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HSC 2014 MX2 Marathon ADVANCED (archive) (1 Viewer)
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Re: HSC 2014 4U Marathon - Advanced Level
} \\ =\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-?
Ooh,
How do you guys know if its 0+ or 0-?
Last edited:
seanieg89
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Re: HSC 2014 4U Marathon - Advanced Level
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.
Yeah it tends to zero.Ooh,
How do you guys know if its 0+ or 0-?
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.
VBN2470
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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
Can someone please explain the following question? It would preferable if someone did this question using Gaussian elimination (matrix row reduction)..
Thanks
Shadowdude
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Re: HSC 2014 4U Marathon - Advanced Level
go from there
If they intersect, then you can find values for the parameters such that it works.
go from there
Carrotsticks
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Re: HSC 2014 4U Marathon - Advanced Level
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/
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/
seanieg89
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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.
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.
Last edited:
awesome-0_4000
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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
Expand [cos(theta) + isin(theta)]^5 using De Moivre's Theorem for the LHS and Binomial Theorem on the RHS. Equate the real parts
awesome-0_4000
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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
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
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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
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
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Sy123
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Re: HSC 2014 4U Marathon - Advanced Level
 = 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.
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.
Sy123
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Re: HSC 2014 4U Marathon
 + (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
Ifand
are real numbers such that
find the maximum possible difference between any two of
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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