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Re: HSC 2013 4U Marathon I see, well I changed the question so you know if you got it right or not.

Re: HSC 2013 4U Marathon Not quite. I am not sure how you arrived at your observation in the second line. EDIT: I will make it a 'show that' question so you can check the answer.

Re: HSC 2013 4U Marathon $Given: $ \sum_{i=1}^j i^2 = \frac{j}{6}(j+1)(2j+1) $Show that $ \sum_{r=1}^n (\left \sum_{k=1}^r k)\right = \frac{n}{6}(n+1)(n+2) \\ \\ $Hence prove that the product of any three consecutive integers is divisible by 6$

Re: HSC 2013 4U Marathon I would say: 'A circle with radius 1, centre (1,0) with an empty circle at the origin due to z=/=0' I would make a tiny sketch to show what I mean and that is all. The question is actually originally sketch the locus, but since I cant really ask that on the internet...

Re: HSC 2013 4U Marathon Yes it is the 'mindless algebra', think carefully about what z can be

Ah, I see there. Nice solution, thank you.

Thank You

What raw mark out of 70 in this year's Extension 1 Maths exam would people expect to get a 95 Examination mark?

Re: HSC 2013 4U Marathon I am hoping people have learnt Binomial Theorem from Extension 1: $Expand $ (1+i)^{4n} \\ \\ $Show that $ \binom{4n}{0}-\binom{4n}{2}+\binom{4n}{4}-\binom{4n}{6}+...+\binom{4n}{4n} = (-4)^n

Re: HSC 2013 4U Marathon Not quite

Re: HSC 2013 4U Marathon The stupid flaw with my idea is that I counted the averages of the classes as a whole one unit when calculating total average. So for example average of class 1 for females is 90 males 80, average for class 2 females 95 males 85 Then total average for females is...

Re: HSC 2013 4U Marathon $The roots of the equation $ z^n=r $ are plotted on the Argand diagram. The points are then connected by straight lines$ $i) What polygon is formed with the connection of all the roots?$ $ii) By considering the arguments of each of the complex numbers...

Re: HSC 2013 4U Marathon Ah yes this is true. Is this explanation good enough? All roots of z can be represented with: z_k=\cos \frac{2\pi}{k}+i\sin \frac{2\pi}{k} \ \ \ \ (=cis\frac{2\pi}{k}) Now when any of these values undertake z (EXCEPT the trivial root when k=0, which is covered...

Re: HSC 2013 4U Marathon Factorise: z^n-1=0 \\ \\ (z-1)(1+z+z^2+...+z^{n-1})=0 By null factor law: 1+z+z^2+...+z^(n-1)=0 But in order for z^n-1=0, z=1 as well. Hence the other solution is 1+1+1^2+...+1^(n-1)=n Hence two values that it can take, 0 and n

Re: HSC 2013 4U Marathon An easier one: \text{If } z $ is a complex number on the Argand diagram$ \\ \\ \text{Describe the locus } \frac{1}{z}+\frac{1}{\bar{z}}=1 (remember to post a question after you answer)

Re: HSC 2013 4U Marathon Alright I figured it out, some of them I arrived at constructively (not any of the deduce ones) And I will not get into the working out in depth, some of it is quite tedious Now: a) \ \ i) z_1, z_2, \bar{z_1}, \bar{z_2} \ \ \rightarrow $roots$ \therefore...

Re: HSC 2013 4U Marathon Call the complex number P: z Hence, since OP is perpendicular to OR (and half its size) OR:\frac{iz}{2} To find the midpoint S, we add and divide by 2: S:\frac{z}{4}(2+i) Define z=4+iy S=\frac{4+iy}{4}(2+i) \\ \\ S=\frac{8-y}{4}+i(\frac{y+2}{2}) Finding in...

This is a very good paper Demento, well done :) It has inspired me to create my own topic test for 4U I might add that people might be confused to see the notation m ^\circ n(5) in question 5

(\sqrt{x})^2=x \ \ \ x>0

Now lets establish the need that in order to for two people to be sociable, then they must be both social right? You say that most intelligent people you meet are non-sociable. I say most intelligent people I meet are sociable Hence, I can make the conclusion that since you do not find as many...