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March 22, 2026, 07:19:52 am

Author Topic: VCAA 2009 Exam 2  (Read 3093 times)  Share 

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Ilovemathsmeth

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Re: VCAA 2009 Exam 2
« Reply #15 on: October 16, 2010, 01:48:08 pm »
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That's pretty nasty of them, out of all the trial exams I did in 2008 I never came across this. How weird. I haven't seen what the graph looks like though. Which question is this?
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bomb

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Re: VCAA 2009 Exam 2
« Reply #16 on: October 16, 2010, 02:06:57 pm »
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So if it says "strictly", you don't incude TPs?
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jasoN-

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Re: VCAA 2009 Exam 2
« Reply #17 on: October 16, 2010, 02:51:22 pm »
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I honestly doubt they'd expect you to know that.
Most likely or would be accepted, don't stress
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thushan

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Re: VCAA 2009 Exam 2
« Reply #18 on: October 16, 2010, 05:48:51 pm »
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Hopefully, but VCAA is sadistic, so I'd play safe
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TrueTears

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Re: VCAA 2009 Exam 2
« Reply #19 on: October 16, 2010, 06:13:43 pm »
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lol definitions in VCAA exams, making it look like real analysis

http://en.wikipedia.org/wiki/Monotonic_function

so yes 9 IS included, from a mathematician point of view, if you didn't include 9 you'd be wrong (and no it's not because VCAA is pedantic or anything, it's coz including 9 is the correct answer! It is analogous to how mathematicians define 1+1 = 2 and if you had 1+1 = 3 then it is wrong (well not wrong in different modulo classes lol) not because we are pedantic but it is because you did not follow the fixed definitions)

It is convention in analysis that strictly decreasing is defined how it is as per the wiki article, thus we can differentiate between correct answers and incorrect answers. If this was not a conventionally defined term, eg, how you define "faces" in graph theory, then there would be some leeway in accepting different answers (as long as you state which definition you are using)
« Last Edit: October 16, 2010, 06:34:56 pm by TrueTears »
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kenhung123

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Re: VCAA 2009 Exam 2
« Reply #20 on: October 16, 2010, 06:42:59 pm »
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Trial exams are emphasising this concept and I find it so confusing. When is TP included when increasing/decreasing? I don't understand the definition a<b and f(b)<f(a) :S