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Nooblet: Thanks, I usually make silly errors when I try things like that. Carrot: I remember getting a good 4U student to prove that formula for regions of a certain type, it can be deduced from syllabus techniques. Will post a question on it shortly.

One more thing, i) rules out the possibility of the two real roots coinciding. (If the two real roots had the same sign then the product of roots would be positive). Hence the polynomial has exactly two distinct real roots.

96pi/5, may be wrong though...do not have paper at hand.

Whilst I am pretty sure Carrot's reasoning works, I think it is actually a little tedious to prove that the suspicion is correct. The most anal of HSC markers may object to such a proof by pictures as it is indeed quite a jump from the related (and commonly used in 4U) argument: f(a)<0, f(b)>0...

Seems silly to require it to be done by induction when de moivre's is a course theorem...

Dude it's just a number. No where near as important to life/success as people make it out to be. Most people don't even remember what they got a few years after high school. Regarding ATAR cutoffs, there are a million different ways of getting into the course you want besides going straight...

Re: 2012 HSC MX2 Marathon When thinking about whether shells or annuli will be more efficient, consider the resulting integrands. There will generally be a sort of inverse relation between our integrands (as we are integrating in perpendicular directions for these two methods), and we prefer to...

A hint for the first two number theory questions I posted on the previous page. Consider the prime factors of squares and try to use properties of prime numbers.

Yep. Alas Green's theorem is actually not so easy to prove properly... I have been studying vector calculus/calculus on manifolds again for the last couple of weeks, because I realised how little I actually understood the big theorems in undergrad (beyond being able to blindly apply them in...

Isn't the isoperimetric inequality a problem of essentially equal difficulty? By assuming it it seems we are circumventing the main bulk of the problem...Also, showing that a circle in particular attains the bound that the isoperimetric inequality asserts does not guarantee that ANY such maximal...

If you change the ab=ac=bc=27 to ab+ac+bc=27 you get a problem that makes sense and is considerably more difficult that the one obtained by replacing 27 with 25. If we let P(x) be the monic polynomial with roots at a,b,c, then P(x)=x^3-15x^2+27x+c for some real c. The problem amounts to finding...

Then trivially, a=b=c=5, abc=125.

If the equality signs in the second condition are changed to addition signs, then the question is more interesting. The answer is: -13 =< abc =< 243 using the cubic discriminant.

Question is vacuous, no triple of real numbers (a, b, c) can satisfy the two given conditions.

And a quick sketch of my solution to the question I posted earlier, I might tidy it up a bit later.

\sum_{n=1}^N T_n^{-1}=2\sum_{n=1}^N \frac{1}{n(n+1)}=2\sum_{n=1}^N \frac{1}{n}-\frac{1}{n+1}=2-\frac{2}{N+1}\rightarrow 2\\ \\$as $N\rightarrow \infty New number theory questions, ordered roughly by increasing difficulty: 1. Prove that the product of 5 consecutive positive integers cannot be...

T_n+T_{n-1}=n^2\\ \Rightarrow T_n=(-1)^nT_0+\sum_{k=0}^{n-1} (-1)^k(T_{n-k}+T_{n-k-1})=\sum_{k=0}^{n-1}(-1)^k(n-k)^2

Proving that z>2 is trivial, but proving z<4 is not...Assume solutions with z>3 exists and look for a contradiction. I will post a solution tomorrow.

Hint: There are no solutions if z>3. Once you have proven this, the problem is in two variables.

Good start :). This question is reasonably difficult.