HSC 2014 MX2 Marathon (archive) (2 Viewers)

nrlwinner · Aug 24, 2014

Re: HSC 2014 4U Marathon

Here is a great question from the HSC a while back.

faisalabdul16 · Aug 24, 2014

Re: HSC 2014 4U Marathon

nrlwinner said:
Here is a great question from the HSC a while back.

View attachment 30819

What year?

Axio · Aug 24, 2014

Re: HSC 2014 4U Marathon

a.

$(\sqrt{x}-\sqrt{y})^{2}\geq 0\\x-2\sqrt{xy}+y\geq 0\\x-2\sqrt{n}+y\geq 0\\\therefore\frac{x+y}{2}\geq \sqrt{n}$

b.

$xy=n\\as, x> \sqrt{n}, y< \sqrt{n}\\\therefore x> y\\2x> x+y\\x> \frac{x+y}{2}$

Therefore (x+y)/2 is a closer approximation to root n than x is.

sepseminar · Aug 26, 2014

Re: HSC 2014 4U Marathon

Prove that

$\int_\frac{1}{b}^b \frac{\tan^{-1}(x)}{x} = \frac{\pi \cdot \mathrm{ln}(b)}{2}$ .

dunjaaa · Aug 26, 2014

Re: HSC 2014 4U Marathon

Screen shot 2014-08-26 at 4.48.33 PM.png

sepseminar · Aug 26, 2014

Re: HSC 2014 4U Marathon

Yep, that's correct, Dunjaaa. There is a slightly faster solution in which you make the substitution u=1/x. But all the deft moves are in your solution. First, the use of a substitution to equate the integral with a similar looking integral. Then the use of some basic properties to relate the two integrals together. In this case it was the fact that the inverse tan of a number and the inverse tan of the number's reciprocal is constant.

Fade1233 · Sep 4, 2014

Re: HSC 2014 4U Marathon

Fibonacci Based Strong Induction.

Edit: Forgot to put up the condition: U(n)= U(n-1) + U(n-2)

dunjaaa · Sep 6, 2014

Re: HSC 2014 4U Marathon

Screen shot 2014-09-06 at 9.59.45 PM.png

faisalabdul16 · Sep 10, 2014

Re: HSC 2014 4U Marathon

Second part? Need to check my answer...

RealiseNothing · Sep 10, 2014

Re: HSC 2014 4U Marathon

faisalabdul16 said:
View attachment 30897

Second part? Need to check my answer...

$1-\binom{9}{3}$

Kurosaki · Sep 11, 2014

Re: HSC 2014 4U Marathon

RealiseNothing said:
$1-\binom{9}{3}$

Whut
Second bit is 420 I think

Carrotsticks · Sep 11, 2014

Re: HSC 2014 4U Marathon

Kurosaki said:
Whut
Second bit is 420 I think

Blaze it.

Also I think he just mixed up probability with permutations.

RealiseNothing · Sep 11, 2014

Re: HSC 2014 4U Marathon

Carrotsticks said:
Blaze it.

Also I think he just mixed up probability with permutations.

Yer wtf was i doing.

Also on phone so 9C3 are in decreasing order. Not in decreasing should then be 5*9C3.

mreditor16 · Sep 14, 2014

Re: HSC 2014 4U Marathon

I personally enjoyed this question

mreditor16 · Sep 14, 2014

Re: HSC 2014 4U Marathon

btw above question was in fort street 2010

only Q3....

mreditor16 · Sep 15, 2014

Re: HSC 2014 4U Marathon

bumpity bump!

dunjaaa · Sep 19, 2014

Re: HSC 2014 4U Marathon

Since nobody has answered it, I might as well do so

Screen shot 2014-09-19 at 2.42.52 PM.png

. Strong induction in disguise

Sy123 · Sep 21, 2014

Re: HSC 2014 4U Marathon

$\\ $Let$ \ a,b,c \ $be positive real numbers such that$ \ ab + ac + bc = 1 \\ \\ $ Prove that$ \ \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{a+c} \geq \sqrt{3} + \frac{ab}{a+b} + \frac{ac}{a+c} + \frac{bc}{b+c} \\ \\ $and show where equality holds$$

mreditor16 · Sep 21, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
$\\ $Let$ \ a,b,c \ $be positive real numbers such that$ \ ab + ac + bc = 1 \\ \\ $ Prove that$ \ \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{a+c} \geq \sqrt{3} + \frac{ab}{a+b} + \frac{ac}{a+c} + \frac{bc}{b+c} \\ \\ $and show where equality holds$$

slightly too hard for normal 4u marathon thread, imo. should be in advanced thread imo.

Fizzy_Cyst · Sep 21, 2014

Re: HSC 2014 4U Marathon

A non-real number, w, is such that |w| = 1.

If z = (1+w)/(1-w) find the locus of z as w moves on the complex number plane

HSC 2014 MX2 Marathon (archive) (2 Viewers)

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