CSSA MX2 Trial (1 Viewer)

[math man]

Saw the questions for the MX2 CSSA.

The last question was pretty nice, except not quite happy about the fact that to complete one of the questions, you have to swap an integral and a summation sign under the assumption that it's allowed.

At least say something like "You may freely interchange a summation and the integral, without proof".

But overall this is something I would expect to be HSC difficulty, maybe a wee bit more difficult for the last question.

i agree with the assumed uniform convergence thing, today when i went through the cssa trial in my class i had
to tell my students to just swap the integral and sum notation and that you cant justify it, you just have to do it...

 

[asianese]

I've got everything except the last 2 marks. Any ideas?

 

[Carrotsticks]

 

[asianese]

AH! Thankyou.

Why is it always pi^2/6...HSC/CCSA papers lol

Also carrot what is that typeset from? (program?)

 

[Carrotsticks]

BTW a small typo: The bottom limit for 'O' should be 0, not 1.

AH! Thankyou.

Why is it always pi^2/6...HSC/CCSA papers lol

Also carrot what is that typeset from? (program?)

Math Type + Microsoft Word. Couldn't be bothered TeXing.

And it's asked often because it's the first converging p-series (assumed integer exponent).

 

[asianese]

Yes I realised it should be 0, to get the '1' at the front.

Thankyou!

 

[seanieg89]

On the issue of rigour, it should be noted that the manipulation in the last question is only valid because the series in question is absolutely convergent.

 

[tywebb]

does anyone remember the last question exactly?

Here is question 16b with the solution and Boo Rim Choe's 1987 article upon which it is based:

http://4unitmaths.com/cssa16b.pdf

 

[jezzamon]

i agree with the assumed uniform convergence thing, today when i went through the cssa trial in my class i had
to tell my students to just swap the integral and sum notation and that you cant justify it, you just have to do it...

I just did the question now (not in the test, unfortunately)

Well, you can swap the sum and integral because you can split integrals if the things inside are just added, like we do in partial fractions and everything. Then you take the constant out of the integral, because it's a constant

 

[Carrotsticks]

Well, you can swap the sum and integral because you can split integrals if the things inside are just added, like we do in partial fractions and everything. You can then take the constant out of the integral... not too difficult

That is true for a FINITE number of terms in the sum, but not necessarily for an INFINITE number of terms.

 

[cutemouse]

The pi^2/6 result can also be proved using Complex Analysis. Quite neat I think.

 

[Carrotsticks]

The pi^2/6 result can also be proved using Complex Analysis. Quite neat I think.

Essentially any convergent infinite series can be proven via complex analysis.

 

[deswa1]

Just finished it. All up I thought it was good, a bit harder than our school's paper and the MC was challenging I thought. I liked the last question- I love it when you do all this random integration etc. and then end up with neat results that have almost nothing to do with what you started with. Do you guys know the answers to:

13 d ii) The one about find the coordinates of C if ABCD is a square
16 a ii) Find the values of k and c-> the one where f(x) has line symmetry around x=c

I'm not that confident with my answers on these two

 

[billym]

Just finished it. All up I thought it was good, a bit harder than our school's paper and the MC was challenging I thought. I liked the last question- I love it when you do all this random integration etc. and then end up with neat results that have almost nothing to do with what you started with. Do you guys know the answers to:

13 d ii) The one about find the coordinates of C if ABCD is a square
16 a ii) Find the values of k and c-> the one where f(x) has line symmetry around x=c

I'm not that confident with my answers on these two

13 d ii)

x - value of C = (1+2^(1/2))^(1/2)

16 a ii)

k = (AB)^(1/2)

c = (1/2)ln(B/A)


Pardon the horrible typesetting, not very good with LaTex

 

[deswa1]

13 d ii)

x - value of C = (1+2^(1/2))^(1/2)

16 a ii)

k = (AB)^(1/2)

c = (1/2)ln(B/A)


Pardon the horrible typesetting, not very good with LaTex

Yay got 13 right but my answer for 16 was wrong- it was semi close though- I'll redo it.

Thanks bro

 

[jenslekman]

Just finished it. All up I thought it was good, a bit harder than our school's paper and the MC was challenging I thought. I liked the last question- I love it when you do all this random integration etc. and then end up with neat results that have almost nothing to do with what you started with. Do you guys know the answers to:

13 d ii) The one about find the coordinates of C if ABCD is a square
16 a ii) Find the values of k and c-> the one where f(x) has line symmetry around x=c

I'm not that confident with my answers on these two

is that what you do in class?

 

[deswa1]

is that what you do in class?

Nah I was at home because I'm sick. But yeah, this is sorta what we do in maths lol

 

[jenslekman]

Nah I was at home because I'm sick. But yeah, this is sorta what we do in maths lol

lol doing maths while being sick! get well soon wow.. didnt notice that you were away =/

 

[kayven]

Could I get a copy of the CSSA 2012 trial paper or no?

 

[Nooblet94]

I had a go at it on the weekend. I skipped the ABCD square question because I had no idea how to do it. Got everything else though, just silly errors here and there. Full marks on the last question, which I was very happy about (although I did have to think for a while about the last part of a)).

I thought the multiple choice was surprisingly hard. All my school exams had fairly easy questions, so it came as a bit of a shock. Overall though, it was easier than I'd expected. There was nothing that was really "out there", so to speak.