HSC 2016 Maths Marathon (archive) (1 Viewer)

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leehuan

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Re: HSC 2016 2U Marathon

I can't be bothered Googling but I am slightly confused about the domain for an absolute function.

Would the below setting be right for the domain of a function over that interval?

Yes.

And for the above, [0, ...?] check the pi/2. Also by arbitrary points I do mean random points, but in this situation I mean variable points on the unit circle.

Also, are your points really points on the unit circle?
 
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davidgoes4wce

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Re: HSC 2016 2U Marathon

So variable points in your question could be
A( 1/ sqrt(2), 1/sqrt(2)) B( sqrt(3)/2, 1/2)

We treat it like a parametric equation for 'x' and 'y'.
 

InteGrand

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Re: HSC 2016 2U Marathon

I can't be bothered Googling but I am slightly confused about the domain for an absolute function.

Would the below setting be right for the domain of a function over that interval?

"Domain" refers to the set of all input values where the function is defined. For the absolute value function you posted, the maximal domain the set of all real numbers ().
 

leehuan

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Re: HSC 2016 2U Marathon

So variable points in your question could be
A( 1/ sqrt(2), 1/sqrt(2)) B( sqrt(3)/2, 1/2)

We treat it like a parametric equation for 'x' and 'y'.
Pretty much but leave the parameters in there, in which case a and b here.
"Domain" refers to the set of all input values where the function is defined. For the absolute value function you posted, the maximal domain the set of all real numbers ().
Lol I immediately forgot he said domain.
 

davidgoes4wce

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Re: HSC 2016 2U Marathon

Maybe what I should have asked was , 'is it the right set of x values upon which the interval lies'? (Not sure if there is a specific term in maths to refer to where the inequality exists x<=1 or x>1
 

calamebe

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Re: HSC 2016 2U Marathon

Ok so I haven't gone far in integration, so I may do a little bit of research but here is my answer to first bit: https://m.imgur.com/u9s89zo
I think I may have skipped a bit in the 2nd to 3rd line of ii), so let me explain what I did. So as cos(90-x)=sin(x), cos(270-x)=sin(x), sin(90-x)=cos(x), and sin(270-x)=-cos(x), I just subbed all of those values in to get an expression for that in terms of cos(a), cos(b), sin(a) and sin(b). Also the white out because I used the compound angle formula, something I had to prove later lol.
 

leehuan

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Re: HSC 2016 2U Marathon

Ok so I haven't gone far in integration, so I may do a little bit of research but here is my answer to first bit: https://m.imgur.com/u9s89zo
I think I may have skipped a bit in the 2nd to 3rd line of ii), so let me explain what I did. So as cos(90-x)=sin(x), cos(270-x)=sin(x), sin(90-x)=cos(x), and sin(270-x)=-cos(x), I just subbed all of those values in to get an expression for that in terms of cos(a), cos(b), sin(a) and sin(b). Also the white out because I used the compound angle formula, something I had to prove later lol.
Note that this is a 2U marathon, and the compound angle formula is therefore NOT assumed under normal circumstances.

But otherwise, I like your working. Lol, I accidentally made C(0,1) instead of (1,0) and complicated the question, but use of ASTC is acceptable without explanation for this question as otherwise it's too much explanation.
______________________________
Tbh, the second part is easier than the entire rest-of-the-question, believe it or not. It can be done very easily using part (a) and the table of standard integrals.
 
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leehuan

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Re: HSC 2016 2U Marathon

NEXT QUESTION:

 

Drsoccerball

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Re: HSC 2016 2U Marathon

NEXT QUESTION:

I think we should post questions as they do the topics. Log's was something we learnt towards the end and it can be not valued as much as it should be. We should stick with calculus questions for a while.
 

flavurr

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Re: HSC 2016 2U Marathon

a^o=X => loga(X)=o
a^g= Y => loga(Y)=g
a^l=XY => loga(XY)=l

a^l = a^o * a^g (when multiplying by same base add exponents)

a^l = a^o+g (same base, exponents must equal each other)

l = o+g (substitute back into log form)

loga(XY) = loga(X)+loga(Y)

Im going to get roasted for not using latex
 

DatAtarLyfe

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Re: HSC 2016 2U Marathon

a^o=X => loga(X)=o
a^g= Y => loga(Y)=g
a^l=XY => loga(XY)=l

a^l = a^o * a^g (when multiplying by same base add exponents)

a^l = a^o+g (same base, exponents must equal each other)

l = o+g (substitute back into log form)

loga(XY) = loga(X)+loga(Y)

Im going to get roasted for not using latex
its k, join the club.
Next question, for 2016ers ONLY (wanna see how many can get it, found it relatively easy)
Untitled.png
 

leehuan

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Re: HSC 2016 2U Marathon

I think we should post questions as they do the topics. Log's was something we learnt towards the end and it can be not valued as much as it should be. We should stick with calculus questions for a while.
My bad, I forgot logs was a later topic in 2U.

its k, join the club.
Next question, for 2016ers ONLY (wanna see how many can get it, found it relatively easy)
View attachment 32641
Aiya...
 

DatAtarLyfe

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Re: HSC 2016 2U Marathon

was in 2015 2u paper but wanted one of the 2016ers to do it (that didn't accelerate)
 

calamebe

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Re: HSC 2016 2U Marathon

its k, join the club.
Next question, for 2016ers ONLY (wanna see how many can get it, found it relatively easy)
View attachment 32641
Didn't get time to do it until now, though someone else would have done it by now. Anyway here's my stuff:

https://imgur.com/a/TUf8b

lol also the album is the wrong way around, just look at the second pic first then the 1st
 

leehuan

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Re: HSC 2016 2U Marathon

Didn't get time to do it until now, though someone else would have done it by now. Anyway here's my stuff:

https://imgur.com/a/TUf8b

lol also the album is the wrong way around, just look at the second pic first then the 1st
Did you perform an argument with areas for (i)? Because similar triangles was a more tidier method.
_________________________________________________
NEXT QUESTION:
 
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calamebe

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Re: HSC 2016 2U Marathon

Did you perform an argument with areas for (i)? Because similar triangles was a more tidier method.
Yeah, I could tell there was a better way, just thought that if I've already proved it there's no reason to do it again.
 

Green Yoda

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Re: HSC 2016 2U Marathon

Did you perform an argument with areas for (i)? Because similar triangles was a more tidier method.
_________________________________________________
NEXT QUESTION:
a) 1/36
b) bit confused here as the game will end as proposed in the question so my answer of sky winning is 1/2 as there are 2 players. (might be wrong).
 

leehuan

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Re: HSC 2016 2U Marathon

a) 1/36
b) bit confused here as the game will end as proposed in the question so my answer of sky winning is 1/2 as there are 2 players. (might be wrong).
Nope, it's not a symmetry probability as Sky has the advantage of going first.
 
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