25 Replies to “WitCH 50: No Partial Credit”

  1. Ugh! indeed.

    This is a really stupid way of testing partial fractions. Some speculations on the motivation behind the question:

    1. The writer was trying to test partial fractions in a ‘different’ way. Why? Stupid. Just test partial fractions in Exam 1 in the usual way.

    2. The writer was trying to test partial fractions in a way that thwarted directly using a CAS. Stupid. If you don’t want students using a CAS, test the skill in Exam 1. Memo to VCAA: You don’t try to test skills that are trivialised using CAS technology in Exam 2. Because it leads either to trivialisation or to stupid questions.

    3. The writer thought that s/he was being clever (linked to 1). Wrong. You’re being *stupid* not clever.

    Verdict: There is no reasonable justification for this question. It’s stupid to test partial fractions in Exam 2, no matter how ‘cleverly’ you think you can do it. And the most stupid thing is that partial fractions *was* tested in Exam 1. So why have this question in Exam 2?? Don’t the writers of each exam communicate with each other? Isn’t there someone with a global perspective?

    1. Following on from point 2, even if that was the intent, the question failed, since one can still use CAS to get the partial fraction form by appending the restriction b<0 to an expand command (on the TI).

      1. Hi SRK. Indeed. I thought the same as I typed. I figured the writer probably thought this extra ‘layer’ of ‘CAS code’ was sufficient to make the question more challenging. I suppose it might and for some students maybe it did …

        The writer probably expected students to do this question using a CAS, which is quite ridiculous since it’s more efficient to do it ‘by hand’. (As an aside, one of my biggest bugbears are commercial trial exams where the solutions consist predominantly of screenshots of a CAS. Stupid people). The question ultimately tests nothing except pushing buttons or writing simple Mathematica code.

        Whenever I see things like “where b < 0 my first move is to substitute \displaystyle b = - \beta^2 and go from there. This is what I tell students to do.

  2. Thanks, JF and SRK. It’s obviously a contrived and weird and pointless question. But it’s also as clumsy as all hell. As you point out, JF, having \boldsymbol{b<0} is ridiculous, leading to the ridiculous and hellishly ugly \boldsymbol{\sqrt{|b|}}. It’s just dumb, and it’s not the only such dumbness in the question.

    1. Well, I don’t see the point in having the \displaystyle a in the \displaystyle \frac{A}{ax} term in Options A, B and E. That’s dumb too.

      And having Option A containing a term with a quadratic denominator and immediately following the statement “with linear denominators” is dumb too.

      So by reckoning the question is triple dumb. An impressive effort by VCAA.

      1. JF, that’s sort of the other issue that bugged me. It’s reasonable enough for a multiple choice question to offer dumb or weird choices. But, what should be done with that factor \boldsymbol{a} in the denominator?

        1. I thought of that, namely if A and a are both constants then \frac{A}{a} is a constant, too and really only needs one letter.

          But then I wondered if it wasn’t meant to imply a was a common feature, somehow…

          …then I gave up because it is a VCAA exam and there is no knowing what (if?) they are actually thinking.

          1. RF, it’s reasonable for the question to include the factor \boldsymbol{a} in the denominator. But how does one naturally deal with that factor, and how does that relate to the correct and incorrect answers on offer?

            1. I guess what I am saying (in a round about way) is that I would be quite annoyed if \frac{A}{x} was marked as wrong but \frac{A}{a x} were marked as correct.

              On the whole I found the question stupid. It is a “by hand” skill and testing these on Paper 2 examinations just leads to all kinds of crazy bull$h!t

                1. In a multiple choice question, there is normally an “intended answer” but, as we have seen here repeatedly, sometimes other answers are technically correct because of issues such as this.

                  As to what is marked correct or not, we may never know when it comes to VCAA exams.

                  My comment was more general in nature and not responding to how this particular question was/is marked.

                  1. Yeah but the question is about this question above. How does this question handle the factor a, and is it that correct and reasonable?

                    1. My view (see my comment below) is yes.

                      But it’s a qualified yes. Yes, I think it’s correct and reasonable. BUT no, I don’t think it’s correct and reasonable to do it for all options except the correct one. I think that’s misleading for students and stupid.

                      It should be one for all and all for one.

        1. That’s the exact point I’ve been raising in a couple of these blogs.

          And having anything in Exam 2 that is tying to test ‘by hand’ skills is very stupid. This happens because Exam 1 is too skinny and Exam 2 is too fat.

  3. The correct answer is D and there’s not an ‘a’ to be seen. So I don’t know why ALL the other options have an ‘a’. It’s stupid and will potentially confuse some students in the future.

    I thinks it’s helpful to consider TM’s example \displaystyle{\frac{1}{2x(x^2-9)}}:

    Would we prefer students to decompose it as

    \displaystyle \frac{A}{2x} + \frac{B}{x - 3} + \frac{C}{x+3}, or

    \displaystyle \frac{A}{x} + \frac{B}{x - 3} + \frac{C}{x+3} ?

    I actually prefer the former because students make too many stupid mistakes with the latter when calculating the values of A, B, C. But in the abstract VCAA MCQ I don’t know why all their options except for the correct one do this …?

    And would encouraging the former create problems if something like \displaystyle{\frac{1}{(2x+1)(x^2-9)}} is presented, whereby some students might either:

    1) think it decomposes as \displaystyle \frac{A}{x+1} + \frac{B}{x - 3} + \frac{C}{x+3}, or

    2) try to decompose it as \displaystyle \frac{A}{x+\frac{1}{2}} + \frac{B}{x - 3} + \frac{C}{x+3} and get themselves into a tangle?

    1. Thanks, John. Picking on Terry’s example helps here (although not on the negative \boldsymbol{b} nonsense).

      Clearly the factor \boldsymbol{a} is intended to test/confuse/trick students (take your pick). I’m ok with that in principle, although I don’t think it’s great. The larger problem for me is that I don’t like the correct answer. It’s not what I would do. Neither of your suggested approaches to Terry’s specific example is what I would do.

      (Your “Which would we prefer” question made think of how to answer whether I prefer Mr. Potato Head or ScoMoFo as prime minister. Sometimes it’s better to take the Fifth.)

      1. So Marty, what *would* you do? I don’t know about the fifth, but here’s the third … Would you treat it as \displaystyle \frac{1}{2} {\frac{1}{x(x^2-9)}} and then focus on decomposing \displaystyle {\frac{1}{x(x^2-9)} as \displaystyle \frac{A}{x} + \frac{B}{x - 3} + \frac{C}{x+3} ?

            1. Conceptually, the \boldsymbol{2} has nothing particularly to do with the factor \boldsymbol{x}, so why make a denominator \boldsymbol{2x}? What would you do with the 2 if the whole denominator happened to be \boldsymbol{2\left(x^2-9\right)}?

              Computationally, taking out the 1/2 and partialling the rest, you increase the odds of avoiding fractions.

              1. Better yet, what would I do with the 2 if the denominator had 2(x-1) …?

                Well, in both cases (yours and mine), I’d take the 2 out. But if the denominator had 2x, I wouldn’t.

                Re: “Computationally, taking out the 1/2 and partialling the rest, you increase the odds of avoiding fractions.”

                I’m not sure about that. I’ll have to think about what that probability calculation might look like. I’ve got a feeling it will make no difference.

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