HSC 2012 MX1 Marathon #2 (archive) (8 Viewers)

Nooblet94 · May 20, 2012

Re: HSC 2012 Marathon

barbernator said:
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 · May 20, 2012

Re: HSC 2012 Marathon

$Let the roots be $\alpha , \frac{1}{\alpha }$ and $\beta \\ $i) Using product of roots,$\\ -b= \frac{\alpha }{\alpha }\cdot \beta \\ \therefore \beta = -b\\ $ii) Sum of roots two at a time,$ \\ 1+\frac{-b}{\alpha }-b\alpha = 0 \\ $dividing by b $\Rightarrow \frac{1}{b}-\frac{1}{\alpha }-\alpha = 0 \\ \therefore \frac{1}{b} = (\frac{1}{\alpha }+\alpha )\\ $Sum of roots,$ -a=\alpha + \frac{1}{\alpha } -b \\ $Multiplying by -1 $\Rightarrow a = b - (\alpha +\frac{1}{\alpha }) \\ $But $\alpha +\frac{1}{\alpha }=\frac{1}{b}\\ \therefore a=b-\frac{1}{b} \\ \\ iii) $Not quite sure, but I'll give it a go$ \\ -\frac{1}{2}\leq b\leq \frac{1}{2} \\ $Taking the reciprocal$ \Rightarrow -2\leq \frac{1}{b}\leq 2 \\ $But $\frac{1}{b}=\frac{1}{\alpha }+\alpha \\ \therefore -2\leq \frac{1}{\alpha }+\alpha \leq 2 \\ $Thus the two roots are real$

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

Sidekickk · May 20, 2012

Re: HSC 2012 Marathon

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

Sanjeet · May 20, 2012

Re: HSC 2012 Marathon

How can you take the discriminant for a polynomial of degree 3?

Carrotsticks · May 21, 2012

Re: HSC 2012 Marathon

Sanjeet said:
How can you take the discriminant for a polynomial of degree 3?

You can have discriminants for polynomials of degree greater than 2.

Sanjeet · May 21, 2012

Re: HSC 2012 Marathon

Is it in the 4 unit course?

asianese · May 21, 2012

Re: HSC 2012 Marathon

You can't take the reciprocal of

$-\frac{1}{2} \leq b \leq \frac{1}{2}$ .

Imagine the curve $y=\frac{1}{x}$ , over the domain $-\frac{1}{2} \leq x \leq \frac{1}{2}$ the curve goes off to $\pm \infty$ so the domain is undefined. (Part of Harder X1 Inequalities) That being said I'm not sure how to approach the question.

Nooblet94 · May 21, 2012

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 · May 21, 2012

Re: HSC 2012 Marathon

Nooblet94 said:
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.

barbernator · May 21, 2012

Re: HSC 2012 Marathon

sum1 else post some questions

Carrotsticks · May 21, 2012

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.

barbernator · May 21, 2012

Re: HSC 2012 Marathon

(1 - pi/4) units squared

Carrotsticks · May 21, 2012

Re: HSC 2012 Marathon

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

Sanjeet · May 21, 2012

Re: HSC 2012 Marathon

Carrotsticks said:
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.

Are we assuming the ball is a perfect circle?

Carrotsticks · May 21, 2012

Re: HSC 2012 Marathon

Sanjeet said:
Are we assuming the ball is a perfect circle?

Of course.

barbernator · May 21, 2012

Re: HSC 2012 Marathon

|k|-arctan(|k|)

celerysticks, does this cover all bases?

Nooblet94 · May 21, 2012

Re: HSC 2012 Marathon

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

$\frac{1}{2}(k-\tan ^{-1} k)$

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

barbernator · May 21, 2012

Re: HSC 2012 Marathon

Nooblet94 said:
$\frac{1}{2}(k-\tan ^{-1} k)$ x2

nice!

Nooblet94 · May 21, 2012

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 · May 21, 2012

Re: HSC 2012 Marathon

Nooblet94 said:
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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