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April 28, 2025, 05:13:41 pm
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Author Topic: Heffernan 2008 trial exam 1  (Read 6975 times)  Share 

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Ken

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Heffernan 2008 trial exam 1
« on: October 31, 2008, 09:23:35 pm »
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does anyone think that Q4 is a bit crazy?
« Last Edit: October 31, 2008, 09:43:38 pm by Ken »
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ed_saifa

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Re: Heffernan 2008 trial exam
« Reply #1 on: October 31, 2008, 09:28:45 pm »
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which exam?
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Ken

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Re: Heffernan 2008 trial exam
« Reply #2 on: October 31, 2008, 09:43:06 pm »
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oops forgot to put it in, it's exam 1
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ed_saifa

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Re: Heffernan 2008 trial exam
« Reply #3 on: October 31, 2008, 09:44:09 pm »
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It could have been worse. I'm just glad they didn't put any complex terms into it.
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"It's not a community effort"
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Ken

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Re: Heffernan 2008 trial exam 1
« Reply #4 on: October 31, 2008, 09:46:13 pm »
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so ur saying that it's possible for VCAA to put in such a question?
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ed_saifa

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Re: Heffernan 2008 trial exam 1
« Reply #5 on: October 31, 2008, 09:47:22 pm »
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Yeah, they've done it before
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"It's not a community effort"
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"How can you expect to do well when
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-your lemons come with Cu2+ ions built in" - Dwyer
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Ken

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Re: Heffernan 2008 trial exam 1
« Reply #6 on: October 31, 2008, 10:00:08 pm »
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i dont understand the solutions properly, can someone pls explain it to me? thanks
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ed_saifa

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Re: Heffernan 2008 trial exam 1
« Reply #7 on: October 31, 2008, 10:08:06 pm »
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http://en.wikipedia.org/wiki/Complex_conjugate_root_theorem
This explain the conjugate root theorem which makes this question a lot easier.
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"How can you expect to do well when
-you draw a lemon as having two half-cells connected with a salt bridge
-your lemons come with Cu2+ ions built in" - Dwyer
"Why'd you score so bad?!" - Zotos
"Your arguments are seri

fredrick

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Re: Heffernan 2008 trial exam 1
« Reply #8 on: October 31, 2008, 10:39:07 pm »
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Basically any polynomial with real coefficents has its complex roots occur in conjugate pairs.(ie if is a soln then so is )

Q)If one of the roots of is . Find all other roots.

--roots theorm. Notice the common term (z-1), let it equal a if you cant see it.



So is one quadratic factor.

Now

Equate coefficents:


and


So
Now just quadratic equation or CTS and you get

there the right answers i hope
« Last Edit: October 31, 2008, 11:33:58 pm by fredrick »
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shinny

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Re: Heffernan 2008 trial exam 1
« Reply #9 on: October 31, 2008, 11:11:41 pm »
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Yeh this is a very likely style of question to be given by VCAA so make sure you know how to do it. It seems like a ridiculous question until you know the conjugate root theorem which makes things quite simple.
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AppleXY

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Re: Heffernan 2008 trial exam 1
« Reply #10 on: October 31, 2008, 11:24:31 pm »
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That's easy. Just long divide the polynomial lol

only takes a few minutes lool
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fredrick

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Re: Heffernan 2008 trial exam 1
« Reply #11 on: October 31, 2008, 11:31:29 pm »
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oh rite better fix that *slaps head*
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Ken

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Re: Heffernan 2008 trial exam 1
« Reply #12 on: October 31, 2008, 11:54:32 pm »
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i didn't understand the part where u equate the coefficients
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shinny

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Re: Heffernan 2008 trial exam 1
« Reply #13 on: October 31, 2008, 11:59:58 pm »
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Just use long division if you're not familiar with that technique.
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Re: Heffernan 2008 trial exam 1
« Reply #14 on: November 01, 2008, 12:27:23 pm »
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How about question 2.
It asks for the value of D if the box is 'not at the point of moving across the floor'. Should the answer be R \ {0.5g}?
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