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September 19, 2025, 09:33:57 pm

Author Topic: Methods (CAS) 2009 exam 2 Suggested Solutions.  (Read 22602 times)  Share 

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jasrulz63

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Re: Methods (CAS) 2009 exam 2 Suggested Solutions.
« Reply #30 on: November 09, 2009, 04:33:44 pm »
I need to fix up the pdf to add in TT's amendments, then I'll upload it (assuming it's ok with him)

Over9000

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Re: Methods (CAS) 2009 exam 2 Suggested Solutions.
« Reply #31 on: November 09, 2009, 04:34:16 pm »
thus B is false
There is no stationary point at x=0, its undefined, did u differentiate the function?
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simpak

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Re: Methods (CAS) 2009 exam 2 Suggested Solutions.
« Reply #32 on: November 09, 2009, 04:34:37 pm »
I totally can't remember if I accidentally put D for the one about 'how many solutions'.
I either put E or D.
If I put D, I'll die D:
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TrueTears

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Re: Methods (CAS) 2009 exam 2 Suggested Solutions.
« Reply #33 on: November 09, 2009, 04:34:41 pm »
there is a stationary pt at (0,2) try it on your calculator
B is true.
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TrueTears

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Re: Methods (CAS) 2009 exam 2 Suggested Solutions.
« Reply #34 on: November 09, 2009, 04:34:59 pm »
I need to fix up the pdf to add in TT's amendments, then I'll upload it (assuming it's ok with him)
Yeah go for it.
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Crazometer

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Re: Methods (CAS) 2009 exam 2 Suggested Solutions.
« Reply #35 on: November 09, 2009, 04:35:16 pm »
Errr, not beating a dead horse or anything but isn't the domain for f(g(x)) the intersection of the domain of f(x) and range of g(x) ?

TrueTears

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Re: Methods (CAS) 2009 exam 2 Suggested Solutions.
« Reply #36 on: November 09, 2009, 04:35:45 pm »
Errr, not beating a dead horse or anything but isn't the domain for f(g(x)) the intersection of the domain of f(x) and range of g(x) ?
No it's the domain of g(x).
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Over9000

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Re: Methods (CAS) 2009 exam 2 Suggested Solutions.
« Reply #37 on: November 09, 2009, 04:35:49 pm »
Errr, not beating a dead horse or anything but isn't the domain for f(g(x)) the intersection of the domain of f(x) and range of g(x) ?
Thats for f(x) + or X g(x).
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hyperblade01

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Re: Methods (CAS) 2009 exam 2 Suggested Solutions.
« Reply #38 on: November 09, 2009, 04:36:24 pm »
With regards with 1e)i finding the rule

For f(g(x)) to exist , ran g needs to be a subset or equal to dom f right?

But dom f [0, infinity) whilst ran g is R?




EDIT: But I assumed it was possible otherwise that whole page woulda been blank
« Last Edit: November 09, 2009, 04:39:33 pm by hyperblade01 »
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hard

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Re: Methods (CAS) 2009 exam 2 Suggested Solutions.
« Reply #39 on: November 09, 2009, 04:37:09 pm »
OMGGGGGG WTFFFFFFFFF SOME OF THE Q'S  ON THE EXAM YOU POSTED TT WAS DIFFERENT TO THE QS ON MY EXAM LIKE Q1 MC AND THERE WAS NO GRAPH FOR Q1 OF SHORT ANSWER

WTFFFFFFFFFFFFFFFF

hyperblade01

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Re: Methods (CAS) 2009 exam 2 Suggested Solutions.
« Reply #40 on: November 09, 2009, 04:37:30 pm »
OMGGGGGG WTFFFFFFFFF SOME OF THE Q'S  ON THE EXAM YOU POSTED TT WAS DIFFERENT TO THE QS ON MY EXAM LIKE Q1 MC AND THERE WAS NO GRAPH FOR Q1 OF SHORT ANSWER

WTFFFFFFFFFFFFFFFF

It's CAS, you mighta done non-CAS
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almostatrap

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Re: Methods (CAS) 2009 exam 2 Suggested Solutions.
« Reply #41 on: November 09, 2009, 04:37:57 pm »
thanks TT!

lost 6 marks. would have only lost one but in my spare time i went back to EA 1 and fucked it all up lol
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TrueTears

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Re: Methods (CAS) 2009 exam 2 Suggested Solutions.
« Reply #42 on: November 09, 2009, 04:38:25 pm »
With regards with 1e)i finding the rule

For f(g(x)) to exist , ran g needs to be a subset or equal to dom f right?

But dom f [0, infinity) whilst ran g is R?




EDIT: But I assumed it was possible otherwise that whole page woulda been blank
Range g is .
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anti

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Re: Methods (CAS) 2009 exam 2 Suggested Solutions.
« Reply #43 on: November 09, 2009, 04:38:47 pm »
I checked the amended version, but I still think cii. is wrong unless I'm overlooking something which is very possible XD

You're told that you have a tin of balls, so can't you presume that X<68.5 and use 1 minus the result of ci as the probablity?

Thanks for the solutions by the way.

ash.w

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Re: Methods (CAS) 2009 exam 2 Suggested Solutions.
« Reply #44 on: November 09, 2009, 04:38:53 pm »
Quote
Nah the domain is R.
the domain is R+ U {0}