Search results

Have you tried the Stars and Bars method?

Re: HSC 2018 MX2 Marathon P(√x)=0 has roots α², β² Simplify the equation to form a standard quadratic equation. x - (a+1) = (a-2)√x x² - 2(a+1)x + (a+1)² = (a-2)²x x² - 2(a+1)x + (a+1)² = (a²-4a+4)x x² - (a²-4a+4+2a+2)x + (a+1)² = 0 x² - (a²-2a+6)x + (a+1)² = 0 α²+β² = a²-2a+6 = (a-1)²+5...

Alternatively you can differentiate sec³x and then rearrange both sides and re-integrate. However this is essentially the same as Integration By parts.

HSC 2018 MX2 Integration Marathon may as well just jump ahead and multiply by \frac{\sqrt{x} + \sqrt{x+1}}{\sqrt{x} + \sqrt{x+ 1}}

Re: HSC 2018 MX2 Integration Marathon is anyone else seeing a bunch of extra addition symbols lodged between every character or is that just me

Re: HSC 2018 MX2 Marathon \sqrt{x+ i y} = \pm \left( \sqrt{\frac{x+\sqrt{x^2+y^2}}{2}} + i \, \text{sgn}(y) \sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}} \right)

Re: HSC 2018 MX2 Marathon Find separate expressions for the real and imaginary parts of \sqrt{x+i y} Hence determine \sqrt{x+i y} as a sum of independent real and imaginary parts

Re: HSC 2018 MX2 Integration Marathon but in the hsc such a question would not be likely to appear, all that work just for a simple integral....

Re: HSC 2018 MX2 Integration Marathon using a highly unmotivated proof, sure.... xd

Re: HSC 2018 MX2 Marathon x² − y² = y & 2xy = x x=0, y = -1,0 or y=½, x = ±½√3 Sent from my iPhone using Tapatalk

Consider the graph when x and y are very very large. (x+y)(x²-xy+y²)=1 ⇒ x+y = 1/(x²-xy+y²) ≈ 0 ⇒ x≈-y

Re: HSC 2018 MX2 Marathon That does not follow. You can verify it for yourself by converting the conjugates to reciprocals and clearing them.

Re: HSC 2018 MX2 Marathon Alternatively, you can use the fact that the Imaginary part of 1 is 0 to do this: $\noindent$ S \equiv \Im{\left( \sum_{k=0}^n z^k \right)} \equiv \Im{\left( \frac{z^{n+1}-1}{z-1}\right)} \equiv \Im{\left(z^{\frac{n}{2}} \frac{z^{\frac{n+1}{2}} -...

Re: HSC 2018 MX2 Integration Marathon not without assuming the standard series expansion of e

Re: HSC 2018 MX2 Marathon Polar bash (mod-arg for the high schoolers) makes this problem trivial...

http://community.boredofstudies.org/14/mathematics-extension-2/344856/hsc-2016-mx2-integration-marathon-archive-14.html#post7092990

Re: HSC 2018 MX2 Integration Marathon sik MIT steal

$\noindent$ 4 + \sin^2{x} - 4\sin{x}\cos{x} + 3\cos^2{x} - 4 \equiv 4 - 3\sin^2{x} - 4\sin{x}\cos{x} - \cos^2{x} \\\\ \equiv 4-(\sin{x}+\cos{x})(3\sin{x}+\cos{x})

$\noindent$ \int_0^1 (-\log(x))^n \log{(-\log{x})} \, \text{d}x $ where $ n $ is a non-negative integer$ \\\\ \int_a^1 \frac{\sin^{-1}{x}}{\sqrt{x^2-a^2}} \, \text{d}x \\\\ \int_a^\infty \frac{x}{(x^2+b^2)\sqrt{x^2-a^2}} \, \text{d}x \\\\ \int_a^\infty...