Search results

i should say in Q2 the first part was use substitution t = tan x/2to evaluate (same integrals and values) 1/(2 + cosx) dx and i thinkkk that equaled 0 or smething im not sure though i forgot....

1. mmmm i prob drew my graph wrong ... silly... 2. but with T results why r u gonna do wiht the u?????/ 3. indeed. 4. its catholic trial... 2003 fucking arnolds

Question 1 Find the values of K such that the equation x/(1-x²) = kx has three distinct real roots normally these sort of questions r pretty easy except i cant find graphically any values of K that give 3 distinct roots.. I can find values of K with 2 distinct roots. answered...

OK i did all that until this step... but i would have never thought of that step... I dont think this quesiton will ever be in the hsc without us first getting to do this step in like i.

i prefer locus/parabola/conics from like application of calculus... like 80% of the locus questions are the same... with enough practice it should become automatic.

hey, I have a question but before i post it.. I really find the way u guys solve integration in this forum weird, like i can solve almost all of the questions posted but never the way u guys do. Also its so hard to understand them with the poor formatting so please do your best to format the...

can u use mod in maths? i thought it was just used in programming.. :s

question 6 i think... its weird cause Q6 had a question like this.. and Q8 had 5 mark 2u differentiation and liek 10 mark easy inverse function question.. I dont think they really went with Question numbers. yea i thought of what u did...omg i just realized what i was doing wrong... i...

yea thats definatly in the syllabus there is a whole chapter on it aswell.. last chapter in integration in cambridge.

this was in 2003 trail hsc from our school... i dont really know where to start... if a 3 digit number contains the letter a for its hunderds and letter b for its tens and letter c for its units and a + b + c is divisible by 3 .. then prove that the 3 digit number is also divisible by 3...

i think u forgot to read the last sentence in his post he was talking about ranks not marks

sorry Richiee but these are so bad...

hey if u go to the BUY and SELL page there r hunderds of people from last yr selling these. btw if i was u i would get the cambridge, and maybe fitzpatric then the rest past hsc's n past trials with worked solutions...

this is a very common question that a lot of people get stuck in... its bout the only one that u cant see instantly... just gotta memorize what to do i guess

this is a really famous question....

a + b = m^2 + n^2 (by definition) ab = mn (as above) 3m^2 + 3n^2 = 10mn <i>m = n/3 or m=3n</i> 3m^2 + 3n^2 - 10mn = 0 put in n/3 for m n^2/3 + 3n^2 - 10n^2/3 = 0 as required m = 3n = 0 as required therefore m=n/3 or 3n you can also do this by factorizing or using...

i like 4u better than 3u... maybe that would answer some questions

10.4*1.015= [] expand the [] 10.4*1.015 = 10.8160p + 10 - 10p 10.4*1.015 - 10 = p (10.8160 -10) p = (10.4*1.015 - 10)/0.8160 hope that helps

i know a whole maths class who has 20 people + that all failed.. highest mark was 46%

Thank you.