WitCH 11: Impartial

The following WitCH comes from the (CAS permitted) 2018 Specialist Mathematics Exam 2:

The Examiners’ Report indicates that about half of the students gave the intended answer of D, with about a third giving the incorrect answer B. The Report notes:

Option B did not account for common factors and its last term is not irreducible, so should not have Dx in the numerator.

13 Replies to “WitCH 11: Impartial”

  1. I’ve got three points that I would love to make, but I don’t want to spoil the fun for other crappers. So I’ll start off with just one.

    The problem with option D is that they have cancelled out common factors, which is particularly problematic because the original expression was undefined when x=-1, whereas it is defined for option D.

    1. That’s very funny. I didn’t even see that. Excellent point, Damo. (p.s. Perhaps “crapologist” is a more inviting term than “crapper”.)

      1. Crapologist is certainly more inviting, not to mention more accurate. However, I would argue that the fact that ‘crapper’ feels a bit looser and dirtier is what gives it its appeal. I’m part of a community of crappers. Maybe it’s just me….

    2. I’ll simply make the point that there is absolutely no mathematical understanding required in this question. None at all. Just press the buttons on the CAS calculator(or in Mathematica’s case, simply command Apart[…] like Moses). Furthermore, it encourages a student to have no mathematical understanding, to simply press some buttons and move on.

      In fact, it actually penalises students who attempt to use some mathematical knowledge.

      It’s a poor question that has no place on a CAS-Active Exam. What bugged me the most was the number of ‘copycat’ questions that appeared on commercial trial exams the following year ….

      By the way, Damo. I love your comment. The so-called correct option is not even equal to the original function …. So none of the options are correct.

      1. This terrifies me, John. The justification of CAS was that it would enable more ‘deep thinking’ by reducing the need for students to spend large amounts of time on calculations. This kind of questions exposes the lie of that justification. Students are rewarded for knowing how to push buttons, as opposed to rewarding mathematical knowledge. Unfortunately, I can’t see this situation improving – indeed, in light of the recent ‘review’, I am considering alternatives to teaching.

  2. There is a problem with their note on option B. I’ll do my best to unpick it, although I’m not sure I’ve quite nailed it. It is not that the fraction should not have Dx in the numerator because the denominator is not irreducible, but rather: when one of the fractions has a quadratic denominator, then the numerator should be Dx+E and; because the denominator is not irreducible, we should be writing this fraction as the sum of two fractions, both with linear denominators (in this case).

    As John said, I think that the big problem with this question is that the correct partial fraction is not there. I would suggest that it should be:

    A/(2x+1) + B/(2x+1)^2 + C/(2x+1)^3 + D/(x+1) + E/(x-1)

    1. Thanks, Damo, and I agree. Even on the Examiners’ own terms, their comment on Option B is way off key. People have pretty much nailed the awfulness of this question, and I’ll update when I next come up for air. But, I think there’s a little more awfulness to isolate.

    2. That’s a correct option VCAA could have had which would have at least required more than simple button pushing. But that boat sailed the moment they specified that A, B, etc were non-zero real numbers.

      1. Specifying that A, B etc were non-zero is an unnecessary addition to the problem – and potentially quite destructive, as students reading this problem may assume that this is generally true for partial fractions – but I don’t think it invalidates the above solution.
        I get A= 34/9 B = -11/3 C = 1 D = 1/9 and E = -2

        1. There might be a slip in your calculations, Damo (or mine) – specifically, I get A = 0 (I haven’t checked the other numerators of the fractions that would otherwise ‘cancel’):

          Assuming (2x^2 + 3x + 1)/(2x+1)^3(x^2-1) = A/(2x + 1)^3 + B/(2x + 1)^2 + …. + E/(x + 1) I get

          2x^2 + 3x + 1 = A(x^2 – 1) + B(x^2 – 1)(2x + 1) + C(x^2 – 1)(2x + 1)^2 + D(2x + 1)^3(x + 1) + E(2x + 1)^3(x – 1)

          Substitute x = -1/2: 1/2 – 3/2 + 1 = -3A/4 => A = 0.

          I think the other numerators corresponding to fractions of the fractions that would otherwise ‘cancel’ (C and E) will also be zero. So I think the non-zero statement is necessary to force the option that VCAA want.

          1. Ha, yes. I was scratching my working on the back of an envelope, and at some point mistook the expression on the original numerator to be 2x^3+3x+1. Which it is not. As you were.

        2. The amazing thing about these WitCHes is they are often worse than I thought when I posted. You guys commonly point out aspects that I hadn’t seen. And here you’re both correct: requiring the coefficients be non-zero absolutely kills any reasonable non-CAS approach.

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