Witch 113: Smoothing Over the Cracks

This one is a combo WitCH. The main concern is a multiple choice question from last week’s Methods Exam 2. The question may not be an “error” in the newspaper sense, but it is bad. To appreciate some (but far from all) of its badness, however, we need to see VCAA’s solution. We won’t likely see that solution, however however, for months, if ever; transparency is not VCAA’s strong suit (Section 7).

To deal with this, we’ve teamed up last week’s MCQ with a similar MCQ from the 2021 Exam 2, together with VCAA’s solution to that earlier question from the exam report. Last week’s question appears first. Continue reading “Witch 113: Smoothing Over the Cracks”

WitCH 43: Period Piece

This one comes courtesy of a smart VCE student, the issue having been flagged to them by a fellow student. It is a multiple choice question from the 2009 Mathematical Methods, Exam 2; the Examination Report indicates, without comment, that the correct answer is D.

UPDATE (08/12/23)

The exam report was amended on 18/09/2020, after this post appeared. The report now includes a note, with no indication that the note was added a decade later:

B was also accepted as it leads to an equivalent expression.

If true, then the original exam report was consciously deceptive.

WitCH 33: Below Average

We’re not actively looking for WitCHes right now, since we have a huge backlog to update. This one, however, came up in another context and, after chatting about it with commenter Red Five, there seemed no choice. The following 1-mark multiple choice question appeared in 2019 Exam 2 (CAS) of VCE’s Mathematical Methods. The problem was to determine Pr(X > 0), the possible answers being

A. 2/3      B. 3/4      C. 4/5      D. 7/9      E. 5/6

Have fun.

Update (04/07/20)

Who writes this crap? Who writes such a problem, who proofreads such a problem, and then says “Yep, that’ll work”? Because it didn’t work, and it was never going to. The examination report indicates that 27% of students gave the correct answer, a tick or two above random guessing.
 
We’ll outline a solution below, but first to the crap. The main awfulness is the double-function nonsense, defining the probability distribution \boldsymbol{f} in terms of pretty make the same function \boldsymbol{p}. What’s the point of that? Well, of course \boldsymbol{f} is defined on all of \boldsymbol{R} and \boldsymbol{p} is only defined on \boldsymbol{[-a,b]}. And, what’s the point of defining \boldsymbol{f} on all of \boldsymbol{R}? There’s absolutely none. It’s completely gratuitous and, here, completely ridiculous. It is all the worse, and all the more ridiculous, since the function \boldsymbol{p} isn’t properly defined or labelled piecewise linear, or anything; it’s just Magritte crap. 
 
To add to the Magritte crap, commenter Oliver Oliver has pointed out the hilarious Dali crap, that the Magritte graph is impossible even on its own terms. Beginning in the first quadrant, the point \boldsymbol{(b,b)} is not quite symmetrically placed to make a 45^{\circ} angle. And, yeah, the axes can be scaled differently, but why would one do it here? But now for the Dali: consider the second quadrant and ask yourself, how are the axes scaled there? Taking a hit of acid may assist in answering that one.
 
Now, finally to the problem. As we indicated, the problem itself is fine, its just weird and tricky and hellishly long. And worth 1 mark. 
 
As commenters have pointed out, the problem doesn’t have a whole lot to do with probability. That’s just a scenario to give rise to the two equations, 
 
1) \boldsymbol{a^2 \ +\ \frac{b}{2}\left(2a+b\right) = 1} \qquad      \mbox{(triangle + trapezium = 1).}
 
and
 
2) \boldsymbol{a + b = \frac43} \qquad           \mbox{( average = 3/4).}
 
The problem is then to evaluate
 
*) \boldsymbol{\frac{b}2(2a + b)} \qquad \mbox{(trapezium).}
 
or, equivalently, 
 
**) \boldsymbol{1 - a^2 \qquad} \mbox{(1 - triangle).}
 
 
The problem is tricky, not least because it feels as if there may be an easy way to avoid the full-blown simultaneous equations. This does not appear to be the case, however. Of course, the VCAA just expects the lobotomised students to push the damn buttons which, one must admit, saves the students from being tricked.
 
Anyway, for the non-lobotomised among us, the simplest approach seems to be that indicated below, by commenter amca01. First multiply equation (1) by 2 and rearrange, to give
 
3) \boldsymbol{a^2 + (a + b)^2 = 2}.
 
Then, plugging in (2), we have 
 
4) \boldsymbol{a^2 = \frac29}.
 
That then plugs into **), giving the answer 7/9. 
 
Very nice. And a whole 90 seconds to complete, not counting the time lost making sense of all the crap.