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you lost a factor of cos(x) spiral.

Yeah its not a polynomial equation in z...

I wouldn't risk using it in the HSC. The fact that exponentials dominate polynomials which dominate logarithms is given to you without proof. (In one of the Cambridge 3U books I think.)

The correct answer is: z=0,\textrm{cis}\left(\frac{k\pi}{3}\right)\quad{k=0,1,2,3,4,5}

The point (2,3) also satisfies the given conditions. So the correct solution is the circle of radius 1 about (-2,-3) and the point (2,3).

$Case I: $x\geq -2$.\\ $x+2=|x+2|>3x+7\Rightarrow -2\leq x < -5/2$ which has no solutions.\\ \\ Case II: $x<-2$.\\$-(x+2)=|x+2|>3x+7\Rightarrow x<-9/4\; \; \&\;\; x<-2\Rightarrow x<-9/4$.\\ So our overall solution is $x<-9/4.$ $

x < -9/4.

I used to study to metal through a sick pair of headphones. Find it too distracting these days though.

Re: 2012 HSC MX2 Marathon Don't know what you are expecting us to assume, as arbitrary real powers are not defined in the HSC. But we can make sense of the question using the integral definition of the logarithm, and the log laws/differentiation laws that follow. $Let $x_n:=(1+x/n)^n$. \\...

I strongly recommend Cambridge as it is written by an actual mathematician (Dr Pender). The exposition is clear and the questions are excellent.

Yep, a clever proof by contradiction would be the nicest way of doing it. Alternatively, an ugly coordinate geometry approach will work. Hint: Standard techniques involving congruence/similarity are surprisingly useless here...try playing with inequalities instead.

Prove that any triangle with two internal angle bisectors equal in length is isosceles.

(To prove the upper bound in d, bound the integrand above separately on the intervals [0,1/sqrt(2)] and [1/sqrt(2),1].)

Options a, b and d are all correct. And none is "more correct" than any other. Perhaps the question asked you to single out the incorrect statement?

Mine are like carrots but the horizontal segments are rarely equal in length or parallel unless I am making an effort to write something up to show other people / submit for something. When I am figuring stuff out for myself the working jumps all over the page and isn't particularly neat.

Agree with you obviously, but don't really see how the rearrangement of conditionally convergent series relates to this...

Hilberts 'paradox' doesn't really demonstrate the nonexistence of infinity as a real object, it simply provides an example of a counterintuitive mathematical property that infinite sets (as a formal mathematical object) can have: namely that an infinite set can have proper subsets of the same...

Why?. (Not that I disagree with you. I have stated my philosophical view.)

This is just a formal peculiarity of the way we write down numbers: The decimal representation of a real number is not necessarily unique. For similar reasons, 0.1111...=1 in binary. And regarding the existence of infinity 'in reality' one may pose the question: "do real numbers exist?"...

If the Gamma function is used in the question, surely its definition as an integral can be assumed? From there its one substitution and a couple of lines...