# Search results

1. ### 2015 HSC Mathematics Solutions

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.
2. ### 2015 HSC Mathematics Solutions

Typed up solutions to Maths 2U 2015 :)
3. ### Distance from Ellipse to Point Q

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...
4. ### HSC 2015 MX2 Marathon (archive)

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...
5. ### HSC 2015 MX2 Integration Marathon (archive)

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 \\...
6. ### HSC 2015 MX2 Integration Marathon (archive)

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} \\...
7. ### Parametric Equations Cambridge Ex 9J

Solve for the intersection of OQ and the parabola for point Q.
8. ### HSC 2014 MX2 Marathon (archive)

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
9. ### cube roots of unity question

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
10. ### 2014 HSC Mathematics Solutions

Thus the bearing = 360 - 27.5 = 333 (nearest degree) which is the same thing as (360 + 11) - ACB
11. ### 2014 HSC Mathematics Solutions

I remember learning that in year 9 or 10 ;)
12. ### 2014 HSC Mathematics Solutions

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...
13. ### 2014 HSC Mathematics Solutions

Don't know if it's a trick question or not haha. (doesn't look like it though)
14. ### 2014 HSC Mathematics Solutions

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.
15. ### 2014 HSC Mathematics Solutions

yeah i forgot to save the change while checking :P its changed now :)
16. ### 2014 HSC Mathematics Solutions

Here it is! Solutions to Maths 2U 2014 if you're interested :)
17. ### State Rank Mark

Ooops.. tbh I would think much more will get high because relatively this paper is not too hard
18. ### State Rank Mark

And reckon only about 10 people will be >= 98 ?
19. ### State Rank Mark

What is the top raw mark you guys reckon? Would 100% be possible?
20. ### Need help with Simple Harmonic Question

\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...