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I don't think I've ever done a test where 25 Raw seemed good What have you done Carrot...what have you done

Re: HSC 2013 4U Marathon Proving by mathematical induction on n, for integer m m \neq n $Step 1$ \ \ n=1 \frac{(2m)! \times 2}{(m+1)! m!} = \frac{2}{m+1} \binom{2m}{m} = 2\left(\binom{2m}{m} - \binom{2m}{m+1} \right) = $integer$ $Step 2$ \ \ n=k $Assume for integer$ \ \ p p =...

The main problem was time, if I was a little faster I reckon I could of done a lot more, I got 0/5 for conics (the good outcome of this was that I learnt that something called the 'reflection property of ellipse' is actually in the syllabus), 0/3 for probability missed the 4 mark volumes, but...

Multiply top and bottom by sec^2 x \int_0^{\pi/2} \frac{a \sec^2 x + b \sec x \tan x}{(b \sec x + a\tan x)^2} \ dx Substitution then

Permutations/Combinations

'toned it down' yeeaaaaaaaaaaaaaaaaaaa m8 edit: I meant the 4U paper, this is the wrong thread

Re: HSC 2013 4U Marathon Yep well done. Can anyone post a question please? (preferably probability)

bumping because I almost missed this message also thank goodness it changed

Here are solutions for Q14, also part (c) is exactly the same as a recent 4U marathon question by coincidence Part (iii) the 'average' is part c last part of Carrot's paper Q14a: http://i.imgur.com/aQDd6Z1.jpg Q14b: http://i.imgur.com/3HKJD9B.jpg Q14c: http://i.imgur.com/hncIgI0.jpg

I reckon 3U last year was harder than 4U last year... So if this 3U was harder than last 3U....... w0t

Thank you both, I think I get it

Re: HSC 2013 4U Marathon The inequality came about due to the condition that y > 0 at x = d, thus if we take a quadratic in terms of tan theta, it will describe all possible tan theta in terms of V such that y > 0 at x=d (i.e. the conditions we need for the question), so yeah that's another method

Ah alright, can you explain how to get seanieg's answer?

lol well that was another fail again on my part I'm so bad at probability

Ohh I see, is this similar to what you had in mind: P_n(A) = \frac{1}{2} P_{n-1} (E) P_n(B) = \frac{1}{2} P_{n-2} (E) P_n (C) = \frac{1}{2} P_{n-3} (E) P_n (D) = \frac{1}{2} P_{n-4} (E) + \frac{1}{2} P_{n-1} (E) P_n (E) = \frac{1}{2} P_{n-5} (E) + \frac{1}{2} P_{n-2} (E) Add them all...

I get the first part, but what do you mean by what is bolded?

Also can anyone verify my answers to Problem 4: (a) \frac{1}{2} (b) \frac{9}{16} --------------------------- Also does anyone have copies of the previous UNSW maths competition papers (preferably with solutions), it says its the 51st competition and yet they only put 1 on their site :/

http://www.maths.unsw.edu.au/sites/default/files/final_senior_51.pdf Problem 1, Scenario 1. What I was planning to do was, to recursively find the probability of finding people at E on the first night. So, P_{n}(E) = $Probability that on the$ \ nth \ $night the salesman is at$ \ \ E...

Re: HSC 2013 4U Marathon Yeah I think that makes sense, well done. My way was more straightforward, by subbing in t=d/Vcos a, and then y > 0 We arrive at: V^2 > \frac{gd}{2} (\tan \theta + \cot \theta) Since (a + 1/a) > 2, put in a = than theta to get, \tan \theta + \cot \theta \geq 2

Re: HSC 2013 4U Marathon $A projectile is fired from the origin with a velocity of$ \ V \ \ $to hit a wall$ \ \ d \ \ $units away from the origin$ $Ignore air resistance, take gravitational acceleration as$ \ \ g $Show that if it can reach the wall above ground level if and only if$ V >...