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uhh no that doesn't work but come to think of it you are right that there should be more than just the circle. it works for the entire real axis except for -1 <= x <= 0

i did the 2003 paper on thursday so when i saw it in BOS trials it was a good moment

11.e) was actually the exact opposite of question 2.e) in HSC 2003. http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2003exams/pdf_doc/mathematics_ext2_03.pdf

this is essentially what i was expecting to hear. cheers!

yes... i said |z| = 1

i constructed a quadrilateral using the points z, z+1, 1, and 0. and since it's a parallelogram, where the diagonal bisects the base angle, it must be a rhombus -> |z| = 1 obviously with exception z = -1 because then it's not a quadrilateral

aligned mark 99 is state rank

got slapped as in got 1/8 (it wasn't hard either) i go to penrith

got slapped by inequals in HY and blanked during assessment 3, forgot how to prove |PS - PS'| = 2a on a hyperbola, which cost me 3 marks

well i didn't really try early on so it's my fault

depends whether the question says "Hence"

this is my problem. i'm averaging 94-95% on past papers but i'm rank 4 internally

what type of raw marks do MX2 state rankers actually get? i want something concrete here like if someone knows a state ranker who paid to get their paper back

OH RIGHT uhm i answered it but i got it wrong

i didn't quite get the absolute last part, but i thought the question was very neat. do you mean the greatest term is m+1/n question? i did that one.

i realised, because it didn't say that O lies on AM, but i didn't get it

yes i did both! for the binomial one i just proved LHS = 2^n - 1

q11 is probably harder than hsc q16

(n^{2}+an)(n^{2}+an+2) + 1 \\ = (n^{2}+an)^{2} + 2(n^{2}+an) + 1 \\ = (n^{2}+an + 1)^{2} \\ \text{Therefore, } 2011*2012*2013*2014 + 1 = (2011^{2} + 3*2011 + 1)^{2}

3u was much much much harder and 4u felt very similar