Cssa Trial Exam (1 Viewer)

[haboozin]

i know an early question....

but the area under f(x) and f^-1(x) = 3u^2 ??

 

[FinalFantasy]

i know an early question....

but the area under f(x) and f^-1(x) = 3u^2 ??

hey i did question 1 last and rushed it
but i got area is 3 units² also haha

 

[Antwan23q]

yeh i reckon i went alright, but i lost heaps of marks! like the last question i forgot the surface area rule of a cone. that was 8 marks. and question 5. i fucked that one up. but the rest i went fair good, im lookin at a mark of 70-80

 

[FinalFantasy]

was question 5 the volumes and the trig?
the tan2@ and stuff?

 

[~ ReNcH ~]

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

What FinalFantasy said...I just can't remember the exact form but I remember that it was a fraction with tan(x/2) in the denominator somewhere...and looking at FinalFantasy's working, it seems as though the answer you got was the one I got, as I remember having to integrate 2/(1+t)² now

 

[~ ReNcH ~]

hey i did question 1 last and rushed it
but i got area is 3 units² also haha

I just left that...I didn't know how to approach it. Did you actually have to find the equation of f^-1(x)? Coz when I figured that I had to do that, then I just thought I'd leave it...

 

[~ ReNcH ~]

was question 5 the volumes and the trig?
the tan2@ and stuff?

The whole paper was posted up here in an earlier post in this thread...maybe page 2 or 3...

 

[Dumsum]

I just left that...I didn't know how to approach it. Did you actually have to find the equation of f^-1(x)? Coz when I figured that I had to do that, then I just thought I'd leave it...

You can do the area of the square minus twice the area under f(x) from sqrt(3) to 2, because of symmetry.

A = 4 - 2 int{sqrt3 -> 2} (x^3 - 3x)dx
= 4 - 2[x^4 / 4 -3/2 x^2] _sqrt3²
= 4 - 2(4 - 6 - (9/4 - 9/2))
= 3.5 sq units.

I think that's right.

 

[Stefano]

I just got my mark back: 40% (Equal 2nd in my school)
First place was 69%

FAAAAAARRRRKKKK IM ANNOYED!!

 

[fantasy27]

wow you got it back so quickly.. i dont even think our teacher has started marking it, let alone saw our solutions yet... hmmm.. so quick!

 

[Stefano]

Well, theres only 3 of us you see.

 

[EATAPIE]

I just got my mark back: 40% (Equal 2nd in my school)
First place was 69%.
Well, theres only 3 of us you see.

hahaha, good on ya stefano
haha

 

[David_O]

You can do the area of the square minus twice the area under f(x) from sqrt(3) to 2, because of symmetry.

A = 4 - 2 int{sqrt3 -> 2} (x^3 - 3x)dx
= 4 - 2[x^4 / 4 -3/2 x^2] _sqrt3²
= 4 - 2(4 - 6 - (9/4 - 9/2))
= 3.5 sq units.

I think that's right.

That's right.
I managed to get it wrong by saying 2 times 2 is 3 instead of 4.
How hideous.

 

[ishq]

You can do the area of the square minus twice the area under f(x) from sqrt(3) to 2, because of symmetry.

A = 4 - 2 int{sqrt3 -> 2} (x^3 - 3x)dx
= 4 - 2[x^4 / 4 -3/2 x^2] _sqrt3²
= 4 - 2(4 - 6 - (9/4 - 9/2))
= 3.5 sq units.

I think that's right.

Exactly what I did - hooray!!!
Except - I think my limits are wrong..I did zero to sqrt 3....
DAMN!!!!

 

[Dumsum]

That's right.
I managed to get it wrong by saying 2 times 2 is 3 instead of 4.
How hideous.

Reminds me of this time in yr 11 2u where i did 2/2 = 0... suffice to say, the only thing I did wrong in that exam

 

[~ ReNcH ~]

You can do the area of the square minus twice the area under f(x) from sqrt(3) to 2, because of symmetry.

A = 4 - 2 int{sqrt3 -> 2} (x^3 - 3x)dx
= 4 - 2[x^4 / 4 -3/2 x^2] _sqrt3²
= 4 - 2(4 - 6 - (9/4 - 9/2))
= 3.5 sq units.

I think that's right.

Aha...that makes sense.
Coz I was thinking about finding the equation of the inverse and doing it that way, but I opted against it and worked on further questions...ah well, 2 marks lost there then.

 

[LaCe]

Well i did the 4u Trial last Wednesday and a couple of questions came from the Independent Trial and CSSA. Altogether it wasnt anything ridiculously hard but then again there werent too many "gimme" marks.

 

[yook]

Well i did the 4u Trial last Wednesday and a couple of questions came from the Independent Trial and CSSA. Altogether it wasnt anything ridiculously hard but then again there werent too many "gimme" marks.

isnt that... not allowed?!?! even "parts" of the CSSA??

 

[~ ReNcH ~]

isnt that... not allowed?!?! even "parts" of the CSSA??

You would think so...that's how Cheltenham got into trouble i.e. for releasing whole CSSA exams and even parts of it.

 

[fantasy27]

Aha...that makes sense.
Coz I was thinking about finding the equation of the inverse and doing it that way, but I opted against it and worked on further questions...ah well, 2 marks lost there then.

yeh i tried, but it got really messy and abadoned it several minutes later and worked on something else ><