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Problem with Question 10 in 2009 HSC paper (1 Viewer)

Stuzul

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Hey
so yeah pretty much im struggling to understand a couple of the supplied answers

id post the questions as a pic but dont know how, and you kind of have to look at a, b and first part of c to kind of get it
its Q10c part ii -> i answered q10 correctly until here and dont understand te proof as such

links are
http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2009exams/pdf_doc/2009-hsc-exam-mathematics.pdf
and
http://www.boardofstudies.nsw.edu.a...s/pdf_doc/mathematics-sample-answers-2009.pdf

any help would be greatly appreciated. thanks :)
 

RishBonjour

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which part of the proof don't you understand?

basically, f dash x - g dash x is greater than or equal to zero for all x>=0, which is given, you get that from the previous question, and therefore f dash x >= g dash x?
 

Stuzul

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got the first bit but struggling to make the link between the first bit and the 2nd bit of what you said, so i get that f dash x - g dash x is always greater than zero, but why does that justify that fdashx is greater than gdashx?
 

RishBonjour

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lets put it this way:


10 - 3= 7
now 7 > 0 therefore 10 > 3
thats basically what they did. makes sense?
 

Stuzul

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haha oh wow, i feel like a moron. cheers
 

Stuzul

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The other question i can't understand is Question 6a from 2008

(a) Solve 2 sin^2 (x/3)= 1 for –π ≤ x ≤ π .

answers

(a)
solution.png

my questions are;
1. why does it change to x/3 rather than x - and does it always do this when the fucntion is x/3, for example?
2. what goes on between line 2 and line 3????
 

jet

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Hopefully you know that


So between lines two and three they are just taking the square root and you need to account for the fact that both the negative square root and the positive square root are possible.

I'm not sure what you mean by 1.
 

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