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HSC 2015 MX1 Marathon (archive) (2 Viewers)

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InteGrand

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Re: HSC 2015 3U Marathon

was this how it was done?
y-x >= 0
(y-x)^2 >= 0
y^2 + x^2 >= 2xy
(x^2 + y^2)/2x >= y
r >= y
First line is unnecessary, since the second line is true always, as $y$ and $x$ are real numbers. But otherwise, yes. (Basically AM-GM Inequality) :)
 

Crisium

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Re: HSC 2015 3U Marathon

Using the x = 4sin^2(O) substitution you should get

[ cos^2(O) - sin^2(O) ] / sin(O)cos(O)

It's in the form of f'(x) / f(x)

So it should result in log(sin(O)cos(O)) and sub in the limits of integration - But these limits have to change to account for the substitution
 

rand_althor

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Re: HSC 2015 3U Marathon

Using the x = 4sin^2(O) substitution you should get

[ cos^2(O) - sin^2(O) ] / sin(O)cos(O)

It's in the form of f'(x) / f(x)

So it should result in log(sin(O)cos(O)) and sub in the limits of integration - But these limits have to change to account for the substitution
You've made a mistake in the bold part. Perhaps you have incorrectly substituted?
 
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Zlatman

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Re: HSC 2015 3U Marathon

Find the least possible value of 4x^2 + 1/x when x is positive.












EDIT: a bit of extra info may be needed.

As x approaches 0, y approaches 1/x, which tends towards positive infinity as x approaches 0.
As x approaches positive infinity, y approaches 4x^2, which tends towards positive infinity as x approaches positive infinity.

Since there are no vertical asymptotes for positive x, there must be a local minimum point between the two places the curve approaches positive infinity.
 
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integral95

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Re: HSC 2015 3U Marathon

Using the symmetrical properties, find


 
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