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HSC 2015 MX2 Marathon (archive) (1 Viewer)

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braintic

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Re: HSC 2015 4U Marathon

NEXT QUESTION

A Geogebra sketch tells me that the locus defined by that equation is a pair of ellipses centred at the origin whose major axes are perpendicular to each other, and that the approximate answer is 8.47
I haven't done the algebra, so I don't know the exact answer.
 

Drsoccerball

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Re: HSC 2015 4U Marathon

A Geogebra sketch tells me that the locus defined by that equation is a pair of ellipses centred at the origin whose major axes are perpendicular to each other, and that the approximate answer is 8.47
I haven't done the algebra, so I don't know the exact answer.
Well that's great didn't even learn it yet :/
 

braintic

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Re: HSC 2015 4U Marathon

Well that's great didn't even learn it yet :/
That Geogebra sketch was just for me to get a visual on whether the answers offered so far are correct.
I still haven't attempted the question, so I can't rule out the possibility that there is another way to do the question without knowing that locus.
 

Ekman

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Re: HSC 2015 4U Marathon

A Geogebra sketch tells me that the locus defined by that equation is a pair of ellipses centred at the origin whose major axes are perpendicular to each other, and that the approximate answer is 8.47
I haven't done the algebra, so I don't know the exact answer.
Well your approximations are very accurate:

 

braintic

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Re: HSC 2015 4U Marathon

Well your approximations are very accurate:

That actually makes a lot of sense in terms of my diagram.

Your triangle inequality in line 2 will turn into an equality when z lies on the real axis (outside -2 and 2), and that is where the points of maximum modulus were located.
 

Drsoccerball

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Re: HSC 2015 4U Marathon

Can we please not let this thread die
 

Kaido

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Re: HSC 2015 4U Marathon

calm down boy, this thread wont die as long as im here
 

Kaido

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Re: HSC 2015 4U Marathon

cos4θ=Re ∑k=0->4 (4Ck)(cosθ)^k(isinθ)^(4−k)=8cos^4θ-8cos^2θ+1
let cosθ=x and then cos4θ=1/2
so, 8x^4-8x^2+1=1/2
16x^4-16x^2+1=0

solve cos4θ=1/2
therefore: +/- cospi/12, +/- cos5pi/12
 

Kaido

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Re: HSC 2015 4U Marathon

In the meantime, someone try solving this (just curious):
x^5+x=1, find real x
 

Drsoccerball

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Re: HSC 2015 4U Marathon

cos4θ=Re ∑k=0->4 (4Ck)(cosθ)^k(isinθ)^(4−k)=8cos^4θ-8cos^2θ+1
let cosθ=x and then cos4θ=1/2
so, 8x^4-8x^2+1=1/2
16x^4-16x^2+1=0

solve cos4θ=1/2
therefore: +/- cospi/12, +/- cos5pi/12
Makes sense the question says 3 though so in that case you'd make Thanks
 
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