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HSC 2017 MX2 Marathon (archive) (1 Viewer)

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leehuan

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Re: HSC 2017 4U Marathon

Had to make sure lol
In sciences a LINE of best fit can be a linear line of best fit or a curve of best fit
Actually, I don't even know why a line of best fit can be a curve. I would just call a curve of best fit its own seperate thing. (Polynomial of best fit, blah blah)
 

seanieg89

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Re: HSC 2017 4U Marathon

Had to make sure lol
In sciences a LINE of best fit can be a linear line of best fit or a curve of best fit
That's just hazy high school science definitions. There are various precise definitions of lines/curves of best fit (regression analysis, interpolating polynomials etc), and whenever such an object is named a line it is definitely a line in the usual sense of the word (at least as far as I have seen).
 

mini8658

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Re: HSC 2017 4U Marathon

Is there a way to expand it so the cis2kpi/5's cancel out, leaving only the z^5-1?
 

Green Yoda

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Re: HSC 2017 4U Marathon

Quick question: When we are finding the roots of unity (or any complex number) and leaving it in mod-arg form do we have to change the argument to the principle range or just leave it as it as (e.g. for k=5 arg = 15pi/8). The answers in past papers leave it as it is...so not sure what to do.
 

calamebe

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Re: HSC 2017 4U Marathon

Quick question: When we are finding the roots of unity (or any complex number) and leaving it in mod-arg form do we have to change the argument to the principle range or just leave it as it as (e.g. for k=5 arg = 15pi/8). The answers in past papers leave it as it is...so not sure what to do.
I always converted it to the principle range, technically you'd be right, but it's better to be safe.
 

pikachu975

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Re: HSC 2017 4U Marathon

Quick question: When we are finding the roots of unity (or any complex number) and leaving it in mod-arg form do we have to change the argument to the principle range or just leave it as it as (e.g. for k=5 arg = 15pi/8). The answers in past papers leave it as it is...so not sure what to do.
I like to use -2, -1, 0, 1, 2 instead of 0, 1, 2, 3, 4 because it'll always give you arguments in the principal argument range.

Just a side note it's principal argument not principle
 

InteGrand

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Re: HSC 2017 4U Marathon






 
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integral95

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Re: HSC 2017 4U Marathon

This is probably a vague explanation but...


In the case where both roots are real.
The sum of roots = -b/a, which is negative if a,b are either both positive or both negative. This implies that at least one of the roots is negative.
Product of roots = c/a, if the roots are both negative, then the product must be positive therefore a,c must be both positive or both negative.

If the roots are complex, they must be in conjugate pairs due to real coefficients, this means that the real parts must be the same.
So if the real part is negative, then the sum of roots must be negative, and the product of roots must be positive.
 

Kingom

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Re: HSC 2017 4U Marathon

here's a nice binomial followed by a harder binomial
133.PNG
134.PNG
 

stupid_girl

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Re: HSC 2017 MX2 Marathon

This geometry question doesn't involve any advanced knowledge but requires a bit of creativity.:tongue:

ABCD is a quadrilateral with three equal sides AB,BC and CD. Show that the mid-point of AD lies on a circle with diameter BC if and only if the area of ABCD is a quarter of the product of its diagonals.
 

frog1944

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Re: HSC 2017 MX2 Marathon

A question posted in the 2016 MX2 marathon by Paradoxica that wasn't answered:
 

si2136

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Re: HSC 2017 MX2 Marathon

A question posted in the 2016 MX2 marathon by Paradoxica that wasn't answered:
It approaches infinity.

lim x--> 0, sin x / (1-cos x)

= lim x--> 0 (1+cos x) / sin x

Apply l'hopital, sin^2 x, sin^4 x, all approach infinity.

Therefore the limit approaches infinity.

Could someone check my answer please? Thanks.
 

InteGrand

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Re: HSC 2017 MX2 Marathon

It approaches infinity.

lim x--> 0, sin x / (1-cos x)

= lim x--> 0 (1+cos x) / sin x

Apply l'hopital, sin^2 x, sin^4 x, all approach infinity.

Therefore the limit approaches infinity.

Could someone check my answer please? Thanks.
 

seanieg89

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Re: HSC 2017 MX2 Marathon

Suppose you have n points on a circle such that no three distinct chords coincide at any single point.

How many regions do the nC2 chords divide the interior of the circle into?

(You don't have to provide rigorous proof for this question if you can guess the answer correctly).
 

dan964

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Re: HSC 2017 MX2 Marathon

Suppose you have n points on a circle such that no three distinct chords coincide at any single point.

How many regions do the nC2 chords divide the interior of the circle into?

(You don't have to provide rigorous proof for this question if you can guess the answer correctly).
2m where m is the number of chords. (2*nC2??)

Sorry thats not right, hold on.
 

Sy123

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Re: HSC 2017 MX2 Marathon

Part (b) is a little long and requires good understanding of the problem, so keep that in mind






---------------------------------









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Paradoxica

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Re: HSC 2017 MX2 Marathon

Suppose you have n points on a circle such that no three distinct chords coincide at any single point.

How many regions do the nC2 chords divide the interior of the circle into?

(You don't have to provide rigorous proof for this question if you can guess the answer correctly).
something along the lines of the sum of three binomial coefficients of even lower argument...

I'll leave the details to the interested reader...
 

dan964

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Re: HSC 2017 MX2 Marathon

Just a reminder this thread is mainly for Q1-15 material, and should be accessible to all Ext 2 students, not just the really talented ones.

I have moved the Lagrange multiplier & IMO question to the Advanced Marathon thread here...
http://community.boredofstudies.org...-2/360463/hsc-2017-mx2-marathon-advanced.html

If you have questions for help not related to questions posted here, it is advisable to post a separate thread rather than posting here in the marathon thread thanks. Although feel free to post a question here of course...
I have moved some of those also out of this thread.

here is another question to attempt
abbotsleigh20034u_q7b.PNG
 
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