HSC 2014 MX2 Marathon ADVANCED (archive) (1 Viewer)

Chlee1998 · Nov 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

seriously if it was that easy i wouldn't have posted it here

SilentWaters · Nov 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Axio said:
Could you explain to me how you get that first line to fit the AM-GM inequality?

Sum the components of the product and divide it by the number of terms so that the geometric mean is less than the arithmetic one:

$\\ \frac{(1+m)+1+1+1+...}{n}$

Notice we are adding $n-1$ 1's to $1+m,$ (the latter counting for 1 term).

Chlee1998 · Nov 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

perhaps Glitter/ Sy/Frank can tackle the Question (the one I posted)

FrankXie · Nov 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

glittergal96 said:
$(1+m)^{1/n}=((1+m)\cdot 1^{n-1})^{1/n}<\frac{m+n}{n}$

by AM-GM. (Noting that not all terms are equal, so we cannot have equality.)

Similarly, we have

$(1+n)^{1/m}<\frac{m+n}{m}.$

Invert these inequalities and sum them to complete the proof.

well done, so smart! Here is my proof:

$$ let $ (1+m)^{\frac1n}=1+\alpha, (1+n)^{\frac1m}=1+\beta, $ so $ \alpha>0, \beta>0 $ and the desired result is $ \frac1{1+\alpha}+\frac1{1+\beta}>1, $ which is equivalent to $ 1+\alpha+1+\beta>(1+\alpha)(1+\beta), $ which is equivalent to $ \alpha\beta<1$

$$ Now $ 1+m=(1+\alpha)^n=1+n\alpha+$ positive terms $>1+n\alpha, $ so $ m>n\alpha$

$similarly $ n>m\beta. $ Multiplying these two results gives $ mn>mn\alpha\beta $ so $ \alpha\beta<1 $ Done!$

Sy123 · Nov 30, 2014

Re: HSC 2014 4U Marathon - Advanced Level

glittergal96 said:
$(1+m)^{1/n}=((1+m)\cdot 1^{n-1})^{1/n}<\frac{m+n}{n}$

by AM-GM. (Noting that not all terms are equal, so we cannot have equality.)

Similarly, we have

$(1+n)^{1/m}<\frac{m+n}{m}.$

Invert these inequalities and sum them to complete the proof.

Yep that was my method when seanieg89 posted this question near the start of the year

(yes I am guilty of recyling some questions)

FrankXie · Dec 1, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Chlee1998 said:
perhaps Glitter/ Sy/Frank can tackle the Question (the one I posted)

Now that you nominated me, i just have my try, not an inspiring method though.

$$ First of all i'll base my solution on the facts that $ \cos72^\circ=\frac{\sqrt5-1}{4}, \cos36^\circ=\frac{\sqrt5+1}{4}, $ which (by themselves are good classroom exercises) can be derived neatly by drawing an isosceles triangle with interior anlges being $ 72^\circ, 72^\circ $ and $ 36^\circ.$

$Now $ \frac{1}{\sin6^\circ}-\frac{1}{\sin66^\circ}=\frac{\sin66^\circ-\sin6^\circ}{\sin6^\circ\sin66^\circ}=\frac{2\cos36^\circ\sin30^\circ}{\frac12(\cos60^\circ-\cos72^\circ)}=\frac{2\cos36^\circ}{\frac12-\cos72^\circ}=\frac{\frac{\sqrt5+1}{2}}{\frac12-\frac{\sqrt5-1}{4}}=\frac{2(\sqrt5+1)}{3-\sqrt5} \qquad $ (1)$

$\frac{1}{\sin78^\circ}-\frac{1}{\sin42^\circ}=\frac{\sin138^\circ-\sin78^\circ}{\sin78^\circ\sin42^\circ}=\frac{2 \cos108^\circ\sin30^\circ}{\frac12(\cos36^\circ-\cos120^\circ)}=\frac{2\cos108^\circ}{\frac12+\cos36^\circ}=\frac{-\frac{\sqrt5-1}{2}}{\frac12+\frac{\sqrt5+1}{4}}=\frac{-2(\sqrt5-1)}{3+\sqrt5} \qquad $ (2)$$

Finally adding equations (1) and (2) completes the proof.

Sy123 · Dec 5, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$\\ $Solve among the reals$ \ x^2 + (4x^3 - 3x)^2 = 1$

glittergal96 · Dec 5, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ $Solve among the reals$ \ x^2 + (4x^3 - 3x)^2 = 1$

Since we must have $|x|\leq 1$ we make the substitution $x=\cos \theta$ . (This is also motivated by the fact that we are obviously looking for points on the unit circle satisfying a certain condition.)

Then

$\cos^2 3\theta=(4\cos^3\theta-3\cos\theta)^2=1-\cos^2\theta=\sin^2\theta=\cos^2(\pi/2-\theta).$

So $\theta=\pi(4k+1)/4,\pi(4k-1)/8.$

The 6 distinct solutions are

$x=\pm \cos(\pi/4),\pm\cos(\pi/8),\pm\cos(3\pi/8).$

Sy123 · Dec 6, 2014

Re: HSC 2014 4U Marathon - Advanced Level

_____________________________________________

NEW YEAR 2015

______________________________________________c

Sy123 · Dec 6, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Trebla said:
Post any questions within the scope and level of Mathematics Extension 2 mainly targeting Q16 difficulty in the HSC.

Any questions beyond the scope of the HSC syllabus should be posted in the Extracurricular Topics forum:
http://community.boredofstudies.org/238/extracurricular-topics/

Once a question is posted, it needs to be answered before the next question is raised.

I will get the ball rolling:

Prove that

$\dfrac{3}{8}+\dfrac{4}{16}+\dfrac{5}{32}+\dfrac{6}{64}+....= 1$

$1 + x + x^2 + \dots + x^{n-1} = \frac{1-x^n}{1-x}$

$x + 2x^2 + 3x^3 + \dots + (n-1)x^{n-1} = x \cdot \frac{d}{dx} \left(\frac{1-x^n}{1-x} \right)$

$\sum_{n=3}^{\infty} \frac{n}{2^n} = \lim_{n \to \infty} \left(x \cdot \frac{d}{dx} \left(\frac{1-x^n}{1-x}\right) \right) \left( \frac{1}{2} \right) - \frac{1}{2} - \frac{1}{2}$

$\therefore \sum_{n =3}^{\infty} \frac{n}{2^n}= 1$

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Please mods keep this thread, just change the title, I'll change my OP

Axio · Dec 6, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$1 + x + x^2 + \dots + x^{n-1} = \frac{1-x^n}{1-x}$

$x + 2x^2 + 3x^3 + \dots + (n-1)x^{n-1} = x \cdot \frac{d}{dx} \left(\frac{1-x^n}{1-x} \right)$

$\sum_{n=3}^{\infty} \frac{n}{2^n} = \lim_{n \to \infty} \left(x \cdot \frac{d}{dx} \left(\frac{1-x^n}{1-x}\right) \right) \left( \frac{1}{2} \right) - \frac{1}{2} - \frac{1}{2}$

$\therefore \sum_{n =3}^{\infty} \frac{n}{2^n}= 1$

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Please mods keep this thread, just change the title, I'll change my OP

I agree with this.

HSC 2014 MX2 Marathon ADVANCED (archive) (1 Viewer)

Chlee1998

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Sy123

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