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(Finalised) Official BOS Mathematics Trial Examinations. (1 Viewer)

umm what

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I have no idea where Parra Library is, could someone meet me at the station and we will go together?
 

seanieg89

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PM me the paper as it stands now so I can attempt it tomorrow p00nsticks :). (The MX2 one.)
 

seanieg89

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100% for you already mate
Haha it was more to check for typos / errors / potentially confusing wording in questions, have been providing feedback on a couple of drafts so I know most of what's in it.
 

chriss95

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So is the daily schedule still the same. ie 2 hours for MX1, 3 for MX2... then a few hours for marking where we do whatever, and then feedback/tips?
 

Carrotsticks

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So is the daily schedule still the same. ie 2 hours for MX1, 3 for MX2... then a few hours for marking where we do whatever, and then feedback/tips?
Not anymore, I figured it would be quite impractical because I will need to write a full set of marking guidelines etc, which will take far too long. Also, too difficult to find tutors who understand how the marking processes work in the HSC, they exist in very small numbers. I will do the marking myself and release a thread on the day of the exam. The more able students will be able to explain some of the things in the examination.

Whilst this is happening, I will be marking the papers and providing input into the thread if needed.

Perhaps afterwards, around the 6th or 8th October, I will organise to have another 'Feedback day' where you get all your marks back and I have a seminar going through the harder problems.
 

Carrotsticks

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I will be uploading some more scrapped questions soon.

Also, I will be uploading the front cover of your writing booklets and your test paper, which will include the winning cover design =)
 

deswa1

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Too straightforward, and didn't have enough space for it.

I don't get why arg (z+k) has to be greater than zero. Lets say z is 0.5-0.5i which has modulus 0.5 which is less than one which is what the restriction requires. Obviously though the arg of z is less than zero (being -pi/4 in this case). So when k is any real number, z+k will always remain below the real axis and hence the argument will always be negative.

Am I missing something simple and being retarded?

But yes, I've never seen a question like this before so I'm getting even more excited haha
 

Carrotsticks

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Ahhh I missed out on some (rather crucial) details.

1. Im(z) > 0

2. k > 1
 
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RealiseNothing

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I don't get why arg (z+k) has to be greater than zero. Lets say z is 0.5-0.5i which has modulus 0.5 which is less than one which is what the restriction requires. Obviously though the arg of z is less than zero (being -pi/4 in this case). So when k is any real number, z+k will always remain below the real axis and hence the argument will always be negative.

Am I missing something simple and being retarded?

But yes, I've never seen a question like this before so I'm getting even more excited haha
hehehehhe.
 

nightweaver066

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Is this question right? I ended up with unless i messed up somewhere..

Also with the complex number question, what if |z| > k? Wouldn't this mean the max arg would be pi? Not too sure about this one

Edit: Nvm inequality doesn't necessarily correspond to max, but just less than lol
 
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