no it's right keep at itAm 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.
PM'edNeed to see the paper
Please someone send me a link as soon as possible
this happened to me for a while. in the end i used a totally different method.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.
How did you go from less than 3 to less than 1this 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?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.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)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?
wtf major derp then editing nowHow did you go from less than 3 to less than 1
YEAH JUST REALISEDHow 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.
My method seems right too right?YEAH JUST REALISED
wtf i didn't even do that in the exam idk where that just came from
I don't really follow your third line?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 havingwell 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 just looked at your post and basically the same thing I think.I don't really follow your third line?
But you end up with > 1 instead of < 1 (you fudged a little without realising).I just looked at your post and basically the same thing I think.
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.
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 1.14bii.
found the value of theta for which Fi was minimum not maximum. 0/2