HSC 2014 MX2 Marathon ADVANCED (archive) (4 Viewers)

Trebla · Mar 31, 2014

Re: HSC 2014 4U Marathon - Advanced Level

braintic said:
Surely its zero? Or better: 0+ if m>n, 0- if m<n

Not sure how to show working though (at least not without making the same sort of assumptions I have made in my head already).

rationalise the numerator

Ikki · Apr 1, 2014

Re: HSC 2014 4U Marathon - Advanced Level

RealiseNothing said:
$\lim_{x \to \infty} \sqrt{x+m} - \sqrt{x+n}$

Ooh,
$= \lim_{x \rightarrow \infty}\frac{x+m-x-n}{(\sqrt{x+m}+\sqrt{x+n})} \\ =\lim_{x \rightarrow \infty}\frac{m-n}{(\sqrt{x+m}+\sqrt{x+n})} \\ $So does it$ \rightarrow 0?$
How do you guys know if its 0+ or 0-?

seanieg89 · Apr 1, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Ikki said:
Ooh,
$= \lim_{x \rightarrow \infty}\frac{x+m-x-n}{(\sqrt{x+m}+\sqrt{x+n})} \\ =\lim_{x \rightarrow \infty}\frac{m-n}{(\sqrt{x+m}+\sqrt{x+n})} \\ $So does it$ \rightarrow 0?$
How do you guys know if its 0+ or 0-?

Yeah it tends to zero.

Depends on whether m or n is larger.

I would just write 0 though. The +/- notation is more commonly used when talking about one-sided limits.

Ikki · Apr 2, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Ah k cool thanks

VBN2470 · Apr 4, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Can someone please explain the following question? It would preferable if someone did this question using Gaussian elimination (matrix row reduction)..

Thanks

Shadowdude · Apr 4, 2014

Re: HSC 2014 4U Marathon - Advanced Level

VBN2470 said:
Can someone please explain the following question? It would preferable if someone did this question using Gaussian elimination (matrix row reduction)..
View attachment 30215

Thanks

If they intersect, then you can find values for the parameters such that it works.

go from there

Carrotsticks · Apr 4, 2014

Re: HSC 2014 4U Marathon - Advanced Level

VBN2470 said:
Can someone please explain the following question? It would preferable if someone did this question using Gaussian elimination (matrix row reduction)..
View attachment 30215

Thanks

Find the equation of their normal vectors.

If the normal vectors are parallel, then so are the planes.

Also, these university questions should be directed here

http://community.boredofstudies.org/1003/maths/

Sy123 · Apr 6, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$\\ $Prove via squeeze theorem$ \ \ e^x = \lim_{n \to \infty} \left(1 + \frac{x}{n} \right)^n$

Someone else post a question please

seanieg89 · Apr 6, 2014

Re: HSC 2014 4U Marathon - Advanced Level

The squeeze theorem isn't really necessary if we are working from MX2 knowledge, where e^x refers to a (not entirely rigorously defined) function which we know how to differentiate, and whose inverse we know how to differentiate.

Here is my proof:

Fix x real and let g(t)=(1+tx)^(1/t) for positive t. It suffices to show that log g(t) tends to x as t tends to zero.

log g(t)=log(1+tx)/t = (h(t)-h(0))/t where h(s)=log(1+sx).

So letting t->0 we get log g(t) -> h'(0)=x.

gahyunkk · Apr 6, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Can anyone explain this to me?

awesome-0_4000 · Apr 6, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Expand [cos(theta) + isin(theta)]^5 using De Moivre's Theorem for the LHS and Binomial Theorem on the RHS. Equate the real parts

gahyunkk · Apr 6, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Yea i did but i cant get those value with solving 16x^5-20x^3+5x-1=0 especially bcuz of -1 at the end

awesome-0_4000 · Apr 6, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Cos(5theta) = cos(5x2pi/5) = cos(2pi) = 1
So substituting back into the RHS of the original equation we get 16cos^5(theta) -20cos^3(theta) + 5cos(theta) = 1
Then subtract the 1

gahyunkk · Apr 6, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Yes yes im up to this bit
But after this i have to find the x value of this equation and compare to the 5 roots i got from cos(5theta)=1 to deduce exact value of cos2/5pi and cos4/5pi.

But how do i find the roots from 16x^5-20x^3+5x-1=0?
Normally we use quadratic formula but this case doesnt work bcuz of -1

Sy123 · Apr 6, 2014

Re: HSC 2014 4U Marathon - Advanced Level

gahyunkk said:
Yes yes im up to this bit
But after this i have to find the x value of this equation and compare to the 5 roots i got from cos(5theta)=1 to deduce exact value of cos2/5pi and cos4/5pi.

But how do i find the roots from 16x^5-20x^3+5x-1=0?
Normally we use quadratic formula but this case doesnt work bcuz of -1

I answered in the thread you posted

VBN2470 · Apr 7, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Can someone please answer the following

:

Thanks.

Sy123 · Apr 7, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$\\ $The function$ \ g \ $is continuous at$ \ x = a \ $if$ \ \lim_{x \to a} g(x) = g(a) \\ $Taking the first inequality$ \ \ \lim_{x \to a} f(x) \leq \lim_{x \to a} g(x) \leq \lim_{x \to a} h(x) \\ $Since$ \ f \ $and$ \ h \ $are continuous, then$ \\ \Rightarrow \ f(a) \leq \lim_{x \to a} g(x) \leq h(a) \\ $But$ \ f(a) = h(a) = g(a) \\ \therefore \ f(a) \leq \lim_{x \to a} \leq f(a) \\ \Rightarrow \ g(a) \leq \lim_{x \to a} g(x) \leq g(a) \ \ \ \ $then by squeeze theorem$ \\ \lim_{x \to a} g(x) = g(a) \\ $Hence$ \ g \ $is continuous$$

But next time please post your question to the Extracurricular Section, the rigorous definition of continuity is not covered in the 4U HSC course, and hence it is not appropriate for this question to be asked here.

VBN2470 · Apr 8, 2014

Re: HSC 2014 4U Marathon

If $m$ and $n$ are positive integers such that $\frac{m+n}{m^2+mn+n^2}=\frac{4}{49}$ find the value of $m+n$ .

VBN2470 · Apr 8, 2014

Re: HSC 2014 4U Marathon

If $x, y$ and $z$ are real numbers such that $x^2+y^2+z^2-xy-yz-zx=8$ find the maximum possible difference between any two of $x, y$ and $z$

Sy123 · Apr 8, 2014

Re: HSC 2014 4U Marathon

VBN2470 said:
If $x, y$ and $z$ are real numbers such that $x^2+y^2+z^2-xy-yz-zx=8$ find the maximum possible difference between any two of $x, y$ and $z$

$z^2 - z(x+y) + (x^2+y^2-xy - 8) = 0 \ \ \triangle \geq 0 \ \ $since$ \ x,y,z \ $real$ \\ \\ (x+y)^2 - 4(x^2+y^2-xy-8) \geq 0 \\ 32 - 3(x-y)^2 \geq 0 \Rightarrow \ \sqrt{\frac{32}{3}} \geq |x -y|$

Also if you're question if Q16 difficulty or beyond, please post it in the Advanced Level marathon. This marathon is for easy questions

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