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HSC 2014 Maths Marathon (archive) (1 Viewer)

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hi im trash

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Re: HSC 2014 2U Marathon

How would you the first one using 2u method only?
i think its just a 1 mark question, because 1/(xlnx) can be seen as (1/x)/(lnx), hence, ln(lnx)+c.

Also the answer to this question http://postimg.org/image/id86iq0qb/ = ((pi^pi)-1)/ln(pi). (working, you just move the lnpi from the rhs to lhs and bingo)

also guys... how do you guys find these questions?! :OOO
 

hi im trash

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Re: HSC 2014 2U Marathon

Assuming it's 1 dx instead of two, it would eventually turn out like 1/x/lnx/ln (lnx) which integrated would be ln (ln (lnx))) + C. If it's two 2dx I'd have no clue
 

dunjaaa

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Re: HSC 2014 2U Marathon

yeah basically the same idea, [1/xln(x)]/[ln(ln(x))] and integrate
 

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Re: HSC 2014 2U Marathon

Sorry. A better question: Find the volume of the solid of rotation when the region bounded by the curve y=(x-4)(8-x) and the lines x=4, x=8 and y=10 is rotated about the x-axis.
 

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Re: HSC 2014 2U Marathon

I) it's a gp with r = -x
ii) integrate x^2n+1/1+x = 1-1/2+1/3-1/4.....+1/2n+1 - ln2
iii) since n approaches infinity, 1 /2(n+1) approaches 0. And since I is less than 1/2(n+1), hence I < or equal to 0
Iv) I don't really know how to do this one, but I feel like it's fishy cause I don't think k can = 0
 
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dunjaaa

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Re: HSC 2014 2U Marathon

Yeah, was a typo. It was meant to be k=1. Hint for (iv), use part (ii) and (iii).
 

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Re: HSC 2014 2U Marathon

Yeah, was a typo. It was meant to be k=1. Hint for (iv), use part (ii) and (iii).
have already tried assuming k=1, ended up getting a result that was opposite in sign compared to II so ))))): sadday
 

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Re: HSC 2014 2U Marathon

Sorry another typo :(. It's meant to be (-1)^(k+1)
 

dunjaaa

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Re: HSC 2014 2U Marathon

Nothing is wrong with the question, I promise haha :D
 
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