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Trebla, If you can get hold of a copy of the MANSW solutions book (lilac) prior to 2000, there are raw HSC stats in the back, which seems to confirm that in 1993 only 1 person scored 120/120. cheers

probably on his own... see MANSW Student Services

Hey Trebla, I don't think solving solving y=-2x, y= 4-x and y=2x, y=4-x to get the points of intersection is exactly rocket science <G>

a useful approach for quests like (ii) is to graph y=2|x| and y= |x-2|+ 2 on the same diagram and solve 2|x| > |x-2|+2

ur right about the 93 paper.. it had, at 46.7 the lowest raw mean in that era, but the was the only time a perfect score was achieved; 120/120.... the student was Anthony Henderson, Sydney Grammar.... google the name to see what he's achieved since

vol= Int (0,2)[ 2(pi)xy] dx= 2(pi) Int (0,2)[x(4x^2-x^4)dx= 32(pi)/3

I(sqrt(x^2+1)dx=x.sqrt(x^2+1)-I(x^2/sqrt(x^2+1)dx =x.sqrt(x^2+1)-I((x^2+1-1)/sqrt(x^2+1) dx =x.sqrt(x^2+1)-I(sqrt(x^2+1)dx+I(1/sqrt(x^2+1) dx so I(sqrt(x^2+1)dx=1/2(x.sqrt(x^2+1) +1/2. ln(x+sqrt(x^2+1)) The orig quest then...

From Int 2(1-u^2)/(1+u^2)(1-u^2+2u), by partial fractions u have Int[(u-1)/(u^2-2u-1)-(u-1)/u^2+1) so have 0.5 ln(u^2-2u-1)-0.5 ln(u^2+1) + arctan(u) and then put back u= tan(x/2)

the answer is missing an x between the 1 and 3 .... for some reason its not printing right <x> <x></x></x>

1/((x-1)(x-3))+1<0 so (x-2)^2/((x-1)(x-3))<0 only soln for this is 1<x<3

u=exp(x)>>du = exp(x)dx I=Int[ 1/(u-1)-1/(u+1)]du = 1/2[ln(u-1)-ln(u+1)] etc...

24. V(y)= 0 when x=30 and t = VsinA/10 which gives 300 = V*V sinA cosA ......(1) Also subst y = 26.25 and t=VsinA/10 into equation y= -5t*t + VtsinA +15 have finally 225 = V*V sinA*sinA ..... (2) from (1) and (2) tan A = 225/300...

Nothing wrong with the equation... Its your integration skill that's lacking: x=int(v/(g-kv^2))dv etc...