Answer to 10b? (1 Viewer)

[jtyler]

I should not have made a common denominator in part i, haha crap it would've been so much easier. I have been a fool.

You have no idea how retarded I feel... I found a common denominator once I'd differentiated (mindlessly), and the resultant expression was impossible to solve. The entire time I was staring at it (and it was a long time believe me, I found the previous 9 questions VERY easy, even compared to previous HSC's), the thought of cross multiplying didn't cross my mind once.

I feel so stupid.

 

[sonson fonson]

dammit! why the hell didn't i crossmultiply!????? grrr. oh well hopefully got 2/4 or even 1.

 

[ando24]

i got to the differentation of it to find the minimum and then simplified it for 2 pages of working only to get stuck.....

 

[Fish Sauce]

i got to the differentation of it to find the minimum and then simplified it for 2 pages of working only to get stuck.....

I did about a page of simplifying then noticed a tiny mistake further up the page so I just gave up and went back to do questions I'd missed. I got a lot more marks that way I must say.

 

[Originality]

I think i wouldve given up since it looks really complex, i would have thought it was wrong.

 

[pattersonst]

mitchbazza...
i got that too!
scoooore

 

[Poad]

I should not have made a common denominator in part i, haha crap it would've been so much easier. I have been a fool.

This. Is. SO. True. =\

 

[karentao77]

i gave up in the first 5 minutes of thinking about it...

 

[Fish Sauce]

This. Is. SO. True. =\

Someone feels my pain!

Seriously, the numerator I had in my differentiated equation was three lines long. It was ridiculous. I probably could have done it if I hadn't simplified the thing in part 1

 

[me121]

Yea. I got the same.
P = L[1]/x^2+L[2]/(m-x)^2

dP/dx = -2*L[1]/x^3+2*L[2]/(m-x)^3

Then by making this equal to zero,

x = m*L[1]^(1/3)/(L[1]^(1/3)+L[2]^(1/3))

d^2*P/dx^2 = 6*L[1]/x^4+6*L[2]/(m-x)^4

As the denominators are to the power of 4, they are always +, and as L1 and L2 are >0, the second deriv is +, therefore it is a local minimum.

 

[Steth0scope]

it didnt say L1 and L2 > 0 although they may pay it im not too sure... i simplified my fractions to end up with one fraction with everything to the power of 4/3 or to the power of 4 .. which means the fraction has to be positive because the cube root of a qaurtic (x^4/3) is always positive too... i dunno i may be wrong too