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Yep! I have changed 14a ii (didnt read the question lol) But for 13b ii) - the first inequality will include the rectangular bit below the semicircle. Pick (0, -1) and put that in the inequality -- -1 <= sqrt(9-0^2) which is true.

Typed up solutions to Maths 2U 2015 :)

For a\geq 2 shortest distance is \left | a-2 \right | Let a point on the ellipse to be \left ( x_{1},y_{1} \right ). Distance from \left ( a,0 \right ) to the point on ellipse is given by \sqrt{(x_{1}-a)^2+y_{1}^2}. We can use the equation of the ellipse to replace y_{1} by x_{1}. We...

Re: HSC 2015 4U Marathon $Let $z = x + iy,\,\,z \ne 0.\\\\\,\,\frac{1}{{x + iy}} + \frac{1}{{x - iy}} = 1\\\frac{{x - iy + x + iy}}{{\left( {x + iy} \right)\left( {x - iy} \right)}} = 1\\\\2x = {x^2} + {y^2}\\{\left( {x - 1} \right)^2} + {y^2} = 1 Circle with centre (1,0), radius 1, excluding...

Re: MX2 2015 Integration Marathon \,\,\,\,\int{\frac{1}{1+{{\sin }^{2}}x}dx} \\ \text{=}\int{\frac{1}{{{\cos }^{2}}x+2{{\sin }^{2}}x}dx} \\ =\int{\frac{{{\sec }^{2}}x}{1+2{{\tan }^{2}}x}dx} \\ \\ \text{Let}\,u=\tan x,\,\,\frac{du}{dx}={{\sec }^{2}}x \\...

Re: MX2 2015 Integration Marathon \,\,\,\int_{0}^{\tfrac{\pi }{6}}{{{\sin }^{4}}x\ {{\cos }^{3}}x\,\,dx} \\ =\int_{0}^{\tfrac{\pi }{6}}{{{\sin }^{4}}x\left( 1-{{\sin }^{2}}x \right)\cos x\,\,dx} \\ =\int_{0}^{\tfrac{\pi }{6}}{\left( {{\sin }^{4}}x-{{\sin }^{6}}x \right)\cos x\,\,dx} \\...

Solve for the intersection of OQ and the parabola for point Q.

Re: HSC 2014 4U Marathon (ii) a\sqrt{1-b^2}=1-b\sqrt{1-a^2}\\a^2(1-b^2)=1-2b\sqrt(1-a^2)+b^2(1-a^2)\\2b\sqrt{1-a^2}=[1-(a^2-b^2)]^2\\4b^2(1-a^2)=1-2(a^2-b^2)+(a^2-b^2)^2\\0=1-2(a^2+b^2)+(a^2+b^2)^2\\a^2+b^2=\frac{2\pm \sqrt{2^2-4}}{2}=1

Yes. w has an imaginary part (thus non-real) and so it cannot be 1. If w is 1 , then 1 + \omega +\omega ^2 = 3 \neq 0

Thus the bearing = 360 - 27.5 = 333 (nearest degree) which is the same thing as (360 + 11) - ACB

I remember learning that in year 9 or 10 ;)

I always think that in maths as long as your method is valid and you get the correct result, they should award you the mark because there is always more than one possible solution. Even though writing one line to explain how the result is obtained (as in my solutions) seems like inadequate for...

Don't know if it's a trick question or not haha. (doesn't look like it though)

Hmm I see your point. Didn't pick it up while I was rushing through it. It could be an open circle on the graph (i.e. approaching 0) so there might only be two solutions.

yeah i forgot to save the change while checking :P its changed now :)

Here it is! Solutions to Maths 2U 2014 if you're interested :)

Ooops.. tbh I would think much more will get high because relatively this paper is not too hard

And reckon only about 10 people will be >= 98 ?

What is the top raw mark you guys reckon? Would 100% be possible?

\frac{\mathrm{d} ^2x}{\mathrm{d} t^2}=-\frac{k}{x^2}\\\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{1}{2} v^2 \right )=-\frac{k}{x^2}\\\frac{1}{2} v^2 =\frac{k}{x} + c\\\\$When $x$ = $10^9 $ m, $v=0$:$\\c = -\frac{k}{10^9}\\\\$When $x$ = $6.4\times 10^6 $ m, $v$:$\\\frac{1}{2} v^2...