 2020 at 9:35 am
      
      Suppose we are told that $\boldsymbol{f(x) = \sqrt{(x+k)}}$ where $\boldsymbol{k\in\mathbb N}$ . (That is, $\boldsymbol{k}$ is a positive integer, but we don’t know which.) Then, we are asked:
      
      A possible domain for $\boldsymbol{f(x)}$ is
      
      A. $\boldsymbol{[-2,\infty]}$ .
      
      B. $\boldsymbol{[2,\infty]}$ .
      
      How would you respond?
      
      Reply
      1. John Friend says:
        
        December 6, 2020 at 9:56 am
        
        Marty, I see what you’re trying to do here but it’s not the same. In your example both A and B are correct if $k \geq 2$ . In Q20 there is only one value of the period that gives a correct answer. For any other period there is no correct answer:
        
        If there is a correct answer then period = 1 and if period = 1 then the correct answer is B.
        
        For your question, if I *had* to choose either A or B (it’s a test and I’m told there is only one correct answer for each question and I can’t argue that the question is defective) I would choose B because it’s correct for all values of k.
        
        My beef with Q20 is that the ‘minimal-period’ assumption is expected and required.
        
        Reply
        
        marty says:
        
        December 6, 2020 at 10:06 am
        
        But you agree that somehow B is “correct” or “more correct” or “more likely correct” or something? That seems to me inarguable, and that’s the problem.
        
        The natural interpretation when told $\boldsymbol{k}$ is an integer is that this is all $\boldsymbol{all}$ we know about $\boldsymbol{k}$ . The natural interpretation is to look for an answer that works for all $\boldsymbol{k}$ , not to assume a simple $\boldsymbol{k}$ and then fish.
        
        The more mathematical you think, the more MCQ20 screws you up.
        
        Reply
        
        John Friend says:
        
        December 6, 2020 at 10:45 am
        
        Yes, but in an MCQ the over-riding interpretation/assumption must be that a correct option exists. This is data that must be used …
        
        Once this data is used in Q20 (and what choice do we have?) then k =1 follows and then the mathematics takes over.
        
        I’m not saying I agree with this, but that’s the insane VCAA logic that I guarantee will be seen in solutions from The Bizarro Twilight Zone:
        
        k = 1 because a correct option exists.
        
        For your example, we must assume that a *single* correct option exists. So we must choose the option that is true for all k: B.
        
        Again, I’m NOT saying I agree with this. A maths question on a maths exam should be able to be answered using purely mathematics. But in VCAA’s case we must apply an insane brand of fuzzy logic.
        
        I’ll say again for the record: Where I think the question is fucked is that the ‘minimal-period’ assumption is expected and required. (Because it follows from the assumption that a correct option exists).
        
        Reply
        
        marty says:
        
        December 6, 2020 at 11:00 am
        
        Sure. Again I’m not arguing with the pragmatics, with coming up with some interpretation and then the plausible strategies to deal with the question thus interpreted. That’s the job of teachers like you, to shepherd your students through the jungles of idiocy, and I respect and admire the job you guys do.
        
        But, that is not my (self-appointed) job, and that is not (primarily) what this blog is about. My job is to consider what is said and written and proposed – not what was intended to be said or written or proposed – and to critique it all, fairly but thoroughly, from a mathematician’s perspective.
        
        Reply
        
        John Friend says:
        
        December 6, 2020 at 11:17 am
        
        Indeed. “the pragmatics, … coming up with some interpretation and then the plausible strategies to deal with the question thus interpreted … [which is] the job of teachers … to shepherd [their] students through the jungles of idiocy”
        
        gives me the absolute shits. We must do this because of imbeciles at VCAA and their cretinous goons.
        
        Reply
    2. John (No) Friend (of VCAA) says:
      
      December 6, 2020 at 9:42 am
      
      Marty, don’t think for 1 minute I’m precluding. I absolutely agree with you.
      
      What makes me angry is that we don’t get a chance to hold the writers or vettors accountable for this shit. We don’t get the chance to interrogate or debate or engage with them. There is no accountability. You get a pissant on-line opportunity to provide feedback that just gets ignored. It’s the teachers and students who end up suffering the consequences of this shit.
      
      The MAV should be facilitating accountability. Instead, it runs the weak-as-piss Meet the VCAA Stooges where you don’t get an answer to anything. And VCAA publishes a mealy-mouthed Report that is silent on the many issues plaguing its exams. I’m not advocating a public flogging, but if there *was* robust accountability, I think standards would rise. There are too many VCAA stooges and people within VCAA that get away with this because there is no accountability.
      
      What makes me really angry is the “necessity of quickly clawing out some plausible answer to an exam question” when the question is fucked.
      
      Reply
      1. marty says:
        
        December 6, 2020 at 9:53 am
        
        Of course we agree. The whole game is insane, the VCAA doesn’t give a shit and the MAV is a fucking lapdog, and that’s the main point.
        
        But the specific point on this post is, to what extent is the exam question screwed and, precisely, how? My square root question and my question about it being “possible” that $\boldsymbol{k=1}$ are intended to tease that out.
        
        Reply
Red Five says:

December 6, 2020 at 12:15 pm

If I hear you all (Marty, JF and Glen in particular) if this weren’t a multiple choice question, are you suggesting that no answer is possible, even an answer in terms of $a$ ?

If that is your suggestion, and if that suggestion is correct (which my brief modal logic training tells me is a valid conclusion, albeit based on a couple of hidden premises) then…

Reply
1. marty says:
  
  December 6, 2020 at 1:22 pm
  
  I wish you hadn’t used the word “possible”.
  
  My natural interpretation of such a question is that, since you don’t know what k is, you are looking for an answer that works for all relevant k. The MCQ is then tangled by the issue of whether k should be general, combined with the “possible” in the actual question. I think JF and others are more willing than me to permit k to have a specific value that will work.
  
  Again, this is all secondary. What is primary is that the question is insane.
  
  Reply
  1. John Friend says:
    
    December 6, 2020 at 2:11 pm
    
    Yes, that’s the heart of the matter.
    What if the question had said “A possible interval for D $\displaystyle \text{if f has the smallest possible period}$ is”
    
    Reply
    1. John Friend says:
      
      December 10, 2020 at 10:23 am
      
      *maximum possible period
      
      Reply
2. John Friend says:
  
  December 6, 2020 at 1:31 pm
  
  There are an infinite number of answers corresponding to each of the infinite number of possible values of $a$ . Q20 narrows this down to a single answer corresponding to a single value of $a$ . In general:
  
  The period is $\displaystyle \frac{1}{n}$ where $n$ is a positive integer, thus $a = 2 n \pi$ and so $f(x) = \cos{(2 n \pi x)}$ .
  
  Apply the range constraint on $g$ :
  
  $\displaystyle -1 \leq \log_{2} (\cos{(2 n \pi x)}) \leq 0 \implies \frac{1}{2} \leq \cos{(2 n \pi x)} \leq 1$
  
  and so the maximal domain is the union of all intervals of the form
  
  $\displaystyle \left[ \frac{6m-1}{6n}, m\right] \cup \left[ m, \frac{6m+1}{6n}\right]$
  
  where $m$ is an integer.
  
  So for Q20, if you take $n = 1$ (so that the period is 1 and $a = 2 \pi$ ) you get option B from $\displaystyle \left[ m, \frac{6m+1}{6}\right]$ with $m = 1$ .
  
  None of the other options in Q20 are an interval with either of these two required forms for any other value of $n$ .
  
  So if this was not a MCQ, but simply a question that asked what is a possible domain, then my answer would be
  
  A possible domain is any interval of the form
  
  $\displaystyle \left[ \frac{6m-1}{6n}, m\right]$ or $\displaystyle \left[ m, \frac{6m+1}{6n}\right]$
  
  where $a = 2 n \pi$ , $n$ is a positive integer and $m$ is an integer, or any union of such intervals.
  
  Reply
  1. John Friend says:
    
    December 10, 2020 at 2:48 pm
    
    For those who are interested, here is a typed version (with a couple of errata corrected) of the above. The only thing missing is a critical evaluation of the question.
    
    2020mm-exam 2-Section A Q20-solution and discussion
    
    Reply
    1. marty says:
      
      December 10, 2020 at 3:54 pm
      
      Thanks, JF.
      
      Reply
      1. John Friend says:
        
        December 10, 2020 at 6:10 pm
        
        You’re welcome, Marty. In fact, I’ve edited that document a bit so that it’s much better and now even contains a very brief evaluation of the question (as well as the correction of a careless typo). I promise there will be no further edits and uploads!!
        
        Aside 1: It’s this sort of discussion that I think should appear in commercially available VCAA exam solutions. This would create a non-trivial point of difference between expensive solutions and free solutions and actually make the expensive solutions worth buying. Such discussion did appear in some Bizarro Twilight Zone solutions prior to 2019, but then ‘VCAA assessors’ started writing ALL the solutions. Now there’s no danger of any discussion or critical evaluation (when appropriate). I don’t see the point in paying big money for a product when you can get something just as good for *free* … (you’re simply paying for the brand rather than the content and getting sucked in by misleading marketing).
        
        Aside 2: If schools are trying to save money as a result of 2020, they could do worse things (and unfortunately some are …) than refuse to pay big money for:
        
        1) VCAA solutions (since just-as-useful solutions can be obtained for free on-line, the expensive solutions have no non-trivial point-of-difference despite marketing to the contrary),
        
        2) Trial exams (unnecessary because there are over 30 freely available VCAA exams plus over 30 free trial exams plus whatever commercial trial exams a school already has),
        
        3) meeting useless VCAA stooges,
        
        4) Commercial SACs (use of which is against VCAA regulations and leads to audit failure),
        
        5) Useless PD delivered by organisations and people with snouts in the trough. (You don’t have to attend paid PD to satisfy the PD requirements of irrelevant blood-sucking vampires. Schools have plenty of staff with plenty of experience to share as internal free PD).
        
        By my calculation this would save every school many thousands of dollars …
        And if you look at public schools and multiply this saving per school by the number of schools, you are potentially saving many, many hundreds of thousand of dollars – money that could be much better spent.
        
        2020mm-exam 2-Section A Q20-solution and discussion
        
        Reply
Glen says:

December 6, 2020 at 2:34 pm

Cool discussion.

Yes the question is bananas. Convoluted in the extreme. It is also way too long. Don’t forget I fucked it up at the start!

I see what Marty is arguing in general. For this specific question, however, I believe we are being told that $f$ is $1$ -periodic, and nothing further. Nothing about the principal (minimal) period. So it could be 1/2, 1/3, whatever, periodic. Regardless, it is definitely $1$ -periodic.

We are asked what a “possible” interval is, and so, given that there is an interval that works when setting the other parameters to admissible values, there is a correct answer.

Marty is welcome to tell me why I’m wrong :).

Reply
1. marty says:
  
  December 6, 2020 at 3:23 pm
  
  Glen, I wouldn’t go as far as to say that you (and JF and S-T) wrong: I think the question is made sufficiently ambiguous by the “possible” that there is no definitive answer. But I think, as suggested by my square root example, that mine is the preferred reading.
  
  Reply

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