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yep. 84 - any silly mistakes

diagram.....

possibly trapezoidal rule but im not sure how you could get rid of the triangle at the top of each

yeah my answer agrees with yours (though i didnt use integration) that does not add to that you have forgotton all the internal terms (a+b+c)(x+y+z) =/= ax + by + ca it = ax + ay + az + bx + by + bz + cx + cy + cz

simple solution. all 4 uniters smash the 3 unit and we all get scaled to heaven :D

na he means 93 (average mark 47ish). ran out of time :( hopefully scrape >100/120 but i dunno. not so much hard just time-consuming ie. i didnt see the elegant ways to do things

a good nights sleep is the best study :)

4(250) + 3 =1003 > 1000 that should help

agreed, just post em up in the board. new posts every minute anyway :)

looking at the suggested answers to Q3)a)i) from 2003 HSC 1/undefined remains undefined. However this could be because they do not explicitly give you the equation of the graph.

yeah it probably should :p ta for picking that up

maybe, but would you have to find 1st, 2nd derivitives, intercepts etc? as well as controlling n and x :s

as 1 + X > 1 1/(1+X)n < 1 hence (1 + X)-n < 1 instead 1/(1-nX) - (1+X)n = [ 1 - (1+X)n + nX(1+X)n] / ( 1-nX) * common denominator using the first inequality (1+X)n > 1 + nX it becomes > [1 - (1+nX) + nX(1 + nX)] / (1-nX) > nX/1-nX > 0 as 1-nX > 0 and nX >0 hence...

looking at all of your 15 posts you use expletives in at least 10 and are abusive at other members. One wonders why you bother posting in these forums.

haha thats alright, i see your doing your HSC next year so its okay :) a few lines before my first to explain it a bit better P(x)/(x2+1) = Q(x) + (x - 2)/(x2+1) hence P(x) = (x2+1)Q(x) + (x - 2)

P(x) = (x2-1)Q(x) + x - 2 where Q(x) is some other polynomial P(-1) = 0 - 1 - 2 &there4; remainder = -3

woops should just be x = 0, &pi;/2, 2&pi;

hmm RHS = sin2x +2sinxcosx + cos2x = (sinx + cosx)2 &there4; (sinx + cosx)2 - (sinx + cosx) = 0 (sinx + cosx)(sinx + cosx - 1) = 0 sinx + cosx = 0 x = 3&pi;/4 7&pi;/4 or sinx + cosx = 1 x = 0, &pi;/2, &pi;, 3&pi;/2, 2&pi; late night solution so ive prolly done...

i have the same integral. yet get my answer :S edit: goodo i can still add this close the hsc :P

woops forgot to square. my revised answer = (27a3&pi; )/20