HSC 2012 MX1 Marathon #2 (archive) (7 Viewers)

RealiseNothing · Jun 17, 2012

Re: HSC 2012 Marathon

barbernator said:
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)]

barbernator · Jun 17, 2012

Re: HSC 2012 Marathon

RealiseNothing said:
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)]

cmon, change it to sum to product as well

RealiseNothing · Jun 17, 2012

Re: HSC 2012 Marathon

barbernator said:
cmon, change it to sum to product as well

I proved above that:

$sin(x+y) + sin(x-y) = 2sin(x)cos(y)$

Now let $x + y = \alpha ~and~ x - y = \beta$

So we have $sin\alpha + sin\beta = 2sin(x)cos(y)$

Using simultaneous equations of $x + y = \alpha ~and~ x - y = \beta$ we arrive at the expressions:

$2x = \alpha + \beta ~and~ 2y = \alpha - \beta$

Hence:

$x = \frac{\alpha + \beta}{2} ~and~ y = \frac{\alpha - \beta}{2}$

Sub these into $sin\alpha + sin\beta = 2sin(x)cos(y)$ and we end up with:

$sin\alpha + sin\beta = 2sin(\frac{\alpha + \beta}{2})cos(\frac{\alpha - \beta}{2})$

Timske · Jun 17, 2012

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 · Jun 17, 2012

Re: HSC 2012 Marathon

Timske said:
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.

$1. \ \ \ddot{x}=18x^3+27x^2+9x \\\left(\frac{1}{2}v^2\right)=18x^3+27x^2+9x \\ \frac{1}{2}v^2=18\frac{x^4}{4}+27\frac{x^3}{3}+9\frac{x^2}{2}+C \\ \\ x=-2 \Rightarrow v=-6 \\ \frac{1}{2}(-6)^2=18\frac{(-2)^4}{4}+27\frac{(-2)^3}{3}+9\frac{(-2)^2}{2}+C \\ \\ C=0 \\ \\ \therefore v^2=9x^4+18x^3+9x^2 \\ \\ v^2=9x^2(x^2+2x+1) \\ v^2=9x^2(1+x)^2$
$2. \ \ v=-3x(1+x) \\ \\ \text {Since} \ \ \ v<0 \\ \frac{dx}{dt}=-3x(1+x) \\ \frac{dt}{dx}=\frac{1}{-3x(1+x)} \\ \therefore t=\int \frac{-1}{3}\frac{1}{x(1+x)} \\ \therefore -3t=\int \frac{1}{x(1+x)}$
$3. \ \ t=0 \Rightarrow x=-2 \\ \\ \therefore \ln{(1+\frac{1}{-2})}=C \\ \therefore C=\ln{(\frac{1}{2})} \\ \\ \ln{1+\frac{1}{x}}=3t+\ln{0.5} \\ \\ \frac{1}{x}=e^{3t+\ln{0.5}}-1 \\ \\ \therefore x=\frac{1}{e^{3t+ln{0.5}}-1}$

I will edit one in

Timske · Jun 17, 2012

Re: HSC 2012 Marathon

Check your working for 3 there's a mistake

Sy123 · Jun 17, 2012

Re: HSC 2012 Marathon

Hmm, I cant find a mistake there

Timske · Jun 17, 2012

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 · Jun 17, 2012

Re: HSC 2012 Marathon

Find two squares such that their difference is 2011.

Sy123 · Jun 17, 2012

Re: HSC 2012 Marathon

Do they have to be integers?
If not, then a possible answer is $\sqrt{2013} \ \ \ \ and \ \ \ \ \sqrt{2}$

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

I may have forgotten the brackets in $\ln{1+\frac{1}{x}}=3\ln{0.5}$

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

nightweaver066 · Jun 17, 2012

Re: HSC 2012 Marathon

RealiseNothing said:
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

RealiseNothing · Jun 17, 2012

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.

Timske · Jun 17, 2012

Re: HSC 2012 Marathon

Sy123 said:
Do they have to be integers?
If not, then a possible answer is $\sqrt{2013} \ \ \ \ and \ \ \ \ \sqrt{2}$

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

I may have forgotten the brackets in $\ln{1+\frac{1}{x}}=3\ln{0.5}$

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 · Jun 17, 2012

Re: HSC 2012 Marathon

RealiseNothing said:
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 · Jun 17, 2012

Re: HSC 2012 Marathon

RealiseNothing said:
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 · Jun 17, 2012

Re: HSC 2012 Marathon

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

Given that:

$I_n = n! - (n-1)!$

Prove that:

$I_n = (n-1)I_{n-1} + (n-1)!$

barbernator · Jun 17, 2012

Re: HSC 2012 Marathon

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

Given that:

$I_n = n! - (n-1)!$

Prove that:

$I_n = (n-1)I_{n-1} + (n-1)!$

thats not too bad, typing up working now. ok i wont lol

nightweaver066 · Jun 17, 2012

Re: HSC 2012 Marathon

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

Given that:

$I_n = n! - (n-1)!$

Prove that:

$I_n = (n-1)I_{n-1} + (n-1)!$

Mind if i go backwards? lol

$I_n = (n -1)I_{n - 1} + (n - 1)!$

$= (n - 1)[(n - 1)! - (n - 2)!] + (n - 1)!$

$= (n - 1)^2 (n - 2)! - (n - 1)! + (n - 1)!$

$= (n - 1)(n - 1)!$

$= \frac{(n - 1)n!}{n}$

$= n(n - 1)! - (n - 1)!$

$= n! - (n - 1)!$

RealiseNothing · Jun 17, 2012

Re: HSC 2012 Marathon

barbernator said:
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 $I_n$ in terms of (n-1) only, so that no other term appaears in the expression but some form of (n-1).

RealiseNothing · Jun 17, 2012

Re: HSC 2012 Marathon

nightweaver066 said:
Mind if i go backwards? lol

$I_n = (n -1)I_{n - 1} + (n - 1)!$

$= (n - 1)[(n - 1)! - (n - 2)!] + (n - 1)!$

$= (n - 1)^2 (n - 2)! - (n - 1)! + (n - 1)!$

$= (n - 1)(n - 1)!$

$= \frac{(n - 1)n!}{n}$

$= n(n - 1)! - (n - 1)!$

$= n! - (n - 1)!$

I probably shouldn't of told you what it was equal to so you couldn't go backwards lol.

It would of been much harder working it out going forwards.

HSC 2012 MX1 Marathon #2 (archive) (7 Viewers)

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