HSC 2013 MX2 Marathon (archive) (3 Viewers)

RealiseNothing · Mar 19, 2013

Re: HSC 2013 4U Marathon

Exodia said:
$I get this:$ $\left ( x^{2}+\frac{1}{x^{2}} \right )+a\left ( x+\frac{1}{x} \right )+b=0$
$x+\frac{1}{x}=\frac{-a\pm \sqrt{a^{2}-4b}}{2}$

Check your working out, it should be:

$x+\frac{1}{x} = \frac{-a \pm \sqrt{a^2-4b+8}}{2}$

RealiseNothing · Mar 19, 2013

Re: HSC 2013 4U Marathon

Lieutenant_21 said:
What is the fastest way to prove the AM:GM inequality? My teacher uses induction + calculus which is a tedious method.

$(\sqrt{a} - \sqrt{b})^2 \geq 0$

$a+b-2\sqrt{ab} \geq 0$

$a+b \geq 2\sqrt{ab}$

$\frac{a+b}{2} \geq \sqrt{ab}$

You can also prove it using a really quick circle geo construction. (I'm assuming you mean this one and not the more general one).

RealiseNothing · Mar 19, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
$Find the minimum possible value of $a^2+b^2$ where $a$ and $b$ are two real numbers such that the polynomial:\\ \\ $x^4+ax^3+bx^2+ax+1$\\ \\ has at least one real root.$

Dividing by $x^2$ gives:

$x^2 + ax + b + \frac{a}{x} + \frac{1}{x^2} = 0$

Grouping appropriate terms:

$x^2 + \frac{1}{x^2} + 2 + ax + \frac{a}{x} + (b-2) = 0$

$(x+\frac{1}{x})^2 + a(x+\frac{1}{x}) + (b-2) = 0$

Now using the quadratic formula:

$x + \frac{1}{x} = \frac{-a \pm \sqrt{a^2 - 4b + 8}}{2}$

Now let $x+\frac{1}{x} = U$

We want to find the values of U such that it has atleast one real root:

$x^2 - Ux + 1 = 0$

$U^2 - 4 \geq 0$

$U \geq |2|$

So $x+\frac{1}{x} \geq |2|$

Therefore going back to our quadratic equation and the roots:

$\frac{-a \pm \sqrt{a^2 - 4b + 8}}{2} \geq |2|$

$-a \pm \sqrt{a^2 - 4b + 8} \geq |4|$

Continuing.....

Homo Sapien · Mar 20, 2013

Re: HSC 2013 4U Marathon

Prove that square root of two is irrational (3 marks), hence prove that cos of 1 degree is irrational (2 marks).

Homo Sapien · Mar 20, 2013

Re: HSC 2013 4U Marathon

To make the question easier you can assume that square root of two is irrational

But I think proving the irrationality of square root of two isn't very difficult, not sure if it is within the MX2 syllabus though?

seanieg89 · Mar 20, 2013

Re: HSC 2013 4U Marathon

Differences to products implies that

cos(nx) = 2*cos((n-1)x)*cos(x) + cos((n-2)x)

applying this formula inductively with x = 1 under the assumption that cos(1 degree) is rational implies the assertion that cos(m) is rational for EVERY integer m.

But we know that cos(45)=sqrt(2)/2 is irrational, so we have a contradiction.

Hence our assumption was led ad absurdum and cos(1 degree) is irrational.

Homo Sapien · Mar 20, 2013

Re: HSC 2013 4U Marathon

Nice solution!
Out of curiosity, what are you working on in your PhD?

seanieg89 · Mar 20, 2013

Re: HSC 2013 4U Marathon

Semiclassical analysis, microlocal analysis and harmonic analysis mostly.

Homo Sapien · Mar 20, 2013

Re: HSC 2013 4U Marathon

Wow, a lot of analysis. Sounds very interesting!

RealiseNothing · Mar 20, 2013

Re: HSC 2013 4U Marathon

Homo Sapien said:
To make the question easier you can assume that square root of two is irrational
But I think proving the irrationality of square root of two isn't very difficult, not sure if it is within the MX2 syllabus though?

It's MX1.

RealiseNothing · Mar 20, 2013

Re: HSC 2013 4U Marathon

Assume it is rational, so $\sqrt{2}$ can be expressed as a fraction where the numerator and denominator have no common factors:

$\sqrt{2} = \frac{p}{q}$

$q\sqrt{2} = p$

$2q^2 = p^2$

So now we know that $p^2$ is even and hence $p$ is even. So now we let $p=2k$ :

$p^2 = 4k^2$

$2q^2 = 4k^2$

$q^2 = 2k^2$

So now we know that $q^2$ is even, and so $q$ is even.

However if both $p$ and $q$ are even, then they have a common factor of $2$ . This is a contradiction and so our assumption is wrong, that is, $\sqrt{2}$ is not rational - it is irrational.

Lieutenant_21 · Mar 21, 2013

Re: HSC 2013 4U Marathon

Got my 4U half yearly results today (which was on all topics except volumes, mech and harder 3U) and I am disagreeing with the teacher about 1 mark

There was a really easy question, find i^{2013} and I did (i^{3})^{671}=i but my teacher said it is wrong because I didn't show that I understand why i^{3}=-i (I should have written i^{3}=iXi^{2}=iX-1=-i and then continue. My teacher is obsessed with understanding of why things happen and wants us to demonstrate our understanding of the concepts in the test which is good but not for this question... as far as I know they aren't allowed to do this in the HSC? Lost 1 mark in the whole paper because of this (wasn't the hardest paper ever lol).

asianese · Mar 21, 2013

Re: HSC 2013 4U Marathon

How many marks for the question?

You perhaps could have avoided this by saying:

$i^{2013} = i^{2012}\times i\\=(i^4)^{503}\times i \\ = i $ since $ i^4=1.$

In my opinion you did skip a step. If it was /2 marks you would lose 1, but if it was /1, then it is reasonable that you lose that 1 mark.

RealiseNothing · Mar 21, 2013

Re: HSC 2013 4U Marathon

Lieutenant_21 said:
Got my 4U half yearly results today (which was on all topics except volumes, mech and harder 3U) and I am disagreeing with the teacher about 1 mark
There was a really easy question, find i^{2013} and I did (i^{3})^{671}=i but my teacher said it is wrong because I didn't show that I understand why i^{3}=-i (I should have written i^{3}=iXi^{2}=iX-1=-i and then continue. My teacher is obsessed with understanding of why things happen and wants us to demonstrate our understanding of the concepts in the test which is good but not for this question... as far as I know they aren't allowed to do this in the HSC? Lost 1 mark in the whole paper because of this.

Your teacher is being silly, but an easier way to do that question is just:

$i^{2013} = i\times i^{2012} = i$ as $i$ to the power of any multiple of 4 is 1 (2012 is a multiple of 4).

Lieutenant_21 · Mar 21, 2013

Re: HSC 2013 4U Marathon

1

Lieutenant_21 · Mar 21, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
Your teacher is being silly, but an easier way to do that question is just:

$i^{2013} = i\times i^{2012} = i$ as $i$ to the power of any multiple of 4 is 1 (2012 is a multiple of 4).

Should have done this way

RealiseNothing · Mar 21, 2013

Re: HSC 2013 4U Marathon

Lieutenant_21 said:
1

If it's worth one mark, then that mark always comes from a correct answer (using the marking guidelines of the HSC) and nothing to do with the working out.

However your teacher could have some seriously fucked up marking criteria.

Lieutenant_21 · Mar 21, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
If it's worth one mark, then that mark always comes from a correct answer (using the marking guidelines of the HSC) and nothing to do with the working out.

However your teacher could have some seriously fucked up marking criteria.

He does. In his marking criteria it says you can even lose marks for untidy work lol (don't know if that is true for the HSC) but I will argue with him and ask him to change the criteria. If he doesn't then I will be careful next time. It won't affect my rank anyway

iBibah · Mar 21, 2013

Re: HSC 2013 4U Marathon

Lieutenant_21 said:
Got my 4U half yearly results today (which was on all topics except volumes, mech and harder 3U) and I am disagreeing with the teacher about 1 mark
There was a really easy question, find i^{2013} and I did (i^{3})^{671}=i but my teacher said it is wrong because I didn't show that I understand why i^{3}=-i (I should have written i^{3}=iXi^{2}=iX-1=-i and then continue. My teacher is obsessed with understanding of why things happen and wants us to demonstrate our understanding of the concepts in the test which is good but not for this question... as far as I know they aren't allowed to do this in the HSC? Lost 1 mark in the whole paper because of this (wasn't the hardest paper ever lol).

How did you go all up? (if you don't want t osay here pm?)

RealiseNothing said:
If it's worth one mark, then that mark always comes from a correct answer (using the marking guidelines of the HSC) and nothing to do with the working out.

However your teacher could have some seriously fucked up marking criteria.

+1

Lieutenant_21 said:
He does. In his marking criteria it says you can even lose marks for untidy work lol (don't know if that is true for the HSC) but I will argue with him and ask him to change the criteria. If he doesn't then I will be careful next time. It won't affect my rank anyway

In the HSC, as long as they can read it and understand your working, they mark it.

Lieutenant_21 · Mar 21, 2013

Re: HSC 2013 4U Marathon

iBibah said:
How did you go all up? (if you don't want t osay here pm?)

I am pretty sure that was the only mark I lost (hopefully, as I haven't gotten the whole paper back yet). The test wan't really that hard, was very similar to questions found in Cambridge.
I never used Cambridge before but I used it to study for this test because my teacher uses it and I think Cambridge and Terry Lee complement each other very well for 4U.

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