Fake HSC paper (1 Viewer)

[lyounamu]

Go to page 11/14
Q10 (i)

This question is ridiculous. People will agree with me on this.

Better method to solve the question than just simply solving it?

 

[Aerath]

Where's the paper? =\

 

[lyounamu]

Where's the paper? =\

Here we go, sorry!

http://www.boredofstudies.org/courses/maths/2u/1152574997_2006_Mathematics_HSC.pdf

 

[Just.Snaz]

lol.. nah, there's no other way.. the question is basically just testing your algebraic skills.. easy two marks I reckon..

 

[lolokay]

Go to page 11/14
Q10 (i)

This question is ridiculous. People will agree with me on this.

Better method to solve the question than just simply solving it?

what's so ridiculous about it?

 

[conics2008]

thats old...

 

[lyounamu]

what's so ridiculous about it?

Because it's only two marks. I had to write at least 10 lines to work that out.

 

[agirlinatutu]

HSC exam 4002 spoof – please note this is not an actual Board of Studies endorsed HSC examination – only a parody of a 2 unit HSC
examination, created by live fast @ Bored of Studies

Namu are you aiming for bloody genuis?

 

[Advv]

Because it's only two marks. I had to write at least 10 lines to work that out.

It takes a minute or two to do. It shouldn't be worth more than two.

 

[lyounamu]

It takes a minute or two to do. It shouldn't be worth more than two.

Really? Show me the working out that takes a minute or two. (not trying to be rude here, I am serious)

 

[Advv]

I'd scan it but I don't have a scanner. I just differentiated them both using -v'/v^2 and then used common denominator. Then I factorised the 2 out and voila. I didn't time myself or anything but it didn't take very long.

 

[shaon0]

Go to page 11/14
Q10 (i)

This question is ridiculous. People will agree with me on this.

Better method to solve the question than just simply solving it?

What's so ridiculous about it?

 

[monstylez]

lynamou, I don't say this often but stop being a little bitch. The last few questions are just long tedious algebra to be the discminator between band 5 and 6 students.

It's easy.

 

[lyounamu]

lynamou, I don't say this often but stop being a little bitch. The last few questions are just long tedious algebra to be the discminator between band 5 and 6 students.

It's easy.

Seriously, what did I do to deserve such a blame?

I only wanted to see if there is a better method to solve the question. If there isn't, there isn't and if there is, there is. I don't also say this often but piss off.

And if this is as easy as you make it out to be, why not provide a few line solution to back your statement up.

 

[lolokay]

what's so ridiculous about it?

What's so ridiculous about it?

lol

just do normal chain rule differentiation to each part, let u = (x+8), so du/dx = 1
you get d(b² + u²)^-1/du
= -2u(b² + u²)^-2

and likewise for the other part, then add them

 

[lyounamu]

lol

just do normal chain rule differentiation to each part, let u = (x+8), so du/dx = 1
you get d(b² + u²)^-1/du
= -2u(b² + u²)^-2

and likewise for the other part, then add them

Hahaha...that's more like it.

Thanks.

 

[cheney31]

TO ANSWER PARTS II AND III YOU MAY ASSUME THE FOLLOWING FACTORIZATION GIVEN BY A COMPUTER PACKAGE


lololololol............

 

[Dota55]

anyone got the answers for parts (ii) or (iii)

or at least a method of dealing with them?

 

[Dota55]

Seriously, what did I do to deserve such a blame?

I only wanted to see if there is a better method to solve the question. If there isn't, there isn't and if there is, there is. I don't also say this often but piss off.

And if this is as easy as you make it out to be, why not provide a few line solution to back your statement up.

Actually...it IS easy. Yeah, it looks very daunting to begin with...but if you know how to differentiate and have decent algebra skills (which you do, i know you do ) then you can do it.

(i) that is. The rest...i don't understand

 

[Advv]

anyone got the answers for parts (ii) or (iii)

or at least a method of dealing with them?

Uhh. My guess would be to sub x=15 then let dI/dx = 0 and find any turning points. There should be a maximum turning point at x=0.

Then use the sign of y'.So on the left of 0, the curve would be increasing and on the right, it'd be decreasing? o_o;

Iono.