HSC 2012 MX1 Marathon #2 (archive) (1 Viewer)

seanieg89 · Jul 15, 2012

Re: HSC 2012 Marathon

Sanjeet said:
asianese's solution made complete sense to me.

Doesn't make it any more legal unfortunately...

Sanjeet · Jul 16, 2012

Re: HSC 2012 Marathon

seanieg89 said:
Doesn't make it any more legal unfortunately...

What part of his solution would a 3u student not be expected to produce? Just curious

deswa1 · Jul 16, 2012

Re: HSC 2012 Marathon

seanieg89 said:
This is the problem with your argument Sy123, you cannot treat such a series "term by term". There isn't really a proof of this equality that a 3U student could be expected to produce...I will post a sandwich argument a bit later.

You can prove this identity with entirely 3U methods. I remember an extension question in Cambridge went through it (it used the upper and lower bound rectangles)- I'll try and find the question.

Carrotsticks · Jul 16, 2012

Re: HSC 2012 Marathon

Sanjeet said:
What part of his solution would a 3u student not be expected to produce? Just curious

Everything he did was within the scope of the Syllabus for Extension 1, so a good Extension 1 student would be able to 'understand' everything he did.

HOWEVER, he assumed that the behaviour of the binomial expansion is the same for infinitely large values of n, which CANNOT be assumed. It has to be justified, refer to seanieg89's post about treating an infinite number of terms.

Also deswa1, the proof for it has been in the HSC beforehand (or were you referring to the series expansion? If so, it's in the BOS MX2 Seminar booklet):

seanieg89 · Jul 16, 2012

Re: HSC 2012 Marathon

deswa1 said:
You can prove this identity with entirely 3U methods. I remember an extension question in Cambridge went through it (it used the upper and lower bound rectangles)- I'll try and find the question.

Yep, the methods are 3U but I more meant that we cannot expect a 3U student to answer this kind of question without being led through a specific argument...there are too many stumbling blocks.

Sy123 · Jul 18, 2012

Re: HSC 2012 Marathon

Time for another question I guess.

$$Using the fact that$ \ \ (1+x)^{-n}=\frac{1}{(1+x)^n} \\ \\ $Show that$ \ \ \ -n 2^{-n-1}=-\left(\frac{\sum_{r=1}^{n} r \binom{n}{r}}{\left(\sum_{r=0}^n \binom {n}{r}}\right)^2 \right)$

Also Im trusting the fact that the values for the summation notation, are beside the sigma because of how I put them in a fraction?

Ok its all fine, you can now attempt the question, the typo was corrected.

Also please post a question after your done...

Sanjeet · Jul 18, 2012

Re: HSC 2012 Marathon

Sy123 said:
Time for another question I guess.

$$Using the fact that$ \ \ (1+x)^{-n}=\frac{1}{(1+x)^n} \\ \\ $Show that$ \ \ \ -n \times 2^{-n-1}=-1 \times \left(\frac{\sum^n _{r=1} r \binom{n}{r}}{\left(\sum^n _{r=0} \binom {n}{r}}\right)^2 \right)$

Also Im trusting the fact that the values for the summation notation, are beside the sigma because of how I put them in a fraction?

Ok its all fine, you can now attempt the question, the typo was corrected.

Also please post a question after your done...

$\sum_{r=1}^{n}$
I'm guessing you wanted it like this? If so the tex code is \sum_{r=1}^{n}

Sy123 · Jul 18, 2012

Re: HSC 2012 Marathon

Thats pretty much what I did, except Im guessing that because I condensed it so much, the latex editor decided to put the values there.

OMGITzJustin · Jul 22, 2012

Re: HSC 2012 Marathon

Sy123 said:
Thats pretty much what I did, except Im guessing that because I condensed it so much, the latex editor decided to put the values there.

Integral of secx

SpiralFlex · Jul 22, 2012

Re: HSC 2012 Marathon

OMGITzJustin said:
Integral of secx

$=\int \sec x * \frac{\sec x + \tan x}{\sec x+\tan x} \; dx$

$=\int \frac{\sec^2 x + \sec x \tan x}{\sec x+\tan x} \;dx$

$=\ln |\sec x + \tan x|+C$

SpiralFlex · Jul 22, 2012

Re: HSC 2012 Marathon

$ABC is a triangle with sides a,b,c and a perimeter of p.$

$$Prove$\; \cos (\frac{A}{2})=\frac{1}{2}\sqrt{\frac{p(p-2a)}{bc}}$

OMGITzJustin · Jul 22, 2012

Re: HSC 2012 Marathon

integral of (a+(1/a))^n da, a is a complex number (over C - unit circle)

barbernator · Jul 22, 2012

Re: HSC 2012 Marathon

OMGITzJustin said:
integral of (a+(1/a))^n da, a is a complex number (over C - unit circle)

dis is the extension 1 marathon.

seanieg89 · Jul 22, 2012

Re: HSC 2012 Marathon

OMGITzJustin said:
integral of (a+(1/a))^n da, a is a complex number (over C - unit circle)

Haha how is this anywhere near HSC level?

Assuming the integration is counterclockwise, the answer is:

I=0 if n is an even non-negative integer
I=2*pi*i*nC_{(n+1)/2} if n is an odd non-negative integer.

Aesytic · Jul 22, 2012

Re: HSC 2012 Marathon

SpiralFlex said:
$ABC is a triangle with sides a,b,c and a perimeter of p.$

$$Prove$\; \cos (\frac{A}{2})=\frac{1}{2}\sqrt{\frac{p(p-2a)}{bc}}$

Using cosine rule,
cosA = [b^2 + c^2 - a^2]/2bc
2cos^2(A/2) - 1 = [b^2 + c^2 - a^2]/2bc
2cos^2(A/2) = [b^2 + c^2 - a^2]/2bc + 1
= [b^2 + c^2 - a^2 + 2bc]/2bc
= [(b+c)^2 - a^2]/2bc
= [(b+c+a)(b+c-a)]/2bc
= [p((p-a)-a)]/2bc
= [p(p-2a)]/2bc
cos^2(A/2) = [p(p-2a)]/4bc
cos(A/2) = +- sqrt{[p(p-2a)]/4bc}
= +- 1/2*sqrt{[p(p-2a)]/bc}
but 0 < A < 180
0 < A/2 < 90
.'. cos(A/2) > 0
.'. cos(A/2) = 1/2*sqrt{[p(p-2a)]/bc}

barbernator · Jul 22, 2012

Re: HSC 2012 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=cos(A)=\frac{b^{2}@plus;c^{2}-a^{2}}{2bc}\\ 2cos^{2}(\frac{A}{2})-1=\frac{b^{2}@plus;c^{2}-a^{2}}{2bc}\\ cos^{2}(\frac{A}{2})=\frac{b^{2}@plus;c^{2}-a^{2}@plus;2bc}{4bc}\\ cos(\frac{A}{2})=\sqrt{\frac{b^{2}@plus;c^{2}-a^{2}@plus;2bc}{4bc}}~(taking~@plus;ve)\\ ~~~~~~~~~~~=\sqrt{\frac{(b@plus;c)^{2}-a^{2}}{4bc}}\\ ~~~~~~~~~~~=\sqrt{\frac{(b@plus;c-a)(b@plus;c@plus;a)}{4bc}}\\ ~~~~~~~~~~~=\frac{1}{2}\sqrt{\frac{(p)(p-2a)}{bc}}~as~required\\" target="_blank"><img src="http://latex.codecogs.com/gif.latex?cos(A)=\frac{b^{2}+c^{2}-a^{2}}{2bc}\\ 2cos^{2}(\frac{A}{2})-1=\frac{b^{2}+c^{2}-a^{2}}{2bc}\\ cos^{2}(\frac{A}{2})=\frac{b^{2}+c^{2}-a^{2}+2bc}{4bc}\\ cos(\frac{A}{2})=\sqrt{\frac{b^{2}+c^{2}-a^{2}+2bc}{4bc}}~(taking~+ve)\\ ~~~~~~~~~~~=\sqrt{\frac{(b+c)^{2}-a^{2}}{4bc}}\\ ~~~~~~~~~~~=\sqrt{\frac{(b+c-a)(b+c+a)}{4bc}}\\ ~~~~~~~~~~~=\frac{1}{2}\sqrt{\frac{(p)(p-2a)}{bc}}~as~required\\" title="cos(A)=\frac{b^{2}+c^{2}-a^{2}}{2bc}\\ 2cos^{2}(\frac{A}{2})-1=\frac{b^{2}+c^{2}-a^{2}}{2bc}\\ cos^{2}(\frac{A}{2})=\frac{b^{2}+c^{2}-a^{2}+2bc}{4bc}\\ cos(\frac{A}{2})=\sqrt{\frac{b^{2}+c^{2}-a^{2}+2bc}{4bc}}~(taking~+ve)\\ ~~~~~~~~~~~=\sqrt{\frac{(b+c)^{2}-a^{2}}{4bc}}\\ ~~~~~~~~~~~=\sqrt{\frac{(b+c-a)(b+c+a)}{4bc}}\\ ~~~~~~~~~~~=\frac{1}{2}\sqrt{\frac{(p)(p-2a)}{bc}}~as~required\\" /></a>

OMGITzJustin · Jul 22, 2012

Re: HSC 2012 Marathon

oh sorry boys, wrong thread hahahaha (i usually type in "marathon" in my search bar) saw seanieg and typed away

lets take a break from the harder ones, find the integral of sin^2 (1/2x)

barbernator · Jul 22, 2012

Re: HSC 2012 Marathon

OMGITzJustin said:
oh sorry boys, wrong thread hahahaha (i usually type in "marathon" in my search bar) saw seanieg and typed away

lets take a break from the harder ones, find the integral of sin^2 (1/2x)

again, assuming u didn't type out the question wrongly, this is 3 unit. did you mean the integral of sin^2(x/2)?

Timske · Jul 22, 2012

Re: HSC 2012 Marathon

lets do some 3d pm

Sy123 · Jul 22, 2012

Re: HSC 2012 Marathon

$$a) Show that$ \ \ \frac{d}{dx} x\sqrt{r^2-x^2}=\sqrt{r^2-x^2}-\frac{x^2}{\sqrt{r^2-x^2}} \\ \\ $b) Show that$ \frac{x^2}{\sqrt{r^2-x^2}}=\frac{x^2-r^2}{\sqrt{r^2-x^2}}+\frac{r^2}{\sqrt{r^2-x^2}} \\ \\ $Hence show that$ \ \ 2 \times \int_{-r}^{r} \sqrt{r^2-x^2} \ dx=\pi r^2$

Enjoy.

Typo has been fixed.

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