HSC 2012 MX1 Marathon #2 (archive) (1 Viewer)

[Nooblet94]

Re: HSC 2012 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=tan(\theta ) = -4 @plus; 3\sqrt{3}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?tan(\theta ) = -4 + 3\sqrt{3}" title="tan(\theta ) = -4 + 3\sqrt{3}" /></a>

Barb please say yes

That's it. Can't believe everyone struggled on that one so much! LOL

 

[barbernator]

Re: HSC 2012 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=tan(\theta ) = -4 @plus; 3\sqrt{3}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?tan(\theta ) = -4 + 3\sqrt{3}" title="tan(\theta ) = -4 + 3\sqrt{3}" /></a>

Barb please say yes

YES WOOOOOOOOOO!! now you post a question

 

[Timske]

Re: HSC 2012 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=Show ~ \frac{T_{k@plus;1}}{T_{k}} = \frac{n - k @plus; 1}{k} * \frac{x}{a} ~ , ~ for~ (a@plus;x)^n" target="_blank"><img src="http://latex.codecogs.com/gif.latex?Show ~ \frac{T_{k+1}}{T_{k}} = \frac{n - k + 1}{k} * \frac{x}{a} ~ , ~ for~ (a+x)^n" title="Show ~ \frac{T_{k+1}}{T_{k}} = \frac{n - k + 1}{k} * \frac{x}{a} ~ , ~ for~ (a+x)^n" /></a>

Medium/hard

 

[Timske]

Re: HSC 2012 Marathon

That's it. Can't believe everyone struggled on that one so much! LOL

mis read Q :S

 

[barbernator]

Re: HSC 2012 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=Show ~ \frac{T_{k@plus;1}}{T_{k}} = \frac{n - k @plus; 1}{k} * \frac{x}{a} ~ , ~ for~ (a@plus;x)^n" target="_blank"><img src="http://latex.codecogs.com/gif.latex?Show ~ \frac{T_{k+1}}{T_{k}} = \frac{n - k + 1}{k} * \frac{x}{a} ~ , ~ for~ (a+x)^n" title="Show ~ \frac{T_{k+1}}{T_{k}} = \frac{n - k + 1}{k} * \frac{x}{a} ~ , ~ for~ (a+x)^n" /></a>

Medium/hard

oh god i cbfd latexing for that haha, ill let some other latex leet do that one

 

[Timske]

Re: HSC 2012 Marathon

oh god i cbfd latexing for that haha, ill let some other latex leet do that one

hahahaha

 

[Nooblet94]

Re: HSC 2012 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\\ T_{k@plus;1}=\frac{n!}{k!(n-k)!}\cdot a^k\cdot x^{a-k}\\ T_{k}=\frac{n!}{(k-1)!(n-k@plus;1)!}\cdot a^{k-1}\cdot x^{a-k@plus;1}\\ ~\\ \therefore \frac{T_{k@plus;1}}{T_k}=\frac{n!}{k!(n-k)!}\cdot \frac{(k-1)!(n-k@plus;1)!}{n!}\cdot a^k\cdot a^{1-k} \cdot x^{a-k} \cdot x^{k-a-1}\\ =\frac{n-k@plus;1}{k}\cdot \frac{a}{x}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\ T_{k+1}=\frac{n!}{k!(n-k)!}\cdot a^k\cdot x^{a-k}\\ T_{k}=\frac{n!}{(k-1)!(n-k+1)!}\cdot a^{k-1}\cdot x^{a-k+1}\\ ~\\ \therefore \frac{T_{k+1}}{T_k}=\frac{n!}{k!(n-k)!}\cdot \frac{(k-1)!(n-k+1)!}{n!}\cdot a^k\cdot a^{1-k} \cdot x^{a-k} \cdot x^{k-a-1}\\ =\frac{n-k+1}{k}\cdot \frac{a}{x}" title="\\ T_{k+1}=\frac{n!}{k!(n-k)!}\cdot a^k\cdot x^{a-k}\\ T_{k}=\frac{n!}{(k-1)!(n-k+1)!}\cdot a^{k-1}\cdot x^{a-k+1}\\ ~\\ \therefore \frac{T_{k+1}}{T_k}=\frac{n!}{k!(n-k)!}\cdot \frac{(k-1)!(n-k+1)!}{n!}\cdot a^k\cdot a^{1-k} \cdot x^{a-k} \cdot x^{k-a-1}\\ =\frac{n-k+1}{k}\cdot \frac{a}{x}" /></a>


I'm really, really bored.

(Also, wut. captcha? since when?)

 

[asianese]

Re: HSC 2012 Marathon

Is the limit computable by HSC MX1? I had to use LHopitals..

 

[Carrotsticks]

Re: HSC 2012 Marathon

Is the limit computable by HSC MX1? I had to use LHopitals..

You can't use L'Hopital's Principle unless the limit is in the form infinity/infinity.

You would acquire the answer by rationalising the numerator.

 

[bleakarcher]

Re: HSC 2012 Marathon

^ multiply numerator and denominator by the conjugate of the numerator.

 

[Timske]

Re: HSC 2012 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\inline \dpi{150} \lim_{x \rightarrow - 5}~ \frac{\sqrt{20 - x} - 5}{5 @plus; x} ~ * ~ \frac{\sqrt{20 - x} @plus; 5}{\sqrt{20 - x} @plus; 5}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\inline \dpi{150} \lim_{x \rightarrow - 5}~ \frac{\sqrt{20 - x} - 5}{5 + x} ~ * ~ \frac{\sqrt{20 - x} + 5}{\sqrt{20 - x} + 5}" title="\inline \dpi{150} \lim_{x \rightarrow - 5}~ \frac{\sqrt{20 - x} - 5}{5 + x} ~ * ~ \frac{\sqrt{20 - x} + 5}{\sqrt{20 - x} + 5}" /></a>

 

[asianese]

Re: HSC 2012 Marathon

Oh...0/0 doesn't work? Hm ok.

 

[Carrotsticks]

Re: HSC 2012 Marathon

Oh...0/0 doesn't work? Hm ok.

Sometimes it does ie:

 

[Nooblet94]

Re: HSC 2012 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\\ ~~~~\lim_{x \rightarrow -5} \frac{\sqrt{20-x}-5}{5@plus;x}\\ =\lim_{x \rightarrow -5} \frac{20-x-25}{(5@plus;x)(\sqrt{20-x}@plus;5)}\\ =\lim_{x \rightarrow -5} \frac{-(x@plus;5)}{(5@plus;x)\sqrt{20-x}@plus;5}\\ =\lim_{x \rightarrow -5} \frac{-1}{\sqrt{20-x}@plus;5}\\ =-\frac{1}{10}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\ ~~~~\lim_{x \rightarrow -5} \frac{\sqrt{20-x}-5}{5+x}\\ =\lim_{x \rightarrow -5} \frac{20-x-25}{(5+x)(\sqrt{20-x}+5)}\\ =\lim_{x \rightarrow -5} \frac{-(x+5)}{(5+x)\sqrt{20-x}+5}\\ =\lim_{x \rightarrow -5} \frac{-1}{\sqrt{20-x}+5}\\ =-\frac{1}{10}" title="\\ ~~~~\lim_{x \rightarrow -5} \frac{\sqrt{20-x}-5}{5+x}\\ =\lim_{x \rightarrow -5} \frac{20-x-25}{(5+x)(\sqrt{20-x}+5)}\\ =\lim_{x \rightarrow -5} \frac{-(x+5)}{(5+x)\sqrt{20-x}+5}\\ =\lim_{x \rightarrow -5} \frac{-1}{\sqrt{20-x}+5}\\ =-\frac{1}{10}" /></a>

Sometimes it does ie:

I thought L'Hopitals applied to all indeterminate limits... so it should work for all limits of the form 0/0, not just sometimes. (I may be wrong, you're the one studying second year advanced maths after all)

 

[barbernator]

Re: HSC 2012 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\\ ~~~~\lim_{x \rightarrow -5} \frac{\sqrt{20-x}-5}{5@plus;x}\\ =\lim_{x \rightarrow -5} \frac{20-x-25}{(5@plus;x)(\sqrt{20-x}@plus;5)}\\ =\lim_{x \rightarrow -5} \frac{-(x@plus;5)}{(5@plus;x)\sqrt{20-x}@plus;5}\\ =\lim_{x \rightarrow -5} \frac{-1}{\sqrt{20-x}@plus;5}\\ =-\frac{1}{10}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\ ~~~~\lim_{x \rightarrow -5} \frac{\sqrt{20-x}-5}{5+x}\\ =\lim_{x \rightarrow -5} \frac{20-x-25}{(5+x)(\sqrt{20-x}+5)}\\ =\lim_{x \rightarrow -5} \frac{-(x+5)}{(5+x)\sqrt{20-x}+5}\\ =\lim_{x \rightarrow -5} \frac{-1}{\sqrt{20-x}+5}\\ =-\frac{1}{10}" title="\\ ~~~~\lim_{x \rightarrow -5} \frac{\sqrt{20-x}-5}{5+x}\\ =\lim_{x \rightarrow -5} \frac{20-x-25}{(5+x)(\sqrt{20-x}+5)}\\ =\lim_{x \rightarrow -5} \frac{-(x+5)}{(5+x)\sqrt{20-x}+5}\\ =\lim_{x \rightarrow -5} \frac{-1}{\sqrt{20-x}+5}\\ =-\frac{1}{10}" /></a>

well done, now post a new question

 

[Nooblet94]

Re: HSC 2012 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\\ $By rationalising the numerator, prove that$ \sqrt{n@plus;1} -\sqrt{n} > \frac{1}{2\sqrt{n@plus;1}}. $\\ ~\\ Hence prove by induction that$ 1@plus;\frac{1}{2}@plus;\frac{1}{3}...\frac{1}{n} < \sqrt{n} $ for $n\geq 7" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\ $By rationalising the numerator, prove that$ \sqrt{n+1} -\sqrt{n} > \frac{1}{2\sqrt{n+1}}. $\\ ~\\ Hence prove by induction that$ 1+\frac{1}{2}+\frac{1}{3}...\frac{1}{n} < \sqrt{n} $ for $n\geq 7" title="\\ $By rationalising the numerator, prove that$ \sqrt{n+1} -\sqrt{n} > \frac{1}{2\sqrt{n+1}}. $\\ ~\\ Hence prove by induction that$ 1+\frac{1}{2}+\frac{1}{3}...\frac{1}{n} < \sqrt{n} $ for $n\geq 7" /></a>

 

[Carrotsticks]

Re: HSC 2012 Marathon

I thought L'Hopitals applied to all indeterminate limits... so it should work for all limits of the form 0/0, not just sometimes. (I may be wrong, you're the one studying second year advanced maths after all)

I thought the exact same thing too and asked my lecturer after we first learnt it (couple weeks into 1st Year).

My motive for asking is because sometimes we can use limit transformations:



To turn say x -> infinity to x -> 0 and change the form of the functions f(x) and g(x), so it may be possible to turn a 0/0 to a infinity/infinity form.

He told me that there are some families of curves where using L'Hopital's Principle for 0/0 doesn't work.

Naturally I asked him "Okay.. can you give me an example?", to which he replied something along the lines of "But you don't know enough Maths to understand the example, so just keep it in mind.".

I later emailed him for an example anyway so maybe some day when I 'know enough Maths', I would be able to understand (of course I didn't actually say that in the email) and to this day, I am still waiting for a reply.

 

[Sanjeet]

Re: HSC 2012 Marathon

Naturally I asked him "Okay.. can you give me an example?", to which he replied something along the lines of "But you don't know enough Maths to understand the example, so just keep it in mind.".

I later emailed him for an example anyway so maybe some day when I 'know enough Maths', I would be able to understand (of course I didn't actually say that in the email) and to this day, I am still waiting for a reply.

lol'd

 

[Sy123]

Re: HSC 2012 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\\ $By rationalising the numerator, prove that$ \sqrt{n@plus;1} -\sqrt{n} > \frac{1}{2\sqrt{n@plus;1}}. $\\ ~\\ Hence prove by induction that$ 1@plus;\frac{1}{2}@plus;\frac{1}{3}...\frac{1}{n} < \sqrt{n} $ for $n\geq 7" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\ $By rationalising the numerator, prove that$ \sqrt{n+1} -\sqrt{n} > \frac{1}{2\sqrt{n+1}}. $\\ ~\\ Hence prove by induction that$ 1+\frac{1}{2}+\frac{1}{3}...\frac{1}{n} < \sqrt{n} $ for $n\geq 7" title="\\ $By rationalising the numerator, prove that$ \sqrt{n+1} -\sqrt{n} > \frac{1}{2\sqrt{n+1}}. $\\ ~\\ Hence prove by induction that$ 1+\frac{1}{2}+\frac{1}{3}...\frac{1}{n} < \sqrt{n} $ for $n\geq 7" /></a>

Here goes first use of latex,

Ok here goes second part:

Therefore, the statement is true for n=7, If the statement is true for n=k, then it is also true for n=k+1. Therefore the statement is true for all integers, n greater than or equal to 7

There may be a few working lines missing, but that was my first real attempt at using latex, and it worked out alright. For some reason my reasoning for the induction was missing, but I filled it in, under normal text above.

Ok Here is my question.

(Fitzpatrick 3U Ex23(c) Question 36)

It is in the induction excercise, so you will have to use induction

How do you deatach items?

 

[Timske]

Re: HSC 2012 Marathon

Difficulty: Easy

<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{1}{\left | x \right |} < 5" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{1}{\left | x \right |} < 5" title="\frac{1}{\left | x \right |} < 5" /></a>