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Yep thats what I did. Haha i shouldn't say yes, I have a thesis to complete within the next three weeks that I've been majorly procrastinating. Where are these from though? Are they easier olympiad questions? Or hard 4U questions?

Yes the proof of the Banach-Tarski Paradox relies crucially on AC. I know that you cannot construct non-Lebesgue measurable sets without AC, as for non-constructive proofs of their existence I am not too sure. In any case the vast majority of mathematicians accept AC since there are several very...

Okay, i misunderstood you then :). Where that 'paradox' comes from is the fact that it is impossible to define a measure satisfying all the desirable properties on the entire power set of R^3. My point is that there still exists a nice measure defined on the class of sets called Lebesgue...

What exactly do you mean by this? There certainly exists something called a measure on R^3, which satisfies all the reasonable properties you might expect of a "volume". Or are you referring to the fact that in the HSC course, you calculate "volumes" without them being properly defined?

As a test of solutions, when n=1, we are forced to have a_1=2. So F(2)=2/5. I think your expression comes out as 1/5 at 1. Your method sounds reasonable though, so im not really sure why somethings going wrong.

The solution is fairly similar, except two transformations were required for this one. Don't really have time to write down a complete solution but the end result is: \\F(n+1)=\frac{(n+1)!(n^2+3n+1)!+(2n+1)!(n+1)^2!}{(n+1)^2(n+1)!(n^2+3n+1)!}.\\ Some clever manipulations with binomial...

Q(k)=(k+1)P(k)-k=0$ for $ k=0,1,...,n\\ \Rightarrow Q(x)=K\prod_{k=0}^n (x-k) \quad $ for some constant $K$, since $ \deg Q=n+1\\ \Rightarrow 1=Q(-1)=K\prod_{k=0}^n (-1-k)=K\prod_{k=0}^n (1+k)=(n+1)!K\\$where the penultimate equality is from the fact that $n$ is odd. Hence $K=1/(n+1)!$. This...

$A standard definition of concavity is that a curve is concave up if the line segment joining any two points on the curve lies entirely above the curve. Hence the point:$ \left(\frac{a+b}{2},\frac{f(a)+f(b)}{2}\right)$ which bisects such a segment will lie above the corresponding point on the...

$By annuli: $V=\pi\int_0^1 e^{2x}-1\, dx.\\ $By cylindrical shells: $V=2\pi\int_1^e y(1-\log{y})\, dy.\\$Obviously the first integral is easier to evaluate,$\\V=\frac{\pi(e^2-3)}{2}.

Assume that the given equation holds and try to deduce that this forces the three points to be vertices of an equilateral triangle. It isn't easy, but I will wait for more people to attempt it before i give any big hints.

Thankyou comeeatmebro and good work arj, although you have only proven the 'if' statement, not the 'only if' statement. The latter is slightly harder.

$Prove that the complex numbers $p,q,r$ satisfy:\\$p^2+q^2+r^2=pq+pr+qr$\\ if and only if they correspond to the vertices of an equilateral triangle in the complex plane.$

yes

$You won't be finding any formulae, since functions of that form do not necessarily have elementary primitives. For example, $f(x):=a\sqrt{e^{x^2}+1} $ does not. When questions of this form are asked in the HSC, you either have a factor of $f'(x)$ in the numerator making it a trivial...

If p/q is a rational root of a polynomial with integer coefficients, then p must divide the constant term and q must divide the leading coefficient. For the monic case this reduces to the integer root condition you state above. Also note that the 'p/q' here has to be in simplest form, i.e. p and...

$From division algorithm: $P(x)=Q(x)(x^2+1) + Ax+B\\ $Hence by evaluation: $ P(i)=Ai+B,\quad P(-i)=-Ai+B\\$Solving simultaneously for $A,B$ yields the desired result.\\ For odd $P$ we have that $P(i)+P(-i)=0$, and hence our remainder reduces to $R(x)=P(i)x/i$.\\ Try this question: Set...

$For Q6, is the limiting sum $S_{\infty}=\theta^2-\sin^2(\theta)$?$

The Cambridge book has a proof based on coordinate geometry and several leadup questions. There is a shorter proof based on the (equal sum of focal lengths) property of the ellipse, along with the triangle inequality. I wrote it up formally for a student once before but this is the outline: 1...

$Are you sure you got the correct solutions to the quartic? Judging by your use of the symmetric polynomials in the roots it seems you have assumed that $ \cos (\pi/8)$ and $\cos(5\pi/8)$ are double roots. This is not the case. The roots of the polynomial are $\cos(n\pi/8)$ for $n=1,3,5,7$. Now...

yes, because x=a/e is the equation of the right directrix.