HSC 2012 MX1 Marathon #2 (archive) (1 Viewer)

Nooblet94

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Re: HSC 2012 Marathon :)

Two of the roots of the equation x^3 + ax^2 + b = 0 are reciprocals of each other (a and b are real).
i) show that the third root is equal to -b
ii) show that a= b - 1/b
iii) show that the two roots, which are reciprocals, will be real if -0.5 < = b < = 1/2
Got it, nice question :)

I'll type up my solution after school tomorrow if nobody else has by then.
 

Sanjeet

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Re: HSC 2012 Marathon :)



The last part is probably wrong, but that's all I could think of lol (4 unit concept?)
 

Sidekickk

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Re: HSC 2012 Marathon :)

^ If i remember correctly, you take discriminant for part iii)
 

Sanjeet

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Re: HSC 2012 Marathon :)

How can you take the discriminant for a polynomial of degree 3?
 
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Re: HSC 2012 Marathon :)

You can't take the reciprocal of

.

Imagine the curve , over the domain the curve goes off to so the domain is undefined. (Part of Harder X1 Inequalities) That being said I'm not sure how to approach the question.
 

Nooblet94

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Re: HSC 2012 Marathon :)

For the last part:

<a href="http://www.codecogs.com/eqnedit.php?latex=\alpha @plus;\frac{1}{\alpha}=\frac{1}{b}\\ \alpha ^2@plus;1=\frac{\alpha}{b}\\ \alpha ^2-\frac{\alpha}{b}@plus;1=0\\ $For $\alpha$ to be real, $\Delta \geq 0\\ \left (-\frac{1}{b} \right )^2-4\cdot1\cdot1\geq 0\\ \frac{1}{b^2}\geq 4\\ b^2\leq \frac{1}{4}\\ -\frac{1}{2}\leq b\leq \frac{1}{2}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\alpha +\frac{1}{\alpha}=\frac{1}{b}\\ \alpha ^2+1=\frac{\alpha}{b}\\ \alpha ^2-\frac{\alpha}{b}+1=0\\ $For $\alpha$ to be real, $\Delta \geq 0\\ \left (-\frac{1}{b} \right )^2-4\cdot1\cdot1\geq 0\\ \frac{1}{b^2}\geq 4\\ b^2\leq \frac{1}{4}\\ -\frac{1}{2}\leq b\leq \frac{1}{2}" title="\alpha +\frac{1}{\alpha}=\frac{1}{b}\\ \alpha ^2+1=\frac{\alpha}{b}\\ \alpha ^2-\frac{\alpha}{b}+1=0\\ $For $\alpha$ to be real, $\Delta \geq 0\\ \left (-\frac{1}{b} \right )^2-4\cdot1\cdot1\geq 0\\ \frac{1}{b^2}\geq 4\\ b^2\leq \frac{1}{4}\\ -\frac{1}{2}\leq b\leq \frac{1}{2}" /></a>
 

barbernator

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Re: HSC 2012 Marathon :)

For the last part:

<a href="http://www.codecogs.com/eqnedit.php?latex=\alpha @plus;\frac{1}{\alpha}=\frac{1}{b}\\ \alpha ^2@plus;1=\frac{\alpha}{b}\\ \alpha ^2-\frac{\alpha}{b}@plus;1=0\\ $For $\alpha$ to be real, $\Delta \geq 0\\ \left (-\frac{1}{b} \right )^2-4\cdot1\cdot1\geq 0\\ \frac{1}{b^2}\geq 4\\ b^2\leq \frac{1}{4}\\ -\frac{1}{2}\leq b\leq \frac{1}{2}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\alpha +\frac{1}{\alpha}=\frac{1}{b}\\ \alpha ^2+1=\frac{\alpha}{b}\\ \alpha ^2-\frac{\alpha}{b}+1=0\\ $For $\alpha$ to be real, $\Delta \geq 0\\ \left (-\frac{1}{b} \right )^2-4\cdot1\cdot1\geq 0\\ \frac{1}{b^2}\geq 4\\ b^2\leq \frac{1}{4}\\ -\frac{1}{2}\leq b\leq \frac{1}{2}" title="\alpha +\frac{1}{\alpha}=\frac{1}{b}\\ \alpha ^2+1=\frac{\alpha}{b}\\ \alpha ^2-\frac{\alpha}{b}+1=0\\ $For $\alpha$ to be real, $\Delta \geq 0\\ \left (-\frac{1}{b} \right )^2-4\cdot1\cdot1\geq 0\\ \frac{1}{b^2}\geq 4\\ b^2\leq \frac{1}{4}\\ -\frac{1}{2}\leq b\leq \frac{1}{2}" /></a>
dats it. would rep but must spread first.
 

Carrotsticks

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Re: HSC 2012 Marathon :)

Consider the curve y=|x|.

A ball of radius 1 is 'dropped' into the V part of it.

Find the area bounded by the ball and the curve.
 

Carrotsticks

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Re: HSC 2012 Marathon :)

Generalise the problem to y=|kx| for some real number k.
 

barbernator

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Re: HSC 2012 Marathon :)

|k|-arctan(|k|)

celerysticks, does this cover all bases?
 
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Nooblet94

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Re: HSC 2012 Marathon :)

Generalise the problem to y=|kx| for some real number k.


1/2 shouldn't be there.. hmm...

Edit 2: Worked it out, I had simplified it to half of the area to make the working out easier and forgot to multiply by 2 again :p
 
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Nooblet94

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Re: HSC 2012 Marathon :)

That was a nice question Carrot - looks hard, but when you actually work out HOW to do it, the working out itself is easy.

Next question, do it in 3d (i.e. a sphere being dropped into a cone)
 

Carrotsticks

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Re: HSC 2012 Marathon :)

That was a nice question Carrot - looks hard, but when you actually work out HOW to do it, the working out itself is easy.

Next question, do it in 3d (i.e. a sphere being dropped into a cone)
Might have to be a bit more specific about the equation of the cone. Best to use spherical coordinates but then that would make it out of syllabus.

So the best way to properly define the question is probably to give us the cross sectional angle of the cone.
 

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