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Re: HSC 4U Integration Marathon 2017 61root(2)/840 right? What I did was break up the sin part, get three different integrals, one of which is easy to work out, then for the other two I created a recursion formula for the integral of sec(x)^n from 0 to pi/4, and added/subtracted the three...

Yeah, I'll do that, thanks.

Oh wait, so if I only got 3 band sixes (I got band sixes for both 3U and 4U this year) for my next HSC I wouldn't get an All-Rounders Award? That sucks.

Sleeping and doing two assignments

Well let the roots be cisx and cisy, because they're easy to write. So we know that cisx+cisy = -b/a and cisx*cisy=cis(x+y)=c/a. Squaring the first equation, b^2/a^2 = cis(x)^2 +2cis(x)*cis(y) + cis(y)^2 = cis(2x) + 2cis(x+y) + cis(2y). So as cis(x+y) = c/a: (b^2/a^2) - 2c/a = cis(2x) +...

Firstly, just consider the asymptotes of the hyperbola x^2/64-y^2/36=1. This would just be y = ±3x/4. Now, to find the asymptotes of (x-3)^2/64 - (y+1)^2/36 = 1, notice that it is just the hyperbola x^2/64-y^2/36=1 but translated 3 to the right and 1 down. So just replace x with x-3 and y with...

Thomas Hassall went from around 400th to 136th haha.

98 for both Extension 1 and Extension 2.

Re: HSC 4U Integration Marathon 2017 www.imgur.com/nGjEjX8 www.imgur.com/O3sZNpH EDIT: That final c should be swapped with a k, but I can't be bothered to edit it and upload another photo.

Yeah sure. So all we need in this question is the argument. arg((z3-z2)/z3)=arg(z3-z2)-arg(z3), or arg(BC)-arg(OC). Now, as vector BC is perpendicular to vector OC, this means that if we translate them both to the origin, there will be an angle of ± pi/2 between them. In other words...

For i), note that z3-z2 is vector BC, and z3 is vector OC. As the radius is perpendicular to the tangent, angle OCB will be pi/2. Thus, (z3-z2)/z3 is imaginary, and similarly so is (z2-z1)/z2. ii) This is just tangents from a point have equal lengths. iii) Here, I would prove that OB and AC are...

Re: HSC 2017 4U Marathon I always converted it to the principle range, technically you'd be right, but it's better to be safe.

And if anyone wants to know why that works, just use the cosine formula and note that the cosine of any angle is less than or equal to 1.

Hey Red_of_Head, did you see this?