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Re: HSC 2015 2U Marathon The line is a tangent to the curve.....

Nope that's how I'd do it in the exam. (well no way I would expand that lol)

Let's just say I had nothing better to do hahaah. how did you get it wrong?

For simplicity you can let u = x+4 (you don't have to) u^4-u^3-2u^2 = u^2(u^2-u-2) = \\ \\ u^2(u-2)(u+1) = 0 \\ \\ (x+4)^2(x+4-2)(x+4+1) = 0 \\ \\ (x+4)^2(x+2)(x+5) = 0 \\ \\ \\ \therefore x = -4,-2,-5

da hell i have no idea.... I mean wouldn't they send you a zmail or something to inform you of such absence? (obviously I didn't get any of that)

Re: HSC 2015 4U Marathon By riemann sum we get \int_{0}^{1} 2^x dx = \frac{1}{ln(2)}

What's A and B?

Lol Premutations Anyway, clearly you can't have a 4 digit number since that's always bigger than 100. Same goes for 3 digit numbers So you could have any 2 digit number since that's always less than 100 and there are 4P2 possible 2 digit numbers made from digits 1 to 4 then you have to...

lol i would choose a) since it asked for y. hehehehe

Re: HSC 2015 3U Marathon $Let y = sin^{-1}x$ \\ \\ \\ $i) Explain why$\ \ cosy\geq0 \\ \\ \\ ii) \ \ \ $Hence show that$ \ \ \ cosy = \sqrt{1-x^2}\\ \\ \\ iii)$Show that$ \ \ \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}

If you can get the questions right then who cares :P But I reckon that's the fastest way to pull it out if you somehow forgot them.

well if you're doing really well in your other 5 subjects, drop it.

Re: HSC 2015 4U Marathon - Advanced Level Well I dunno if this is too hard for the integration thread, but i'll post it here anyway. From UNSW integration bee. \int \frac{1}{x(x+1)(x+2)...(x+n)}\ \ dx

Re: HSC 2015 3U Marathon I get the feeling @Drsoccorball is not familiar with binomial theorem yet.

\int \frac{1}{\sqrt{x}}=\int x^{\frac{-1}{2}} = 2x^{\frac{1}{2}} = 2\sqrt{x}

Not that I know of sorry.

Lol I remember someone else asked the same question last year. i) Rearrange the equation to get z1-z2 = z4-z3 which implies AB and DC are parallel and equal in length. Rearrange again to get z1-z4 = z2-z3, meaning AD and DC are also parallel to each other and equal. This makes ABCD a...

Re: HSC 2015 3U Marathon \frac{d}{dx}sinx = \lim_{h \to 0}\frac{sin(x+h)-sinx}{h} Expand and simply and use \ \ \ \lim_{x \to 0}\frac{sinx}{x} = 1\\ \\ \\ \lim_{x \to0}\frac{cosx-1}{x}=0 To get the answer

GOOOOO yongggggg

ahhhh crap you're right the hyperbola isn't even defined at x=3