HSC 2013 MX2 Marathon (archive) (1 Viewer)

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$

HSC 2013 MX2 Marathon (archive) (1 Viewer)

Well-Known Member

Active Member

Cult of Personality

Well-Known Member

Member

Active Member

This too shall pass

Active Member

Heroic!

what is that?It is Cowpea

Well-Known Member

Member

This too shall pass

Premium Member

This too shall pass

what is that?It is Cowpea

This too shall pass

This too shall pass

what is that?It is Cowpea

what is that?It is Cowpea

Users Who Are Viewing This Thread (Users: 0, Guests: 1)