HSC 2012 MX2 Marathon (archive) (1 Viewer)

nightweaver066 · May 19, 2012

Re: 2012 HSC MX2 Marathon

I like this question.

$$The polynomial equation $ x^4 + 4x^3 - 2x^2 - 12x - 3 = 0 $ has roots $ \alpha, \beta, \gamma~and~\delta. $ Find the polynomial equation whose roots are $ \alpha + 1, \beta + 1, \gamma + 1, \delta +1$

$$Hence or otherwise, Solve $ x^4 + 4x^3 - 2x^2 - 12x - 3 = 0$

asianese · May 19, 2012

Re: 2012 HSC MX2 Marathon

$\text{Let}\, y=x+1\\ \therefore x=y-1.\\ \begin{align*}\therefore (y-1)^4+4(y-1)^3-2(y-1)^2-12(y-1)-3&=0\\ y^4-8y^2+4&=0\\(y^2-2y-2)(y^2+2y-2)&=0\\\\ \therefore (x+1)^2-2(x+1)-2&=0\\ x^2 &=3\\\therefore x&=\pm \sqrt{3}.\\\\(x+1)^2+2(x+1)-2&=0\\(x+2)^2&=3\\ \therefore x&=-2\pm \sqrt{3}\end{align*}$

I've spared the messy algebra...

The addition/product of roots would work also I think.

Inference · May 22, 2012

Re: 2012 HSC MX2 Marathon

Why don't we add some pure maths questions in here,

here's a fave one of mine, combinatorics

A permutation $\{x_1, \ldots, x_{2n}\}$ of the set $\{1,2, \ldots, 2n\}$ where $\mbox{n}$ is a positive integer, is said to have property $T$ if $|x_i - x_{i + 1}| = n$ for at least one $i \in \{1,2, \ldots, 2n - 1\}$ . Show that, for each $\mbox{n}$ , there are more permutations with property $T$ than without.

Hint: Indicator functions will help, and think about counting sets

Carrotsticks · May 22, 2012

Re: 2012 HSC MX2 Marathon

Inference said:
Why don't we add some pure maths questions in here,

here's a fave one of mine, combinatorics

A permutation $\{x_1, \ldots, x_{2n}\}$ of the set $\{1,2, \ldots, 2n\}$ where $\mbox{n}$ is a positive integer, is said to have property $T$ if $|x_i - x_{i + 1}| = n$ for at least one $i \in \{1,2, \ldots, 2n - 1\}$ . Show that, for each $\mbox{n}$ , there are more permutations with property $T$ than without.

Hint: Indicator functions will help, and think about counting sets

If you want some Pure Maths questions unrelated to the HSC, then feel free to make your own thread in the 'Extracurricular Maths' section: http://community.boredofstudies.org/forumdisplay.php?f=238

Inference · May 22, 2012

Re: 2012 HSC MX2 Marathon

Thanks, but it is related to the HSC albeit a fair extension.

You guys do combinatorics in the HSC right? The above question can be solved purely combinatorially.

barbernator · May 22, 2012

Re: 2012 HSC MX2 Marathon

Inference said:
Thanks, but it is related to the HSC albeit a fair extension.

You guys do combinatorics in the HSC right? The above question can be solved purely combinatorially.

we learn factorial notation for combs and perms and how to apply them, that is about all lol.

Inference · May 22, 2012

Re: 2012 HSC MX2 Marathon

Ah right I see, okay I guess it is a fair stretch then, maybe a bit of wishful thinking can lead somewhere

httton · May 27, 2012

Re: 2012 HSC MX2 Marathon

Hey guys just two questions

1/ what are some ways to determine whether a question should be done with shells or slices
2/ if I'm given a shape like x^2(6-x^2) find area enclosed in 1st quadrant rotated about the y axis how do I go about it?

bleakarcher · May 27, 2012

Re: 2012 HSC MX2 Marathon

^ cylindrical shells would be easier there

Carrotsticks · May 27, 2012

Re: 2012 HSC MX2 Marathon

httton said:
Hey guys just two questions

1/ what are some ways to determine whether a question should be done with shells or slices
2/ if I'm given a shape like x^2(6-x^2) find area enclosed in 1st quadrant rotated about the y axis how do I go about it?

Generally, rotation around a line parallel to the Y axis is more easily done using Shells.

X axis is more easily done using Slicing.

But of course both will yield correct answers (but one will yield a harder integral than the other).

Example: Rotate the area bounded by the curve y=-x(x-1) and the X axis, around the line x=2.

If you use Slicing, you will need to eventually use the Quadratic formula. and then integrate that expression.

httton · May 27, 2012

Re: 2012 HSC MX2 Marathon

haha i went out to have a good think and came back with the same idea!

for the others, essentially try to have your slice correspond to 1 x to 1 y values each

seanieg89 · May 28, 2012

Re: 2012 HSC MX2 Marathon

When thinking about whether shells or annuli will be more efficient, consider the resulting integrands. There will generally be a sort of inverse relation between our integrands (as we are integrating in perpendicular directions for these two methods), and we prefer to have things like tan(x) in our integrand rather than tan^{-1}(x).

Trebla · May 30, 2012

Re: 2012 HSC MX2 Marathon

This one might look like it is beyond the course but it can be solved using methods within the scope of Ext2 level.

Suppose that f(x) is an even function. Show that

$\displaystyle\int_{-a}^a f(x)(\cos x + i\sin x)\,dx = 2\displaystyle\int_{0}^a f(x)\cos x\,dx$

Hence evaluate

$\displaystyle\int_{0}^{\pi} \left|\dfrac{\pi}{2} - x\right|(\sin x + i\cos x)\,dx$

Fus Ro Dah · May 31, 2012

Re: 2012 HSC MX2 Marathon

Trebla said:
This one might look like it is beyond the course but it can be solved using methods within the scope of Ext2 level.

Suppose that f(x) is an even function. Show that

$\displaystyle\int_{-a}^a f(x)(\cos x + i\sin x)\,dx = 2\displaystyle\int_{0}^a f(x)\cos x\,dx$

Hence evaluate

$\displaystyle\int_{0}^{\pi} \left|\dfrac{\pi}{2} - x\right|(\sin x + i\cos x)\,dx$

First part split the integral as two terms then the integral with the i in it cancels out due to the sine function and we are left with 2x integral of f(x)cos(x). Second part use the property f(x) = f(a-x), add the two integrals, the icos(x) term cancels and the result follows.

I am a bit skeptical regarding this question being within Ext2 level without proper justification because we are taking the Riemann integral of a complex-valued function. It may require justification that it works in this case because $\mathbb{C}$ is a vector space. Other students might get confused about having an i in the integral.

OMGITzJustin · Jun 3, 2012

Re: 2012 HSC MX2 Marathon

Fus Ro Dah said:
First part split the integral as two terms then the integral with the i in it cancels out due to the sine function and we are left with 2x integral of f(x)cos(x). Second part use the property f(x) = f(a-x), add the two integrals, the icos(x) term cancels and the result follows.

I am a bit skeptical regarding this question being within Ext2 level without proper justification because we are taking the Riemann integral of a complex-valued function. It may require justification that it works in this case because $\mathbb{C}$ is a vector space. Other students might get confused about having an i in the integral.

yeah I agree with with you, the question seems like it isnt in the scope of the ext 2 course

Trebla · Jun 3, 2012

Re: 2012 HSC MX2 Marathon

Okay made it heaps easier...

Suppose that f(x) is an even function and k is a constant. Show that

$\displaystyle\int_{-a}^a f(x)(\cos x + k\sin x)\,dx = 2\displaystyle\int_{0}^a f(x)\cos x\,dx$

Hence evaluate

$\displaystyle\int_{0}^{\pi} \left|\dfrac{\pi}{2} - x\right|(\sin x + k\cos x)\,dx$

bleakarcher · Jun 3, 2012

Re: 2012 HSC MX2 Marathon

can you treat 'i' as a constant?

funnytomato · Jun 3, 2012

Re: 2012 HSC MX2 Marathon

bleakarcher said:
can you treat 'i' as a constant?

why not?

bleakarcher · Jun 3, 2012

Re: 2012 HSC MX2 Marathon

wasnt sure..

seanieg89 · Jul 12, 2012

Re: 2012 HSC MX2 Marathon

Might as well bring this back to life.

Prove that the polynomial

$p(x)=\sum_{k=0}^{2n} \frac{x^k}{k!}$

has no real roots. (Where n is a non-negative integer.)

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