HSC 2012 MX2 Marathon (archive) (1 Viewer)

SpiralFlex · Jan 17, 2012

Re: 2012 HSC MX2 Marathon

IamBread said:
Yeah that's probably true.. I just really liked mechanics

You can post some integration questions next week if you like. I am still consolidating my first topics.

RealiseNothing · Jan 17, 2012

Re: 2012 HSC MX2 Marathon

My school's finished complex numbers and done half of graphs.

Is this a reasonable pace? Will we have enough time to finish the course in-depth.

nightweaver066 · Jan 17, 2012

Re: 2012 HSC MX2 Marathon

RealiseNothing said:
My school's finished complex numbers and done half of graphs.

Is this a reasonable pace? Will we have enough time to finish the course in-depth.

It should be enough time.

Some schools don't start teaching 4U until this year like Newington and Sydney Boys.

largarithmic · Jan 17, 2012

Re: 2012 HSC MX2 Marathon

IamBread said:
Yeah that's probably true.. I just really liked mechanics

Hey your thing says HSC 2011, are you doing science / advanced maths this year and if so at which uni?

Also may as well post a question. Dunno how suitable thing one is but its neat:

Part (A). Trapezium ABCD is given with AB parallel to CD. Let diagonals AC and BD intersect at X. Prove that triangles ADX and BCX have the same area.

Part (B). A triangle and any point on one of its sides are given. Show how to construct, with straight edge and compass, a line through this given point that bisects the triangle into two regions of equal area.

IamBread · Jan 18, 2012

Re: 2012 HSC MX2 Marathon

largarithmic said:
Hey your thing says HSC 2011, are you doing science / advanced maths this year and if so at which uni?

I was tossing up between Advanced Math and Advanced Science at UNSW, but in the end I picked science (major in physics). I was looking at the handbooks, and it still had a fair bit of maths in it, as well as the physics so I went with that.

Carrotsticks · Jan 18, 2012

Re: 2012 HSC MX2 Marathon

IamBread said:
I was tossing up between Advanced Math and Advanced Science at UNSW, but in the end I picked science (major in physics). I was looking at the handbooks, and it still had a fair bit of maths in it, as well as the physics so I went with that.

=(

IamBread · Jan 18, 2012

Re: 2012 HSC MX2 Marathon

Carrotsticks said:
=(

=( It's ok, I'm doing as much math as possible

IamBread · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

Guess it's time for a question, don't want this thread to die!

$$If $\int^6_0f(x)dx=42$

$$and $\int^3_0f(x)dx=12,$

$$Find $\int^6_32f(x)-3dx$

nightweaver066 · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

$\int^6_0 f(x)dx - \int^3_0 f(x)dx = 30$

$\int^6_3 f(x)dx = 30$

$\int^6_3 2f(x)dx = 60$

$\int^6_3 2f(x) - 3dx = 60 - [3x]^6_3$

$= 60 - (3^6 - 3^3)$

$= -642$

Unsure lol.

IamBread · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

nightweaver066 said:
$\int^6_0 f(x)dx - \int^3_0 f(x)dx = 30$

$\int^6_3 f(x)dx = 30$

$\int^6_3 2f(x)dx = 60$

$\int^6_3 2f(x) - 3dx = 60 - [3x]^6_3$

$= 60 - (3^6 - 3^3)$

$= -642$

Unsure lol.

It's all right until the last bit

$\int^6_3 2f(x) - 3dx = 2\int^6_3f(x)dx - \int^6_3 3dx$

$= 60 - [3x]^6_3$

$= 60 - (18 - 9)$

$=51$

Have you done integration yet?

nightweaver066 · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

IamBread said:
It's all right until the last bit

$\int^6_3 2f(x) - 3dx = 2\int^6_3f(x)dx - \int^6_3 3dx$

$= 60 - [3x]^6_3$

$= 60 - (18 - 9)$

$=51$

Have you done integration yet?

LOL. My bad, i misread it as power.....

I've finished 2U and 3U integration but haven't done 4U integration yet.

IamBread · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

nightweaver066 said:
LOL. My bad, i misread it as power.....

I've finished 2U and 3U integration but haven't done 4U integration yet.

Haha that makes sense. 4U integration is much better

Here's another one:

$$Prove $ \int \frac{1}{1+x^2}dx = tan^{-1}x + c$

Not much harder then 3U, except you have to figure out the substitution

nightweaver066 · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

IamBread said:
Haha that makes sense. 4U integration is much better

Here's another one:

$$Prove $ \int \frac{1}{1+x^2}dx = tan^{-1}x + c$

$$Let $ y = tan^{-1}x$

$x = tany$

$Differentiating with respect to x,$

$1 = sec^2y\frac{dy}{dx}$

$\frac{dy}{dx} = \frac{1}{sec^2y}$

$= \frac{1}{1 + tan^2y}$

$= \frac{1}{1 + x^2}$

$\therefore \int \frac{1}{1 + x^2}dx = tan^{-1}x + c$

SpiralFlex · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

Or.

Let $x=\tan \theta$

$dx=\tan \theta \: d\theta$

$=\int \frac{1}{1+\tan^2 \theta}*\sec^2 \theta \: d\theta$

$=\int 1 \:d\theta$

$= \theta + C$

$=\tan^{-1}x+C$

IamBread · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

nightweaver066 said:
$$Let $ y = tan^{-1}x$

$x = tany$

$Differentiating with respect to x,$

$1 = sec^2y\frac{dy}{dx}$

$\frac{dy}{dx} = \frac{1}{sec^2y}$

$= \frac{1}{1 + tan^2y}$

$= \frac{1}{1 + x^2}$

$\therefore \int \frac{1}{1 + x^2}dx = tan^{-1}x + c$

That works, can you do it the other way though?

IamBread · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

SpiralFlex said:
Or.

Let $x=\tan \theta$

$dx=\tan \theta d\theta$

$=\int \frac{1}{1+\tan^2 \theta}*\sec^2 \theta d\theta$

$=\int 1 d\theta$

$=\tan^{-1}x+C$

That's it.

Next one.

$$Show that $\int \frac{1}{\sqrt{a^2 - x^2}}dx = sin^{-1}\frac{x}{a} + c$

SpiralFlex · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

Let $x=a\sin \theta$

$dx=a\cos \theta \:d\theta$

$=\int \frac{1}{a\cos \theta}*a\cos \theta\: d\theta$

$=\int 1 \:d\theta$

$=\theta+C$

$=\sin^{-1}\frac{x}{a}+C$

IamBread · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

$$Show that $\int \frac{1}{\sqrt{a^2 + x^2}}dx = ln|x +\sqrt{a^2 + x^2}| + c$

Then

$\int \frac{1}{\sqrt{x^2 - a^2}}dx = ln|x +\sqrt{x^2 - a^2}| + c$

SpiralFlex · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

IamBread said:
$$Show that $\int \frac{1}{\sqrt{a^2 + x^2}}dx = ln|x +\sqrt{a^2 + x^2}| + c$

Then

$\int \frac{1}{\sqrt{x^2 - a^2}}dx = ln|x +\sqrt{x^2 - a^2}| + c$

Let $x=a\tan \theta$

$dx=a\sec^2 \theta\: d\theta$

$=\int \sec \theta \: d\theta$

$=\int \frac{\sec \theta (\sec \theta+\tan \theta)}{sec \theta+\tan \theta}\: d\theta$

$=\int \ln|\sec \theta +\tan \theta|+C_{1}$

$=\int \ln|\frac{\sqrt{x^2+a^2}}{a}+\frac{x}{a}|+C_{1}$

$=\int \ln|\sqrt{x^2+a^2}+x|+C$

STOP POSTING OR I CAN'T GO TOILET!

Let $x=a\sec \theta$

$dx=a\sec \theta \tan \theta\: d\theta$

$=\int\frac{1}{a\tan \theta}*a\sec \theta \tan \theta\: d\theta$

$=\int \sec \theta \:d\theta$

$=\ln|\sec \theta+ \tan \theta|+C_{1}$

$=\ln|\frac{x}{a}+\frac{\sqrt{x^2-a^2}}{a}|+C_{1}$

$=\ln|x+\sqrt{x^2-a^2}}|+C$

IamBread · Jan 19, 2012

Re: 2012 HSC MX2 Marathon

SpiralFlex said:
STOP POSTING OR I CAN'T GO TOILET!

Haha ok, I'll let you go to the toilet

Plus I'm out of standard integrals to post lol.

HSC 2012 MX2 Marathon (archive) (1 Viewer)

Well-Known Member

what is that?It is Cowpea

Well-Known Member

Member

Member

Retired

Member

Member

Well-Known Member

Member

Well-Known Member

Member

Well-Known Member

Well-Known Member

Member

Member

Well-Known Member

Member

Well-Known Member

Member

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