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AH Maths 2006 Exam Post Mortem


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Poll: How many marks (out of 100) do you think you got? (37 member(s) have cast votes)

How many marks (out of 100) do you think you got?

  1. Less than 30 (6 votes [16.22%])

    Percentage of vote: 16.22%

  2. 30-50 (4 votes [10.81%])

    Percentage of vote: 10.81%

  3. 50-60 (5 votes [13.51%])

    Percentage of vote: 13.51%

  4. 60-70 (11 votes [29.73%])

    Percentage of vote: 29.73%

  5. 70-80 (5 votes [13.51%])

    Percentage of vote: 13.51%

  6. 80-90 (2 votes [5.41%])

    Percentage of vote: 5.41%

  7. 90-100 (4 votes [10.81%])

    Percentage of vote: 10.81%

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#121 Jason Bourne

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Posted 20 May 2006 - 01:27 PM

i think i got x=1 y=1 z=0
Nothing!!!!!!!!

#122 south lanarkshire jag

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Posted 20 May 2006 - 01:30 PM

QUOTE(Jason Bourne @ May 20 2006, 02:27 PM) View Post

i think i got x=1 y=1 z=0



im pretty sure im right cos i put them back into the equations and got the right answer

#123 Jonny-Sensei

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Posted 20 May 2006 - 01:44 PM

Both your answers appear to be right though?

therefore infinite solutions???
Roll on ze gap year !!

#124 south lanarkshire jag

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Posted 20 May 2006 - 01:47 PM

QUOTE(PinkFloyd @ May 20 2006, 02:44 PM) View Post

Both your answers appear to be right though?

therefore infinite solutions???




oooo i get it now

cos the gaussiam elimination was of the sytle that it had an infinite amount of solutions then you were supposed to say that it was infact redundant

can anyone back up what im saying?

#125 bob753

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Posted 20 May 2006 - 02:09 PM

yes i can as lots of people here are getting results and if are put back into the equation then they will get the right answer but this is because the equations are redundent i.e infinite number of solutions
so every one will be thinking they are right when in fact they are wrong as there solotion is only one of many


#126 south lanarkshire jag

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Posted 20 May 2006 - 02:12 PM

QUOTE(bob753 @ May 20 2006, 03:09 PM) View Post

yes i can as lots of people here are getting results and if are put back into the equation then they will get the right answer but this is because the equations are redundent i.e infinite number of solutions
so every one will be thinking they are right when in fact they are wrong as there solotion is only one of many



thats kinda crap cos not many people will relise you will get loads of results

#127 Jason Bourne

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Posted 20 May 2006 - 02:34 PM

so is the answer: there are infinite solutions.

For 5 marks? huh.gif
Nothing!!!!!!!!

#128 south lanarkshire jag

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Posted 20 May 2006 - 02:40 PM

QUOTE(Jason Bourne @ May 20 2006, 03:34 PM) View Post

so is the answer: there are infinite solutions.

For 5 marks? huh.gif



no

you need to solve the equation and show there are infinite solutions

#129 bob753

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Posted 20 May 2006 - 03:18 PM

when you solve the equations the bottom line of the matrix should be

0 0 0 0

when there are 4 zeros that meens there are infinite solutions

#130 waitingforan_alibi

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Posted 20 May 2006 - 04:00 PM

you have to show that there are infinite solutions and then find the line that they intersect on

#131 YIC

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Posted 20 May 2006 - 04:08 PM

I got parametric equations

I ain't sure if you had to do that, but hey

#132 bigyun

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Posted 20 May 2006 - 04:31 PM

did any1 get e^x(sin2x) for the differential equation

or sumthin alog those lines?

#133 Jonny-Sensei

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Posted 20 May 2006 - 05:01 PM

Yeah that looks pretty familiar. I believe the answer was in the form  y=e{^-^x}(Asinx + Bcosx)

then the proper substitution gave B=0

differentiating gave:  dy/dx = e{^-^x}(Acosx - Bsinx) - e{^-^x}(Asinx + Bcosx)

substituting gave A=2

Back substitution :

 y=e{^-^x}((2)sinx + (0)cosx)
 y = 2e{^-^x}sinx

not 100% sure if thats right, but that's what I did.
Roll on ze gap year !!

#134 marcinkania7

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Posted 20 May 2006 - 05:41 PM

I would say (and I'm pretty confident in these results) that the Gaussian elimination's solution is
x = 1
and
y = 2z + 1

and the differential equation:
y = e^(-x)sin(x)

because when you differentiate it and substitute it it works so it should be correct..

#135 Pringles

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Posted 20 May 2006 - 07:22 PM

QUOTE
I would say (and I'm pretty confident in these results) that the Gaussian elimination's solution is
x = 1
and
y = 2z + 1

and the differential equation:
y = e^(-x)sin(x)

because when you differentiate it and substitute it it works so it should be correct..


I got the same answer for the gaussian elimination but realised I made a mistake on the value of x in the exam. The differential equation I got y = 2e^-1(sinx)

#136 Jason Bourne

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Posted 20 May 2006 - 10:08 PM

Heres the question that everybody loves rolleyes.gif wacko.gif stuart.gif w00t.gif cool2.gif weeeeeeeeeeeee....


A. Show that

 \int sin^2x cos^2x dx = \int cos^2x dx - \int cos^4x dx

B. By writing  cos^{4}x= cosxcos^3x and using integration by parts, show that

 \int_{0}^{\frac {\pi} {4}}cos^4x dx = \frac {1} {4} + 3  \int_{0}^{\frac {\pi} {4}}sin^2xcos^2x dx

C. Show that  \int_{0}^{\frac {\pi} {4}}cos^2x dx = \frac {\pi + 2} {8}

D. Hence using the above results show that

 \int_{0}^{\frac {\pi} {4}} cos^4x dx = \frac {3\pi + 8} {32}

Nothing!!!!!!!!

#137 Jonny-Sensei

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Posted 21 May 2006 - 01:10 AM

AAAAAAAAAAAAAAAAAAAARGH IT BURNS!!!!!!!!!
Roll on ze gap year !!

#138 st-and Paul

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Posted 21 May 2006 - 10:57 AM

QUOTE(Jason Bourne @ May 20 2006, 11:08 PM) View Post

Heres the question that everybody loves rolleyes.gif wacko.gif stuart.gif w00t.gif cool2.gif weeeeeeeeeeeee....


A. Show that

 \int sin^2x cos^2x dx = \int cos^2x dx - \int cos^4x dx

B. By writing  cos^{4}x= cosxcos^3x and using integration by parts, show that

 \int_{0}^{\frac {\pi} {4}}cos^4x dx = \frac {1} {4} + 3  \int_{0}^{\frac {\pi} {4}}sin^2xcos^2x dx

C. Show that  \int_{0}^{\frac {\pi} {4}}cos^2x dx = \frac {\pi + 2} {8}

D. Hence using the above results show that

 \int_{0}^{\frac {\pi} {4}} cos^4x dx = \frac {3\pi + 8} {32}


Do you reckon you got at least a B in the exam? You will regret it if you get a C, it will affect your 1st semester module choices for Maths at St Andrews if you get anything lower. And from somone who did it, you dont want to do MT1001 Introductory Maths... complete waste of time. Waste of a semester. Complete with a lecturer who is so bad that he can make you forget how to do Integration...

#139 albidalbibab

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Posted 21 May 2006 - 01:52 PM

1

#140 sacraficialangel7

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Posted 21 May 2006 - 02:27 PM

OMG that paper was horrid! i dunno how anyone managed to solve the gaussian. i got x=1 and that was about it. used about 4 pages looking for y and z. nope! tried everything but noooo. don't think i even got 35 marks. that paper was outta 90 btw. some ppl in ma mate's school counted them all. i did all the questions in a random order, but wtf was Q17? and it didn't help that i revised all the wrong stuff. if i pass that, it'll be a miracle. glad it's only a fallback! eek!



Mod edit: Ouch. My eyes.







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