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#1 punto

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Posted 20 May 2006 - 03:53 PM

Any chance someone can post full solution for 2006 adv exam???
Thanks

#2 Jason Bourne

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Posted 20 May 2006 - 10:10 PM

if someone knows the answers!? tongue.gif
Nothing!!!!!!!!

#3 john g 22

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Posted 04 August 2006 - 05:22 PM

lol i no this was posted ages ago so maybe sum1 has found a link by now .. if any has please post it up ! would be intrseting to see the right answers

#4 ad absurdum

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Posted 05 August 2006 - 11:39 AM

Also, does anyone have a copy of the paper?
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#5 waitingforan_alibi

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Posted 06 August 2006 - 10:30 AM

I still have a copy of the paper so i might post solutions to some of the questions if i can be bothered doing the questions - i cant remember the answers anymore but i didnt find the paper as hard as everybody made it out to be.

#6 john g 22

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Posted 06 August 2006 - 05:53 PM

QUOTE(ad absurdum @ Aug 5 2006, 12:39 PM) View Post

Also, does anyone have a copy of the paper?


13 through to 16 would be nice .. up to you though man .. cheers

QUOTE(waitingforan_alibi @ Aug 6 2006, 11:30 AM) View Post

I still have a copy of the paper so i might post solutions to some of the questions if i can be bothered doing the questions - i cant remember the answers anymore but i didnt find the paper as hard as everybody made it out to be.



13 though to 16 would be nice man ... it doesnt REALLY matter though .. am just curious .. cheers

#7 Mr H

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Posted 07 August 2006 - 07:42 PM

You can get answers here: http://www.madras.fife.sch.uk/maths/answer...nswersAH06.html if that's any help?

H tends 2 infinity

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Never argue with an idiot. They drag you down to their level then beat you with experience.


#8 waitingforan_alibi

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Posted 08 August 2006 - 10:36 AM

looks like my work here is done biggrin.gif

i ll try and post the proofs if i can be bothered at some point


#9 ad absurdum

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Posted 09 August 2006 - 09:50 AM

QUOTE(ad absurdum)
Also, does anyone have a copy of the paper?
Anyone? I'll love you forever....
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#10 dfx

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Posted 09 August 2006 - 01:23 PM

QUOTE(ad absurdum @ Aug 9 2006, 10:50 AM) View Post

QUOTE(ad absurdum)
Also, does anyone have a copy of the paper?
Anyone? I'll love you forever....


You have PM. smile.gif

#11 ad absurdum

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Posted 09 August 2006 - 02:36 PM

QUOTE(dfx @ Aug 9 2006, 02:23 PM) View Post

QUOTE(ad absurdum @ Aug 9 2006, 10:50 AM) View Post

QUOTE(ad absurdum)
Also, does anyone have a copy of the paper?
Anyone? I'll love you forever....


You have PM. smile.gif
Thank you very much biggrin.gif

Since the answer sheet did not give proofs I've done the first couple of them.

For question 7.a):

Assume the statement is true. n3 - n is divisible by 6 then it must be divisible by both 2 and 3.

First consider the case of divisibility by 2:
n^3 - n \equiv 0 (mod 2) \\
n(n^2-1) \equiv 0 (mod 2) \\
now the factors are n and (n2-1). if one of these is odd, the other is even hence the product is even and the congruence in mod 2 holds true hence the product is divisible by 2.

Now consider the case of divisibility by 3:
Fermat's little theorem:
a^p \equiv a (mod p)
where a is an integer and p is prime
So:
n^3 \equiv n (mod 3) \\
n^3 - n \equiv 0 (mod 3)\\
hence it is divisible by 3. Since it is divisible by both 2 and 3 it is divisible by 6. Quod erat demonstrandum.

and for 7.b)

This is not true, n=5 is a counterexample.

and for 11.

1 + \cot^2x = \csc^2x \\
1 + \frac{\cos^2x}{\sin^2x} = \frac{1}{\sin^2x} \\
\sin^2x + cos^2x = 1
quod erat demonstrandum.

I'll need to think about the matrix one and the integration one for a bit. I'm not sure I have enough knowledge to tackle these yet to be honest sad.gif

Edit - made a stupid mistake. I fixed it now though, so hopefully this is right.
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#12 George

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Posted 09 August 2006 - 06:15 PM

QUOTE(ad absurdum @ Aug 9 2006, 03:36 PM) View Post

For question 7.a):

Another way to tackle it is:

n^3 - n = n(n^2 - 1) = (n+1)n(n-1)

Since it is a product of three consecutive integers, one is guaranteed to be divisible by 2, and another by 3. Thus 6 will always be a factor.

I think someone posted that solution just after the exam, and quoted a theorem that a number which is the product of k consective integers has k! as a factor.

Your approach was nice though smile.gif Congruences aren't really dealt with in AH if I remember correctly, and also Fermat's little theorem only holds if p is prime (which is fine in this question).

#13 ad absurdum

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Posted 09 August 2006 - 09:38 PM

QUOTE(George @ Aug 9 2006, 07:15 PM) View Post
Another way to tackle it is:

n^3 - n = n(n^2 - 1) = (n+1)n(n-1)

Since it is a product of three consecutive integers, one is guaranteed to be divisible by 2, and another by 3. Thus 6 will always be a factor.

I think someone posted that solution just after the exam, and quoted a theorem that a number which is the product of k consective integers has k! as a factor.

Your approach was nice though smile.gif Congruences aren't really dealt with in AH if I remember correctly, and also Fermat's little theorem only holds if p is prime (which is fine in this question).
Ah, I would never have thought of that. I also didn't know about the theorem you mentioned, I'll look into that smile.gif

I didn't realise that congruences weren't dealt with in AH. It seems a bit strange actually, I thought that they would get introduced at some point at school. I'm sure I'll learn the ins and outs of the AH course in due time though smile.gif

I really need to be more careful when I'm doing maths...in my last post I wrote that odd*even=odd but then finished off pretending I had wrote the right thing. I'm always dropping marks for stupid things like this sad.gif
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#14 waitingforan_alibi

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Posted 11 August 2006 - 06:06 PM

17a)



\begin{align*}\int {sin^2{x}cos^2{x}\, dx} \end{align*}

but since 

\begin{align*} sin^2{x}= 1 - cos^2{x}\end{align*}


\begin{align*}\int {(1-cos^2{x})cos^2{x}\, dx}\end{align*}


\begin{align*}\int {cos^2{x}\,dx} - \int {cos^4{x}\, dx}\end{align*}

b)



\begin{align*} \int_{0}^{\pi /4}{cos{x}cos^3{x} \, dx}\end{align*}

let 

\begin{align*} u = cos^3{x}\end{align*}



\begin{align*}v' = cos{x} \\ \end{align*}


\begin{align*} u' = -3cos^2{x}sin{x} \end{align*}



\begin{align*}v = sin{x}\\\end{align*}
using integration by parts,


\begin{align*} = \frac{1}{4} + 3 \int_{0}^{\pi /4} {cos^2{x}sin^2{x}\,dx} \end{align*}


thats the first two parts
i ll post c and d tomorrow since i have them on paper (this is my first time using LaTex so it gets very tiring)

#15 ad absurdum

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Posted 11 August 2006 - 10:04 PM

QUOTE(waitingforan_alibi @ Aug 11 2006, 07:06 PM) View Post

17a)



\begin{align*}\int {sin^2{x}cos^2{x}\, dx} \end{align*}

but since 

\begin{align*} sin^2{x}= 1 - cos^2{x}\end{align*}


\begin{align*}\int {(1-cos^2{x})cos^2{x}\, dx}\end{align*}


\begin{align*}\int {cos^2{x}\,dx} - \int {cos^4{x}\, dx}\end{align*}
I am officially an idiot for not noticing that tongue.gif

Thanks for posting them though.
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#16 waitingforan_alibi

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Posted 12 August 2006 - 06:15 PM

17 c)



\begin{align*} \int_{0}^{\pi /4} {cos^2{x} \, dx} \end{align*}
i did this a really weird way in the exam - i used integration by parts, but this is how i did it at the time:

let


\begin{align*}u=cosx\end{align*}


\begin{align*}v'=cosx\end{align*}


\begin{align*}u'=-sinx\end{align*}


\begin{align*}v=sinx\end{align*}



\begin{align*}\int_{0}^{\pi /4} {cos^2{x} \, dx} = [cos{x}sin{x}]_{0}^{\pi /4} + \int_{0}^{\pi /4} {sin^2{x} \, dx} \end{align*}

but (sinx)^2 = 1-(cosx)^2, so



\begin{align*} \int_{0}^{\pi /4} {cos^2{x} \, dx} = 1/2 + \int_{0}^{\pi /4} {1 \, dx} - \int_{0}^{\pi /4} {cos^2{x} \, dx}\end{align*}

=>

\begin{align*}2 \int_{0}^{\pi /4} {cos^2{x} \, dx} = \frac{1}{2} + \int_{0}^{\pi /4} {1 \, dx}\end{align*}



\begin{align*}2 \int_{0}^{\pi /4} {cos^2{x} \, dx} = \frac{1}{2} + \frac{\pi}{4}\end{align*}



\begin{align*}2 \int_{0}^{\pi /4} {cos^2{x} \, dx} = \frac{\pi + 2}{4}\end{align*}



\begin{align*} \int_{0}^{\pi /4} {cos^2{x} \, dx} = \frac{\pi + 2}{8}\end{align*}

17 d) when i ve had a break from all this 'texing

#17 waitingforan_alibi

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Posted 12 August 2006 - 06:31 PM

17 d)
using 17 b)


\begin{align*}\int_{0}^{\pi /4} {cos^4{x} \, dx} = \frac{1}{4} +3\int_{0}^{\pi /4} {cos^2{x}sin^2{x} \, dx} \end{align*}

using 17 a) on the above,


\begin{align*}\int_{0}^{\pi /4} {cos^4{x} \, dx} = \frac{1}{4} +3(\int_{0}^{\pi /4} {cos^2{x} \, dx} - \int_{0}^{\pi /4} {cos^4{x} \, dx}) \end{align*}

so


\begin{align*}4\int_{0}^{\pi /4} {cos^4{x} \, dx} = \frac{1}{4} +3\int_{0}^{\pi /4} {cos^2{x} \, dx} \end{align*}

using the result from 17c)


\begin{align*}4\int_{0}^{\pi /4} {cos^4{x} \, dx} = \frac{1}{4} +\frac{3\pi + 6}{8} \end{align*}



\begin{align*}4\int_{0}^{\pi /4} {cos^4{x} \, dx} = \frac{3\pi + 8}{8} \end{align*}



\begin{align*}\int_{0}^{\pi /4} {cos^4{x} \, dx} = \frac{3\pi + 8}{32} \end{align*}

-fin-
there might have been better ways to do 17 but this is how i did it in the exam at the time

#18 waitingforan_alibi

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Posted 13 August 2006 - 10:49 AM

17 was the question that most people seemed to have problems with, but if there was a specific question you want done to see teh full working - just ask





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