Cssa Trial Exam (2 Viewers)

[fantasy27]

the locus with the circle, drawing that stupid graph took me 3-4 min,, caz we have to use bloody compasses, or we lose marks... does ur school make u guys do that? and i heard from my tutor in hsc its ok to freehand as long as u put radius, centre, and maybe provide a eqn'.

yeh we have to draw "perfect" and half page diagrams, if we dont.. possible deduction of marks.. but i dont think its ever happened!

 

[~ ReNcH ~]

Hmm...
Let's see. I didn't get:
Q1c(ii), Q2e(i), Q3b(ii), Q4c(ii), Q5b(ii), Q6c(iii)/(iv) (lost 2 marks), Q8c(i)/(ii) (did some but ended up with a maximum volume slightly different to the one given) - Q7 was a replacement for me.

Did you guys have to use t-formulae for the first integration question?
I fudged Q5b(iii) but got the answers and it didn't seem like I could 3/3 for the few lines of working I actually did...so I'm slightly pessimistic about that one.

And with Q4c(iii): was the locus y=1 with a restriction of x=/0?

 

[FinalFantasy]

^ yea that integration q
int. (1-sinx)\cos²x dx=int. (1-sinx)\(1-sinx)(1+sinx) dx
then a t formula

what was that Q5biii)? was it the tan2@ one?

 

[David_O]

x<0, x>= 1

damn I am bored.

edit: integ: sec^2x - secxtanx

 

[FD3S-R]

^ yea that integration q
int. (1-sinx)\cos²x dx=int. (1-sinx)\(1-sinx)(1+sinx) dx
then a t formula

what was that Q5biii)? was it the tan2@ one?

wtf ∫ (1-sinx) / cos²x dx
= ∫1/ cos²x- secxtanx dx
= tanx - secx + c
???

thats what i did

 

[fantasy27]

Did you guys have to use t-formulae for the first integration question?
I fudged Q5b(iii) but got the answers and it didn't seem like I could 3/3 for the few lines of working I actually did...so I'm slightly pessimistic about that one.

And with Q4c(iii): was the locus y=1 with a restriction of x=/0?

yeh i use t-formulae for that one.. and i got locus of y=1 as well.. but i thought it was wrong and couldnt establish a 'domain or restriction... so i left it.. damn!

 

[ishq]

^ yea that integration q
int. (1-sinx)\cos²x dx=int. (1-sinx)\(1-sinx)(1+sinx) dx
then a t formula

what was that Q5biii)? was it the tan2@ one?

I put inverse tan sqrt sinx

 

[~ ReNcH ~]

wtf ∫ (1-sinx) / cos²x dx
= ∫1/ cos²x- secxtanx dx
= tanx - secx + c
???

thats what i did

True...well, at least my method isn't wrong. It's just another way of expressing it.

On that note...even if your method is different, thus yielding a different result to the one in the criteria, you would still get full marks wouldn't you?

 

[Boz_87]

wtf ∫ (1-sinx) / cos²x dx
= ∫1/ cos²x- secxtanx dx
= tanx - secx + c
???

thats what i did

I got the same mate im pretty sure its right coz 1/cos^2 X is sec^2 X=tanX etc did u use substution for the sinX/cos^2 X?

 

[lum]

yup, lots of intergrals can be intergrated in more than one way, doesn't matter which method you use, as long as it's corret

 

[FinalFantasy]

wat?? no one did it my way?
so which one's right?

 

[~ ReNcH ~]

yup, lots of intergrals can be intergrated in more than one way, doesn't matter which method you use, as long as it's corret

That's cool...particularly with trig. which can have so many variations.

 

[~ ReNcH ~]

wat?? no one did it my way?
so which one's right?

They're both right afaik...I did it your way and ended up with something like 1/[2+tan(x/2)]

 

[FinalFantasy]

cool
-----------

 

[noah]

u sound like you did well, what schools u guys go to?

the locus with the circle, drawing that stupid graph took me 3-4 min,, caz we have to use bloody compasses, or we lose marks... does ur school make u guys do that? and i heard from my tutor in hsc its ok to freehand as long as u put radius, centre, and maybe provide a eqn'.

for probability, did anyone else get p = 2/7 or 3/7? it's something on 7.

im pretty sure it was 4/7 or 7/4 or something

 

[lum]

They're both right afaik...I did it your way and ended up with something like 1/[2+tan(x/2)]

what? i did it wif t method too, but i got -2/( tanx/2 +1)

 

[FinalFantasy]

int. (1-sinx)\cos²x dx=int. 1\(1+sinx) dx
let t=tan (x\2)
dx=2dt\(1+t²)
int. 1\(1+2t\(1+t²) ) * 2dt\(1+t²)
=int. 2dt\(1+t²+2t)=int. 2dt\(1+t)²
=-2\(1+t)
=-2\(1+tan (x\2) )

rench said "something like...." so maybe he just didn't type it properly here

 

[tim_dawborn]

why not just say that it was a pretty shite paper all rounded

 

[ishq]

The marks wil tell

 

[lum]

nah, i arldy know how i went, it can only become worse with time