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Re: HSC 2013 4U Marathon Well the really hard ones should probably not be. For example my triangular numbers question should be in this thread imo (or 3U one)

Re: HSC 2013 4U Marathon Very good, a different approach to what I had in mind for the second one however. The third one is straightforward when b=3a then there is a horizontal point of inflection at x=1. Can someone else post a question now? (preferably Complex and Polynomials, (or also...

Re: HSC 2013 4U Marathon $Given the polynomial $ S(x)=ax^3+bx^2 + bx+a $The polynomial has 3 real roots all of which have a multiplicity of 2 or less.$ $i) Factorise the polynomial as one linear and one quadratic factor$ $ii) Show that if the polynomial has 3 real roots then...

Re: HSC 2013 4U Marathon Alright haha. Ill post another one tomorrow (I will try to make it a Complex Numbers/Polynomials one)

Re: HSC 2013 4U Marathon Nice work. I will post another question if I can think of one (unless someone else can post one?)

Re: HSC 2013 4U Marathon Apologies, will fix it now

Re: HSC 2013 4U Marathon You are given: 0<a_0, a_1, a_2, a_3 .... ,a_n < 10 \ \ \ \ \ \ \ \ \ a_0, a_1, a_2, .... , a_n \in \mathbb{N} a_0+a_1+a_2+...+a_n=3p \ \ \ \ \ p \in \mathbb{N} $Prove by process of mathematical induction or otherwise$ 10^n...

Re: HSC 2013 4U Marathon Ah my second question was not built nicely. Sorry about that. (Was supposed to lead on from part i) It should of been. x^3-2x^2+5x+2012 Will think of another question soon.

Re: HSC 2013 4U Marathon Note: \sum_{k=1}^r k = \frac{r}{2}(r+1) (Arithmetic series) \sum_{r=1}^n \frac{r}{2}(r+1) \\ \\ = 0.5 \sum_{r=1}^n r^2 +r \\ \\ = 0.5 [\sum_{r=1}^n r^2 + \sum_{r=1}^n r] \\ \\ = 0.5[\frac{n}{6}(n+1)(2n+1)+\frac{n}{2}(n+1)] \\ \\ = 0.5(n/6 (n+1) (2n+1+3) =...

Re: HSC 2013 4U Marathon Well that was an excellent question. Solution I feel like my solution is unnecessarily long however.

Look at their answers, what do they imply? That their own x is x=5\sqrt{5}+5-5 y=4\sqrt{5} Lets multiply to check if it meets 200 xy=4\sqrt{5}(5\sqrt{5}) \neq 200 \therefore $Textbook is wrong$ (I did it and you are correct)

Re: HSC 2013 4U Marathon I havent done the question yet, but do you mean Im(z) >= -1 Im(z) <= 1 Because it is impossible to have complex inequalities.

0.9999...=0.9+0.09+0.009+... \\ \\ = 9 \cdot \frac{1}{10}+ 9 \cdot \frac{1}{100} + 9 \cdot \frac{1}{1000}+... \\ \\ = 9(\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+...) In the bracket we arrive at an infinite converging geometric series. 9(0.1+0.01+0.001+...)=9 \cdot \frac{0.1}{1-0.1} = 9 \cdot...

Of course it equals one 0.9999... is the same thing as: \lim_{x->1^-}x=1 (I am pretty sure)

1. \ \ \ y=(x-1)^2+2 \\ \pm \sqrt{y-2}=x-1 \\ x \geq 1 \\ \therefore x=1+\sqrt{y-2} \\ f^{-1}(x)=1+\sqrt{x-2} 2. Go by inverse functions logic. f^{-1}(f(a))=a Not sure what its asking here 3. To find h^{-1}(0) \rightarrow h(x)=0 0=x^3-3x \\ x=0, x=\sqrt{3}, x=-\sqrt{3}

http://community.boredofstudies.org/showthread.php?t=294872&p=6178572&viewfull=1#post6178572 t ; ) legal19 kekeke

Ah of course, very silly oversight of mine

Re: HSC 2013 4U Marathon Oh no not this again haha, I still only half understand this concept.

Why does: \lim_{x \to \infty} \frac{0}{x} =0

Re: HSC 2013 4U Marathon Heh, I believe you are over complicating the question a bit. In your original solution you made an observation that you could of just applied into the expression. Apply the given sum and you are done.