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My sentiments exactly. I'm not going to debate the fact (once again) that for HSC Maths, one does not require tertiary level mathematics. That's why it is HSC, not Math3031.

No, I believe the answer is: P(x)=-2x^3-3x^2+12x+5. But i could be wrong.

My ATAR was 96.3. only need around ~93 with bonus pts for my course as its cut-off will probably be 98 next yr (this year it was 97.2).

I wouldn't risk losing 2 marks...the markers are testing your knowledge on IBP not how to use Euler's formula. You'll probably get 0 marks for that in the HSC unless you rigorously prove Euler's formula.

It's Shaon0, i've gotten something like Affinity but with x,y=/=0. It's obviously a hyperbola not a line, unless i'm making a major mistake in my working.

2[sec(2x)]^2/tan(2x) =2*cos(2x)/[sin(2x)(cos(2x))^2] =2/[sin(2x)cos(2x)] =2csc(2x)sec(2x)

Let z=x+iy: arg((x-1)+iy)+arg((x+1)+iy)=pi/3 atan(y/(x-1))+atan(y/(x+1))=pi/3 tan(pi/3)=(y/(x-1)+y/(x+1))/(1-y^2/(x^2-1)) sqrt(3)=2xy/(x^2-1-y^2) Affinity has already graphed it.

1981? It may have been but 2010 onwards, they'll never ask it but either way it's very easy to derive. The proofs easier using polar coordinates and a lot quicker.

Nice, i was wondering when a 10er would answer this question. btw, it'll never be tested ie Non-uniform circular motion won't ever be tested.

I= S sqrt(x)e^(xsqrt(x)) dx = S sqrt(x)e^(x^(3/2)) dx Let u=x^(3/2):du=3/2 x^(1/2) dx I= 2/3 S e^u du =2/3 e^u +C =2/3 e^(xsqrt(x)) +C

Yeah

Do you know why its plus instead of +-?

Yeah, you made a mistake. 3rd line, theres no way i'm wrong.

Let x=y: x=4y+1/y xy=4y^2+1 4y^2-xy+1=0 y=x+-sqrt(x^2-16)/8 [Using quadratic formula]

I=S e^(1/x)/x^2 dx =-e^(1/x)+C

Yeah, wats everyone talking about? You're correct.

Answer is 3/5!!

They won't ask you a general case in HSC. They'll ask you to maybe solve an easy case using rational root theorem, factor theorem and later maybe conjugate root theorem. OR they might give a symmetric quartic poly to be solved ie where A=E,B=D eg: x^4+4x^3+6x^2+4x+1.