HSC 2016 Maths Marathon (archive) (1 Viewer)

Paradoxica · Feb 7, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
Okay question, do I have to use this Euclid's lemma thing

Nope, you can always assume unique factorisation exists, down to the order of the factors.

Nailgun · Feb 7, 2016

Re: HSC 2016 2U Marathon

Paradoxica said:
Nope, you can always assume unique factorisation exists, down to the order of the factors.

I don't know what this means lol
are the answers x=2,200

Paradoxica · Feb 7, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
I don't know what this means lol
are the answers x=2,200

no, I said they were integers, I never said they weren't naturals.

$$a.k.a. $(-1)\times(-1)=1$

Nailgun · Feb 7, 2016

Re: HSC 2016 2U Marathon

Ah okay lol I get it

x = (0,-198) or (2,200)

Nailgun · Feb 7, 2016

Re: HSC 2016 2U Marathon

$$Let roots of the equation be $ (\alpha ,\beta ) \\ \frac { -b }{ a } =-p=\alpha +\beta ,\quad \frac { c }{ a } =q=\alpha \beta \quad \\ p=-(\alpha +\beta ),\quad q=\alpha \beta \\ p+q=198\\ -\alpha -\beta +\alpha \beta =198\\ \alpha \beta -\alpha -\beta +1=199\\ (\alpha -1)(\beta -1)=199\\ $Since 199 is a prime number $ \\ \alpha -1=1,\quad \beta -1=199\quad \\ OR\quad \alpha -1=-1,\quad \beta -1=-199\\ \therefore (\alpha ,\beta )=(2,200)\quad or\quad (0,-198)$

EDIT: By the way, can someone explain this to me anyways lol, it seems interesting, but not entirely useful

Nailgun said:
Okay question, do I have to use this Euclid's lemma thing

Paradoxica · Feb 7, 2016

Re: HSC 2016 2U Marathon

Euclid's Lemma lets us determine solutions to diophantine equations (see the elementary mathematics marathon) when we are certain that one product term divides a certain expression. Euclid's Lemma is also valid for the ring of polynomials with integer co-efficients. That is to say, that if we have a product of small polynomials on one side, and the other side is a product very large and messy polynomials, we can be certain that the smaller terms divide the larger terms of the polynomial we are interested in.

Here's an example using a root of unity polynomial.

$$\noindent$ P(z) \equiv x^{15}-1 \equiv (x^3 -1)(x^{12}+x^9+x^6+x^3+1) \equiv (x^5-1)(x^{10}+x^5+1) \\P(z) \equiv (x-1)(x^2+x+1)(x^{12}+x^9+x^6+x^3+1) \equiv (x-1)(x^4+x^3+x^2+x+1)(x^{10}+x^5+1) \\ $Since $x^4+x^3+x^2+x+1$ is irreducible over the integers, and $(x-1)$ is common to both sides, it follows by Euclid's Extended Lemma that $\\(x^2+x+1)|(x^{10}+x^5+1)\\$Indeed, $x^{10}+x^5+1 \equiv (x^2+x+1)(x^8-x^7+x^5-x^4+x^3-x+1)$

davidgoes4wce · Feb 8, 2016

Re: HSC 2016 2U Marathon

Wondering if someone could confirm with me if there was a mistake in Fitzpatrick and Bob Aus textbook for Exercise 14.1 e.g 2

Everything was right except for the x-values not existing for x=0 and x=1. The book has x=-1.

InteGrand · Feb 8, 2016

Re: HSC 2016 2U Marathon

davidgoes4wce said:
Wondering if someone could confirm with me if there was a mistake in Fitzpatrick and Bob Aus textbook for Exercise 14.1 e.g 2

Everything was right except for the x-values not existing for x=0 and x=1. The book has x=-1.

$\noindent The denominator of the original expression (the one they asked us to simplify) is $\frac{1}{x}-x$. This is undefined iff $x = 0$ and equals $0$ iff $x =\pm 1$ (and the numerator is defined for all real $x$ except $x =0$.) Hence for the original expression, we have $x \neq 0, \, \pm 1$.$

leehuan · Feb 8, 2016

Re: HSC 2016 2U Marathon

Paradoxica said:
$$\noindent$ f(x) = \frac{x^2-6x+6}{2x-4} \,\,\, g(x) = \frac{ax^2+bx+c}{x-d} \\ $The following information is known about $g(x). \\ g(x) $ and $ f(x) $ have the same vertical asymptotes.$ \\ $The oblique asymptotes are perpendicular, and intersect on the $y$-axis. $ \\ f(x) = g(x) $ has a solution at $x = -2. \\ $Find $g(x)$ and hence find the other solution to $f(x) = g(x)$

Oblique asymptotes are not a part of 2unit.

The hardest rational function you can give a 2u student and let them figure it out is f(x)/g(x) where deg f(x) = deg g(x) = 1. I.e. a linear over a linear.

(And even then a 2u kid misses the +1 -1 trick but you're allowed to assume they know that though)

Drsoccerball · Feb 10, 2016

Re: HSC 2016 2U Marathon

Drsoccerball said:
New Question:

$$ Find : $ \frac{d}{dx} (\pi^{\sin^n{x}}+e^{\sin^n{x}})^n \cdot \cos{(nx)}$

$If you're bothered find the double derivative.$

$Hence or otherwise find:$

$\int{\cos{(2x)}{(\pi ^{\sin ^2 x}+ e ^{\sin ^2 x})^2}}dx$

$$ Given that $ \int{\pi ^{2u} (1+e^{2u})}du $ can be expressed as $ \int{(\pi ^{2 \sin ^2x}+ \pi ^{2 \sin ^2 x} e^{2 \sin ^2 x}) \cos{2x}}dx $ where $ u= \sin ^2 x$

davidgoes4wce · Feb 12, 2016

Re: HSC 2016 2U Marathon

Just confirming there is a mistake with this question:

leehuan · Feb 12, 2016

Re: HSC 2016 2U Marathon

davidgoes4wce said:
Just confirming there is a mistake with this question:

I plugged the equation into GeoGebra and it is showing an x-intercept at x=2, not x=1.2.

So it would see that none of the graphs are correct

Bestintheworld · Feb 12, 2016

Re: HSC 2016 2U Marathon

Factorise x^2+6x+5

Nailgun · Feb 12, 2016

Re: HSC 2016 2U Marathon

Bestintheworld said:
Factorise x^2+6x+5

lellele not sure if serious but (x+5)(x+1)

you know, i remember in like yr9 when stuff like that was actually a legit exam question and i struggled with it lol

Speed6 · Feb 12, 2016

Re: HSC 2016 2U Marathon

Bestintheworld said:
Factorise x^2+6x+5

Really?

Bestintheworld · Feb 12, 2016

Re: HSC 2016 2U Marathon

^its a joke lol

davidgoes4wce · Feb 12, 2016

Re: HSC 2016 2U Marathon

leehuan said:
I plugged the equation into GeoGebra and it is showing an x-intercept at x=2, not x=1.2.

So it would see that none of the graphs are correct

Thanks for confirming leehuan

Nailgun · Feb 12, 2016

Re: HSC 2016 2U Marathon

Drsoccerball said:
$Hence or otherwise find:$

$\int{\cos{(2x)}{(\pi ^{\sin ^2 x}+ e ^{\sin ^2 x})^2}}dx$

$$ Given that $ \int{\pi ^{2u} (1+e^{2u})}du $ can be expressed as $ \int{(\pi ^{2 \sin ^2x}+ \pi ^{2 \sin ^2 x} e^{2 \sin ^2 x}) \cos{2x}}dx $ where $ u= \sin ^2 x$

Is the answer to this just C lol?
Or did I dun goof somewhere

Also, I retried that derivative question again from the first time you posted it, and I could actually do it now lol, and it was fairly easy
I feel so accomplished lelel

InteGrand · Feb 12, 2016

Re: HSC 2016 2U Marathon

Nailgun said:
Is the answer to this just C lol?
Or did I dun goof somewhere

Also, I retried that derivative question again from the first time you posted it, and I could actually do it now lol, and it was fairly easy
I feel so accomplished lelel

$\noindent The answer can't just be $C$, because that would be equivalent to saying that the integrand is identically 0, but this is clearly not the case. One easy way to see that it is not the case is to sub. $x =0$ into the integrand.$

Nailgun · Feb 12, 2016

Re: HSC 2016 2U Marathon

ok redoing it lol

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