HSC 2012 MX2 Marathon (archive) (2 Viewers)

shaon0

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Re: 2012 HSC MX2 Marathon

d(f'(x))=∂f'/∂x dx =df'/dx dx = f"(x) dx
 

mnmaa

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Re: 2012 HSC MX2 Marathon

fun question to do for anyone interested-
Suppose that f(1) = 2, f(4) = 5, f '(1) = 5, f '(4) = 4, and f '' is continuous. Find the value of the following integral.
the answer is 8
 

Rezen

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Re: 2012 HSC MX2 Marathon

Prove that, for any triangle with sides a, b, c and area A,

a^2 + b^2 + c^2 >= 4sqrt(3)A
 

deswa1

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Re: 2012 HSC MX2 Marathon

Hey guys can you implicitly differentiate this:

<a href="http://www.codecogs.com/eqnedit.php?latex=(\frac{y}{x})^{\frac{1}{2}}@plus;(\frac{x}{y})^{\frac{1}{2}}=9" target="_blank"><img src="http://latex.codecogs.com/gif.latex?(\frac{y}{x})^{\frac{1}{2}}+(\frac{x}{y})^{\frac{1}{2}}=9" title="(\frac{y}{x})^{\frac{1}{2}}+(\frac{x}{y})^{\frac{1}{2}}=9" /></a>

I did it but I want to confirm with someone. Thanks :)
 

Carrotsticks

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Re: 2012 HSC MX2 Marathon

Hey guys can you implicitly differentiate this:

<a href="http://www.codecogs.com/eqnedit.php?latex=(\frac{y}{x})^{\frac{1}{2}}@plus;(\frac{x}{y})^{\frac{1}{2}}=9" target="_blank"><img src="http://latex.codecogs.com/gif.latex?(\frac{y}{x})^{\frac{1}{2}}+(\frac{x}{y})^{\frac{1}{2}}=9" title="(\frac{y}{x})^{\frac{1}{2}}+(\frac{x}{y})^{\frac{1}{2}}=9" /></a>

I did it but I want to confirm with someone. Thanks :)
I got y' = y/x.
 

jet

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Re: 2012 HSC MX2 Marathon

Here you go. I did it in as many steps as possible so you don't miss out on anything:

Fairly sure you don't need m < 0 considering that when n is even, n - 1 is odd and you can take odd nth roots of negative numbers.
 

Carrotsticks

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Re: 2012 HSC MX2 Marathon

Fairly sure you don't need m < 0 considering that when n is even, n - 1 is odd and you can take odd nth roots of negative numbers.
Your assertion would be correct if n were allowed to be negative.

However, a polynomial is defined by addition or subtraction powers of increasing non-negative integer exponents of some variable, implying that n must be positive.

Whether n is odd or even, the n-1th root will still be positive.

Hence the sign is determined by m, implying that m must be positive.
 

Carrotsticks

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Re: 2012 HSC MX2 Marathon

^^ sorry, I meant m must be negative in the last line.
 

deswa1

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Re: 2012 HSC MX2 Marathon

Still doesn't work...
 

Carrotsticks

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Re: 2012 HSC MX2 Marathon

It works, but half of the question is cut off haha.
 

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