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0 and 2 I dont understand why they give two marks there for me part (i) looks harder :P

before any simplification, you'll have two sin@'s, move one so u have sin@ on LHS and on RHS when sin@=0, the equation holds when sin@=0, @=0 well probably that explanation is a bit dodgy because if u make the two sin@'s on the LHS then you're dividing by zero.. the best explanation might...

looks like u need some idiotic brain to be able to work out this part :D

this is strange... for me part (iii) was really easy but obviously its not for you guys and the top student in my school couldnt do this part, whilst he did everything else in the whole paper but instead I couldnt figure out part (i) and skipped it @Rorix: what I wrote in the other...

no man my dad works for the Board and he hasn't told me about that So, u LIED! :p

same here :( before I saw ext2 HSC papers (my trial was easy :p)

Oh. I forgot that. But do you know why people say.. "its safe if u got a band 6" or .. "dont worry u r in band 6 or 5"? when band means nothing....

a friend in my school only couldn't do a 1-mark question :D which year??? which part? a or b?

what does getting a certain band give you? for example if I am in the very-very-bottom of band 6, will my mark scales much better than if I was in the very-very-top of band 5? or for extension courses, band E4 and E3. sorry if this is already discussed somewhere...

I was soo stupid!! Regarded v as a constant :(

easy question 8 but I didn't or couldn't do Q2 (d) (complex number), Q4 c (conics), Q5 (b) (ii) (probability, and I dont know if I get part (i) right), Q6 b (ii)-(vi), Q7 (a) (i)-(ii), Q 8 (a) (iii)-(v), Q8 (b) (i), (iii). Noo.. that adds up to 36 marks!!! :( :( in Q6 (b) (ii), is it meant...

whoops.. it was meant to be 1+1=3 haha I typed it really quickly :p

make sure u sleep well tonight I know from experience that this is vitally important otherwise ull end up making stupid mistakes (like 1+1=3 or transcription error) or simply couldnt figure out how to do a simple question

I am not!!! :( :( :( shud've done more practice

well then can u make a triangle with sides 1, 2, and 99999999999 ?? ;) :P or 0.0001, 0.00000000000000001, 9999999999999999999 |a+b| <= |a| + |b| and |a-b| >= ||a|-|b|| is the complete version of triangle inequality, a and b can be complex Cambridge proves that when a and b are real...

you cant just make a triangle with random sides Well I think the third side is equal to the sum of the other two sides when the triangle approaches a straight line :p we did this thing in complex number... But, in the inequality |a+b| <= |a| + |b| and |a-b| >= ||a|-|b|| , a and b real...

what about things like this: integral of sin^3x/cos^2x.dx you get "different" answers using different method... the normal one is sec(x) + cos(x) but if you write the integrand as sec(x)tan(x)*sin^2(x) and integrate by parts, you get tan(x)sin(x) + 2cos(x) of course this simplifies...

true but I dont think they can / are allowed to mark you down

hmm after seeing part (ii) I think you should use De Moivre's ok, so say you're meant to use De Moivre's in part (i) But I solve the quadratic like above, is it ok?

will we still get full marks for this? Say for example, HSC 1998, Q2 (e) (i) By solving the equation z^3 + 1 = 0, fnd the three cube roots of -1. If I do it by solving z^3 = -1, -1 = cis(pi), using De Moivre's theorem, will that be ok? I think you're meant to solve (z+1)(z^2 - z + 1) = 0