HSC 2013-14 MX1 Marathon (archive) (2 Viewers)

asianese · Jun 10, 2013

Re: HSC 2013 3U Marathon Thread

You could also do a t-substitution, but that would be longer and more tedious.

SpiralFlex · Jun 13, 2013

Re: HSC 2013 3U Marathon Thread

Spiral the fluffy bunny with wings flies up in the sky directly above Asianese with a horizontal velocity of 5 meters per second and at a height of 5 metres. Asianese is eating fruit salad. He wants to throw a piece up for Spiral but does not want to hurt him. He throws the grape at some time t, it has e same position, same height and same velocity as Spiral. He throws the piece of fruit at a velocity V and at an angle theta from the horizontal. Assuming air resistance is negligible

a) Should Spiral be in front of Asianese or behind him when the fruit is thrown, and by what distance?

b) At what initial velocity must Asianese throw the fruit, what angle the fruit initially fly out of his hand to ensure Spiral is safe.

Makematics · Jun 13, 2013

Re: HSC 2013 3U Marathon Thread

SpiralFlex said:
Spiral the fluffy bunny with wings flies up in the sky directly above Asianese with a horizontal velocity of 5 meters per second and at a height of 5 metres. Asianese is eating fruit salad. He wants to throw a piece up for Spiral but does not want to hurt him. He throws the grape at some time t, it has e same position, same height and same velocity as Spiral. You throw the piece of fruit at a velocity V and at an angle theta from the horizontal. Assuming air resistance is negligible

a) Should Spiral be in front of Asianese or behind him when the fruit is thrown, and by what distance?

b) At what initial velocity must Asianese throw the fruit, what angle the fruit initially fly out of his hand to ensure Spiral is safe.

man you are bored

SpiralFlex · Jun 14, 2013

Re: HSC 2013 3U Marathon Thread

?

It's a good question

Makematics · Jun 14, 2013

Re: HSC 2013 3U Marathon Thread

SpiralFlex said:
?

It's a good question

i know haha, just saying you must be fairly bored to make/reword such a question.

SpiralFlex · Jun 14, 2013

Re: HSC 2013 3U Marathon Thread

Well if I ever do get the chance to fly I don't want my face to be damaged. I will work as a part time model.

Sy123 · Jun 16, 2013

Re: HSC 2013 3U Marathon Thread

$i) Explain why$

$1-x+x^2-x^3+\dots+(-1)^{n-1}x^{n-1} = \frac{1}{1+x} + (-1)^{n+1} \frac{x^n}{x+1} \ \ \ \ \fbox{1}$

$ii) Hence prove that$

$1 - \frac{1}{2} + \frac{1}{3} - \dots + \frac{(-1)^{n-1}}{n} = \ln 2 + (-1)^{n+1} \int_0^1 \frac{x^n}{x+1} \ dx \ \ \ \ \fbox{2}$

$$iii) Explain why for$ \ \ |x| < 1$

$\frac{x^n}{x+1} \rightarrow 0 \ \ \ $as$ \ \ \ n \to \infty \ \ \ \fbox{1}$

$iv) Hence prove that$

$\ln 2 = \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k} \ \ \ \fbox{1}$

omgiloverice · Jun 18, 2013

Re: HSC 2013 3U Marathon Thread

$$i) Prove that$:$

$\sqrt[x]{x} = e^{\frac{ln{x}}{x}}$

$$ii)Hence show that the maximum value of $f(x)=\sqrt[x]{x} \ $occurs at $ x=e$

Not sure if this belongs to the 2U or 3U so i'll post it here~

Makematics · Jun 18, 2013

Re: HSC 2013 3U Marathon Thread

omgiloverice said:
$$i) Prove that$:$

$\sqrt[x]{x} = e^{\frac{ln{x}}{x}}$

$$ii)Hence show that the maximum value of $f(x)=\sqrt[x]{x} \ $occurs at $ x=e$

Not sure if this belongs to the 2U or 3U so i'll post it here~

$\\\mathbf{i)}~$RHS$=e^\ln(x^\frac{1}{x})$

$\\~~~~~~~~~=x^\frac{1}{x}\\\\~~~~~~~~~=\sqrt[x]{x}\\\\~~~~~~~~~=$LHS$\\\\\mathbf{ii)}~f(x)=e^ \frac{lnx}{x}\\\\f'(x)=e^\frac{lnx}{x}\frac{d( \frac{lnx}{x})}{dx}\\\\~~~~~~~~=e^\frac{lnx}{x} \frac{1-lnx}{x^2}\\\\$To find a maximum value of$~f(x)$, we must solve$ f'(x)=0\\\\lnx=1\\\\\therefore x=e$

Note that the tests for maximum values should be performed, it's just it's getting late and i can't be bothered. also what is up with the lnx on latex? is there any way to get it without those annoying/ugly brackets? Also I think this would be a bit too challenging for 2U :/

seanieg89 · Jun 18, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$i) Explain why$

$1-x+x^2-x^3+\dots+(-1)^{n-1}x^{n-1} = \frac{1}{1+x} + (-1)^{n+1} \frac{x^n}{x+1} \ \ \ \ \fbox{1}$

$ii) Hence prove that$

$1 - \frac{1}{2} + \frac{1}{3} - \dots + \frac{(-1)^{n-1}}{n} = \ln 2 + (-1)^{n+1} \int_0^1 \frac{x^n}{x+1} \ dx \ \ \ \ \fbox{2}$

$$iii) Explain why for$ \ \ |x| < 1$

$\frac{x^n}{x+1} \rightarrow 0 \ \ \ $as$ \ \ \ n \to \infty \ \ \ \fbox{1}$

$iv) Hence prove that$

$\ln 2 = \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k} \ \ \ \fbox{1}$

The key isn't that the integrand tends to zero, we need the whole integral to tend to zero. (Unless we have more sophisticated tools at our disposal for going from the former to the latter.)

Sy123 · Jun 18, 2013

Re: HSC 2013 3U Marathon Thread

seanieg89 said:
The key isn't that the integrand tends to zero, we need the whole integral to tend to zero. (Unless we have more sophisticated tools at our disposal for going from the former to the latter.)

Ah yep, I remember that, oops.

asianese · Jun 18, 2013

Re: HSC 2013 3U Marathon Thread

Nvm == codecogs h8s me

Makematics · Jun 19, 2013

Re: HSC 2013 3U Marathon Thread

yeah well i'm new to latex... how the hell am i supposed to learn all these random things like align and enumerate from LOL? is there any guide you can link me to? :S

asianese · Jun 19, 2013

Re: HSC 2013 3U Marathon Thread

Makematics said:
yeah well i'm new to latex... how the hell am i supposed to learn all these random things like align and enumerate from LOL? is there any guide you can link me to? :S

Yes lol: written by the creator: http://tobi.oetiker.ch/lshort/lshort.pdf

But you just pick up things from this forum or by googling:

Just remember: if you're manually putting in spaces or doing things the 'slow way' there's probably an easier way...

Makematics · Jun 19, 2013

Re: HSC 2013 3U Marathon Thread

asianese said:
Yes lol: written by the creator: http://tobi.oetiker.ch/lshort/lshort.pdf

But you just pick up things from this forum or by googling:

Just remember: if you're manually putting in spaces or doing things the 'slow way' there's probably an easier way...

hmm alright thanks! it isn't that slow but i guess it'll make the solution look neater.

Sy123 · Jun 23, 2013

Re: HSC 2013 3U Marathon Thread

$Prove that$

$x^x \geq \left( \frac{1}{e} \right ) ^{\frac{1}{e}}$

HeroicPandas · Jun 23, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$Prove that$

$x^x \geq \left( \frac{1}{e} \right ) ^{\frac{1}{e}}$

$$\underline{Consider} $ y = x^x = e^{xlnx} \\ \frac{dy}{dx} = x^x(lnx+1) \\ $For \underline{stationary} points, $\frac{dy}{dx} = 0, \\ x^x \neq 0 \\ lnx+1 = 0, x = \frac{1}{e} , y = \left ( \frac{1}{e} \right )^{\frac{1}{e}} \\ $Sign of $\frac{dy}{dx} = lnx+1 ($eliminating positive product terms$) \\ $We draw a normal logarithm curve, x-int of $\frac{1}{e} \\ $From curve, if x$ < \frac{1}{e}, \frac{dy}{dx} < 0 \\ $Similarly, if x$ >\frac{1}{e}, \frac{dy}{dx} >0 \\ $Hence $ \left ( \frac{1}{e}, \left ( \frac{1}{e} \right )^{\frac{1}{e}} \right ) $this point is a \underline{minimum} stationary point$ \\ \therefore $Curve must be above this point, more importantly, the lowest y-value this curve can take is y = \left ( \frac{1}{e} \right )^{\frac{1}{e}}$ \\ \therefore x^x \geq \left ( \frac{1}{e} \right )^{\frac{1}{e}}$

Sy123 · Jun 23, 2013

Re: HSC 2013 3U Marathon Thread

HeroicPandas said:
$$\underline{Consider} $ y = x^x = e^{xlnx} \\ \frac{dy}{dx} = x^x(lnx+1) \\ $For \underline{stationary} points, $\frac{dy}{dx} = 0, \\ x^x \neq 0 \\ lnx+1 = 0, x = \frac{1}{e} , y = \left ( \frac{1}{e} \right )^{\frac{1}{e}} \\ $Sign of $\frac{dy}{dx} = lnx+1 ($eliminating positive product terms$) \\ $We draw a normal logarithm curve, x-int of $\frac{1}{e} \\ $From curve, if x$ < \frac{1}{e}, \frac{dy}{dx} < 0 \\ $Similarly, if x$ >\frac{1}{e}, \frac{dy}{dx} >0 \\ $Hence $ \left ( \frac{1}{e}, \left ( \frac{1}{e} \right )^{\frac{1}{e}} \right ) $this point is a \underline{minimum} stationary point$ \\ \therefore $Curve must be above this point, more importantly, the lowest y-value this curve can take is y = \left ( \frac{1}{e} \right )^{\frac{1}{e}}$ \\ \therefore x^x \geq \left ( \frac{1}{e} \right )^{\frac{1}{e}}$

Yep nice work.
Also I should probably say x > 0 for that question.

omgiloverice · Jun 23, 2013

Re: HSC 2013 3U Marathon Thread

Found this in one of the most underrated textbooks.

$$It is given that$\ A>0,\ B>0,\ $and $n$ is a positive integer.$

$$ a) Divide $A^{n+1}-A^{n}B+B^{n+1}-B^{n}A \ $by $A-B$ and deduce that A^{n+1}+B^{n+1}\geq A^{n}B+B^{n}A$

$$ b) Using (a), show by mathematical induction that $ (\frac{A+B}{2})^{n}\leq \frac{1}{2}(A^{n}+B^{n})$

HeroicPandas · Jun 23, 2013

Re: HSC 2013 3U Marathon Thread

omgiloverice said:
Found this in one of the most underrated textbooks.

$$It is given that$\ A>0,\ B>0,\ $and $n$ is a positive integer.$

$$ a) Divide $A^{n+1}-A^{n}B+B^{n+1}-B^{n}A \ $by $A-B$ and deduce that A^{n+1}+B^{n+1}\geq A^{n}B+B^{n}A$

$$ b) Using (a), show by mathematical induction that $ (\frac{A+B}{2})^{n}\leq \frac{1}{2}(A^{n}+B^{n})$

$A^{n+1} - A^nB + B^{n+1} - B^nA = (A-B)(A^n - B^n)\geq 0 \\($because \dots if A $>B,\therefore A-B>0, \therefore A^n-B^n >0, \therefore (A-B)(A^n - B^n)>),$ same thing happens when A$<B$) \\ $ $ And equality occurs when A = B$ \\ \\ $If ($A-B)(A^n-B^n) \geq 0 \\\\ A^{n+1}- A^nB + B^{n+1} - B^nA \geq 0 \\ \\ \therefore A^{n+1}+ B^{n+1} \geq A^nB+B^nA$

Is this ok?

too lazy to do induction....

here is a very fun question from the greatest textbook:

$(a) $ Prove by induction that cos$\left ( \frac{90^{\circ}}{2^n} \right ) = \frac{1}{2} \underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{\dots + \sqrt{2}}}}}}_\text{n terms} \\ \\ ~~~~(b) $ Find sin $\left ( \frac{90^{\circ}}{2^n} \right ). $ Hence find an expression converging to $\pi,$ and investigate it as a means of approximating \pi.$$

:O i never knew Cambridge 3U uses LaTeX..

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