Maths Extension 2 thoughts (1 Viewer)

leehuan · Oct 21, 2015

Thjv said:
what if n is even, such as if n = 6? Does that still make it 1/2??

If Ekman tested n=4 which I just did in my head it was still 1/2

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t_davis9 · Oct 21, 2015

leehuan said:
If Ekman tested n=4 which I just did in my head it was still 1/2

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That's what I did, just wrote them all out divided by 4!

snail489 · Oct 21, 2015

thank the lord for no circle geo!!

Any of you guys manage to get the lim(n->infin) P_n thing in q16?? was the only question I couldn't figure out how to get

Sy123 · Oct 21, 2015

Anyone with the paper please PM me

RealiseNothing · Oct 21, 2015

Thjv said:
What was the answer for the probability multiple choice question, where 1 is to the left of 2?

1/2 cos its either to the left of 2, or the right of 2.

Only two possibilities.

Librah · Oct 21, 2015

Sy123 said:
Anyone with the paper please PM me

same

Thjv · Oct 21, 2015

RealiseNothing said:
1/2 cos its either to the left of 2, or the right of 2.

Only two possibilities.

Officially confirmed by the BOS master.

-1

Rip in peace.

lita1000 · Oct 21, 2015

can someone upload the paper? Would like to double check to see whether I made any errors

InteGrand · Oct 21, 2015

leehuan said:
I just want to know what lim n->infinity n^p/(npCp) is

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$I'm assuming $p$ is a fixed positive integer. Then $\frac{n^p}{\binom{np}{p}}=\frac{n^p}{\frac{(np)(np-1)\cdots (np -(p -1))}{p!}}=\frac{n^p}{\underbrace{(np)(np-1)\cdots (np -(p-1))}_{p\text{ terms}}}\times p!$. Hence as $n\to \infty$, this quantity approaches $\frac{p!}{p^p}$, since the fraction has the polynomial $n^p$ on the numerator, and a degree $p$ polynomial on the denominator with leading coefficient $p^p$.$

leehuan · Oct 21, 2015

How to do 16b) (iii) and (iv)

(iii) - If you tried solving for 0, you found that all of the roots were cos .........
So you just took the product of roots from there

(iv) x = 1/sqrt2

leehuan · Oct 21, 2015

InteGrand said:
$I'm assuming $p$ is a fixed positive integer. Then $\frac{n^p}{\binom{np}{p}}=\frac{n^p}{\frac{(np)(np-1)\cdots (np -(p -1))}{p!}}=\frac{n^p}{\underbrace{(np)(np-1)\cdots (np -(p-1))}_{p\text{ terms}}}\times p!$. Hence as $n\to \infty$, this quantity approaches $\frac{p!}{p^p}$, since the fraction has the polynomial $n^p$ on the numerator, and a degree $p$ polynomial on the denominator with leading coefficient $p^p$.$

Clever. I would not have considered the leading term.

Ekman · Oct 21, 2015

InteGrand said:
$I'm assuming $p$ is a fixed positive integer. Then $\frac{n^p}{\binom{np}{p}}=\frac{n^p}{\frac{(np)(np-1)\cdots (np -(p -1))}{p!}}=\frac{n^p}{\underbrace{(np)(np-1)\cdots (np -(p-1))}_{p\text{ terms}}}\times p!$. Hence as $n\to \infty$, this quantity approaches $\frac{p!}{p^p}$, since the fraction has the polynomial $n^p$ on the numerator, and a degree $p$ polynomial on the denominator with leading coefficient $p^p$.$

YES OMG YES! I was so unconfident about that answer! Thanks Integrand!

atargainz · Oct 21, 2015

cut-off predictions anyone?

Drsoccerball · Oct 21, 2015

leehuan said:
How to do 16b) (iii) and (iv)

(iii) - If you tried solving for 0, you found that all of the roots were cos .........
So you just took the product of roots from there

(iv) x = 1/sqrt2

i got (-1)^n

I couldn't find the substitution

LOL that moment when you forget that $sec^2x-tan^2x =1$

Lost 3 marks so far lel

InteGrand · Oct 21, 2015

leehuan said:
Clever. I would not have considered the leading term.

$This is the general method of finding the limit as $x\to \infty$ of a ratio of equal degree polynomials. I.e. $\lim _{x\to \infty}\frac{a_N x^N +a_{N-1}x^{N-1} \cdots + a_0}{b_N x^N +b_{N-1}x^{N-1}+ \cdots + b_0}=\frac{a_N}{b_N}$. This is seen by dividing top and bottom of the quotient by $x^N$, and then seeing that all the terms except the leading terms will become fractions with powers of $x$ on the denominator, so all these terms go to 0 as $x\to \infty$, leaving you with $\frac{a_N}{b_N}$.$

Drsoccerball · Oct 21, 2015

That happy moment when seeing $\frac{a+b+c+d}{4} \geq (abcd)^{1/4}$

Zen2613 · Oct 21, 2015

Kaido said:
Rekt, i threw booklets up to 3 tables away and couldnt find where i answered the Q. Its hard to control where 13 booklets go -.-

Same here OMG! 13 booklets and then some slipped away from me, they were scattered all over the ahop.

RealiseNothing · Oct 21, 2015

Drsoccerball said:
That happy moment when seeing $\frac{a+b+c+d}{4} \geq (abcd)^{1/4}$

The inequality was something called the QM-GM which is just the AM-GM extended to the QM-AM-GM-HM but without the AM and HM.

Zen2613 · Oct 21, 2015

Thjv said:
I did not get A. Can anyone confirm if this answer is correct or not?

I did too. Except I found the probability using a specific value for n. I think i let n=5 and got 1/2. so yes A

Kaido · Oct 21, 2015

Wtf is terry lee doing, hurry up and release the solutions LEL

Maths Extension 2 thoughts (1 Viewer)

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