4U Revising Game (3 Viewers)

jet · May 23, 2009

That does look about right....
I dont have the answers btw

GUSSSSSSSSSSSSS · May 23, 2009

jetblack2007 said:
That does look about right....
I dont have the answers btw

HAHAHAHA
..........i mite be able to get a hold of them

Trebla · May 24, 2009

Prove that for a > b > 0:

$\dfrac{1}{a}<\dfrac{\ln a - \ln b}{a - b}<\dfrac{1}{b}$

lolokay · May 24, 2009

Trebla said:
Prove that for a > b > 0:

$\dfrac{1}{a}<\dfrac{\ln a - \ln b}{a - b}<\dfrac{1}{b}$

$consider\ f(x)=x-1-ln(x)\\ f'(x)=1-1/x\\ f'(x)=0\ when\ x=1\\ and\ f''(x)=1/x^2>0\\ so\ x=1,\ f(x)=0\ is\ the\ minimum\\ \therefore f(x)=x-1-ln(x)\geq 0\\ x-1\geq ln(x),\ equality\ if\ x=1\\ \\ let\ x=a/b,a>b>0,\ so\ x\neq 1,a-b>0\\ \frac{a}{b}-1>ln(a/b)= ln(a)-ln(b)\\ 1/b(a-b)>ln(a)-ln(b)\\ \frac{1}{b}>\frac{ln(a)-ln(b)}{a-b}\\ \\ and,\ letting\ x=b/a\\ b/a - 1 > ln(b/a)=ln(b)-ln(a)\\ 1/a(b-a) > ln(b)-ln(a)\\ 1/a(a-b)<ln(a)-ln(b)\\ \frac{1}{a}<\frac{ln(a)-ln(b}{a-b}<\frac{1}{b}\\$

shaon0 · May 24, 2009

lolokay said:
$consider\ f(x)=x-1-ln(x)\\ f'(x)=1-1/x\\ f'(x)=0\ when\ x=1\\ and\ f''(x)=1/x^2>0\\ so\ x=1,\ f(x)=0\ is\ the\ minimum\\ \therefore f(x)=x-1-ln(x)\geq 0\\ x-1\geq ln(x),\ equality\ if\ x=1\\ \\ let\ x=a/b,a>b>0,\ so\ x\neq 1,a-b>0\\ \frac{a}{b}-1>ln(a/b)= ln(a)-ln(b)\\ 1/b(a-b)>ln(a)-ln(b)\\ \frac{1}{b}>\frac{ln(a)-ln(b)}{a-b}\\ \\ and,\ letting\ x=b/a\\ b/a - 1 > ln(b/a)=ln(b)-ln(a)\\ 1/a(b-a) > ln(b)-ln(a)\\ 1/a(a-b)<ln(a)-ln(b)\\ \frac{1}{a}<\frac{ln(a)-ln(b}{a-b}<\frac{1}{b}\\$

nice solution

gurmies · Jul 6, 2009

New MX2 Game

Hey guys, with trial exams looming up ahead, I thought it would be a good idea to revive the MX2 game. Questions should be mainly taken from trial exams (it would be a good idea to quote which school and year) and to make things interesting, try to make them difficult (although this is dependent on perception). Furthermore, to keep things interesting for those who weren't the first to answer the question, add spoiler tags (spoiler) and (/spoiler), but using [] instead. One last thing: If a question has been answered, but you believe you have an alternative method, please post it (irrespective of how strange it may be) because exposure to different approaches will expand our own ways of thinking about problems. To start things off:

Sydney Technical High School 1996

Question 8, Part b)

$i)Explain \,\, why \,\, (6+\sqrt{35})^{1980}+(6-\sqrt{35})^{1980} \,\, is \,\, an \,\, even \,\, number \\\\ ii)Show \,\, that \,\, 0<6-\sqrt{35}<0.1 \\\\ iii)Hence \,\, or \,\, otherwise \,\, prove \,\, that \,\, the \,\, first \,\, thousand \,\, digits \,\, after \,\, the \,\, decimal \,\, point \,\, in \,\, the \,\, value \,\, of \,\, (6+\sqrt{35})^{1980} \,\, are \,\, all \,\, 9's.$

untouchablecuz · Jul 7, 2009

Re: New MX2 Game

nearly done the last bit

Another Q:

Sydney Tech 2001, 8 c)

i) If x1 > 1 and x2 > 2 show that x1 + x2 > sqrt(x1x2)

ii) Use the principle of mathematical induction to show that, for n ≥ 2, if xj > 1 where j = 1, 2, 3, . . . , n then ln(x1+x2+· · ·+xn) > [1/(2n−1)](lnx1 +lnx2+· · ·+lnxn).

(the numeral or pronumeral after "x" is in subscript xj, xn, x1 etc)

azureus88 · Jul 7, 2009

Re: New MX2 Game

(i) [maths](\sqrt{x_1}-\sqrt{x_2})^2\geq 0\\x_1+x_2\geq 2\sqrt{x_1x_2}\geq \sqrt{x_1x_2}[/maths]

(ii) [maths]$for n=2,$\\\ln(x_1+x_2)>\ln(\sqrt{x_1x_2})\\=\frac{1}{2}\ln(x_1x_2)\\>\frac{1}{2(2)-1}\ln(x_1x_2)\\\\$assume n=k is true,\\for n=k+1,$\\\ln((x_1+x_2+...+x_k)+x_{k+1})\\=\ln(x_1+x_2+...+x_k)+\ln(x_{k+1})\\>\frac{1}{2n-1}(\ln x_1+\ln x_2+...+\ln x_n)+\ln( x_{k+1})\\=\frac{1}{2n-1}(\ln x_1+\ln x_2+...+\ln x_n+(2n-1)\ln x_{k+1})\\>\frac{1}{2n-1}(\ln x_1+\ln x_2+...+\ln x_n+\ln x_{k+1})forn\geq 2[/maths]

for the last part of 1st question, cant you just say 0<[6-sqrt(35)]^1980<(0.1)^1980 so [6-sqrt(35)]^1980 must be all 0's for the first 1979 decimal places and since [6-sqrt(35)]^1980 + [6+sqrt(35)]^1980 add up to an even number (a whole number), then [6+sqrt(35)]^1980 must be all 9's for first thousand decimal places.

azureus88 · Jul 7, 2009

Re: New MX2 Game

New Q:

Show that [maths]\sin nt+\sin(nt+\alpha)+\sin(nt+2\alpha)+...+\sin(nt+(k-1)\alpha)=0\,\,where\,\,\alpha =\frac{2\pi}{k}[/maths]

Sums to products might be useful here.

untouchablecuz · Jul 7, 2009

Re: New MX2 Game

azureus88 said:
(i) [maths](\sqrt{x_1}-\sqrt{x_2})^2\geq 0\\x_1+x_2\geq 2\sqrt{x_1x_2}\geq \sqrt{x_1x_2}[/maths]

(ii) [maths]$for n=2,$\\\ln(x_1+x_2)>\ln(\sqrt{x_1x_2})\\=\frac{1}{2}\ln(x_1x_2)\\>\frac{1}{2(2)-1}\ln(x_1x_2)\\\\$assume n=k is true,\\for n=k+1,$\\\ln((x_1+x_2+...+x_k)+x_{k+1})\\=\ln(x_1+x_2+...+x_k)+\ln(x_{k+1})\\>\frac{1}{2n-1}(\ln x_1+\ln x_2+...+\ln x_n)+\ln( x_{k+1})\\=\frac{1}{2n-1}(\ln x_1+\ln x_2+...+\ln x_n+(2n-1)\ln x_{k+1})\\>\frac{1}{2n-1}(\ln x_1+\ln x_2+...+\ln x_n+\ln x_{k+1})forn\geq 2[/maths]

for the last part of 1st question, cant you just say 0<[6-sqrt(35)]^1980<(0.1)^1980 so [6-sqrt(35)]^1980 must be all 0's for the first 1979 decimal places and since [6-sqrt(35)]^1980 + [6+sqrt(35)]^1980 add up to an even number (a whole number), then [6+sqrt(35)]^1980 must be all 9's for first thousand decimal places.

[maths]ln((x_1+x_2+...+x_k)+x_{k+1})\\=\ln(x_1+x_2+...+x_k)+\ln(x_{k+1})[/maths]

this part doesnt follow; try begining with the use of the result a + b > sqrt(ab) from i)

azureus88 · Jul 7, 2009

Re: New MX2 Game

woops, thanks for the correction. 2nd attempt:

[maths](x_1+x_2+...+x_k)+x_{k+1}>\sqrt{(x_1+x_2+...+x_k)x_{k+1}}\\\ln(x_1+x_2+...+x_k+x_{k+1})\\>\frac{1}{2}(\ln(x_1+x_2+...+x_k)+\ln(x_{k+1}))\\>\frac{1}{2}(\frac{1}{2k-1}(\ln x_1+\ln x_2+...+\ln x_k+(2k-1)\ln x_{k+1}))\\>\frac{1}{4k-2}(\ln x_1+\ln x_2+...+\ln x_k+\ln x_{k+1})\\>\frac{1}{2k+1}(\ln x_1+\ln x_2+...+\ln x_k+\ln x_{k+1})forn>2\\=\frac{1}{2(k+1)-1}(\ln x_1+\ln x_2+...+\ln x_k+\ln x_{k+1})[/maths]

untouchablecuz · Jul 7, 2009

Re: New MX2 Game

azureus88 said:
woops, thanks for the correction. 2nd attempt:

[maths](x_1+x_2+...+x_k)+x_{k+1}>\sqrt{(x_1+x_2+...+x_k)x_{k+1}}\\\ln(x_1+x_2+...+x_k+x_{k+1})\\>\frac{1}{2}(\ln(x_1+x_2+...+x_k)+\ln(x_{k+1}))\\>\frac{1}{2}(\frac{1}{2k-1}(\ln x_1+\ln x_2+...+\ln x_k+(2k-1)\ln x_{k+1}))\\>\frac{1}{4k-2}(\ln x_1+\ln x_2+...+\ln x_k+\ln x_{k+1})\\>\frac{1}{2k+1}(\ln x_1+\ln x_2+...+\ln x_k+\ln x_{k+1})forn>2\\=\frac{1}{2(k+1)-1}(\ln x_1+\ln x_2+...+\ln x_k+\ln x_{k+1})[/maths]

nicee

with your Q, i used sum to products, i end up with every term having a 2cos(-alpha/2) as a factor but thats where im stuck

azureus88 · Jul 7, 2009

Re: New MX2 Game

think you gotta consider two cases: when n is even and when n is odd.

so when n is even, you take the sin's in pairs (1st and last, 2nd and 2nd last...) and use sums to products.

shaon0 · Jul 8, 2009

Re: New MX2 Game

gurmies said:
Hey guys, with trial exams looming up ahead, I thought it would be a good idea to revive the MX2 game. Questions should be mainly taken from trial exams (it would be a good idea to quote which school and year) and to make things interesting, try to make them difficult (although this is dependent on perception). Furthermore, to keep things interesting for those who weren't the first to answer the question, add spoiler tags (spoiler) and (/spoiler), but using [] instead. One last thing: If a question has been answered, but you believe you have an alternative method, please post it (irrespective of how strange it may be) because exposure to different approaches will expand our own ways of thinking about problems. To start things off:

Sydney Technical High School 1996

Question 8, Part b)

$i)Explain \,\, why \,\, (6+\sqrt{35})^{1980}+(6-\sqrt{35})^{1980} \,\, is \,\, an \,\, even \,\, number \\\\ ii)Show \,\, that \,\, 0<6-\sqrt{35}<0.1 \\\\ iii)Hence \,\, or \,\, otherwise \,\, prove \,\, that \,\, the \,\, first \,\, thousand \,\, digits \,\, after \,\, the \,\, decimal \,\, point \,\, in \,\, the \,\, value \,\, of \,\, (6+\sqrt{35})^{1980} \,\, are \,\, all \,\, 9's.$

lol, all these MX2/MX1 game's are doomed to fail.

Nice question on sums-to-products from the extension part of yr12 Cambridge 3unit book.

gurmies · Jul 8, 2009

Re: New MX2 Game

shaon0 said:
lol, all these MX2/MX1 game's are doomed to fail.

Nice question on sums-to-products from the extension part of yr12 Cambridge 3unit book.

Yeah man, always a lack of participation! Perhaps people are too engrossed in their trials study =D

x.Exhaust.x · Jul 9, 2009

Re: New MX2 Game

In other words, C B F.

Aerath · Jul 11, 2009

Re: New MX2 Game

And can't be fucked typing it up either.

clintmyster · Jul 11, 2009

Re: New MX2 Game

so you scan like I do =P

Revacious · Jul 15, 2009

Re: New MX2 Game

I shat myself when i saw this question, i mean, it was a diagram that took the whole damn page up.
It wasn't too bad when you did part abcd, but for the fun of it, let's skip to the last part =)

Find the motherfucking volume of that motherfucking solid. (SB THSC 2004)

jet · Jul 15, 2009

Re: New MX2 Game

$\text{Ok, so first, we work out the length } BW \text{ realising that } ABC \text{ is equilateral: } \\ sin 60 = \frac{BW}{2a} \\ BW = \sqrt{3}a \\ \text{Now we work out the value of each parameter as linear functions of } x: KP = a \\ PQ = mx + n \\ x = 0, PQ = HT = a; x = b, PQ = DE = 2a \\ \therefore PQ = \frac{ax}{b}+ a \\ LM = ux + v \\ x = 0, LM = RS = a; x = b, LM = 0 \\ LM = a - \frac{ax}{b} \\ h = sx + t \\ x = 0, h = 0; x = b, h = \sqrt{3}a \\ \therefore h = \frac{\sqrt{3}ax}{b} \\ \text{Thus, the area of the shape } KLMNQP = PQ \times KP + \frac{1}{2}\left (PQ + LM \right ) \times h \\ = a\left (\frac{ax}{b}+ a \right ) + \frac{1}{2}\left( \frac{ax}{b} + a + a - \frac{ax}{b} \right )\times \frac{\sqrt{3}ax}{b} \\ = \frac{a^2x}{b} + a^2 + \frac{\sqrt{3}a^2x}{b} \\ = \frac{a^2}{b}\left(x + b + \sqrt{3}x \right ) \\ \text{Hence the volume of each slice } \Delta V = \frac{a^2}{b}\left(x + b + \sqrt{3}x \right ) \, \Delta x \\$
$\text{and the volume of the solid } V = \lim_{\Delta x \to 0} \sum^{b}_{x = 0} \frac{a^2}{b}\left(x + b + \sqrt{3}x \right ) \, \Delta x \\ = \frac{a^2}{b}\int^{b}_{0} x + b + \sqrt{3}x \, dx \\ = \frac{a^2}{b}\left [\frac{x^2}{2} + bx + \frac{\sqrt{3}x^2}{2} \right ]^{b}_0 \\ = \frac{a^2}{b}\left [\frac{b^2}{2} + b^2 + \frac{\sqrt{3}b^2}{2} \right ] \\ = a^2b\left [\frac{1}{2} + 1 + \frac{\sqrt{3}}{2} \right ] \\ = \frac{a^2b\left (3 + \sqrt{3} \right )}{2} \, \left ( \text{hopefully lol} \right )$

Somebody else ask a question.

4U Revising Game (3 Viewers)

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