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HSC 2013-14 MX1 Marathon (archive) (5 Viewers)

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Axio

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Re: HSC 2013 3U Marathon Thread

:(
 
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dunjaaa

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Re: HSC 2013 3U Marathon Thread

Question: Screen shot 2014-07-12 at 11.37.12 PM.png, just remembered a good question that I did last year in SHM :D
 

Sy123

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Re: HSC 2013 3U Marathon Thread

 
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braintic

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Re: HSC 2013 3U Marathon Thread

As you said 'using the Binomial theorem', I assume you want something different to the previous answer.

Not sure exactly what you're looking for, but perhaps:

x³ + 4x² + 6x + 4 = (x³ + 6x² + 12x + 8) - 2x² - 6x - 4
=(x+2)³ - 2(x² + 3x + 2)
=(x+2)³ - 2(x+2)(x+1)
=(x+2) [ (x+2)² - 2(x+1) ]
=(x+2) (x² + 2x + 2)

So x=-2
 

dunjaaa

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Re: HSC 2013 3U Marathon Thread

I think what Sy meant was this:
x^3+4x^2+6x+4 = 0
Multiplying top and bottom by x,
[(x+1)^4-1]/x = 0
x+1 = -1 (Since x cannot = 0)
Therefore, x=-2 is the only real root of this polynomial.
 

Sy123

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dunjaaa

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Re: HSC 2013 3U Marathon Thread

Find: lim (n->infinity) {[e^(1/n) + e^(2/n) + e^(3/n) + ... + e^(n/n)]/n}
 
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dunjaaa

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Re: HSC 2013 3U Marathon Thread

There was another way that I preferred (3u wise -> differentiation via first principles), but yeah nice work haha :p
 
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Carrotsticks

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Re: HSC 2013 3U Marathon Thread

There was another way that I preferred (3u wise via first principles), but yeah nice work haha :p
First principles? As in differentiation by first principles?
 

dunjaaa

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Re: HSC 2013 3U Marathon Thread

yeah thats what I meant
 

Sy123

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