HSC 2016 Maths Marathon (archive) (4 Viewers)

Drsoccerball · Jan 27, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
Erm not too jump in early
$$Wasn't the LHS $ x^2 -3x +1 $ not $ x^2 -3x + 2$

Yes I mistyped the question. Sorry about that

but +2 makes the question nicer you can do the same thing for +1 as well.

Nailgun · Jan 27, 2016

Re: HSC 2016 2U Marathon

Ohkay I redid it with the same question you used and got it right

Took a while, but I got there in the end lol

Drsoccerball · Jan 27, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
Ohkay I redid it with the same question you used and got it right
Took a while, but I got there in the end lol

Nice. Now try the next question.

Drsoccerball · Jan 30, 2016

Re: HSC 2016 2U Marathon

Drsoccerball said:
$Please past graduates do not help in this next problem. I don't want to waste an amazing question.$

$$ Simplify : $ \frac{\log_{\frac{a}{b}}x \log_{\frac{a}{c}}{x}}{\log_a ^2 x} +\frac{\log_{\frac{b}{c}}x \log_{\frac{b}{a}}{x}}{\log_b^2 x} +\frac{\log_{\frac{c}{a}}x \log_{\frac{c}{b}}{x}}{\log_c ^2 x}$

Bump.

leehuan · Jan 30, 2016

Re: HSC 2016 2U Marathon

Gave that question a bit of thought and here's the hint. It's way easier than it looks.

Nailgun · Jan 31, 2016

Re: HSC 2016 2U Marathon

Drsoccerball said:
Bump.

I don't suppose the answer is;
$(1-\log _{ a }{ c } )(1-\log _{ a }{ b } )+(1-\log _{ b }{ c } )(1-\log _{ b }{ a } )+(1-\log _{ c }{ a } )(1-\log _{ c }{ b } )$

leehuan · Jan 31, 2016

Re: HSC 2016 2U Marathon

$\frac { \log _{ \frac { a }{ b } } x\log _{ \frac { a }{ c } }{ x } }{ \log _{ a } ^{ 2 }x } +\frac { \log _{ \frac { b }{ c } } x\log _{ \frac { b }{ a } }{ x } }{ \log _{ b } ^{ 2 }x } +\frac { \log _{ \frac { c }{ a } } x\log _{ \frac { c }{ b } }{ x } }{ \log _{ c } ^{ 2 }x }$

$\\ =\frac { \left( \frac { \ln { x } }{ \ln { \frac { a }{ b } } } \right) \left( \frac { \ln { x } }{ \ln { \frac { a }{ c } } } \right) }{ \left( \frac { \ln { x } }{ \ln { a } } \right) \left( \frac { \ln { x } }{ \ln { a } } \right) } +\frac { \left( \frac { \ln { x } }{ \ln { \frac { b }{ c } } } \right) \left( \frac { \ln { x } }{ \ln { \frac { b }{ a } } } \right) }{ \left( \frac { \ln { x } }{ \ln { b } } \right) \left( \frac { \ln { x } }{ \ln { b } } \right) } +\frac { \left( \frac { \ln { x } }{ \ln { \frac { c }{ a } } } \right) \left( \frac { \ln { x } }{ \ln { \frac { c }{ b } } } \right) }{ \left( \frac { \ln { x } }{ \ln { c } } \right) \left( \frac { \ln { x } }{ \ln { c } } \right) }$

$\\ =\frac { { \left( \ln { a } \right) }^{ 2 } }{ \ln { \frac { a }{ b } } \cdot \ln { \frac { a }{ c } } } +\frac { { \left( \ln { b } \right) }^{ 2 } }{ \ln { \frac { b }{ c } } \cdot \ln { \frac { b }{ a } } } +\frac { { \left( \ln { c } \right) }^{ 2 } }{ \ln { \frac { c }{ a } } \cdot \ln { \frac { c }{ b } } } \\ =\frac { { \left( \ln { a } \right) }^{ 2 } }{ \left( \ln { a } -\ln { b } \right) \left( \ln { a } -\ln { c } \right) } +\frac { { \left( \ln { b } \right) }^{ 2 } }{ \left( \ln { b } -\ln { c } \right) \left( \ln { b } -\ln { a } \right) } +\frac { { \left( \ln { c } \right) }^{ 2 } }{ \left( \ln { c } -\ln { a } \right) \left( \ln { c } -\ln { b } \right) }$

$\\ $Now: $ \frac { { w }^{ 2 } }{ \left( w-y \right) \left( w-z \right) } +\frac { { y }^{ 2 } }{ \left( y-z \right) \left( y-w \right) } +\frac { { z }^{ 2 } }{ \left( z-w \right) \left( z-y \right) } \\ =\frac { { w }^{ 2 } }{ \left( w-y \right) \left( w-z \right) } +\frac { { y }^{ 2 } }{ \left( z-y \right) \left( w-y \right) } -\frac { { z }^{ 2 } }{ \left( w-z \right) \left( z-y \right) } \\ =\frac { { w }^{ 2 }\left( z-y \right) +{ y }^{ 2 }\left( w-z \right) -{ z }^{ 2 }\left( w-y \right) }{ \left( w-y \right) \left( w-z \right) \left( z-y \right) } \\ =\frac { { w }^{ 2 }z-w^{ 2 }y+{ y }^{ 2 }w-{ y }^{ 2 }z-{ z }^{ 2 }w-{ +z }^{ 2 }y }{ \left( { w }^{ 2 }-yw-wz+yz \right) \left( z-y \right) }$

$\\ =\frac { { w }^{ 2 }z-w^{ 2 }y+{ y }^{ 2 }w-{ y }^{ 2 }z-{ z }^{ 2 }w+{ z }^{ 2 }y }{ { w }^{ 2 }z-wyz-w{ z }^{ 2 }+y{ z }^{ 2 }-{ w }^{ 2 }y+{ y }^{ 2 }w+wyz-{ y }^{ 2 }z } \\ =\frac { { w }^{ 2 }z-w^{ 2 }y+{ y }^{ 2 }w-{ y }^{ 2 }z-{ z }^{ 2 }w+{ z }^{ 2 }y }{ { w }^{ 2 }z-w^{ 2 }y+{ y }^{ 2 }w-{ y }^{ 2 }z-{ z }^{ 2 }w+{ z }^{ 2 }y } \\ =1$

$\\ \therefore $ Let $w=\ln { a } ,y=\ln { b } ,z=\ln { c } : \\ \frac { { \left( \ln { a } \right) }^{ 2 } }{ \left( \ln { a } -\ln { b } \right) \left( \ln { a } -\ln { c } \right) } +\frac { { \left( \ln { b } \right) }^{ 2 } }{ \left( \ln { b } -\ln { c } \right) \left( \ln { b } -\ln { a } \right) } +\frac { { \left( \ln { c } \right) }^{ 2 } }{ \left( \ln { c } -\ln { a } \right) \left( \ln { c } -\ln { b } \right) } \\ =1$

Nailgun · Jan 31, 2016

Re: HSC 2016 2U Marathon

leehuan said:
$\frac { \log _{ \frac { a }{ b } } x\log _{ \frac { a }{ c } }{ x } }{ \log _{ a } ^{ 2 }x } +\frac { \log _{ \frac { b }{ c } } x\log _{ \frac { b }{ a } }{ x } }{ \log _{ b } ^{ 2 }x } +\frac { \log _{ \frac { c }{ a } } x\log _{ \frac { c }{ b } }{ x } }{ \log _{ c } ^{ 2 }x }$

$\\ =\frac { \left( \frac { \ln { x } }{ \ln { \frac { a }{ b } } } \right) \left( \frac { \ln { x } }{ \ln { \frac { a }{ c } } } \right) }{ \left( \frac { \ln { x } }{ \ln { a } } \right) \left( \frac { \ln { x } }{ \ln { a } } \right) } +\frac { \left( \frac { \ln { x } }{ \ln { \frac { b }{ c } } } \right) \left( \frac { \ln { x } }{ \ln { \frac { b }{ a } } } \right) }{ \left( \frac { \ln { x } }{ \ln { b } } \right) \left( \frac { \ln { x } }{ \ln { b } } \right) } +\frac { \left( \frac { \ln { x } }{ \ln { \frac { c }{ a } } } \right) \left( \frac { \ln { x } }{ \ln { \frac { c }{ b } } } \right) }{ \left( \frac { \ln { x } }{ \ln { c } } \right) \left( \frac { \ln { x } }{ \ln { c } } \right) }$

$\\ =\frac { { \left( \ln { a } \right) }^{ 2 } }{ \ln { \frac { a }{ b } } \cdot \ln { \frac { a }{ c } } } +\frac { { \left( \ln { b } \right) }^{ 2 } }{ \ln { \frac { b }{ c } } \cdot \ln { \frac { b }{ a } } } +\frac { { \left( \ln { c } \right) }^{ 2 } }{ \ln { \frac { c }{ a } } \cdot \ln { \frac { c }{ b } } } \\ =\frac { { \left( \ln { a } \right) }^{ 2 } }{ \left( \ln { a } -\ln { b } \right) \left( \ln { a } -\ln { c } \right) } +\frac { { \left( \ln { b } \right) }^{ 2 } }{ \left( \ln { b } -\ln { c } \right) \left( \ln { b } -\ln { a } \right) } +\frac { { \left( \ln { c } \right) }^{ 2 } }{ \left( \ln { c } -\ln { a } \right) \left( \ln { c } -\ln { b } \right) }$

$\\ $Now: $ \frac { { w }^{ 2 } }{ \left( w-y \right) \left( w-z \right) } +\frac { { y }^{ 2 } }{ \left( y-z \right) \left( y-w \right) } +\frac { { z }^{ 2 } }{ \left( z-w \right) \left( z-y \right) } \\ =\frac { { w }^{ 2 } }{ \left( w-y \right) \left( w-z \right) } +\frac { { y }^{ 2 } }{ \left( z-y \right) \left( w-y \right) } -\frac { { z }^{ 2 } }{ \left( w-z \right) \left( z-y \right) } \\ =\frac { { w }^{ 2 }\left( z-y \right) +{ y }^{ 2 }\left( w-z \right) -{ z }^{ 2 }\left( w-y \right) }{ \left( w-y \right) \left( w-z \right) \left( z-y \right) } \\ =\frac { { w }^{ 2 }z-w^{ 2 }y+{ y }^{ 2 }w-{ y }^{ 2 }z-{ z }^{ 2 }w-{ +z }^{ 2 }y }{ \left( { w }^{ 2 }-yw-wz+yz \right) \left( z-y \right) }$

$\\ =\frac { { w }^{ 2 }z-w^{ 2 }y+{ y }^{ 2 }w-{ y }^{ 2 }z-{ z }^{ 2 }w+{ z }^{ 2 }y }{ { w }^{ 2 }z-wyz-w{ z }^{ 2 }+y{ z }^{ 2 }-{ w }^{ 2 }y+{ y }^{ 2 }w+wyz-{ y }^{ 2 }z } \\ =\frac { { w }^{ 2 }z-w^{ 2 }y+{ y }^{ 2 }w-{ y }^{ 2 }z-{ z }^{ 2 }w+{ z }^{ 2 }y }{ { w }^{ 2 }z-w^{ 2 }y+{ y }^{ 2 }w-{ y }^{ 2 }z-{ z }^{ 2 }w+{ z }^{ 2 }y } \\ =1$

$\\ \therefore $ Let $w=\ln { a } ,y=\ln { b } ,z=\ln { c } : \\ \frac { { \left( \ln { a } \right) }^{ 2 } }{ \left( \ln { a } -\ln { b } \right) \left( \ln { a } -\ln { c } \right) } +\frac { { \left( \ln { b } \right) }^{ 2 } }{ \left( \ln { b } -\ln { c } \right) \left( \ln { b } -\ln { a } \right) } +\frac { { \left( \ln { c } \right) }^{ 2 } }{ \left( \ln { c } -\ln { a } \right) \left( \ln { c } -\ln { b } \right) } \\ =1$

Are there questions like that in the HSC?

leehuan · Jan 31, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
Are there questions like that in the HSC?

No LOL

leehuan · Jan 31, 2016

Re: HSC 2016 2U Marathon

$(1-\log _{ a }{ c } )(1-\log _{ a }{ b } )+(1-\log _{ b }{ c } )(1-\log _{ b }{ a } )+(1-\log _{ c }{ a } )(1-\log _{ c }{ b } )\\ =\frac { \ln { a } -\ln { c } }{ \ln { a } } \frac { \ln { a } -\ln { b } }{ \ln { a } } +\frac { \ln { b } -\ln { c } }{ \ln { b } } \frac { \ln { b } -\ln { a } }{ \ln { b } } +\frac { \ln { c } -\ln { a } }{ \ln { c } } \frac { \ln { c } -\ln { b } }{ \ln { c } }$

Your fractions are reciprocated to mine, so somewhere one of us went wrong. But I could've sworn mine is all good.

Drsoccerball · Jan 31, 2016

Re: HSC 2016 2U Marathon

leehuan said:
$\frac { \log _{ \frac { a }{ b } } x\log _{ \frac { a }{ c } }{ x } }{ \log _{ a } ^{ 2 }x } +\frac { \log _{ \frac { b }{ c } } x\log _{ \frac { b }{ a } }{ x } }{ \log _{ b } ^{ 2 }x } +\frac { \log _{ \frac { c }{ a } } x\log _{ \frac { c }{ b } }{ x } }{ \log _{ c } ^{ 2 }x }$

$\\ =\frac { \left( \frac { \ln { x } }{ \ln { \frac { a }{ b } } } \right) \left( \frac { \ln { x } }{ \ln { \frac { a }{ c } } } \right) }{ \left( \frac { \ln { x } }{ \ln { a } } \right) \left( \frac { \ln { x } }{ \ln { a } } \right) } +\frac { \left( \frac { \ln { x } }{ \ln { \frac { b }{ c } } } \right) \left( \frac { \ln { x } }{ \ln { \frac { b }{ a } } } \right) }{ \left( \frac { \ln { x } }{ \ln { b } } \right) \left( \frac { \ln { x } }{ \ln { b } } \right) } +\frac { \left( \frac { \ln { x } }{ \ln { \frac { c }{ a } } } \right) \left( \frac { \ln { x } }{ \ln { \frac { c }{ b } } } \right) }{ \left( \frac { \ln { x } }{ \ln { c } } \right) \left( \frac { \ln { x } }{ \ln { c } } \right) }$

$\\ =\frac { { \left( \ln { a } \right) }^{ 2 } }{ \ln { \frac { a }{ b } } \cdot \ln { \frac { a }{ c } } } +\frac { { \left( \ln { b } \right) }^{ 2 } }{ \ln { \frac { b }{ c } } \cdot \ln { \frac { b }{ a } } } +\frac { { \left( \ln { c } \right) }^{ 2 } }{ \ln { \frac { c }{ a } } \cdot \ln { \frac { c }{ b } } } \\ =\frac { { \left( \ln { a } \right) }^{ 2 } }{ \left( \ln { a } -\ln { b } \right) \left( \ln { a } -\ln { c } \right) } +\frac { { \left( \ln { b } \right) }^{ 2 } }{ \left( \ln { b } -\ln { c } \right) \left( \ln { b } -\ln { a } \right) } +\frac { { \left( \ln { c } \right) }^{ 2 } }{ \left( \ln { c } -\ln { a } \right) \left( \ln { c } -\ln { b } \right) }$

$\\ $Now: $ \frac { { w }^{ 2 } }{ \left( w-y \right) \left( w-z \right) } +\frac { { y }^{ 2 } }{ \left( y-z \right) \left( y-w \right) } +\frac { { z }^{ 2 } }{ \left( z-w \right) \left( z-y \right) } \\ =\frac { { w }^{ 2 } }{ \left( w-y \right) \left( w-z \right) } +\frac { { y }^{ 2 } }{ \left( z-y \right) \left( w-y \right) } -\frac { { z }^{ 2 } }{ \left( w-z \right) \left( z-y \right) } \\ =\frac { { w }^{ 2 }\left( z-y \right) +{ y }^{ 2 }\left( w-z \right) -{ z }^{ 2 }\left( w-y \right) }{ \left( w-y \right) \left( w-z \right) \left( z-y \right) } \\ =\frac { { w }^{ 2 }z-w^{ 2 }y+{ y }^{ 2 }w-{ y }^{ 2 }z-{ z }^{ 2 }w-{ +z }^{ 2 }y }{ \left( { w }^{ 2 }-yw-wz+yz \right) \left( z-y \right) }$

$\\ =\frac { { w }^{ 2 }z-w^{ 2 }y+{ y }^{ 2 }w-{ y }^{ 2 }z-{ z }^{ 2 }w+{ z }^{ 2 }y }{ { w }^{ 2 }z-wyz-w{ z }^{ 2 }+y{ z }^{ 2 }-{ w }^{ 2 }y+{ y }^{ 2 }w+wyz-{ y }^{ 2 }z } \\ =\frac { { w }^{ 2 }z-w^{ 2 }y+{ y }^{ 2 }w-{ y }^{ 2 }z-{ z }^{ 2 }w+{ z }^{ 2 }y }{ { w }^{ 2 }z-w^{ 2 }y+{ y }^{ 2 }w-{ y }^{ 2 }z-{ z }^{ 2 }w+{ z }^{ 2 }y } \\ =1$

$\\ \therefore $ Let $w=\ln { a } ,y=\ln { b } ,z=\ln { c } : \\ \frac { { \left( \ln { a } \right) }^{ 2 } }{ \left( \ln { a } -\ln { b } \right) \left( \ln { a } -\ln { c } \right) } +\frac { { \left( \ln { b } \right) }^{ 2 } }{ \left( \ln { b } -\ln { c } \right) \left( \ln { b } -\ln { a } \right) } +\frac { { \left( \ln { c } \right) }^{ 2 } }{ \left( \ln { c } -\ln { a } \right) \left( \ln { c } -\ln { b } \right) } \\ =1$

This is right

Drsoccerball · Jan 31, 2016

Re: HSC 2016 2U Marathon

$New question$

$$ Find the radius of a circle with arc ABC such that AC the distance of the chord is 2 meters and BD (the perpendicular from AC) is 10 cm.$

Nailgun · Feb 1, 2016

Re: HSC 2016 2U Marathon

$$ Am I correct in assuming that $ arc\quad AB=arc\quad BC ?$

Nailgun · Feb 1, 2016

Re: HSC 2016 2U Marathon

$$ BD bisects CA as $ arcBC = arcAB \\ $Hence $ DC=AD=1\\ CO=AO=r\\ We\quad see\quad that\quad DO+BD=r\\ Hence,\quad r-0.1=DO\\ Use\quad Pythagoreas\quad Theorem\\ (r-\frac { 1 }{ 10 } )^{ 2 }+1=r^{ 2 }\\ r=5.05m\quad$

Drsoccerball · Feb 1, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
$$ BD bisects CA as $ arcBC = arcAB \\ $Hence $ DC=AD=1\\ CO=AO=r\\ We\quad see\quad that\quad DO+BD=r\\ Hence,\quad r-0.1=DO\\ Use\quad Pythagoreas\quad Theorem\\ (r-\frac { 1 }{ 10 } )^{ 2 }+1=r^{ 2 }\\ r=5.05m\quad$

Correct ! New question:

$$ Find $ f'(\pi) $ if $ f(x) = e^{\sin{e^{\cos{x}}}}$

Nailgun · Feb 1, 2016

Re: HSC 2016 2U Marathon

Drsoccerball said:
Correct ! New question:

$$ Find $ f'(\pi) $ if $ f(x) = e^{\sin{e^{\cos{x}}}}$

#logarithmic differentiation #burningladylogrule #brokenpipetterule

$f\left( x \right) ={ e }^{ \sin { e^{ \cos { x } } } }\\ Let\quad g\left( x \right) =\sin { e^{ \cos { x } } } \\ f\left( x \right) ={ e }^{ g\left( x \right) }\\ f'(x)=g'(x).{ e }^{ g\left( x \right) }\\ \\ g\left( x \right) =\sin { e^{ \cos { x } } } \\ Let\quad z(x)=e^{ \cos { x } }\\ g(x)=\sin { z(x) } \\ g'(x)=z'(x).\cos { z(x) } \\ \\ z(x)=e^{ \cos { x } }\\ \ln { z(x) } =\ln { e^{ \cos { x } } } \\ \ln { z(x) } =\cos { x } \\ \frac { d }{ dx } \ln { z(x) } =\frac { d }{ dx } \cos { x } \\ \frac { z'(x) }{ z(x) } =-\sin { x } \\ z'(x)=-\sin { x } .e^{ \cos { x } }\\ \\ g'(x)=z'(x).\cos { z(x) } \\ g'(x)=-\sin { x } .e^{ \cos { x } }.\cos { e^{ \cos { x } } } \\ \\ f'(x)=g'(x).{ e }^{ g\left( x \right) }\\ f'(x)=-\sin { x } .e^{ \cos { x } }.\cos { e^{ \cos { x } } } .e^{ \sin { e^{ \cos { x } } } }\\ \\ f'(\pi )=-\sin { \pi } .e^{ \cos { \pi } }.\cos { e^{ \cos { \pi } } } .e^{ \sin { e^{ \cos { \pi } } } }\\ f'(\pi )=0$

Drsoccerball · Feb 1, 2016

Re: HSC 2016 2U Marathon

I think it's time to explain the burning lady rule. So what happened is on that day we had assembly and there were these dancing people. Just as the show opens the lady puts a ball of flames into her mouth and she screws up and it catches her face on fire... So she was alright afterwards but ojn that day we learnt that rule. And it made sense to call it the burning lady rule as the ladies standing and then she falls because shes on fire to put it out.

$\log_a{b^c}$ When she's standing.

$c \log_a{b}$ When she fell.

And the broken pippet rule happened on the same day that the next year learnt that rule so we can predict the time each year something bad's going to happen lol...

leehuan · Feb 1, 2016

Re: HSC 2016 2U Marathon

Drsoccerball said:
I think it's time to explain the burning lady rule. So what happened is on that day we had assembly and there were these dancing people. Just as the show opens the lady puts a ball of flames into her mouth and she screws up and it catches her face on fire... So she was alright afterwards but ojn that day we learnt that rule. And it made sense to call it the burning lady rule as the ladies standing and then she falls because shes on fire to put it out.

$\log_a{b^c}$ When she's standing.

$c \log_a{b}$ When she fell.

And the broken pippet rule happened on the same day that the next year learnt that rule so we can predict the time each year something bad's going to happen lol...

This made no sense

Drsoccerball · Feb 1, 2016

Re: HSC 2016 2U Marathon

leehuan said:
This made no sense

I tried.

Nailgun · Feb 1, 2016

Re: HSC 2016 2U Marathon

what day was it specifically lol

HSC 2016 Maths Marathon (archive) (4 Viewers)

Well-Known Member

Cole World

Well-Known Member

Well-Known Member

Well-Known Member

Cole World

Well-Known Member

Cole World

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Cole World

Cole World

Well-Known Member

Cole World

Well-Known Member

Well-Known Member

Well-Known Member

Cole World

Users Who Are Viewing This Thread (Users: 0, Guests: 4)