HSC 2012 MX2 Marathon (archive) (5 Viewers)

Nooblet94 · Feb 14, 2012

Re: 2012 HSC MX2 Marathon

Haven't seen many graphing questions, so here we go.

Graph the following parametric equation:
<a href="http://www.codecogs.com/eqnedit.php?latex=\\ x=16\sin^3(t)\\ y=13\cos(t)-5\cos(2t)-2\cos(3t)-\cos(4t)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\ x=16\sin^3(t)\\ y=13\cos(t)-5\cos(2t)-2\cos(3t)-\cos(4t)" title="\\ x=16\sin^3(t)\\ y=13\cos(t)-5\cos(2t)-2\cos(3t)-\cos(4t)" /></a>

rolpsy · Feb 15, 2012

Re: 2012 HSC MX2 Marathon

lol

question:
graph
$x^3 = 1 + xy^2$
(perhaps a tad more accessible than that ^)

Carrotsticks · Feb 15, 2012

Re: 2012 HSC MX2 Marathon

Rolpsy, what program was used to sketch that curve?

rolpsy · Feb 15, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
Rolpsy, what program was used to sketch that curve?

texshop (i.e. latex) with pstricks (and pst-plot)

math man · Feb 15, 2012

Re: 2012 HSC MX2 Marathon

Nooblet94 said:
Haven't seen many graphing questions, so here we go.

Graph the following parametric equation:
<a href="http://www.codecogs.com/eqnedit.php?latex=\\ x=16\sin^3(t)\\ y=13\cos(t)-5\cos(2t)-2\cos(3t)-\cos(4t)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\ x=16\sin^3(t)\\ y=13\cos(t)-5\cos(2t)-2\cos(3t)-\cos(4t)" title="\\ x=16\sin^3(t)\\ y=13\cos(t)-5\cos(2t)-2\cos(3t)-\cos(4t)" /></a>

nooblet what are you trying to imply here?

deswa1 · Feb 16, 2012

Re: 2012 HSC MX2 Marathon

I haven't seen many integration questions... This is from someone at school (you know who you are

)

<a href="http://www.codecogs.com/eqnedit.php?latex=\int sin(lnx)dx" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int sin(lnx)dx" title="\int sin(lnx)dx" /></a>

nightweaver066 · Feb 16, 2012

Re: 2012 HSC MX2 Marathon

deswa1 said:
I haven't seen many integration questions... This is from someone at school (you know who you are )

$\int sin(lnx)dx$

$\int sin(lnx)dx$

$Let u = lnx$

$e^u = x$

$du = \frac{1}{x}dx$

$\therefore \int sin(lnx) dx = \int e^usinudu$

$Applying I.B.P.$

$= e^usinu - \int e^ucosudu$

$= e^usinu - (e^ucosu + \int e^usinudu)$

$\int e^u sinu du= e^usinu - e^ucosu - \int e^usinudu ..... (1)$

$Applying I.B.P. the other way to the original integral$

$= -e^ucosu + \int e^ucosudu$

$\int e^u sinu du = -e^ucosu + e^usinu - \int e^usinudu ..... (2)$

$(1) + (2): 2\int e^u sinu du = 2e^u(sinu - cosu) - 2\int e^u sinu du$

$4\int e^u sinu du = 2e^u(sinu - cosu)$

$\int e^u sinu du = \frac{1}{2}e^u (sinu - cosu)$

$= \frac{x}{2}(sinlnx - coslnx) + c$

SpiralFlex · Feb 16, 2012

Re: 2012 HSC MX2 Marathon

Many methods I can't type as I am on iPhone so I will demonstrate verbally. You can us normal integration by parts then do it again. Then adding both results of the integrate something should cancel out nicely. Second method is to use Eulers formula. Or you can change it to an exponential base and introducE another variable and go from there.

someth1ng · Feb 16, 2012

Re: 2012 HSC MX2 Marathon

Graph: (x^2+y^2-1)^3-x^2*y^3=0

Nooblet94 · Feb 16, 2012

Re: 2012 HSC MX2 Marathon

Patel Ex 5E Q28
"Show that the polynomial $x^n+mx-b$ has a multiple root provided:
$\left( \frac{m}{n} \right)^n+\left( \frac{b}{n-1} \right)^{n-1}=0$ .
Find this root."

I've found the root, I just can't work out how to show that condition. It's pissing me off because it's the only question in the entire chapter I've been unable to do and it'll probably be something really simple that I'm missing.

mnmaa · Feb 16, 2012

Re: 2012 HSC MX2 Marathon

fun question to do for anyone interested-
Suppose that f(1) = 2, f(4) = 5, f '(1) = 5, f '(4) = 4, and f '' is continuous. Find the value of the following integral.

Carrotsticks · Feb 16, 2012

Re: 2012 HSC MX2 Marathon

Nooblet94 said:
Patel Ex 5E Q28
"Show that the polynomial $x^n+mx-b$ has a multiple root provided:
$\left( \frac{m}{n} \right)^n+\left( \frac{b}{n-1} \right)^{n-1}=0$ .
Find this root."

I've found the root, I just can't work out how to show that condition. It's pissing me off because it's the only question in the entire chapter I've been unable to do and it'll probably be something really simple that I'm missing.

Okay I got it, but its a very ugly question. Give me a moment to post up solutions.

Carrotsticks · Feb 16, 2012

Re: 2012 HSC MX2 Marathon

Here you go. I did it in as many steps as possible so you don't miss out on anything:

shaon0 · Feb 16, 2012

Re: 2012 HSC MX2 Marathon

8?

nightweaver066 · Feb 16, 2012

Re: 2012 HSC MX2 Marathon

mnmaa said:
fun question to do for anyone interested-
Suppose that f(1) = 2, f(4) = 5, f '(1) = 5, f '(4) = 4, and f '' is continuous. Find the value of the following integral.

If you want to use Latex in your post, use the opening and closing brackets $[./tex] (without the dot). [tex]\int^4_1 xf''(x) dx$

$= \left[ xf'(x) \right ]^4_1 - \int^4_1 f'(x) dx$

$= 4f'(4) - 4f'(1) - \left [f(x) \right ]^4 _1$

$16 - 8 - f(4) + f(1)$

$= 8 - 5 + 2$

$= 5$

Not too sure ..

Carrotsticks · Feb 16, 2012

Re: 2012 HSC MX2 Marathon

nightweaver066 said:
If you want to use Latex in your post, use the opening and closing brackets $[./tex] (without the dot). [tex]\int^4_1 xf''(x) dx$

$= \left[ xf'(x) \right ]^4_1 - \int^4_1 f'(x) dx$

$= 4f'(4) - 4f'(1) - \left [f(x) \right ]^4 _1$

$16 - 8 - f(4) + f(1)$

$= 8 - 5 + 2$

$= 5$

Not too sure ..

I got 8 as well.

Carrotsticks · Feb 16, 2012

Re: 2012 HSC MX2 Marathon

You made an arithmetic error in the third line:

$4f'(x)-4f'(x) - .....$

The second 4 shouldn't be there, it should be:

$4f'(x)-f'(x) - .....$

Drongoski · Feb 16, 2012

Re: 2012 HSC MX2 Marathon

$I = \int _1 ^4 xdf'(x) = xf'(x) ]^4 _1 - \int _1 ^4 f'(x)dx = \left [ xf'(x) - f(x) \right ] ^4 _1 = 8$

Carrotsticks · Feb 16, 2012

Re: 2012 HSC MX2 Marathon

Drongoski said:
$I = \int _1 ^4 xdf'(x) = xf'(x) - \int _1 ^4 f'(x)dx$

You appear to have forgotten to type out the substitution limits for the:

$xf'(x)$

And there appears to be a 'd' in the middle of your first integral lol.

Drongoski · Feb 16, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
You appear to have forgotten to type out the substitution limits for the:

$xf'(x)$

And there appears to be a 'd' in the middle of your first integral lol.

Yes I wanted to put the limits later - being a slow typist. The 'd' in the 1st integral is absolutely correct - the differential df'(x) = f''(x)dx.

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