Post by: matteh on November 07, 2008, 11:02:02 am
Post by: AppleXY on November 07, 2008, 11:05:33 am
Post by: shadezofemerald on November 07, 2008, 11:07:22 am
Post by: iamdan08 on November 07, 2008, 11:07:34 am
Post by: random.photon on November 07, 2008, 11:08:31 am
Post by: matteh on November 07, 2008, 11:12:16 am
Post by: random.photon on November 07, 2008, 11:12:38 am
Not according to our examiner.. he told us to change it to e x 3x not e^3x. Bloody hell.
=/
Post by: Lulu on November 07, 2008, 11:19:36 am
Post by: iamdan08 on November 07, 2008, 11:24:41 am
Post by: matteh on November 07, 2008, 11:28:09 am
Post by: kurrymuncher on November 07, 2008, 11:34:17 am
THAT FUCKEN PRISM QUESTION PISSED ME OFF
Post by: vce01 on November 07, 2008, 11:55:08 am
Post by: camlawl on November 07, 2008, 11:56:15 am
so yeah, they just told us it was a 3x and said nothing about changing it from e^3x to e3x.....
someone put up answers :knuppel2:
and i hope they dont disclude it. pretty easy for 3 marks. my answer was 1....
Post by: matteh on November 07, 2008, 11:57:24 am
Post by: matteh on November 07, 2008, 11:59:58 am
Quote
1)a) dy/dx=5(6x-5)(3x^2-5x)^4
1)b) f'(0)=1 (lolx at the slight misprint in the question stem!)
2) Negative hyperbolic graph: x-int at (1,0), y-int at (0,-2)
Vertical asymptote x=-1, horizontal asymptote y=2
3) x={-2pi/9, 2pi/9} over [-pi/2,pi/2]
4)a) Show that k=pi/2 by integration such that A=1
4)b) Pr (X<1/4 | x<1/2) = [2-sqrt(2)]/2
5) C= 0.5log(6)
6)a) Mode=3 (0.4 > {0.3,0.2,0.1})
6)b) Probability=0.3*
{*Derived from Binomial Distribution, where n=2, r=2, p=Pr(X=x):
Pr(X=2)=(2 C 2)[Pr(X=x)]^2[1-Pr(X=x)]^0 = [Pr(X=x)]^2
=> Probability = 0.1^2+0.2^2+0.3^2+0.4^2 = 0.3}
7) Markov Chain: Pr(CCD) + Pr(CDC) + Pr(DCC) = 0.336
8)a) dom f' = R/{1,2}
8)b) Modulus graph...stationary points (local maxima) (-1,4), (2/3 , 17/27)
Non-inclusive endpoint at (1,0), flip the given curve such that y>=0
9)a) y=[4000sqrt(3)]/[3(x)^2]
9)b) Show that A=[4000sqrt(3)]/x + [(sqrt(3)x^2)/2]
9)c) Minimum** Surface Area at x=cuberoot(4000) =10cuberoot(4) cm (looking for tangible evidence that VCAA overlooks root simplification...)
{**The graph for A(x) is an oblique curve whose only stationary point is a local minimum at x=10cuberoot(4), hence the minimum surface area occurs at x=10cuberoot(4)}
10)a) Inverse f: (-1,infinity)->R, inverse f(x) = 0.5loge(x+1)
10)b) Graph of y=x, x>1 with non-inclusive endpoint at (-1,-1)
10)c) (-2x)/(2x+1)*** => a=-2,b=2,c=1
{***We need to evaluate f[-inverse f(2x)] => -inverse f(2x) = -0.5loge(2x+1) = 0.5loge[(2x+1)^(-1)] = 0.5loge[1/(2x+1)]
f[-inverse f(2x)] = e^2[0.5loge[1/(2x+1)]]-1 = 1/(2x+1) - 1 = [1-2x-1]/[2x-1] = (-2x)/(2x-1), QED}
Post by: camlawl on November 07, 2008, 12:45:08 pm
Post by: kurrymuncher on November 07, 2008, 12:47:57 pm
would i lose marks for not labelling endpoints of last graph and that mod graph?yeah, i think you would
Post by: camlawl on November 07, 2008, 12:53:47 pm
maybe, but they were only work 1 and 2 marks. and for the mod graph they only asked for turning points.
hmmm
Post by: mskwmskw on November 07, 2008, 01:13:34 pm
Post by: dekoyl on November 07, 2008, 01:54:01 pm
At the start I used (1/2)Base x height though and then spent 10 minutes doing the next part.. then I realised I used the wrong formula.
Post by: onlyfknhuman on November 07, 2008, 02:13:33 pm
thats wat it said on my sheet, it looke dlike it dispite it being blurred a bit.
I thougth he was just clarifying the numbers not the notation, i assumed that he didnt know how to pronounce the notation. LOl
Post by: xox.happy1.xox on November 07, 2008, 04:04:00 pm
Post by: Collin Li on November 07, 2008, 04:21:05 pm
Lol, when I was looking at the Markov chain question,one of the combinations was DCC, and it reminded me of dcc. Lol. :D
hahaha, what would dcc do?
Post by: Glockmeister on November 07, 2008, 04:24:00 pm
Post by: pHysiX on November 08, 2008, 09:29:49 am
Post by: xox.happy1.xox on November 08, 2008, 09:35:55 am
lol...exam 1 was awesome...my only doubt is the last question, question 10 c...i just wrote the thingo as (-2x)/(2x+1)...i didn't state a b or c coz wasn't the question asking for u to express the equation in the form of (ax)/(bx+c)?It wasn't necessary to state a,b and c, but I guess it's always better safe than sorry. You won't lose any marks, so don't worry.
Post by: pHysiX on November 08, 2008, 01:28:07 pm
Post by: danieltennis on November 08, 2008, 01:36:38 pm
Post by: ReVeL on November 08, 2008, 01:38:27 pm
- Left the trig question in the form
- In the last question, forgot to restrict the domain to (-1,infinite) when drawing the graph.
- For the intergration question, made an arithmatic error and ended up with
Oh well.
Thats what, 3 or 4 marks lost?
Post by: onlyfknhuman on November 08, 2008, 02:00:09 pm
Post by: dcc on November 08, 2008, 02:22:54 pm
Lol, when I was looking at the Markov chain question,one of the combinations was DCC, and it reminded me of dcc. Lol. :D
hahaha, what would dcc do?
I laughed when I saw that. I was tempted to write Pr(DCC) = 1337, but my 'not failing exams' instinct overrode any attempts at humour :)
Post by: Captain on November 08, 2008, 03:35:12 pm
How about drawing asymptotes without scattered lines, but just a straight line? Would i lose marks for that?
Yes. Why would you do that?