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Re: HSC 2015 4U Marathon I don't think you've thought this question through.

The speed got better later yesterday, but it seems to have reverted to snail pace again.

If you have your heart set on doing Ext 2 maths (and assuming your school offers it) I would stay. Selective schools typically limit the number of students who can do Ext 2. And many newcomers to a selective school takes a few months to raise their standard to the old-timers in the school, by...

As long as you end up ranking first, you are in full control over whether you get a state rank based on your HSC exam performance, regardless of the marks that saw you come first.

Re: MX2 2015 Integration Marathon Just wondering what is the point of defining a principal square root in the first place? Why should one take precedence over the other, especially when the choice seems completely arbitrary?

Re: HSC 2015 4U Marathon = cis pi . cis pi/2 . cis pi/4 . cis pi/8 . ... = cis (pi + pi/2 + pi/4 + pi/8 + ...) = cis pi/(1-1/2) .... [limiting sum of GP] = cis 2pi = 1

Re: MX2 2015 Integration Marathon Defined by whom?

Re: MX2 2015 Integration Marathon The issue at the moment is not with the answer (we had it), but with how to interpret the answer for x = 0 and for -1/4 < x < 0.

Re: MX2 2015 Integration Marathon Clearly there is something happening there in the limit, because there is no way the answer can't be zero. But fiik what that something is. Perhaps one of the the uni maths freaks can suggest what is happening. Also, it gives a value for values between -1/4...

Re: MX2 2015 Integration Marathon Let the Integrand be f(x). Squaring both sides: f²(x) = x + f(x) f²(x) - f(x) - x = 0 Using the quadratic formula: f(x) = [1+√(1+4x)]/2 (ignoring the -ve since the integrand is clearly +ve) Integrating: x/2 + [ (1+4x)^(3/2) ] / 12

Re: MX2 2015 Integration Marathon Is what necessary?

Re: MX2 2015 Integration Marathon A little easy compared to the standard set here :)

Is it just me, or is this site EXTREMELY slow today?

Re: HSC 2015 4U Marathon Lower boundary is correct, but not the upper boundary. You should find the cartesian equation of the locus defined by that condition, sketch it, then use 2 unit work to finish it off.

Re: HSC 2015 4U Marathon Now that the other question is sorted .... BUMP

Re: Does the Board of Studies actually use the assessment mark submitted by the schoo It's not a silly question if your teachers have fed you that crap. The rank means NOTHING. How your assessment fits in the mark distribution of your cohort, and how your cohort perform in the actual HSC...

Re: HSC 2015 4U Marathon Actually he's done neither ... he met in the middle. But every step is reversible, so all required logic is there.

So what will you do now?

Well ... at least you're not heading down a dead-end path.

Re: HSC 2015 4U Marathon New Question: Find the range of values of arg z for complex numbers z satisfying: | z - (1+i) | = Re [ z - (-1+i) ] (The answer involves simple values, so working is necessary)