Author Topic: 4U Maths Question Thread (Read 665265 times) Tweet Share

beau77bro · « **Reply #1455 on:** October 07, 2017, 10:19:56 pm »

ok so i get the inequality. but wth is m. like is it the coefficient of every single term??? that seems so sus and impossible. how can u factorise it out if its the coeffiecient of each??

RuiAce · « **Reply #1456 on:** October 07, 2017, 10:39:36 pm »

Quote from: beau77bro on October 07, 2017, 10:19:56 pm

ok so i get the inequality. but wth is m. like is it the coefficient of every single term??? that seems so sus and impossible. how can u factorise it out if its the coeffiecient of each??

$M\text{ is the }\textbf{maximum value}\text{ out of }\textbf{ALL}\text{ of the terms }\\ |a_{n-1}|, \dots, |a_0|$
$\text{Because }M\text{ is the }\textbf{max}\\ \textbf{every value}\text{ must be less than or equal to it}\\ \text{Hence, we obtain}\\ |a_{n-1}|\le M, \, |a_{n-2}|\le M,\,\dots, \, |a_1| \le M, \, |a_0|\le M$

Visual example

$\text{So if you had a polynomial}\\ p(x) = x^3 + (1+i)x^2 + (3+i)x + 5$
$\text{The moduli of the coefficients are }1, \sqrt{2}, \sqrt{10}, 5\\ \text{so }M=5\text{, because it is the maximum value OF the moduli of the coefficients.}$

$\text{Also, if you read what I said again}\\ \text{I had not put }M\text{ in there yet.}\\ \text{I had merely pulled out }|a_{n-1}|, \dots, |a_1|, |a_0|$
$\text{Substituting in all the inequalities from above }\textbf{then}\text{ gives our next line.}$

beau77bro · « **Reply #1457 on:** October 08, 2017, 09:08:50 am »

OHHHHH OK THANKYOU RUI.

seventeenboi · « **Reply #1458 on:** October 09, 2017, 11:12:38 am »

hiii i need help with this question

thank you so much!!!

RuiAce · « **Reply #1459 on:** October 09, 2017, 11:15:56 am »

Quote from: seventeenboi on October 09, 2017, 11:12:38 am

hiii i need help with this question

thank you so much!!!

Already addressed in the compilation.

seventeenboi · « **Reply #1460 on:** October 09, 2017, 11:38:16 am »

Hello, sorry I have another question -
what is the effect on a general hyperbola as e approaches infinity

? i know that it approaches a 'pair of vertical lines' (directrices) but i don't understand why that is

thank you

RuiAce · « **Reply #1461 on:** October 09, 2017, 11:47:59 am »

Quote from: seventeenboi on October 09, 2017, 11:38:16 am

Hello, sorry I have another question -
what is the effect on a general hyperbola as e approaches infinity ? i know that it approaches a 'pair of vertical lines' (directrices) but i don't understand why that is

thank you

$\text{Actually, it doesn't approach the directrices.}\\ \text{In fact, the directrices approach the "center" of the hyperbola}$
(The directrices ultimately coincide thanks to the formula $x=\pm \frac{a}{e}$ and that the negative of 0 is also 0.)
$\text{Consider the equation of the hyperbola }\boxed{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1}\\ \text{Using our formula for the eccentricity }b^2=a^2(e^2-1)\\ \text{we see that the equation becomes}\\ \boxed{\frac{x^2}{a^2}-\frac{y^2}{a^2(e^2-1)}=1}$
$\text{We may as well take }a^2\text{ to the other side to obtain}\\ x^2 - \frac{y^2}{e^2-1} = a^2$
______________________
$\text{Let }e\to \infty.$
$\text{As }e\to \infty\text{, we will obviously have }\frac{1}{e^2-1}\to 0\\ \text{So as the eccentricity gets arbitrarily large, the equation of the hyperbola ultimately becomes}\\ x^2 = a^2$
$\text{This becomes }x=\pm a\\ \text{which is a pair of vertical lines.}$

3.14159265359 · « **Reply #1462 on:** October 09, 2017, 09:58:25 pm »

hey! this is my very first post so please excuse me if this doesn't work out. anywho I wanted to ask about 4u. how hard is it? I reallyyy want to try it and I know I'm capable of doing it if I work for it but Ive been hearing things like this subject is killer and when you do it you'll be looking like a zombie from the lack of sleep, which has made me reconsider taking it but what do you recommend?

RuiAce · « **Reply #1463 on:** October 09, 2017, 10:10:01 pm »

Quote from: Mimi5601 on October 09, 2017, 09:58:25 pm

hey! this is my very first post so please excuse me if this doesn't work out. anywho I wanted to ask about 4u. how hard is it? I reallyyy want to try it and I know I'm capable of doing it if I work for it but Ive been hearing things like this subject is killer and when you do it you'll be looking like a zombie from the lack of sleep, which has made me reconsider taking it but what do you recommend?

I may have forgotten to address these exact rumours, but they were basically what I cleared up at the start of the lecture.

Short answer to the validity of the rumours: NO. When people have this sort of problem, they're doing a course that they were never suited for anyway. Many high achieving students in MX2 have never had this problem.

Shlomo314 · « **Reply #1464 on:** October 12, 2017, 04:52:53 pm »

Hey atar team, how would you solve this question?

Cheers

RuiAce · « **Reply #1465 on:** October 12, 2017, 05:33:07 pm »

Quote from: Shlomo314 on October 12, 2017, 04:52:53 pm

Hey atar team, how would you solve this question?

Cheers

$\text{Treat this question the same way you equated coefficients}\\ \text{in binomial theorem proofs in 3U.}$
$\text{Consider the ways in which we combine terms (from each bracket) to create }x^n\\ \text{We take }1\text{ and pair it with }(n+1)x^n\text{ to obtain }(n+1)x^n\\ \text{We take }x\text{ and pair it with }nx^{n-1}\text{ to obtain }nx^n\\ \text{We take }x^2\text{ and pair it with }(n-1)x^{n-2}\text{ to obtain }(n-1)x^n\\ \vdots \\ \text{We take }x^n\text{ and pair it with }1\text{ to obtain }x^n$
$\text{So the term involving }x^n\text{ will be }\\ (n+1)x^n + nx^n + (n-1)x^n + \dots + 1\\ \text{Thus, the coefficient of }x^n\text{ is}\\ (n+1) + n + \dots +1$
$\text{Using the formula for the sum of an AP, this will be }\frac{n+1}{2}[1 + (n+1)] = \frac{(n+1)(n+2)}{2}$
$\text{Put }n=100:\text{ Answer}=5151\,\boxed {C}$

Shlomo314 · « **Reply #1466 on:** October 13, 2017, 02:19:18 pm »

Hi Atar team, Could someone help me with this question please?

Thanks

RuiAce · « **Reply #1467 on:** October 13, 2017, 02:21:56 pm »

Quote from: Shlomo314 on October 13, 2017, 02:19:18 pm

Hi Atar team, Could someone help me with this question please?

Thanks

This is just 1/2, because the other 1/2 is when the number 1 lies somewhere to the right of the number 2.

(Note that two numbers obviously cannot occupy the same position. So we only have two mutually exclusive cases - 1 is to the left of 2, or 1 is to the right of 2.)

itssona · « **Reply #1468 on:** October 16, 2017, 04:22:54 pm »

heey im not sure if I'm getting the right answer for this,
basically I need to now integrate -1/(6 (2x+3)) and 1/(6 (2x+3)) for partial fractions,
is the answer -1/12 ln (2x+3) + 1/12ln (2x-3)

RuiAce · « **Reply #1469 on:** October 18, 2017, 09:27:35 am »

Quote from: sssona09 on October 16, 2017, 04:22:54 pm

heey im not sure if I'm getting the right answer for this,
basically I need to now integrate -1/(6 (2x+3)) and 1/(6 (2x+3)) for partial fractions,
is the answer -1/12 ln (2x+3) + 1/12ln (2x-3)

That looks fine (although you should be using absolute values for ln|2x+3| and ln|2x-3|).

Search the forums now!

We have moved!

Author Topic: 4U Maths Question Thread (Read 665265 times) Tweet Share

beau77bro

Re: 4U Maths Question Thread

RuiAce

Re: 4U Maths Question Thread

beau77bro

Re: 4U Maths Question Thread

seventeenboi

Re: 4U Maths Question Thread

RuiAce

Re: 4U Maths Question Thread

seventeenboi

Re: 4U Maths Question Thread

RuiAce

Re: 4U Maths Question Thread

3.14159265359

Re: 4U Maths Question Thread

RuiAce

Re: 4U Maths Question Thread

Shlomo314

Re: 4U Maths Question Thread

RuiAce

Re: 4U Maths Question Thread

Shlomo314

Re: 4U Maths Question Thread

RuiAce

Re: 4U Maths Question Thread

itssona

Re: 4U Maths Question Thread

RuiAce

Re: 4U Maths Question Thread

Recent Posts

User:
Password: