Search results

You sure are asking a lot of exact value questions! \frac{\sqrt 3}{2} \times \frac{1}{2} - \frac{1}{2} \times \frac{\sqrt 3}{2} = 0 Another way of doing this is recognising the Compound Angle formula for cos(a+b), which gives us cos(30+60)=cos(90), which is equal to 0.

Cool result isn't it? Was trying this question and when I found out that P(x,y,z)=2, I thought it was pretty funny.

Re: HSC 2012 Marathon :) Maybe a 'cooler' one that isn't just calculations and number crunching please? I like proofs more :3

\\ $Let $u=2^y $ such that our expression becomes $ u^2 - 9u + 8 = 0. $ Solving this quadratic of u gives us $ (u-8)(u-1) = 0 $ and therefore $ u=1 $ or $ 8.$ \\\\ 2^y = 1 \Rightarrow y = 0 \\\\ 2^y = 8 \Rightarrow y = 3.

No worries :) Now your function for m in terms of x and y is actually incorrect. What you have there implies that the line goes through the origin, but that is not necessarily the case.

No, that is not the exact value. You just simplified the expression. \frac{1 + \sqrt 3}{1 - \sqrt 3} = \frac{(1+\sqrt 3)^2}{-2} = \frac{4 + 2 \sqrt 3 }{-2} = -(2 + \sqrt 3)

Re: HSC 2012 Marathon :) More questions please =) I'm bored..

\\ \frac{1}{\sqrt 2} \times \frac{\sqrt 3}{2} + \frac{1}{\sqrt 2} \times \frac{1}{2} = \frac{\sqrt 3}{2 \sqrt 2} + \frac{1}{2 \sqrt 2} = \frac{1 + \sqrt 3}{2 \sqrt 2} = \frac{\sqrt 6 + \sqrt 2}{4} $ by rationalising the denominator again.$

Yes it can be equal to zero. Please refer to my example of the case where the line is in the form y=mx and passing through the given point. I answered the question according to what was given, and there was no such specification that x>0 and y>0. If I got this question in an exam and the marker...

Haha hilarious!

\sqrt 3 - \frac{1}{\sqrt 3} = \frac{3 - 1}{\sqrt 3} = \frac{2}{\sqrt 3} = \frac{2}{3} \sqrt 3 $ by rationalising the denominator.$

I don't understand all this confusion. Who cares how ATAR is calculated? Just study your best and let the UAC worry about the ATAR for you. Suppose you knew everything about how the ATAR is calculated, and even knew the scaling formula despite it being kept a secret by the UAC. Does this affect...

Ummm.... the shortest hypotenuse is simply h=0 as per the case when the equation of the line is 8y=243x. Did you mean the positive x and positive y axis?

Oh and I know it's tempting to use Lagrangian Multipliers for max(P(x,y,z)) but please restrain from doing so.

\\ $Define $ P(x,y,z) = \frac{x+1}{xy+x+1} + \frac{y+1}{yz+y+1} + \frac{z+1}{xz+z+1} $ where $ x,y,z \geq 0 $ and $ xyz=1. \\\\ $Find $ \max \left ( P(x,y,z) \right ) $ and $ \min \left ( P(x,y,z) \right ).

Think about where \frac{\sqrt{5}-1}{2} comes from. What degree is the polynomial from which it came? Also I'm not really sure if this disc is the maximal disc of which the solutions exist outside.

Your solution is essentially correct, but I think it's not explained very well. For example when using the Limiting Sum, you could have added that we can use it since |z|<1 rather than jumping directly into it. Also I think proofs should be such that even the average student could understand it...

Here's a fun question I did a couple weeks ago.

To be honest nofate's way is still pretty fast I guess because we can just use the area of a circle formula in place of the trig substitution ^^

Yeah I've lost marks too for small things like forgetting to put a dx when integrating w.r.t x or whatnot. Most recently, I took a bit of a gamble just to see what would happen. At school right now we finished Co-ordinate Geometry and it was examined a few weeks ago for our 2 unit test. The last...