Tag: integration

Witch 96: Root Con All

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February 7, 2023

6:26 pm

50 Comments on Witch 96: Root Con All

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This is the fourth of our three WitCHes on VCAA’s Specialist Mathematics Exam 1 Sample Questions. Yeah, yeah it’s Douglas Adams-ish, but it wasn’t deliberate. We were undecided on this last one, and it’s not as compelling as the others. But, for similar reasons for the other WitCHes, we changed our mind and decided to post it. (There were other close calls, and we’ll soon update this post with brief comments on all the questions.)

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Witch 95: Top of the Pops

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February 7, 2023

6:02 pm

12 Comments on Witch 95: Top of the Pops

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This is the third of our three WitCHes on VCAA’s Specialist Mathematics Exam 1 Sample Questions. Newcomer aps was first to comment on the error, in regard to VCAA’s webinar, where the same question is “solved” (and see the subsequent comments on that post). Once again, it seems worthwhile to encourage a prominent discussion on an unfamiliar topic, and there is more to say than has come up so far.

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WitCH 49: Trigged Again

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December 2, 2020

10:25 pm

14 Comments on WitCH 49: Trigged Again

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The question below is from the second 2020 Specialist exam (not online), and was flagged by commenter John Friend in the discussion here. John has spelled out the problems, but the question is bad enough to warrant its own post, and there’s arguably a little more to be said.

WitCH 44: Estimated Worth

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October 7, 2020

2:27 am

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This WitCH is from Cambridge’s 2020 textbook, Mathematical Methods, Unit 1 & 2. It is the closing summary of Chapter 21A, Estimating the area under a graph. (It is followed by 21B, Finding the exact area: the definite integral.)

We’re somewhat reluctant about this one, since it’s not as bad as some other WitCHes. Indeed, it is a conscious attempt to do good; it just doesn’t succeed. It came up in a tutorial, and it was sufficiently irritating there that we felt we had no choice.

WitCH 36: Sub Standard

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April 21, 2020

8:00 pm

87 Comments on WitCH 36: Sub Standard

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This WitCH is a companion to our previous, MitPY post, and is a little different from most of our WitCHes. Typically in a WitCH the sin is unarguable, and it is only the egregiousness of the sin that is up for debate. In this case, however, there is room for disagreement, along with some blatant sinning. It comes, predictably, from Cambridge’s Specialist Mathematics 3 & 4 (2020).

WitCH 33: Below Average

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February 1, 2020

3:14 pm

24 Comments on WitCH 33: Below Average

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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, it’s 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.