2008 Independent Trial (1 Viewer)

[vds700]

here is the paper

 

[friction]

Does it have answers ??

 

[vds700]

Does it have answers ??

i do have solutions, but i dont have time to scan them, too many pages, though i can do certain q's if people cant do them.

 

[Pwnage101]

thx for the paper

would it be possible to just type, without the full solutions, just the ANSWER for each non-proof question - ie the ANSWER for integrals (definite/indefinite), columes, etc

eg. 1)x)logea
y) pi/3

 

[Forbidden.]

回复: 2008 Independent Trial

Holy fuck have you people from MX2 ever been able to answer every single question in a MX2 paper in time?

 

[tacogym27101990]

Re: 回复: 2008 Independent Trial

Holy fuck have you people from MX2 ever been able to answer every single question in a MX2 paper in time?

thats the main challenge of 4 unit maths

 

[midifile]

Thanks = ]

 

[friction]

Re: 回复: 2008 Independent Trial

I just realised this is the paper my mate did in class (in the aloted time) and got 117 out of 120. I struggled like you wouldnt believe. That is why he is first and i am second i guess.

 

[midifile]

Re: 回复: 2008 Independent Trial

I just realised this is the paper my mate did in class (in the aloted time) and got 117 out of 120. I struggled like you wouldnt believe. That is why he is first and i am second i guess.

117 out of 120?!?! I didnt know that was possible

 

[shaon0]

Re: 回复: 2008 Independent Trial

I just realised this is the paper my mate did in class (in the aloted time) and got 117 out of 120. I struggled like you wouldnt believe. That is why he is first and i am second i guess.

State ranking?

 

[Timothy.Siu]

woah 117 for 4unit thats impressive

 

[Pwnage101]

i do have solutions, but i dont have time to scan them, too many pages, though i can do certain q's if people cant do them.

could u (or anyone else on that note) post up the solution to:

Q2.a.(iv)

Q2.b.(iii) (brief description)

Q2.c.(iii)

Q3.c.(ii)

Q4.b.(ii)

Q4.c.(iii)

Q6.a.(ii) & (iii) - just need to chek answer (just values of theta)

 

[qqmore]

Hey can you post up solutions for 8a) and c) .... its seems tricky, not sure how to approach it.

 

[Pwnage101]

i got 8(c) - yeh need help with 8 (a) ii as well

here, ill post solution for (c) now (this is my solution, dunno if its correct but it sounds right):

a^2+b^2+c^2 = ab + bc + ca
(x2)
2a^2+2b^2+2c^2 = 2ab + 2bc + 2ca
take all on one side with this structure:
(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ca+a^2)=0
factorise:
(a-b)^2+(b-c)^2+(c-a)^2=0....(1)

now (a-b)^2 is greatehr than or equal to 0
as is (b-c)^2 and (c-a)^2 because any real number squared is so

thus, in order for (1) to be true, there is only 1 posiibility - a=b=c, cause if it were any other way one of the three terms would be greater than 0, and thus the sum of them would be greater than 0, as none can be negative, but we derived it must = 0

thus a=b=c

this is the criterion for an equilateral triangle

QED

 

[vds700]

could u (or anyone else on that note) post up the solution to:

Q2.a.(iv)

Q2.b.(iii) (brief description)

Q2.c.(iii)

Q3.c.(ii)

Q4.b.(ii)

Q4.c.(iii)

Q6.a.(ii) & (iii) - just need to chek answer (just values of theta)

2a(iv) arg(z1z2 ^n) = pi/4 - n.pi/6 = (3-2n)pi/12. as (3-2n) cannot be 12 or 0 for integers n, cannot be real (to be real, must have an arg of 0 or pi)
2b(iii)unit circle, argz = pi/4, shows bar z and iz
2c(iii)sub pi/12 into (ii)

3c(ii)h(x)=e^x, h'(x) = e^x
lim[x->0] (e^x-1)/x = lim[x->0] (e^x-e^0)/(x-0) = h'(0) = 1 =f(0) therefore f)(x)is continuous
ill do rest later

 

[Trebla]

8a)i)
The equation of the tangent is y = x + 1 (check by differentiation)
So the length of CK is the y-value of the tangent at x = k which is k + 1
The area of trapezium OACK = k(1 + k + 1)/2 = k(k+2)/2
The area of trapezium OABK = k(e^k + 1)/2
Actual area under the curve is ∫₀^ke^xdx = e^k - 1
area of trapezium OACK < Actual area under the curve < area of trapezium OABK
=> k(k+2)/2 < e^k - 1 < k(e^k + 1)/2
Let k = 1
3/2 < e - 1 < (e + 1)/2
Taking the LHS of inequality: 3/2 < e - 1 => 2.5 < e
Taking the RHS of inequality: e - 1 < (e + 1)/2
=> 2e - 2 < e + 1
=> e < 3
.: 2.5 < e < 3

 

[Pwnage101]

VDS, with 2(c) (iii) if u sub in u get (cospi/12+sinpi/12) on the right as well, which i cant simplify

 

[vds700]

VDS, with 2(c) (iii) if u sub in u get (cospi/12+sinpi/12) on the right as well, which i cant simplify

(cospi/12+sinpi/12)
=rt2((1/rt2)cospi/12+(1/rt2)sinpi/12)
=rt2(cospi/4cospi/12 +sinpi/4sinpi/12) now just simplify using trig identities

 

[Pwnage101]

k thx, just finally - that circle geo's a bitch, can anyone give the sol?

 

[conics2008]

hey guys/girls i can post up worked solutions for that paper...

if you watn just request the specific questi