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you forgot the square root :)

rofl, this is like trying to find a general method to solving quintic equations!

question 32, find lim_x->0 (1-cos2x)/x²

no no, my teacher told me to learn the proof for that one, nothing like newtons root law. T_K+1=nCk a^(n-K)b^k T_K=nC(k-1) a^(n-[K-1])b^(k-1) T_k+1/T_k={nCk a^(n-K)b^k}/{nC(k-1) a^(n-[K-1])b^(k-1)} =[n!b/(n-k)!k!]/[n!a/((n-k+1)!(k-1)!)]...

how hard can harder 2u be? loan repayments what else is there?

tell an ex student to take the exam for you

how much you guys think you need for E4? 70/84?

not really, since 2*7C2= 2*7!/5!*2!= 7x6

mine is binominal theorem, yours?

i don't like question 7b of 2000 paper, took up too much time to do

Q28 show that a particle with displacement x=atan(nt) is not moving in SHM

so with my 62/120, i can get 87ish??

well, i did not find show sin A=2t/(1+t²) straight forward! nor was show that y1=b³/(a²-b²), the e of ellipse is atleast 1/rt 2 straight forward!

you guys are inhuman! 85+ is crazy talk!

coz of no sleep, i stuffed up today! maybe i'm just using it as an excuse...

i found the test harder than expected. nothing like 2004! i think i got 62/120 give or take 5 marks. how about you?

Question 12 show that sinx cosx= 0.5sin2x hence find the exact value of ∫sin²x cos²x dx with boundaries pi/8 and 0

hehe, good job. i managed to close my eyes for 5 hours without falling asleep, world record :)

how did you know it was 4unit? :) 4unit isn't too bad, we get basic questions like square root minus twenty five as well ie. sq rt -25

rofl, this may be a bit late of a reply, but similar triangles. 2 angles common, so 3rd angle must be common, (both share a common angle and both have a 90 degree angle) that is, equalangular. take the proportion of the triangle ie d/200=20/x xd=4000 d=4000/x relationship, as x...