Javascript Disabled Detected

You currently have javascript disabled. Several functions may not work. Please re-enable javascript to access full functionality.

0

Solutions

Started by punto, May 20 2006 03:53 PM

You cannot reply to this topic
Go to first unread post

17 replies to this topic

#1 punto

Showing Improvement

Members
13 posts

Location:Falkirk
Interests:Football n girls
Gender:Male

Posted 20 May 2006 - 03:53 PM

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

Back to top

#2 Jason Bourne

Good Effort

Members
95 posts

Gender:Male

Posted 20 May 2006 - 10:10 PM

if someone knows the answers!?

Nothing!!!!!!!!

Back to top

#3 john g 22

Showing Improvement

Members
10 posts

Gender:Male

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

Back to top

#4 ad absurdum

Top of the Class

Members
390 posts

Location:Cambridge
Interests:Muzak.
Gender:Male

Posted 05 August 2006 - 11:39 AM

Also, does anyone have a copy of the paper?

HMFC - Founded 1874, beefing the Cabbage since 1875

Back to top

#5 waitingforan_alibi

Showing Improvement

Members
37 posts

Gender:Male

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.

Back to top

#6 john g 22

Showing Improvement

Members
10 posts

Gender:Male

Posted 06 August 2006 - 05:53 PM

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

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)

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

Back to top

#7 Mr H

Site Swot

Members
156 posts

Location:Soonagonner Retirement Home
Gender:Male

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

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

Back to top

#8 waitingforan_alibi

Showing Improvement

Members
37 posts

Gender:Male

Posted 08 August 2006 - 10:36 AM

looks like my work here is done

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

Back to top

#9 ad absurdum

Top of the Class

Members
390 posts

Location:Cambridge
Interests:Muzak.
Gender:Male

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....

HMFC - Founded 1874, beefing the Cabbage since 1875

Back to top

#10 dfx

Fully Fledged Genius

Members
1,955 posts

Gender:Male

Posted 09 August 2006 - 01:23 PM

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

QUOTE(ad absurdum)

Also, does anyone have a copy of the paper?

Anyone? I'll love you forever....

You have PM.

Back to top

#11 ad absurdum

Top of the Class

Members
390 posts

Location:Cambridge
Interests:Muzak.
Gender:Male

Posted 09 August 2006 - 02:36 PM

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

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

QUOTE(ad absurdum)

Also, does anyone have a copy of the paper?

Anyone? I'll love you forever....

You have PM.

Thank you very much

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

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

HMFC - Founded 1874, beefing the Cabbage since 1875

Back to top

#12 George

Child Prodigy

Admin
720 posts

Location:West Lothian
Gender:Male

Posted 09 August 2006 - 06:15 PM

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

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

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).

Higher Maths Past Paper Solutions

Back to top

#13 ad absurdum

Top of the Class

Members
390 posts

Location:Cambridge
Interests:Muzak.
Gender:Male

Posted 09 August 2006 - 09:38 PM

QUOTE(George @ Aug 9 2006, 07:15 PM)

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

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

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

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

HMFC - Founded 1874, beefing the Cabbage since 1875

Back to top

#14 waitingforan_alibi

Showing Improvement

Members
37 posts

Gender:Male

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)

Back to top

#15 ad absurdum

Top of the Class

Members
390 posts

Location:Cambridge
Interests:Muzak.
Gender:Male

Posted 11 August 2006 - 10:04 PM

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

17a)

$\begin{align*}\int {sin^2{x}cos^2{x}\, dx} \end{align*}$ but since $\begin{align*} sin^2{x}= 1 - cos^2{x}\end{align*}$
$\begin{align*}\int {(1-cos^2{x})cos^2{x}\, dx}\end{align*}$
$\begin{align*}\int {cos^2{x}\,dx} - \int {cos^4{x}\, dx}\end{align*}$

I am officially an idiot for not noticing that

Thanks for posting them though.

HMFC - Founded 1874, beefing the Cabbage since 1875

Back to top

#16 waitingforan_alibi

Showing Improvement

Members
37 posts

Gender:Male

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

Back to top

#17 waitingforan_alibi

Showing Improvement

Members
37 posts

Gender:Male

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

Back to top

#18 waitingforan_alibi

Showing Improvement

Members
37 posts

Gender:Male

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

Back to top

Back to Mathematics

1 user(s) are reading this topic

0 members, 1 guests, 0 anonymous users