Thoughts on CSSA Mathematics Extension 2 Paper (1 Viewer)

panakap · Jul 31, 2014

To all those who did the Catholic paper today...

Personally I found the test of a reasonable difficulty.

From the multiple choice I struggled a little with Q10 but I think I got it :/

Question 11 to 15 was relatively straight forward except for part (ii) of the polynomial multiplicity and the circle geo (I'm normally really good with circle geo but this one really troubled me... answers anyone?).

HOWEVER, I think this was one of the hardest Q16's in a while.
The first inequality was straight forward.
But that volume and summation question really got to me :/

My school hasn't done mechanics yet so give me your thoughts on those questions...

In all, I'm hoping for an 85% but we shall see

How did you find it?

hecticated · Jul 31, 2014

question 16 was the biggest wtf. but now that i think about the volume question i realised you use the sine rule haha.
and OMG i worked out the circle geo in the last minute and i didnt get to finish writing my answer

you had to use angle at the centre is double the angle at the circumference, and then you used cylcic quadrilaterals, eventually it would work out to be 180 degrees and therefore collinear.
I also struggled with the second part of polynomials.

we havent done mechanics either.

panakap · Jul 31, 2014

From memory I think I do remember using the angle at centre theorem, but not sure about the cyclic quad I think I proved either similarity or congruence don't remember. Cyclic quad does sound like the way to go nice job

Ahh yes the sine rule I never saw that damn that sucks :/
Anyways it is what it is.

[e] I loved that induction question <3

panakap · Jul 31, 2014

Anyone who has done mechanics how were the questions?

faisalabdul16 · Jul 31, 2014

How was the test difficulty in regards to the 2013 CSSA paper?

panakap · Jul 31, 2014

I'm not quite sure my teacher never supplied me with the 2013 paper

hecticated · Jul 31, 2014

I don't think I'd get full marks for the cylcic quad though because I didn't explain it well since I had no time haha.
I explained the volume question to a 3U friend, I said that I was thinking about using triangles but i didn't know the height and she just said - can't you use the sine rule? soo thats awkward. because there was a sine there.

faisalabdul16 · Jul 31, 2014

I have my exam tomorrow!

I don't think mines catholic, probably a mixture of independent, NEAP and old past papers, like old old hahah

panakap · Jul 31, 2014

Haha well good luck then!!

Anyone with questions about the paper I might be able to answer?

Mechanics kids anything?

integral95 · Jul 31, 2014

heh the 4U 2013 paper was probably one of the easiest CSSA papers that it received some criticism (not really).

panakap · Jul 31, 2014

Wow I guess they wanted revenge by killing us with telescoping sums

panakap · Jul 31, 2014

Since I was limited on time on the last question, I could do part (i) but skipped until (iv) which I got :v

[e] PM'd you for the exam my cousin wants to see it

Davo_01 · Jul 31, 2014

Davo_01 said:
$$i) Show that$ \dfrac{x^{2^n}}{1-x^{2^{n+1}}}=\dfrac{1}{1- x^{2^n}}-\dfrac{1}{1-x^{2^{n+1}}}$

$$ii) Hence show that$ \sum_{n=0}^{N} \dfrac{x^{2^n}}{1-x^{2^{n+1}}}=\dfrac{1}{1-x}-\dfrac{1}{1-x^{2^{N+1}}}$

$$iii)Given that$ \lim_{n\rightarrow \infty}\sum_{n=0}^{N} \dfrac{x^{2^n}}{1-x^{2^{n+1}}}=\sum_{n=0}^{\infty} \dfrac{x^{2^n}}{1-x^{2^{n+1}}}\; $and$\; -1<x<1$

$$Hence show that $ \sum_{n=0}^{\infty} \dfrac{x^{2^n}}{1-x^{2^{n+1}}}=\dfrac{x}{1-x}$

$$iv)Evaluate$ \sum_{n=0}^{\infty} \dfrac{1}{2014^{2^n}-2014^{-2^n}}$

Solution:
$$i)\dfrac{x^{2^n}}{1-x^{2^{n+1}}}=\dfrac{x^{2^n} (1-x^{2^{n}})}{(1-x^{2^{n+1}})(1-x^{2^{n}})}=\dfrac{(1-x^{2^{n+1}})-(1-x^{2^{n}})}{(1-x^{2^{n+1}})(1-x^{2^{n}})}$
$=\dfrac{1}{1- x^{2^n}}-\dfrac{1}{1-x^{2^{n+1}}}$

$$ii) Hence show that$ \sum_{n=0}^{N} \dfrac{x^{2^n}}{1-x^{2^{n+1}}}=\dfrac{1}{1-x}-\dfrac{1}{1-x^{2^{N+1}}}$
$$Telescoping sum:$ \sum_{n=0}^{N}(a_n-a_{n+1})=a_0-a_{N+1}$

$$Substituting$\; a_n=\dfrac{1}{1- x^{2^n}}\; $gives the required result$

$$iii) Since$\; -1<x<1\; $then$ \lim_{n\rightarrow \infty} x^{2^n}=0$

$\sum_{n=0}^{\infty} \dfrac{x^{2^n}}{1-x^{2^{n+1}}}=\dfrac{1}{1-x}-1=\dfrac{x}{1-x}$

$$iv)$ \sum_{n=0}^{\infty} \dfrac{1}{2014^{2^n}-2014^{-2^n}}=\sum_{n=0}^{\infty} \dfrac{2014^{-2^n}}{1-2014^{-2^{n+1}}}=\dfrac{2014^{-1}}{1-2014^{-1}}=\dfrac{1}{2013}$

panakap · Jul 31, 2014

Personally I don't understand the maths behind parts (ii) and (iii)... :/

Davo_01 · Jul 31, 2014

panakap said:
Since I was limited on time on the last question, I could do part (i) but skipped until (iv) which I got :v

[e] PM'd you for the exam my cousin wants to see it

you mean the whole exam? i dont have it, safety period...

I remember most of the exam though

panakap · Jul 31, 2014

Ahh ok sorry a misunderstanding...

How did you show that 4p^3=qr^2 in the roots of multiplicity I think I got most of the way there..

Davo_01 · Jul 31, 2014

panakap said:
Personally I don't understand the maths behind parts (ii) and (iii)... :/

For the telescoping sum, what happens is that the "middle terms" cancel out:

$\sum_{n=0}^{N}(a_n-a_{n+1})=(a_0-a_1)+(a_1-a_2)+(a_2-a_3)+...+(a_{n}-a_{n+1})=a_0-a_{n+1}$

As for part 3, because the magnitude of x is less than 1, the larger we "power it" the smaller it becomes. If we raise it to the largest number possible it approaches a very small number, zero!

so if you look at part 2 the 2^{n+1} approaches zero leaving x/(1-x)-1

panakap · Jul 31, 2014

Davo_01 said:
For the telescoping sum, what happens is that the "middle terms" cancel out:

$\sum_{n=0}^{N}(a_n-a_{n+1})=(a_0-a_1)+(a_1-a_2)+(a_2-a_3)+...+(a_{n-1}-a_{n+1})=a_0-a_{n+1}$

Ahh I see yes that makes sense.
And is this known as a telescoping sum?

Davo_01 · Jul 31, 2014

panakap said:
Ahh I see yes that makes sense.
And is this known as a telescoping sum?

Yes

panakap · Jul 31, 2014

I understand. Wish I had known that for the exam that would've been easy haha

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