HSC 2014 MX2 Marathon ADVANCED (archive) (3 Viewers)

abecina · Sep 12, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$\lim_{x\rightarrow \infty} \left( \tan \frac{\pi x}{2x+1}\right)^{\frac{1}{x}}\textrm{ Hint: uses the definition } e^x = \lim_{n\rightarrow \infty} \left(1+\frac{x}{n}\right)^n$

dunjaaa · Sep 12, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Is the definition even needed? I ended up getting 1 just by inspection because [tan(pi/2)]^0 =1

glittergal96 · Sep 12, 2014

Re: HSC 2014 4U Marathon - Advanced Level

dunjaaa said:
Is the definition even needed? I ended up getting 1 just by inspection because [tan(pi/2)]^0 =1

I don't think you can raise infinity to the power of zero to make conclusions like that lol.

I do think it's 1, but I suspect abecina might have typo'd the question, because with different exponents or something, the exponential could definitely get involved.

(If the question is correct, I will type out my proof of convergence to 1, which is fairly basic.)

dunjaaa · Sep 12, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Yeah, I know it's indeterminate. It was just a hunch lol.

abecina · Sep 12, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Its not a typo. And yeah, the answer is 1. I had to use exponents to do it, though happy to be shown an easier way

glittergal96 · Sep 13, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Okay, I don't think mine is elegant or quick, it is just that I don't think we don't need to know anything more detailed about the definition/properties of the exponential than is taught in the syllabus.

First note that

$\sin(\frac{\pi x}{2x+1})^{1/x}\rightarrow 1$

trivially, because base tends to 1 and exponent tends to 0. So it suffices to consider the expression

$\cos(\frac{\pi x}{2x+1})^{1/x}=\sin(\frac{\pi}{4x+2})^{1/x}$

If we divide and multiply this by

$(\frac{\pi}{4x+2})^{1/x}$

we are left with the product of something that tends to 1 from knowledge of the sin(x)/x as x -> 0 limit and

$(\frac{\pi}{4x+2})^{1/x}.$

For sufficiently large x this quantity is smaller that 1 (as the inside is smaller than 1 and we are raising to a positive power). For sufficiently large x it is also bounded below by

$(\frac{\pi}{5x})^{1/x}.$

The constants are obviously irrelevant as the exponentiation sends them to 1 in the limit as x-> inf.

Finally

$\log(x^{1/x})=\log(x)/x \rightarrow 0 \Rightarrow x^{1/x} \rightarrow 1.$

where the first limit is a HSC assumption (*) for graphing purposes I am pretty sure. (Is it in the 3U book somewhere?) The idea is that logs grow much slower than any positive power.

Putting this all together finishes the proof by the squeeze law.

Just for the sake of completeness, I will include a sketch of a proof of assumption (*):

The assumption is equivalent to showing that $\frac{x}{e^x}\rightarrow 0.$

But

$e^x \geq 1+x+x^2/2$

by integrating

$e^x - 1 \geq 0$

twice from 0 to x. That a quadratic expression dominates a linear one is clear from dividing numerator and denominator by x.

dan964 · Sep 14, 2014

Re: HSC 2014 4U Marathon - Advanced Level

next q:
View attachment 30910

Carrotsticks · Sep 14, 2014

Re: HSC 2014 4U Marathon - Advanced Level

dan964 said:
next q:
View attachment 30910

Interesting questions, much harder than typical ones.

I recognise the Volumes problem from a King's paper (if my memory serves me correctly), from where did you acquire this?

mreditor16 · Sep 14, 2014

Re: HSC 2014 4U Marathon - Advanced Level

dan964 said:
next q:
View attachment 30910

all of them? :/

dan964 · Sep 14, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Carrotsticks said:
Interesting questions, much harder than typical ones.

I recognise the Volumes problem from a King's paper (if my memory serves me correctly), from where did you acquire this?

Q14 a - a sgs trial
Q14 b - some other trial - can't remember think its kincoppal
Q14 c- moriah 2001 made a bit more difficult
Q15-16 - red herring, not from anywhere.
q15a - experimented with the proof behind a rectangular hyperbola and applied it to something else
q15b (ii) is a proof of one of Fermat's theorems using Induction.
q16 - made up, so shouldn't be that hard - just hard slog.

dan964 · Sep 14, 2014

Re: HSC 2014 4U Marathon - Advanced Level

mreditor16 said:
all of them? :/

try 14c, 15 and 16 (hopefully I got all the numbers right in Q16)

Carrotsticks · Sep 14, 2014

Re: HSC 2014 4U Marathon - Advanced Level

dan964 said:
Q14 a - a sgs trial
Q14 b - some other trial - can't remember think its kincoppal
Q14 c- moriah 2001 made a bit more difficult
Q15-16 - red herring, not from anywhere.

Ah! Moriah! I remember that infamous paper.

Axio · Sep 14, 2014

Re: HSC 2014 4U Marathon - Advanced Level

dan964 said:
next q:
View attachment 30910

Is this the solution for 14ai?

$LHS=z^{n}-z^{-n}\\=(cos\theta +isin\theta )^{n}-\frac{1}{(cos\theta +isin\theta)^{n}}\\=(\frac{cos^{2}\theta +2icos\theta sin\theta -sin^{2}\theta -1}{cos\theta +isin\theta })^{n}\\=(\frac{2icos\theta sin\theta -2sin^{2}\theta }{cos\theta +isin\theta })^{n}\\=(\frac{2isin\theta (cos\theta +isin\theta )}{cos\theta +isin\theta })^{n}\\=2isinn\theta$ =RHS

RealiseNothing · Sep 14, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Axio said:
Is this the solution for 14ai?

$LHS=z^{n}-z^{-n}\\=(cos\theta +isin\theta )^{n}-\frac{1}{(cos\theta +isin\theta)^{n}}\\=(\frac{cos^{2}\theta +2icos\theta sin\theta -sin^{2}\theta -1}{cos\theta +isin\theta })^{n}\\=(\frac{2icos\theta sin\theta -2sin^{2}\theta }{cos\theta +isin\theta })^{n}\\=(\frac{2isin\theta (cos\theta +isin\theta )}{cos\theta +isin\theta })^{n}\\=2isinn\theta$ =RHS

Lol De'Moivre's Theorem would have saved you about 5 lines.

Carrotsticks · Sep 14, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Axio said:
Is this the solution for 14ai?

$LHS=z^{n}-z^{-n}\\=(cos\theta +isin\theta )^{n}-\frac{1}{(cos\theta +isin\theta)^{n}}\\=(\frac{cos^{2}\theta +2icos\theta sin\theta -sin^{2}\theta -1}{cos\theta +isin\theta })^{n}\\=(\frac{2icos\theta sin\theta -2sin^{2}\theta }{cos\theta +isin\theta })^{n}\\=(\frac{2isin\theta (cos\theta +isin\theta )}{cos\theta +isin\theta })^{n}\\=2isinn\theta$ =RHS

A much faster way is to simply note that if $z=\cos \theta + i \sin \theta$ , then $\frac{1}{\cos \theta + i \sin \theta}= \cos (-\theta) + i \sin (-\theta)$ and then work from there.

Should take no longer than 2-3 lines.

RealiseNothing · Sep 14, 2014

Re: HSC 2014 4U Marathon - Advanced Level

dan964 said:
next q:
View attachment 30910

I really have to question why 14)d) is there.

dunjaaa · Sep 14, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Screen shot 2014-09-14 at 7.14.31 PM.png

Seeing that there is an oblique rotation question, I may as well post the question we had in 3rd 4u task

faisalabdul16 · Sep 14, 2014

Re: HSC 2014 4U Marathon - Advanced Level

dunjaaa said:
View attachment 30912 Seeing that there is an oblique rotation question, I may as well post the question we had in 3rd 4u task

Oblique is not in the syllabus right? Or is it?

dunjaaa · Sep 14, 2014

Re: HSC 2014 4U Marathon - Advanced Level

It's not, but my teacher is a real big fan of it lol

dan964 · Sep 14, 2014

Re: HSC 2014 4U Marathon - Advanced Level

rotate entire curve using Complex numbers so that you are rotating around a line. done.
btw that's also what q15a in the file I uploaded.

(the hint is that you are given the original unrotated curve)

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