2009 past hsc solutions cant find them >< (1 Viewer)

[fullonoob]

does anyone have the solutions to the 2009 paper, i've looked in search, google, bos everywhere...
if anyone has it may you please post it up
or otherwise please help with these 2 questions

 

[s2 SEductive]

does anyone have the solutions to the 2009 paper, i've looked in search, google, bos everywhere...
if anyone has it may you please post it up
or otherwise please help with these 2 questions

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2009exams/notes/2009-hsc-notes-maths-ext-2.pdf

it's more like notes from marking centre.

 

[Trebla]

For the seond one use substitution x = tan u

 

[Rabbi_Nigger]

For the seond one use substitution x = tan u

oomg i remember this bitch ass question!
I got it wrong roflz, but i got heaps of q8 and 7 right, to make up for it.! hells yea

 

[fullonoob]

thanks once again trebla, how do you know whether to use tanu , sinu or cos u

 

[kcqn93]

http://www.hsccoaching.com/Resources/2009Ext2.pdf

 

[fullonoob]

how do you diff sin^2 x , i got the answer, but dno how
integration by parts seems long

 

[Trebla]

thanks once again trebla, how do you know whether to use tanu , sinu or cos u

How else can you get rid of that square root?

 

[fullonoob]

int cos@ / sin^2@ d@
u = cosec^2@
du = - cot x dx
int - root u du
doesnt work out to what i want it to be...
this is the second question on post
bloody hard for question one

 

[shaon0]

int cos@ / sin^2@ d@
u = cosec^2@
du = - cot x dx
int - root u du
doesnt work out to what i want it to be...
this is the second question on post
bloody hard for question one

Just make the top into 1+x^2-x^2 and you'll get:
I= S [1,sqrt(3)] sqrt(1+x^2)/x^2 dx-S[1,sqrt(3)] 1/sqrt(1+x^2) dx
then let x=tanu. I think you can do the rest

 

[cutemouse]

Or just use long division first. There's less mucking around that way I think.

For the second question you could use x=1/u, but that'd require a bit more work than using the sub. x=secθ