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HSC 2012 MX1 Marathon #2 (archive) (1 Viewer)

RealiseNothing

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Re: HSC 2012 Marathon :)

Ok here is another question. Prove any product-sum or sum-product trig ratio from the startpoint of sin(x+y) and sin(x-y)
sin(x+y) + sin(x-y) = sinxcosy + sinycosx + sinxcosy - sinycosx

sin(x+y) + sin(x-y) = 2sinxcosy

sinxcosy = 1/2[sin(x+y) + sin(x-y)]
 

Timske

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Re: HSC 2012 Marathon :)

For Sy,

<a href="http://www.codecogs.com/eqnedit.php?latex=\textup{A ~particle~ is~ moving ~so~ that} ~\ddot{x}=18x^3@plus;27x^2@plus;9x.~ \textup{Initially ~x=-2~ and~ velocity, ~v~ is ~-6}.\\\\1. ~\textup{Show ~that~} v^2=9x^2(1@plus;x)^2 \\ 2. ~\textup{Hence, or otherwise, show that }\int \frac{1}{x(1@plus;x)} dx = -3t \\3.~\textup{It ~can~ be ~show ~for ~some ~constant ~c},~ \log_{e}(1@plus;\frac{1}{x}) = 3t@plus;c~,\textup{~Using ~this ~equation ~and ~the ~initial~ conditions, ~find~ x ~as ~a~ function ~of ~t.}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\textup{A ~particle~ is~ moving ~so~ that} ~\ddot{x}=18x^3+27x^2+9x.~ \textup{Initially ~x=-2~ and~ velocity, ~v~ is ~-6}.\\\\1. ~\textup{Show ~that~} v^2=9x^2(1+x)^2 \\ 2. ~\textup{Hence, or otherwise, show that }\int \frac{1}{x(1+x)} dx = -3t \\3.~\textup{It ~can~ be ~show ~for ~some ~constant ~c},~ \log_{e}(1+\frac{1}{x}) = 3t+c~,\textup{~Using ~this ~equation ~and ~the ~initial~ conditions, ~find~ x ~as ~a~ function ~of ~t.}" title="\textup{A ~particle~ is~ moving ~so~ that} ~\ddot{x}=18x^3+27x^2+9x.~ \textup{Initially ~x=-2~ and~ velocity, ~v~ is ~-6}.\\\\1. ~\textup{Show ~that~} v^2=9x^2(1+x)^2 \\ 2. ~\textup{Hence, or otherwise, show that }\int \frac{1}{x(1+x)} dx = -3t \\3.~\textup{It ~can~ be ~show ~for ~some ~constant ~c},~ \log_{e}(1+\frac{1}{x}) = 3t+c~,\textup{~Using ~this ~equation ~and ~the ~initial~ conditions, ~find~ x ~as ~a~ function ~of ~t.}" /></a>
 

Sy123

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Re: HSC 2012 Marathon :)

For Sy,

<a href="http://www.codecogs.com/eqnedit.php?latex=\textup{A ~particle~ is~ moving ~so~ that} ~\ddot{x}=18x^3@plus;27x^2@plus;9x.~ \textup{Initially ~x=-2~ and~ velocity, ~v~ is ~-6}.\\\\1. ~\textup{Show ~that~} v^2=9x^2(1@plus;x)^2 \\ 2. ~\textup{Hence, or otherwise, show that }\int \frac{1}{x(1@plus;x)} dx = -3t \\3.~\textup{It ~can~ be ~show ~for ~some ~constant ~c},~ \log_{e}(1@plus;\frac{1}{x}) = 3t@plus;c~,\textup{~Using ~this ~equation ~and ~the ~initial~ conditions, ~find~ x ~as ~a~ function ~of ~t.}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\textup{A ~particle~ is~ moving ~so~ that} ~\ddot{x}=18x^3+27x^2+9x.~ \textup{Initially ~x=-2~ and~ velocity, ~v~ is ~-6}.\\\\1. ~\textup{Show ~that~} v^2=9x^2(1+x)^2 \\ 2. ~\textup{Hence, or otherwise, show that }\int \frac{1}{x(1+x)} dx = -3t \\3.~\textup{It ~can~ be ~show ~for ~some ~constant ~c},~ \log_{e}(1+\frac{1}{x}) = 3t+c~,\textup{~Using ~this ~equation ~and ~the ~initial~ conditions, ~find~ x ~as ~a~ function ~of ~t.}" title="\textup{A ~particle~ is~ moving ~so~ that} ~\ddot{x}=18x^3+27x^2+9x.~ \textup{Initially ~x=-2~ and~ velocity, ~v~ is ~-6}.\\\\1. ~\textup{Show ~that~} v^2=9x^2(1+x)^2 \\ 2. ~\textup{Hence, or otherwise, show that }\int \frac{1}{x(1+x)} dx = -3t \\3.~\textup{It ~can~ be ~show ~for ~some ~constant ~c},~ \log_{e}(1+\frac{1}{x}) = 3t+c~,\textup{~Using ~this ~equation ~and ~the ~initial~ conditions, ~find~ x ~as ~a~ function ~of ~t.}" /></a>
Thanks man.





I will edit one in
 
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Timske

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Re: HSC 2012 Marathon :)

Check your working for 3 there's a mistake
 

Sy123

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Re: HSC 2012 Marathon :)

Hmm, I cant find a mistake there
 

Timske

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Re: HSC 2012 Marathon :)

<a href="http://www.codecogs.com/eqnedit.php?latex=\ln (1 @plus; \frac{1}{x}) = 3t @plus; \ln(\frac{1}{2})" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\ln (1 + \frac{1}{x}) = 3t + \ln(\frac{1}{2})" title="\ln (1 + \frac{1}{x}) = 3t + \ln(\frac{1}{2})" /></a>
 

RealiseNothing

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Re: HSC 2012 Marathon :)

Find two squares such that their difference is 2011.
 

Sy123

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Re: HSC 2012 Marathon :)

Do they have to be integers?
If not, then a possible answer is

Also, my I dont see how my answer has a mistake.

I may have forgotten the brackets in

That 0.5 is 1/2, but I wasnt bothered to put it into fractional form since it was already a long question.
 

nightweaver066

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Re: HSC 2012 Marathon :)

Find two squares such that their difference is 2011.
If they're integers,
Let a and b be two positive integers such that,
a^2 - b^2 = 2011, a > b
(a + b)(a - b) = 2011

But 2011 is a prime number, therefore a + b = 2011, a - b = 1 or a + b = 1, a - b = 2011
Clearly a + b cannot equal to 1 as a, b > 0 and they're both integral
So a + b = 2011
a - b = 1
2a = 2012
a = 1006
b = 1005
 
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RealiseNothing

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Re: HSC 2012 Marathon :)

This may be a bit harder, but find two squares such that their difference is 2012.

And hence find a formula to find two squares whose difference is 4n where n is a positive integer.
 
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Timske

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Re: HSC 2012 Marathon :)

Do they have to be integers?
If not, then a possible answer is

Also, my I dont see how my answer has a mistake.

I may have forgotten the brackets in

That 0.5 is 1/2, but I wasnt bothered to put it into fractional form since it was already a long question.
Oh i see what u did still turns out to be correct..
 

deswa1

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Re: HSC 2012 Marathon :)

This may be a bit harder, but find two squares such that their difference is 2012.
(a+k)^2-a^2=2012 (for a,k integers)
2ak+k^2=2012
k(2a+k)=2012
use 2,1006 as pairs.

k=2, 2a=1004
a=502
Therefore the squares of 502,504 differ by 2012
 

nightweaver066

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Re: HSC 2012 Marathon :)

This may be a bit harder, but find two squares such that their difference is 2012.

And hence find a formula to find two squares whose difference is 4n where n is a positive integer.
Factors of 2012 are 1, 2, 4, ...
Same thing as before, a^2 - b^2 = 2012 = 2(1006)
(a + b)(a - b) = 2(1006)
a + b = 1006, a - b - 2
a = 504, b = 502

(a + b)(a - b) = 4n
a + b = 2n, a - b = 2
a = n + 1, b = n - 1
 

RealiseNothing

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Re: HSC 2012 Marathon :)

Here's a much harder one, but the result is really cool.

Given that:



Prove that:

 

RealiseNothing

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Re: HSC 2012 Marathon :)

thats not too bad
Depends which way you do it tbh. If you weren't given the second part to prove it would be a lot harder. ie - Find in terms of (n-1) only, so that no other term appaears in the expression but some form of (n-1).
 

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