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[Kam!KaZe]

Dopey BOS losers!

Heya! I reckon that was a good 3u exam... Yes true question 7 was abit tricky but ya'll know that if hardly anyone got it out, then we all get the marks! So down with the smart peeps You know, after that 4u maths exam on monday... any exam would do for me HONESTLY, why set exams for when no1 can do them! Those dopes at BOS are just wasting their time! REVENGEEEE!! I say we assasinate the BOS head boss MUHAHHHAHHHAHAHAH!!!! :axedeath:

Anyone prepared for Monday's 3u english exam?

 

[Leo]

English X1 will be easy I reckon. You only have to study for the essay bit, the rest you make it up on the spot. Anyway, no time to prepare, this week is hell alreayd, got two more exams!!! :mad1:

Damn biology tomorrow......... :mad1:

 

[nakata]

i think spice girl got it wrong=)
the 4U test was way harder then the 3U test, as it should be.
And i think i did ok in the 3U one i had enuf time to go throught the questions i was unsure of, and im not even going to mention how i went in 4U=)

The graph in Q5. as as the chick said, is jus a straight y=pi/2 line from 0<x<1 hehe, i had to sub in no. to double check that.

i had half an hour to work Q7 b., and i almost didnt work it out. But im not sure if its right.
so you guys can see, and tell me the answer, i think im wrong.

7b) i). just a matter of binom expansion, multiply everything by x, differentiating, then subbing x=1.

7bii) not sure, but just divided everything in part i by {(n+1)^2}{(n+2)} , then simplified and multiplied everything by (-1)^n. i got {(-1^n)2*n-1}/(n+1)^2 or something like that. Doesnt look right to me.

but im glad maths is over. hate the tests that are comin up though.....argggg

 

[kaseita]

4U was harder than 3U, as it should be.
But 3U was nicely balanced I'd say. Had both easy and hard components, with enough time to work on the harder ones for the 4U students.
I did b)i) a different way, but everyone I know did it your way.
Mine was just longer.

 

[drolle]

I had sort of the opposite experience to josie_is_slut, I was going really fast through the whole paper, skipping the odd question, until I got to the end with 20 minutes left and nothing left to do, so I fiddled around a bit with the questions I had skipped because I had trouble with them. (exactly the same thing that happened to me in 4 unit)

I managed to get another 3 marks by doing Q6 (b) (ii) (It's not as hard as it looks, you just have to split up the inequality from (i) into two equations and look at each one seperately), but other than that I didn't get anywhere.

I found everything to be pretty simple exept for:
Q5(c) (I missed out a 2 somewhere I think, damn calculus tripped me up in 4u as well),
Q7(a) (iii) (WTF? how do you solve x = e^y + e^-y for y? you can't even take out a factor of e^-y, because you still can't do ln(e^2y+1)),
Q7(b)(i) (yeah, yeah, I've worked it out now *slaps self on forehead*)

All up I'm hoping for >80% raw (81/84? you got to be kidding me)

Damn, well that's all the fun exams done , now I'm stuck with physics and chemistry. (Damn social impact! Why did they have to go ruin a perfectly good subject by turning it into a history essay?)

 

[martin]

I liked it a lot, think I could have got close to 100%. Some of it was hard but it was the sort of hard stuff I'm good at. For anyone that's wondering for 7b.
i) differentiate both sides of binomial expansion and then add this to binomial expansion with x=1
ii) integrate binomial expansion twice, evaluating both constants by x=0. Then put x = -1 and it should add up to 1/(n + 2)

 

[wogboy]

I did everything except for Q[6b] (damn inequalities), and Q[7b ii]. I have to mention that I'm the king of careless mistakes, so that will bring down my mark a fair bit. I'm expecting to get hopefully about 70/84 raw. It was a helluva lot easier than the 4U exam though.

 

[Yellow]

yeh i thought the 3u exam was way more easier than th 4u. i got through it the first hour, leaving me with an hour to try and do the questions from 6b) onwards. i'm thinking to got high 80's, low 90's raw for that one.

but 4u though.... probably high 50's - 60's if i'm lucky (i'm talking percentage). i think it was the intimidation factor that got me. from question 2, things were just falling apart.

 

[Dumbarse]

i thought it was pretty alright
couldnt do 6b) but i found out how easy the first part was later
and ran out of time to do 7b)

 

[nakata]

to drolle's question of 7a iii)

it was pretty hard i admit, and it took me a few sec to realise
but its jus a simple quadratic.
all you do is multiple everything by e^y and move everything to one side to get (e^2y) - (xe^y) + 1 i think
then thats a quadratic, and solve for e^y, and then log
then you get an ugly answer i think.

 

[drolle]

people keep mentioning 6(b)(ii), so in case some people are still unsure how to do it, you just look at 1/(n+1) < S dx/x first, which gives
1/(n+1) < ln((n+1)/n)
e^(1/n+1) < 1+1/n
e^(1/n+1)^(n+1) < (1+1/n)^(n+1)
e < (1+1/n)^(n+1)

do a similar thing the other half of the inequality, and you're done

 

[kaseita]

Originally posted by martin

i) differentiate both sides of binomial expansion and then add this to binomial expansion with x=1

alright, maybe I wasn't the only one

 

[spice girl]

Well it was harder than my expected 3U paper.

Anyway, for those who're interested in 7b) WARNING: SPOILER AHEAD!

(i) Consider the polynomial:
a(x) = (c0) + 2(c1)x + 3(c2)x^2 + ... + (n+1)(cn)x^(n)

The integral of a(x) (call this A(x)) is:
A(x) = (c0)x + (c1)x^2 + (c2)x^3 + ... + (cn)x^(n+1)

= x(c0 + (c1)x + (c2)x^2 + ... + (cn)x^n)
= x(1+x)^n

differentiating A(x)
A'(x) = (1+x)^n + nx(1+x)^(n-1)

= (nx + x + 1)(1+x)^(n-1)

a(x) = A'(x) since both are the derivatives of the same function
and (c0) + 2(c1) + 3(c2) + ... + (n+1)(cn) = a(1)
and a(1) = A'(1) = (n+2)*2^(n-1)
Thus (c0) + 2(c1) + 3(c2) + ... + (n+1)(cn) = (n+2)*2^(n-1)

(ii) There's an easier way for (i), but my way takes care of both parts with the same method. The easier way was to consider c0 = cn, c1 = c(n-1), etc.

Anyway...

Consider f(x) = (c0)(x^2)/(1*2) - (c1)(x^3)/(2*3) + (c2)(x^4)/(3*4) - ... + ((-1)^n)(cn)(x^(n+2))/(n+1)(n+2)

the differential, f'(x) is
f'(x) = (c0)x/1 - (c1)(x^2)/2 + (c2)(x^3)/3 - ... + ((-1)^n)(cn)(x^(n+1))/(n+1)

Double differential, f"(x) is
f"(x) = (c0) - (c1)x + (c2)(x^2) - ... + ((-1)^n)(cn)x^n
= (1-x)^n
That's how to get (1-x)^n

Integrating f"(x)
F1(x) = -(1-x)^(n+1)/(n+1) + C1
Compare with f'(x)
f'(0) = 0
F1(0) = -1/n+1
thus C1 = 1/(n+1), if F1(x) = f'(x)

Integrating F1(x):
F2(x) = -(-(1-x)^(n+2))/(n+1)(n+2) + xC1 + C2
= (1-x)^(n+2)/(n+1)(n+2) + x/(n+1) + C2
Compare with f(x)
f(0) = 0
F2(0) = 1/(n+1)(n+2) + C2
C2 = -1/(n+1)(n+2)
So f(x) = (1-x)^(n+2))/(n+1)(n+2) + x/(n+1) - 1/(n+1)(n+2)

The sum we want is conveniently f(1)
Sub 1 into our new expression for f(x)
f(x) = 0 + 1/(n+1) - 1/(n+1)(n+2)
= 1/(n+2)

 

[sango]

question 7

i managed to get question 7 b) without any calculus. no integration and no differentiation.
I basically took the expression (1+x)^(-1) expanded that, and it all worked out. for part ii), i just subbed in x=-2 from what i had worked out in part i) and it worked out to be that. A bit of fudging i guess but it's perfectly valid for a 'show' question.


i think i got max 7. wrong (presuming no mistakes) and if they are lenient around 2 or 3 marks off..

 

[drolle]

ahhh oh course it all makes sense now, thanks nakata

 

[Sourie]

that was the easiest test i have ever come across apart from that son of a bitch last question that i swear cannot be done PORPERLY without binometic expansions......FUCK!!! i spent like half an hour on that damn 7b and in the end i just proved it in the PROPER way through binometrics!!!


i reckon 80/81 if i get them to not count the last question or maybe 79/81 as i am allowing for one careless mistake to be made.......it'd better not stuff up my assessment mark of 47.3!!

damn you Mr BOS for writing that test!!!

cheack the damn test next time so it has questions that can be done using the stuff in the syallabus.....if i dont get at least 79 then some one will suffer biatch!!

 

[drolle]

spice girl, when you say, 'the easier way', did you mean this?
since c0 = cn,
LHS = (n+2)/2*(c0 + c1 + c2 + ... + cn)
= (n+2)/2*(1+1)^n
= (n+2)*2^(n-1)
= RHS

of course I didn't think of this till 15 minutes *after* the exam finished!

 

[Winsux]

too bad you don't find out your raw marks then Sourie. By the way, 7*12 is 84, not 81

i reckon it was a fair test, except of course the end of 6 and 7. but there has to be some hard stuff for the genius' i reckon i got about 80-85% raw

 

[spice girl]

Originally posted by drolle
spice girl, when you say, 'the easier way', did you mean this?

yeah

 

[BlackJack]

Oooh... I found out staight afterwards how to do the two parts I didn't finish (the e^x inequailty which I forgot about and f-1(x))...
sourie was assuming they didn't count the past part worth 3 with evaluate the nCr sum part.
I went slowly thrugh the paper. Theoretical maximum is 81... good enough for me.