HSC 2013 MX2 Marathon (archive) (3 Viewers)

SpiralFlex · Dec 11, 2012

Re: HSC 2013 4U Marathon

Sorry didn't see that substitution.

bleakarcher · Dec 11, 2012

Re: HSC 2013 4U Marathon

seanieg89 said:
Evaluating the definite integral is well within the scope of MX2, the substitution

$x=\frac{1-u}{1+u}$

kills it.

How did you know to use that? It can also be solved using the substitution x=tan(a) and applying a property of integrals.

Shadowdude · Dec 11, 2012

Re: HSC 2013 4U Marathon

SpiralFlex said:
Sorry didn't see that substitution.

To be fair, who would?

seanieg89 · Dec 11, 2012

Re: HSC 2013 4U Marathon

bleakarcher said:
How did you know to use that? It can also be solved using the substitution x=tan(a) and applying a property of integrals.

Wanted a substitution that mapped the interval [0,1] to [0,1] so I could express I in terms of itself and some other terms. (ie I wanted to make a substitution but I didn't want the limits of integration to change.) This was the simplest rational function I could think of satisfying this property and it worked out nicely.

The first way I solved this question though was via a longer, different, less difficult to spot method...differentiation under the integral sign. But this is not MX2 material and it was longer than the substitution anyway.

Synchronised · Dec 12, 2012

Re: HSC 2013 4U Marathon

bleakarcher said:
How did you know to use that? It can also be solved using the substitution x=tan(a) and applying a property of integrals.

The substitution u=arctanx doesn't work if it was an indefinite integral but it works very well here.

bleakarcher · Dec 12, 2012

Re: HSC 2013 4U Marathon

seanieg89 said:
Wanted a substitution that mapped the interval [0,1] to [0,1] so I could express I in terms of itself and some other terms. (ie I wanted to make a substitution but I didn't want the limits of integration to change.) This was the simplest rational function I could think of satisfying this property and it worked out nicely.

The first way I solved this question though was via a longer, different, less difficult to spot method...differentiation under the integral sign. But this is not MX2 material and it was longer than the substitution anyway.

ah okay, I see. nice.

Sy123 · Dec 12, 2012

Re: HSC 2013 4U Marathon

$$Show that$ \ \ \ \ \ \ \ 3=\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6....}}}}}$

$$Similarly, evaluate$ \\ \\ 1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+...}}}}$

bleakarcher · Dec 12, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
$$Show that$ \ \ \ \ \ \ \ 3=\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6....}}}}}$

$$Similarly, evaluate$ \\ \\ 1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+...}}}}$

Let L=sqrt(6+sqrt(6+sqrt(6+...
<=> L^2=6+L
<=> L^2-L-6=0
<=> (L-3)(L+2)=0
=> L=3 since L>0

The continued fractions one should be the golden ratio, [1+sqrt(5)]/2. Sorry I still haven't learnt latex.

HeroicPandas · Dec 12, 2012

Re: HSC 2013 4U Marathon

$Write $\frac{(2-3i)^5}{(3+2i)^3} $ in the form a+ib without expanding at the start$

Difficulty: Easy

RealiseNothing · Dec 13, 2012

Re: HSC 2013 4U Marathon

HeroicPandas said:
$Write $\frac{(2-3i)^5}{(3+2i)^3} $ in the form a+ib without expanding$

Difficulty: Easy

$(2-3i)=-i(3+2i)$

Therefore by substituton:

$\frac{(-i(3+2i))^5}{(3+2i)^3}$

$-i(3+2i)^2$

seanieg89 · Dec 14, 2012

Re: HSC 2013 4U Marathon

Much easier than most of the questions I post here, but still quite a cool result.

$Let $f(x)$ be a continuous function defined on the interval $-1\leq x \leq 1$ which satisfies $|f(x)|\leq 1.$\\ \\i) Prove that there exists an $-1\leq\alpha\leq 1$ with $f(\alpha)=\alpha.$\\ ii) Does the same fact hold true if we replace all inequality signs with strict inequality signs?$

jyu · Dec 14, 2012

Re: HSC 2013 4U Marathon

seanieg89 said:
Much easier than most of the questions I post here, but still quite a cool result.

$Let $f(x)$ be a continuous function defined on the interval $-1\leq x \leq 1$ which satisfies $|f(x)|\leq 1.$\\ \\i) Prove that there exists an $-1\leq\alpha\leq 1$ with $f(\alpha)=\alpha.$\\ ii) Does the same fact hold true if we replace all inequality signs with strict inequality signs?$

i) y=f(x) intersect y=x at least once

ii) no, y=f(x) can be wholely above or below y=x

Sy123 · Dec 14, 2012

Re: HSC 2013 4U Marathon

$What is the sum of all the natural numbers below 1000 that are multiples of 3 or 5?$

AAEldar · Dec 15, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
$What is the sum of all the natural numbers below 1000 that are multiples of 3 or 5?$

Project Euler?

Sy123 · Dec 15, 2012

Re: HSC 2013 4U Marathon

AAEldar said:
Project Euler?

Correct

RealiseNothing · Dec 15, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
$What is the sum of all the natural numbers below 1000 that are multiples of 3 or 5?$

Denote $t_k$ as the k'th triangular number (surprise, surprise).

$999/3 = 333$

So we obtain:

$3t_{333}$ as the sum of the multiples of 3.

$999/5 = 199.8$

So we obtain:

$5t_{199}$ as the sum of multiples of 5.

Now we will double count all the multiples of 15, so using the same logic:

$999/15 = 66.6$

$15t_{66}$ as the sum of multiples of 15.

Now we add and subtract some shit:

$3t_{333} + 5t_{199} - 15t_{66}$

Noting that $t_k = \frac{k}{2}(k+1)$

$\frac{3 \times 333 \times 334}{2} + \frac{5 \times 199 \times 200}{2} - \frac{15 \times 66 \times 67}{2}$

$= 299498$

Sy123 · Dec 15, 2012

Re: HSC 2013 4U Marathon

RealiseNothing said:
Denote $t_k$ as the k'th triangular number (surprise, surprise).

$999/3 = 333$

So we obtain:

$3t_{333}$ as the sum of the multiples of 3.

$999/5 = 199.8$

So we obtain:

$5t_{199}$ as the sum of multiples of 5.

Now we will double count all the multiples of 15, so using the same logic:

$999/15 = 66.6$

$15t_{66}$ as the sum of multiples of 15.

Now we add and subtract some shit:

$3t_{333} + 5t_{199} - 15t_{66}$

Noting that $t_k = \frac{k}{2}(k+1)$

$\frac{3 \times 333 \times 334}{2} + \frac{5 \times 199 \times 200}{2} - \frac{15 \times 66 \times 67}{2}$

$= 299498$

Nice work. I'll try make a question with triangular number properties or something considering how much you like them heh.

Sy123 · Dec 15, 2012

Re: HSC 2013 4U Marathon

$$Define the nth Triangular number as$ \ \ T_n=\sum_{k=1}^{n}k = 1+2+3+...+n$

$$i) Show that $ \ \ \sum_{k=1}^{\infty} \frac{1}{T_k} = 2\sum_{k=1}^{\infty} \frac{1}{k}-\frac{1}{k+1} \ \ \ \ \ \fbox{2}$

$$ii) Hence evaluate $ \ \ \sum_{k=1}^{\infty} \frac{1}{T_k} \ \ \ \ \ \fbox{2}$

RealiseNothing · Dec 15, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
$$Define the nth Triangular number as$ \ \ T_n=\sum_{k=1}^{n}k = 1+2+3+...+n$

$$i) Show that $ \ \ \sum_{k=1}^{\infty} \frac{1}{T_k} = 2\sum_{k=1}^{\infty} \frac{1}{k}-\frac{1}{k+1} \ \ \ \ \ \fbox{2}$

$$ii) Hence evaluate $ \ \ \sum_{k=1}^{\infty} \frac{1}{T_k} \ \ \ \ \ \fbox{2}$

Awesome question. I got the answer but I'll let others have a go first, lets try and have different people answering besides the usual few. If no one answers this by tonight I'll put up my solution though.

I'll try come up with a cool question, but I used most of my ideas in my test lol.

RealiseNothing · Dec 15, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
$$Define the nth Triangular number as$ \ \ T_n=\sum_{k=1}^{n}k = 1+2+3+...+n$

$$i) Show that $ \ \ \sum_{k=1}^{\infty} \frac{1}{T_k} = 2\sum_{k=1}^{\infty} \frac{1}{k}-\frac{1}{k+1} \ \ \ \ \ \fbox{2}$

$$ii) Hence evaluate $ \ \ \sum_{k=1}^{\infty} \frac{1}{T_k} \ \ \ \ \ \fbox{2}$

Hope you don't mind me expanding on this:

$iii) $Deduce that$~\sum_{k=1}^{\infty} \frac{1}{k} - \frac{1}{k+1} = \sum_{k=1}^{\infty} \frac{1}{2^k}$

$iv) $Find the first term of each series$$

$v) $Find a value of 'k' such that$~k(k+1) \neq 2^k$

$vi) $Hence deduce that there is a positive real solution to$~ k(k+1)=2^k ~$other than$~ k=1$

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