2014 HSC Extension 2 Maths Carrotsticks' Solutions (1 Viewer)

[Sy123]

Need to see the paper

Please someone send me a link as soon as possible

 

[mreditor16]

Am I just throwing a big derp, or is Q15 (a) the hardest question in the paper? All my inequality signs are the wrong way around.

no it's right keep at it

 

[mreditor16]

Need to see the paper

Please someone send me a link as soon as possible

PM'ed

 

[aDimitri]

Am I just throwing a big derp, or is Q15 (a) the hardest question in the paper? All my inequality signs are the wrong way around.

this happened to me for a while. in the end i used a totally different method.

since a<c, a^2 < c^2
therefore 5a^2 + 3b^2 + c^2
< 3a^2 + 3b^2 + 3c^2
< 3((a+b+c)^2 - 2(ab+bc+ca)
< 3(a+b+c)^2 (since a, b, c > 0)
< 3
< 1

is this valid?

 

[faisalabdul16]

this happened to me for a while. in the end i used a totally different method.

since a<c, a^2 < c^2
therefore 5a^2 + 3b^2 + c^2
< 3a^2 + 3b^2 + 3c^2
< 3((a+b+c)^2 - 2(ab+bc+ca)
< 3(a+b+c)^2 (since a, b, c > 0)
< 3
< 1




is this valid?

How did you go from less than 3 to less than 1

 

[RealiseNothing]

this happened to me for a while. in the end i used a totally different method.

since a<c, a^2 < c^2
therefore 5a^2 + 3b^2 + c^2
< 3a^2 + 3b^2 + 3c^2
< 3((a+b+c)^2 - 2(ab+bc+ca)
< 3(a+b+c)^2 (since a, b, c > 0)
< 3
< 1

is this valid?

How did you get from < 3 to < 1?

Anyway here is the easiest way:



Since we get and

Sub these two inequalities in to obtain the answer.

 

[billym]

this happened to me for a while. in the end i used a totally different method.

since a<c, a^2 < c^2
therefore 5a^2 + 3b^2 + c^2
< 3a^2 + 3b^2 + 3c^2
< 3((a+b+c)^2 - 2(ab+bc+ca)
< 3(a+b+c)^2 (since a, b, c > 0)
< 3
< 1

is this valid?

Your last step isn't valid. You can't jump from saying it is less than 3, therefore it is less than 1.

 

[IR]

this happened to me for a while. in the end i used a totally different method.

since a<c, a^2 < c^2
therefore 5a^2 + 3b^2 + c^2
< 3a^2 + 3b^2 + 3c^2
< 3((a+b+c)^2 - 2(ab+bc+ca)
< 3(a+b+c)^2 (since a, b, c > 0)
< 3
< 1

is this valid?

well it said expanding so i expanded and got Sum(a^2)+ 2Sum(ab)
Then i used a^2+b^2> 2ab and b^2+c^2>2bc ...... you add these 3 to get 2sum(a^2)> 2sum (ab)
Then you get 3a^2+3b^2+c^2<1 since (a+b +c)=1
Then you do a>b>c and then you get it.

 

[aDimitri]

How did you go from less than 3 to less than 1

wtf major derp then editing now

cbf editing others have posted now

 

[aDimitri]

How did you get from < 3 to < 1?

Anyway here is the easiest way:



Since we get and

Sub these two inequalities in to obtain the answer.

YEAH JUST REALISED
wtf i didn't even do that in the exam idk where that just came from

 

[IR]

YEAH JUST REALISED
wtf i didn't even do that in the exam idk where that just came from

My method seems right too right?

 

[RealiseNothing]

well it said expanding so i expanded and got Sum(a^2)+ 2Sum(ab)
Then i used a^2+b^2> 2ab and b^2+c^2>2bc ...... you add these 3 to get 2sum(a^2)> 2sum (ab)
Then you get 3a^2+3b^2+c^2<1 since (a+b +c)=1
Then you do a>b>c and then you get it.

I don't really follow your third line?

 

[Carrotsticks]

Here is a quick draft!

https://www.dropbox.com/s/dthlsed7gfknygt/HSC 2014 Extension 2 Solutions.pdf?dl=0

Sorry I had to rush through it so much, but I really have to go to class now! Please let me know if you find errors and I'll fix them when I finish teaching.

I hope you can all appreciate how much time it took to make those damn diagrams.

 

[aDimitri]

well it said expanding so i expanded and got Sum(a^2)+ 2Sum(ab)
Then i used a^2+b^2> 2ab and b^2+c^2>2bc ...... you add these 3 to get 2sum(a^2)> 2sum (ab)
Then you get 3a^2+3b^2+c^2<1 since (a+b +c)=1
Then you do a>b>c and then you get it.

nah, this would give > 1 rather than < 1, i think this is the problem carrot is having

 

[IR]

I don't really follow your third line?

I just looked at your post and basically the same thing I think.

 

[RealiseNothing]

I just looked at your post and basically the same thing I think.

But you end up with > 1 instead of < 1 (you fudged a little without realising).

 

[mreditor16]

EDIT - refer to later post

 

[faisalabdul16]

here is a quick draft!

https://www.dropbox.com/s/dthlsed7gfknygt/hsc 2014 extension 2 solutions.pdf?dl=0

sorry i had to rush through it so much, but i really have to go to class now! Please let me know if you find errors and i'll fix them when i finish teaching.

I hope you can all appreciate how much time it took to make those damn diagrams.

thank you legend <3333333

 

[billym]

14bii.
found the value of theta for which Fi was minimum not maximum. 0/2

Depending on your method, you may still get 1. Markers are not supposed to penalise if someone has clearly misread the question and in doing so it doesn't make the question easier. If it is clear that you misread and found the value for a minimum, but you still did some calculus in the process, you could still get 1.

 

[mreditor16]

Carret, for Q11 e), you said by mistake the function is odd

it is even