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2013 CSSA 4U Maths thoughts? This exam was really a disappointment, I thought CSSA could produce better papers than they did today. Hell, the very last question was ripped from a HSC paper.

good luck for CSSA 4U tomorrow everyone

Re: HSC 2013 4U Marathon Subtraction is essentially adding the negative version, so when we minus N, we are first multiplying by -1 then adding side by side. Because for even with subtracting logs, we get stuff like 2 > 1, 5 > 2, ln(2/5) > ln(1/2) which doesn't work. It seems that the only...

Re: HSC 2013 4U Marathon Book my teacher lent me: 'Mathematical Circles (Russian Experience)' by Dimitri Fomin, Sergey Genkin and Ilia Itenberg Its a problem solving book for training and teaching olympiad students. The inequalities chapter uses only uses elementary inequalities such as...

Re: HSC 2013 4U Marathon Yeah that's true :/ Also for anybody wanting to attempt that question, it is a correct result since I got it from a book, so the problem still stands if someone wants to attempt it.

Re: HSC 2013 4U Marathon Yeah I did it again and you are correct Ah yes reading the solution again it is incorrect, what I had in mind was (n- \frac{1}{2} \right )^n \geq (1-x_1)(1-x_2) \dots (1-x_n) ... (1) (n+1/2)^n \geq (1+x_1)(1+x_2) \dots (1+x_n) .... (2) This is done by subbing...

Anything to do with trigonometric functions that require you to calculate something not involving exact ratios i.e. The tidal SHM motion questions. In hindsight do retract my previous statement, whatever the question uses, use that in calculations, (i.e. if the question is using degrees then...

Re: HSC 2013 3U Marathon Thread $A particle is moving at a height of$ \ b \ $above the ground under conditions of gravity$ \ g $The particle is oscillating in simple harmonic motion about $ \ y= 0 \ $parallel to the ground with an amplitude of$ \ b \ $and a period of$ \ \frac{2\pi}{n} $At...

Re: HSC 2013 4U Marathon Yep proving f is increasing is enough, and differentiating with logarithms is the most direct method, but its a little long and you might need to differentiate twice, alternatively: \frac{f(n+1)}{f(n)} = \frac{(2n+1)^{2n+1}}{(2n-1)^n (2n+3)^{n+1}} = \frac{2n+1}{2n+3}...

We all know you are secretly omed dw ; )

Re: HSC 2013 4U Marathon $Prove for some integer$ \ n \binom{n}{0} - \frac{1}{3} \binom{n}{1} + \frac{1}{5} \binom{n}{2} - \dots + \frac{1}{2n+1} \binom{n}{n} = \frac{2^{2n}}{\binom{2n}{n}}

Re: HSC 2013 4U Marathon Solution is correct until the 'let n=1' argument. f(n) = \left(\frac{n - 1/2}{n+1/2} \right )^n which is what you got as the lower bound By subbing in n=1 we do get 1/3, so the next problem then is: $Prove$ \ \left( \frac{2n-1}{2n+1} \right )^n > \frac{1}{3} Or...

Forget about degrees mode Seriously never use it, get into the habit of radians. Especially for MX1 and MX2 you will rarely be using a calculator to compute trigonometric ratios because you should know the common exact ratios already Radians are ultimately superior they give you greater...

The centre of the ellipse is middle of the 2 focii, we know that the focii is (1,3) and (9,3), hence the centre is (5,3) First geometrically interpret the question, z is some number in the complex plane so that its distance from (1,3) (given by |z-1-3i|) plus the distance to (9,3) (given by...

Re: HSC 2013 4U Marathon Yep nice work =========== $You may use the following identity$ \frac{a_1 + a_2 + a_3 + \dots + a_n}{n} \geq \sqrt[n]{a_1 a_2 a_3 \dots a_n} $The sum of the positive real numbers$ \ x_1, x_2, x_3, \dots, x_n \ $is$ \ 1/2 $Prove that$ \frac{1-x_1}{1+x_1} \cdot...

Re: HSC 2013 4U Marathon Yep that is correct. ============ $The product of positive real numbers$ \ a_1, a_2, a_3, \dots , a_n \ $is equal to$ \ 1 $Prove that$ (1+a_1)(1+a_2)\dots (1+a_n) \geq 2^n

Re: HSC 2013 3U Marathon Thread A little different to what I had in mind, integrating x(1-x)^n from 0 to 1 But essentially the same thing ===== Here is question I made: $In this question take$ \phi = \frac{1+\sqrt{5}}{2} , \ \tau = \frac{1-\sqrt{5}}{2} $Consider the identity$...

Re: HSC 2013 3U Marathon Thread You will get on the denominators 2*3, 3*4, 4*5, .... , (n+1)(n+2) while for this one the denominators are only 2, 3, 4, .... , n+2

Re: HSC 2013 3U Marathon Thread Considering the identity: (1+x)^{n+m} = (1+x)^n (1+x)^m Then expanding both sides, equate co-efficient of x^k on both sides knowing that m>n>k , the result comes immediately. ======= $Prove$ \ \ \frac{1}{2}\binom{n}{0} - \frac{1}{3} \binom{n}{1} +...

Re: Australian Maths Competition 2013 Good luck to everyone doing the AMC tomorrow.