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HSC 2015 MX2 Marathon (archive) (1 Viewer)

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Axio

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Re: HSC 2015 4U Marathon




Adding these three inequalities yields:

The strict inequality is true since all three inequalities are zero at different values of x.
Adding only furthers the strictness of the inequality, which leads to:

Which after re-arranging gives the desired inequality.
Now, my question is, how many marks would I get (compared to whatever would be given) in the HSC for doing it this way instead of calculus? (I have nothing against calculus, it's just that I will always try to solve a problem in the most elegant way I can (quickly) think of)
I would have thought that unless they asked you to use calculus or it had 'hence' in the question (and a previous part involved using calculus) that you would get full marks...
 

Drsoccerball

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Re: HSC 2015 4U Marathon

LOOOL my teacher expected us to do this :

From AM-GM






 
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Re: HSC 2015 4U Marathon

i know this is a rlly easy q but i cant seem to get it.
how do u integrate cosec 2x using t formula?
thanks
 

VBN2470

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Re: HSC 2015 4U Marathon

Prove that if is a positive integer such that , then at least one of , or is composite.
 

dan964

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Re: HSC 2015 4U Marathon

Prove that if is a positive integer such that , then at least one of , or is composite.
It surfices to show first, the each prime number bigger than 3 is either 6K + 1 or 6K+5 where K is an integer.
Firstly all numbers can be written in the form 6K+n where K=>0, 0<=n<6
All numbers of the form 6K+2 and 6K+4 are composite, except 2 (divisible by 2) and 6K+3 is also composite, except 3 (divisible by 3).

Thus the question only concerns prime numbers which are of the form 6K+1 or 6K+5 (and not 2 or 3 obviously), as composite numbers already satisfy the result:
If n=6K+1, n+2 = 6K+3 which is composite (divisble by 3)
If n=6K+5, n+4 = 6K+9 which is composite (divisble by 3)
 

VBN2470

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Re: HSC 2015 4U Marathon

Yep, you're method is fine. Here's mine (similiar):

Since we have , we have the following:

Case 1: Let for some positive integer . Then is divisible by 3 and we are done.

Case 2: Let for some positive integer . Then is divisible by 3 and we are done.

Case 3: Let for some positive integer . Then is divisible by 3 and we are done.

Hence, at least one of , or is composite.
 

VBN2470

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Re: HSC 2015 4U Marathon

NEW QUESTION:

Prove that a number is divisible by 3 if and only if the sum of its digits is divisible by 3.
 

Drsoccerball

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Re: HSC 2015 4U Marathon

NEW QUESTION:

Prove that a number is divisible by 3 if and only if the sum of its digits is divisible by 3.
I remember remembering this in year 8 :p dont know how to prove it tho
 
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