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General thoughts on the 2008 CSSA Trial Extension 1 Mathematics Paper (2 Viewers)

davOmeter

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Omfg, that exam was /wrist worthy

Did anyone get the newtons method question?
Like wtf,there was no function?!
 

vds700

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davOmeter said:
Omfg, that exam was /wrist worthy

Did anyone get the newtons method question?
Like wtf,there was no function?!
it was x = 9^(1/3)
x^3 = 9
f(x) = x^3 - 9 = 0
 

davOmeter

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Daaaa ah gaddd, how could I have been so stupid? Least I know 5 other people who didn't get that question haha.
 

lorikeet

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brendanstacey said:
DONT DROP IT!!!

If your getting 50-60 for that exam your basically garanteed a band 6. Remember to compare yourself to the rest of the state, of whom most would fail this badly.
Really? are 50s-60s the kind of raw marks that get bands 5 and 6? I mean, I would love for that to be true, butI thought it was more like 70s?
 

erm

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lorikeet said:
Really? are 50s-60s the kind of raw marks that get bands 5 and 6? I mean, I would love for that to be true, butI thought it was more like 70s?
i would also like to know this...anyone care to make an educated guess at the minimum raw mark for that test that would probably gain a band six?
 

leoyh

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the test was harder than previous years papers but overall it was okay, just a bit pushed for time cause my school took out the Q's we did and instead of replacing it, they put a Q8 and Q9 instead.

and yes, the growth one is 9.6 days
 

proringz

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davOmeter said:
Omfg, that exam was /wrist worthy

Did anyone get the newtons method question?
Like wtf,there was no function?!
That question confused me a bit too but in the end I got it. just minus the 3root9 to the other side so it becomes x - 3root9.
 

DunnyBasher

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proringz said:
That question confused me a bit too but in the end I got it. just minus the 3root9 to the other side so it becomes x - 3root9.
That's what I initially started to do, then realised my logical flaw
You are trying to obtain a value for x=3rt9, so...you cant use 3rt9 in obtaining that answer, otherwise it completely defeats the purpose of using Newton's method.
So, I just cubed and subtracted the 9.
You would have got the correct answer, but whether you will get full marks is questionable.
 

brendanstacey

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Well if you can get 50-60 for this trial which is allways harder than the hsc, youre quite likely to get at least the same in the HSC, and from what my teacher told me, thats all you need.
 

proringz

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DunnyBasher said:
That's what I initially started to do, then realised my logical flaw
You are trying to obtain a value for x=3rt9, so...you cant use 3rt9 in obtaining that answer, otherwise it completely defeats the purpose of using Newton's method.
So, I just cubed and subtracted the 9.
You would have got the correct answer, but whether you will get full marks is questionable.
Oh yeah, I understand your point. I subbed my answer in and it was right, but probably not full marks. But then again if x= 3root9 and if your're finding x, then x-3root9 should = 0.
 

ashraf91

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vds700 said:
it was x = 9^(1/3)
x^3 = 9
f(x) = x^3 - 9 = 0


but y would u do dat, introdcing more solutions wen 9^[1/3] is a constant, it like y^2=x isnt d same as y=x^0.5, please correct me cause i fink im rong
 

davidlee90

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no one's talking about the crazy 3D trig yet... you guys found them easy???
i had 10 degrees and 40 mins as my answer. Anyone else?
 

Zephyrio

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davidlee90 said:
no one's talking about the crazy 3D trig yet... you guys found them easy???
i had 10 degrees and 40 mins as my answer. Anyone else?
Yeah I got this.

It wasn't that hard, was it? I think some people were just put off by the diagram. All you had to do was use the sine rule in the bottom triangle, and use tan in the left-side triangle.
 

Zephyrio

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ashraf91 said:
but y would u do dat, introdcing more solutions wen 9^[1/3] is a constant, it like y^2=x isnt d same as y=x^0.5, please correct me cause i fink im rong
Well if you were that example, y^2 = x,

y = + or - x^0.5. Graph that and you'd get x = y^2.
 

proringz

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Zephyrio said:
Yeah I got this.

It wasn't that hard, was it? I think some people were just put off by the diagram. All you had to do was use the sine rule in the bottom triangle, and use tan in the left-side triangle.
I wasn't put off by the diagram; I was put off by the bearings.
 

DavidTan

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omg, did anyone get the rate of change question on the rocket launching and the camera tracing its path?!
totally looked easy, but stuffed it up badly *sigh* oh well..... :cold:
 

DavidTan

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yeh i know, wat the hell was that all about?!
how in the hell were we supposed to test that?
how the f**k were we supposed to differentiate that to apply newton's thing?
anyone get that either?
 

watzupwitdat

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erm said:
weren't you supposed to start with the other expression and then differentiate it? Then substitute appropriately to get the above expression
I'm not sure, but i was always taught that with a question where they give u a dN/dt and u have to show that N satisfies it, then u have to integrate the dN/dt because they asked u to show, and that differentiating the N would only be a verification.
 

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