HSC 2013 Maths Marathon (archive) (2 Viewers)

Sy123 · May 21, 2013

Re: HSC 2013 2U Marathon

HeroicPandas said:
is this possible in 2U maths? (they dont learn integration by substitution)

Definitely.

nightweaver066 · May 21, 2013

Re: HSC 2013 2U Marathon

edit:accidently posted a 3u question

Sy123 · May 22, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
$$Using the identity$ \ \ \frac{1}{uv} = \frac{\frac{1}{u}}{v} \ \ $find$ \ \ \int \frac{dx}{x\ln x}$

Made it easier.

anomalousdecay · May 23, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
$\int \frac{dx}{x\ln x}$

Bump for this 2-unit question.

HeroicPandas · May 23, 2013

Re: HSC 2013 2U Marathon

anomalousdecay said:
Bump for this 2-unit question.

are u mocking me?

anomalousdecay · May 23, 2013

Re: HSC 2013 2U Marathon

HeroicPandas said:
are u mocking me?

No. I was just trying to encourage 2-unit students to try this out. Even 3-unit students are welcome, as long as they don't use substitution.

But 4-unit students should leave this for others.

omgiloverice · May 23, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
$\int \frac{dx}{x\ln x}$

$\int \frac {\frac {1}{x}} {\ln{x}} \ dx$

$\int \frac {x^{-1}} {\ln{x}} \ dx$

$$Using the glorious standard integrals the answer is:$ \ \ln {\ln {x}} +c$

anomalousdecay · May 23, 2013

Re: HSC 2013 2U Marathon

Yay. We have someone here. You can look at my harder 2-unit question on the 3-unit marathon. You will like it omgiloverice.

Using the standards integrals table, find in simplest possible form:

$\int^1_0\frac{dx}{\sqrt{x^2 +8}}$

omgiloverice · May 25, 2013

Re: HSC 2013 2U Marathon

$[ln {x+\sqrt{x^{2}+8}}]\ _{0}^{1}$

$= ln ({1+\sqrt{3}})-ln {\sqrt{8}}$

$= ln{\frac{4}{2\sqrt{2}}}$

$= ln{\frac{4}{2\sqrt{2}}}$

$= \frac{ln {2}}{2}$

Now somebody answer my question on the 3 unit thread and previous question on here

aerowhale · May 27, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
$Consider a$ \ \ n \ \ $sided regular polygon$

$$There exists a point in this regular polygon, that is equidistant from all other points in the polygon, call this point$ \ \ O$

$$The distance from$ \ \ O \ \ $to any corner of the polygon is given by$ \ \ r$

$$Let$ \ \ K = \frac{1}{2} r^2 \sin \frac{2\pi}{n}$

$i) Prove that the area of the regular polygon$ \ \ A \ \ $is given by$

$A = nK \ \ \ \ \fbox{2}$

$$ii) It is known that, for some small$ \ \ \theta \ \ \sin \theta \approx \theta$

$$From this it follows that$ \ \ \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$

$$By making the substitution$ \ \ n = \frac{1}{m}$

$Prove that$

$\lim_{n \to \infty} A = \pi r^2 \ \ \ \ \fbox{3}$

$$iii) Interpret this result geometrically$ \ \ \ \fbox{1}$

$$Consider a triangle with 2 sides each having a distance of $ r $ and an included angle of $ \theta\\$
$$Its area is $ \frac{1}{2} r^{2} \sin \theta\\$
$$In a regular n-sided polygon as given in the question, there are $ n $ such triangles, each with a central angle of $ \frac{2 \pi}{n}\\$
$$Thus the area of each triangle is $ \frac{1}{2} r^{2} \sin \frac{2 \pi}{n} $, and hence the area of the whole polygon is $ \frac{n}{2} r^{2} \sin \frac{2 \pi}{n} $ which equals $ nK $, as required$.\\$
$LHS= \lim_{n \to \infty } A\\$
$= \lim_{n \to \infty } \frac{n}{2} r^{2} \sin \frac{2 \pi}{n} $ ~~~~~~~~let $ n=\frac{1}{m}\\$
$= \lim_{m \to 0 } r^{2} \frac{ \sin 2 \pi m}{2m}\\$
$= \pi r^{2}\\$
$= RHS\\$
$\therefore \lim_{n \to \infty} A = \pi r^{2}\\$
$$As the number of sides increases toward infinity, the area tends toward $ \pi r^{2} $ i.e. as the shape of a polygon approaches that of a circle, so does its area i.e. the area of a circle is $ \pi r^{2} $, since a circle can be thought of as a polygon with an infinite number of sides.$\\$

omgiloverice · May 27, 2013

Re: HSC 2013 2U Marathon

omg finally......... somebody from 2U, and useful tip use spaces~
$i)\ $Show that$ \ \frac{d}{dx}(ln(sec{x}+tan{x}))= sec{x}$

$ii)\ $Hence find$$ $\int_{0}^{\frac{\pi}{4}}sec{x} \ dx$

aerowhale · May 29, 2013

Re: HSC 2013 2U Marathon

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( \ln \left ( \sec x + \tan x \right ) \right )\\$
$=\frac{\sec x \tan x + \sec^{2} x}{\sec x + \tan x}\\$ $\\$
$=\sec x \times \frac{\sec x + \tan x}{\sec x + \tan x}\\$ $\\$
$=\sec x\\$ $\\$ $\\$
$\int_{0}^{\frac{\pi}{4}}\sec x~dx\\$
$=\int_{0}^{\frac{\pi}{4}}\frac{\mathrm{d} }{\mathrm{d} x}\left ( \ln \left ( \sec x + \tan x \right ) \right ) dx\\$
$=\ln \left ( \sec \frac{\pi}{4} + \tan \frac{\pi}{4} \right ) - \ln \left ( \sec 0 + \tan 0 \right )\\$
$=\ln \left ( \sqrt{2}+1 \right )$

Trebla · May 29, 2013

Re: HSC 2013 2U Marathon

Solve simultaneously:

2a + 3b = 11
2c + 3d = 24
3a - 2b - 3c + 2d = 0
a + 3b + 3c + 6d - ac - bd = 21

The solutions are quite nice if you can handle it.

anomalousdecay · May 31, 2013

Re: HSC 2013 2U Marathon

omgiloverice said:
$[ln {x+\sqrt{x^{2}+8}}]\ _{0}^{1}$

$= ln ({1+\sqrt{3}})-ln {\sqrt{8}}$

$= ln{\frac{4}{2\sqrt{2}}}$

$= ln{\frac{4}{2\sqrt{2}}}$

$= \frac{ln {2}}{2}$

Now somebody answer my question on the 3 unit thread and previous question on here

Correct. Well done.

omgiloverice · Jun 5, 2013

Re: HSC 2013 2U Marathon

It has being a long time....
$$Find the sum of the following series:$ \ 1^{2}-3^{2} + 5^{2} -7^{2} +9^{2} -11^{2} + ... + 10001^{2} - 10003^{2}$

SpiralFlex · Jun 5, 2013

Re: HSC 2013 2U Marathon

$$a) Find the values of$\; A \;$and$\; B \;$such that,\;$ \frac{1}{r(r+2)} \equiv \frac{A}{r}+\frac{B}{r+2}$

$$b) Hence evaluate$\; \sum_{r=1}^{n}\frac{1}{r(r+2)}\; $as$\; n \rightarrow \infty$

omgiloverice · Jun 6, 2013

Re: HSC 2013 2U Marathon

$a)\ \frac{A(r+2)}{r(r+2)}+\frac{Br}{r(r+2)}$

$\frac{A(r+2)+Br}{r(r+2)}$

$\frac{Ar+2A+Br}{r(r+2)}$

$\frac{2A+r(A+B)}{r(r+2)}$

$$\therefore A=\frac{1}{2}, B=-\frac{1}{2}$

$b) \ $Using results in part a we can rearrange$ \ \frac{1}{r(r+2)} \ $into$\ \frac{1}{2}(\frac{1}{r}-\frac{1}{r+2})$

$\therefore \ $the series would like this:$ \ \frac{1}{2}((1-\frac{1}{3})+(\frac{1}{2}-\frac{1}{4})+(\frac{1}{3}+\frac{1}{5})+\cdot \cdot \cdot )$

$$We notice that$ \ \frac{1}{3}$ cancel out, and this repeats to infinity.$

$\therefore \ $We get:$ \ \frac{1}{2}(1+\frac{1}{2})$

$$ Which equals:$ \ \frac{3}{4}$

Good question stumped me for a good few minutes~

braintic · Jun 6, 2013

Re: HSC 2013 2U Marathon

omgiloverice said:
It has being a long time....
$$Find the sum of the following series:$ \ 1^{2}-3^{2} + 5^{2} -7^{2} +9^{2} -11^{2} + ... + 10001^{2} - 10003^{2}$

Factorise each pair as a difference of two squares, take out the common factor of -2, and add the resulting AP.

Carrotsticks · Jun 6, 2013

Re: HSC 2013 2U Marathon

omgiloverice said:
It has being a long time....
$$Find the sum of the following series:$ \ 1^{2}-3^{2} + 5^{2} -7^{2} +9^{2} -11^{2} + ... + 10001^{2} - 10003^{2}$

$\\ $You can also consider the sum $ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} $ and variations of it such as $ \sum_{k=1}^{n} (2k)^2 = \frac{2n(n+1)(2n+1)}{3}$

braintic · Jun 7, 2013

Re: HSC 2013 2U Marathon

For a given function f(x), we know that f(1)=3, f(3)=-2, f(6)=5, f ' (1)=f ' (3)=f ' (6)=0, and f '' (x) > 0 for 2 < x < 4, and -ve elsewhere.

Find the area of the region bounded by y=f ' (x) and the x-axis.

[Do not attempt to fit an equation to f(x) or any of its derivatives]

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