HSC 2016 MX1 Marathon (archive) (1 Viewer)

InteGrand · Feb 4, 2016

Re: HSC 2016 3U Marathon

Shuuya said:
Naww I'm not special enough to have a personalised thread :'(

I think Flop21's thread just got merged with this one too. (Or got moved somewhere...can't find it right now, but when I click on it, it directs me to this thread.)

Edit: Ah, there's a post related to it at the top of this page.

Trebla · Feb 4, 2016

Re: HSC 2016 3U Marathon

Feel free to make separate threads for questions (and strongly encourage you to search past threads before you ask the question), but we really want to avoid a situation where many people start creating their own thread for all their questions.

wu345 · Feb 5, 2016

Re: HSC 2016 3U Marathon

Trebla said:
Feel free to make separate threads for questions (and strongly encourage you to search past threads before you ask the question), but we really want to avoid a situation where many people start creating their own thread for all their questions.

What if we just make one for both Cambridge textbooks? I think a lot of the questions are coming for ext1 are coming from the yr 11 and yr 12 cambridge textbooks (eg. all of mine haha). At the moment there's a thread for yr 11 cambridge but not for yr12

leehuan · Feb 5, 2016

Re: HSC 2016 3U Marathon

$\\ \bf{NEW QUESTION}$

For the actual marathon. To not scare away too many people.

$\\ $Use mathematical induction to prove the formula for the $n$-th partial sum of a geometric progression.$

leehuan · Feb 5, 2016

Re: HSC 2016 3U Marathon

Paradoxica said:
We still have this one.

I left a question for little 2016ers who are still making their way up. I might do your one later.

Trebla · Feb 5, 2016

Re: HSC 2016 3U Marathon

wu345 said:
What if we just make one for both Cambridge textbooks? I think a lot of the questions are coming for ext1 are coming from the yr 11 and yr 12 cambridge textbooks (eg. all of mine haha). At the moment there's a thread for yr 11 cambridge but not for yr12

That is fine as the long as there are only two threads.

davidgoes4wce · Feb 8, 2016

Re: HSC 2016 3U Marathon

Solve for 'x':

leehuan · Feb 8, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
Solve for 'x':

My instinct is that after you use change of base, multiply the equation through by ln(x) [safe because ln(x)≠0] and then it's a quadratic.

Paradoxica · Feb 8, 2016

Re: HSC 2016 3U Marathon

$\noindent a) ${-1,x,x^3}$ is an arithmetic progression.$\\$Find $x$, given $-1$ trivially satisfies the above sequence.$ \\ \phi $ is the positive solution to the above sequence.$ \\ $b) Show by induction or other means that $\phi^{2n}+\phi^{-2n}$ is an integer.$ \\ $c) Similarly, show $\phi^{2n+1}-\phi^{-(2n+1)}$ is an integer.$ \\ $Hint: Manipulate the quadratic equation to your advantage.$

Drsoccerball · Feb 8, 2016

Re: HSC 2016 3U Marathon

Here's a question before I forget :

$Solve:$

$y=x^3+x^2-4x+3---(1)$

$(y+1)(y^2-4)=x-7 ---(2)$

davidgoes4wce · Feb 8, 2016

Re: HSC 2016 3U Marathon

Paradoxica said:
$\noindent a) ${-1,x,x^3}$ is an arithmetic progression.$\\$Find $x$, given $-1$ trivially satisfies the above sequence.$ \\ \phi $ is the positive solution to the above sequence.$ \\ $b) Show by induction or other means that $\phi^{2n}+\phi^{-2n}$ is an integer.$ \\ $c) Similarly, show $\phi^{2n+1}-\phi^{-(2n+1)}$ is an integer.$ \\ $Hint: Manipulate the quadratic equation to your advantage.$

Can an arithmetic progression have a difference of 0?

For part a)

I looked at the difference of the second term and first term D= x-(-1)=x+1
Then I looked at the difference of the 3rd and 2nd term D=x^3-x

equating that

x+1=x^3-x

x^3-2x-1=0

In solving for x, I get x=-1.

Paradoxica · Feb 8, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
Can an arithmetic progression have a difference of 0?

For part a)

I looked at the difference of the second term and first term D= x-(-1)=x+1
Then I looked at the difference of the 3rd and 2nd term D=x^3-x

equating that

x+1=x^3-x

x^3-2x-1=0

In solving for x, I get x=-1.

...do you know what long division is...

or trial factorisation

leehuan · Feb 8, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
Can an arithmetic progression have a difference of 0?

For part a)

I looked at the difference of the second term and first term D= x-(-1)=x+1
Then I looked at the difference of the 3rd and 2nd term D=x^3-x

equating that

x+1=x^3-x

x^3-2x-1=0

In solving for x, I get x=-1.

What. You're meant to factor out (x+1) from that x^3-2x-1 and then solve from THERE.

With or without long division.

davidgoes4wce · Feb 8, 2016

Re: HSC 2016 3U Marathon

Paradoxica said:
...do you know what long division is...

or trial factorisation

Doing the long division was not the hard part, I did it here

I guess being an arithmetic progression , I expected the values to increase by the same 'difference'. Im guessing that you have to do something with the rational number, as there could be 3 values that could be the difference.

leehuan · Feb 8, 2016

Re: HSC 2016 3U Marathon

What am I even typing.

But no, that IS the correct answer. If you try subbing it into x^3 you'll find it pops out.

Speed6 · Feb 8, 2016

Re: HSC 2016 3U Marathon

Are golden ratios an advantage if you know it for 3U?

Drsoccerball · Feb 8, 2016

Re: HSC 2016 3U Marathon

Speed6 said:
Are golden ratios an advantage if you know it for 3U?

There's no use of it in any HSC level. (except maybe general notsure about the syllabus.)

davidgoes4wce · Feb 8, 2016

Re: HSC 2016 3U Marathon

I gather the 3 solutions then are:

AS a bit of a hit and giggle, I also decided then to substitute those rational terms and prove that the difference was indeed the same (sorry its a bit messy):

Paradoxica · Feb 8, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
---

That's unnecessarily complicated.

$$\noindent If $x$ satisfies $ x^2 = bx + c $, then $x^3 = b(x^2) + cx = b(bx+c) + cx = (b^2+c) x + bc \\ $In this case, $b = c = 1$

davidgoes4wce · Feb 8, 2016

Re: HSC 2016 3U Marathon

Paradoxica said:
$\noindent a) ${-1,x,x^3}$ is an arithmetic progression.$\\$Find $x$, given $-1$ trivially satisfies the above sequence.$ \\ \phi $ is the positive solution to the above sequence.$ \\ $b) Show by induction or other means that $\phi^{2n}+\phi^{-2n}$ is an integer.$ \\ $c) Similarly, show $\phi^{2n+1}-\phi^{-(2n+1)}$ is an integer.$ \\ $Hint: Manipulate the quadratic equation to your advantage.$

b) Since we know the solution of part a), by equating n=1, it returns an integer of 3.

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