HSC 2017-2019 MX2 Marathon ADVANCED (1 Viewer)

dan964 · Feb 17, 2017

Post any questions within the scope and level of Mathematics Extension 2 mainly targeting Q16 difficulty in the HSC.

Any questions beyond the scope of the HSC syllabus should be posted in the Extracurricular Topics forum:
http://community.boredofstudies.org...hematics-extension-2/360463/...icular-topics/

Once a question is posted, it needs to be answered before the next question is raised.

I encourage all current students in particular to participate in this marathon.

To get the ball-rolling

Question:
Using MX2 methods & complex numbers, find an expression for $cos\ 36$ (degrees)

Edit 6/3/17: I have moved some of the advanced questions to this forum...

Kingom · Feb 25, 2017

HSC 2017 MX2 Marathon ADVANCED

A nice question

wu345 · Feb 25, 2017

Re: HSC 2017 MX2 Marathon

Kingom said:
A nice question

What paper is that from? Looks interesting

Kingom · Feb 26, 2017

Re: HSC 2017 MX2 Marathon

wu345 said:
What paper is that from? Looks interesting

It's a surprise!

seanieg89 · Feb 27, 2017

Re: HSC 2017 MX2 Marathon

Kingom said:
A nice question

$Partial fractions can be used to decompose the LHS into \\ \\ $7+\sum_{k=0}^6 \frac{\omega^k}{z-\omega^k}$\\ \\ where $\omega=\textrm{cis}(2\pi/7)$\\ \\ Now let $H(z)=(z^7-1)\sum_{k=0}^6 \frac{\omega^k}{z-\omega^k}.\\ \\$ We have $H(\omega^k)=\omega^k\prod_{i\neq k}(\omega^k-\omega^i)=\omega^{nk}\prod_{i\neq 0}(1-\omega^i)\\ \\ = \frac{z^7-1}{z-1}|_{z=1}=7.$ Since $H(z)$ has degree at most $6$, we conclude that $H(z)$ is identically equal to $7$.\\ \\ Hence $LHS\cdot (z^7-1)=7(z^7-1)+2H(z)=7(z^7+1)+2(H(z)-7)=7(z^7+1)$ as required.\\ \\ \\ Then we can rearrange to\\ \\ $2\sum_{k=1}^3 \frac{1}{z^2-2z\cos(2k\pi/7)+1}=\frac{7(z^7+1)}{(z^7-1)(z^2-1)}-\frac{1}{(z-1)^2}=\frac{6z^4+4z^3+8z^2+4z+6}{z^6+z^5+z^4+z^3+z^2+z+1}.$\\ \\ Setting $z=1$ and using the double angle formula yields \\ \\ $\sum_{k=1}^3 \csc^2(k\pi/7)=8.$ \\ \\ Where was it from?$

Kingom · Feb 27, 2017

Re: HSC 2017 MX2 Marathon

seanieg89 said:
$Partial fractions can be used to decompose the LHS into \\ \\ $7+\sum_{k=0}^6 \frac{\omega^k}{z-\omega^k}$\\ \\ where $\omega=\textrm{cis}(2\pi/7)$\\ \\ Now let $H(z)=(z^7-1)\sum_{k=0}^6 \frac{\omega^k}{z-\omega^k}.\\ \\$ We have $H(\omega^k)=\omega^k\prod_{i\neq k}(\omega^k-\omega^i)=\omega^{nk}\prod_{i\neq 0}(1-\omega^i)\\ \\ = \frac{z^7-1}{z-1}|_{z=1}=7.$ Since $H(z)$ has degree at most $6$, we conclude that $H(z)$ is identically equal to $7$.\\ \\ Hence $LHS\cdot (z^7-1)=7(z^7-1)+2H(z)=7(z^7+1)+2(H(z)-7)=7(z^7+1)$ as required.\\ \\ \\ Then we can rearrange to\\ \\ $2\sum_{k=1}^3 \frac{1}{z^2-2z\cos(2k\pi/7)+1}=\frac{7(z^7+1)}{(z^7-1)(z^2-1)}-\frac{1}{(z-1)^2}=\frac{6z^4+4z^3+8z^2+4z+6}{z^6+z^5+z^4+z^3+z^2+z+1}.$\\ \\ Setting $z=1$ and using the double angle formula yields \\ \\ $\sum_{k=1}^3 \csc^2(k\pi/7)=8.$ \\ \\ Where was it from?$

IMO Longlist 1988

seanieg89 · Feb 27, 2017

Re: HSC 2017 MX2 Marathon

Haha that was my suspicion.

Kingom · Feb 27, 2017

Re: HSC 2017 MX2 Marathon

I believe it's accessible to the extension 2 student so i put it on the marathon

seanieg89 · Feb 27, 2017

Re: HSC 2017 MX2 Marathon

Kingom said:
I believe it's accessible to the extension 2 student so i put it on the marathon

Yeah fair enough, they were a lot easier back in the day.

Still should probably put questions like it in the advanced thread (Dan only created it today I believe, but for the last couple of years we have usually split the questions into two marathons, one for routine stuff and one for questions at HSC Q16 difficulty or above. As long as it is still within syllabus of course).

Kingom · Feb 27, 2017

Re: HSC 2017 MX2 Marathon

seanieg89 said:
Yeah fair enough, they were a lot easier back in the day.

Still should probably put questions like it in the advanced thread (Dan only created it today I believe, but for the last couple of years we have usually split the questions into two marathons, one for routine stuff and one for questions at HSC Q16 difficulty or above. As long as it is still within syllabus of course).

Yeh i wanted to put this in the advanced marathon, but none existed at the time it was posted so

frog1944 · Feb 27, 2017

Re: HSC 2017 MX2 Marathon ADVANCED

dan964 said:
Post any questions within the scope and level of Mathematics Extension 2 mainly targeting Q16 difficulty in the HSC.

Any questions beyond the scope of the HSC syllabus should be posted in the Extracurricular Topics forum:
http://community.boredofstudies.org...hematics-extension-2/360463/...icular-topics/

Once a question is posted, it needs to be answered before the next question is raised.

I encourage all current students in particular to participate in this marathon.

To get the ball-rolling

Question:
Using MX2 methods & complex numbers, find an expression for the cos 36.

Note that $(\cos\frac{\pi}{5}+i\sin\frac{\pi}{5})^5=-1$ . Consider $z^5=-1$ whose roots are $z_1=cis\frac{\pi}{5}, z_2 = cis\frac{-\pi}{5}, z_3 = cis\frac{3\pi}{5}, z_4 = cis\frac{-3\pi}{5}, z_5 =-1$ . $z^5 + 1 = (z+1)(z^4-z^3+z^2-z+1) = 0$ Thus the roots of $z^4 - z^3 + z^2 -z +1=0$ are $z_1, z_2, z_3, z_4$ . Note that $z_2, z_4$ are the conjugates of $z_1, z_3$ respectively.
$z^4 -z^3 +z^2-z+1 = (z-z_1)(z-z_2)(z-z_3)(z-z_4) = (z^2 - 2\cos\frac{\pi}{5}z + 1)(z^2 - 2\cos\frac{3\pi}{5}z+1)$
By comparison of coefficients for $z^3, z^2$ we get:
$\cos\frac{\pi}{5} + \cos\frac{3\pi}{5}=\frac{1}{2}$ and $\cos\frac{\pi}{5}\cos\frac{3\pi}{5}=\frac{-1}{4}$
These are the sum of roots and product of roots of the quadratic equation; $4x^2-2x+1=0$ .
$x=\frac{1\pm \sqrt{5}}{4}$ , but $\cos\frac{\pi}{5}>0$ . Thus $\cos\frac{\pi}{5}=\frac{1+\sqrt{5}}{4}$

Mahan1 · Mar 5, 2017

Re: HSC 2017 MX2 Marathon

$\textrm{for} x,y,z \in \mathbb{R}\ xyz(x+y+z) =1$
$\textrm{Find the minimum of the following expression}\ (x+y)(y+z)(x+z)$

Kingom · Mar 6, 2017

Re: HSC 2017 MX2 Marathon

Mahan1 said:
$\textrm{for} x,y,z \in \mathbb{R}\ xyz(x+y+z) =1$
$\textrm{Find the minimum of the following expression}\ (x+y)(y+z)(x+z)$

Lagrange multipliers because why not

Paradoxica · Mar 6, 2017

Re: HSC 2017 MX2 Marathon

Mahan1 said:
$\textrm{for} x,y,z \in \mathbb{R}\ xyz(x+y+z) =1$
$\textrm{Find the minimum of the following expression}\ (x+y)(y+z)(x+z)$

By inspection, the limit is -∞

dan964 · Mar 28, 2017

Re: HSC 2017 MX2 Marathon ADVANCED

$Time for a new question$

$Q1: Prove by induction on $n$ that$
$(fg)^{(n)}=$$\sum_{i=0}^{n}\binom{n}{i}f^{(n-i)}g^{(i)}$

$$Q2: If $ P_{n}(x)=\frac{d^n}{d x^n}[(x^2-1)^n].$ Find $ P_n(1) $ and $ P_n(-1)$
$$where $ f^{(n)} $ is the notation for the n-th derivative$

calamebe · Mar 31, 2017

Re: HSC 2017 MX2 Marathon ADVANCED

dan964 said:
$Time for a new question$

$Q1: Prove by induction on $n$ that$
$(fg)^{(n)}=$$\sum_{i=0}^{n}\binom{n}{i}f^{(n-1)}g^{(i)}$

$$Q2: If $ P_{n}(x)=\frac{d^n}{d x^n}[(x^2-1)^n].$ Find $ P_n(1) $ and $ P_n(-1)$
$$where $ f^{(n)} $ is the notation for the n-th derivative$

For Q1 instead of (n-1) at the top it should just be n-i.

$Anyway for that one in the prove for $k+1$ part (the induction step is it called?) get $(fg)^{(k+1)}$ which will be the derivative of $(fg)^{(k)}$, and this can value can be derived from the induction hypothesis. I derived by breaking the sum into two sums, and using a substitution $r=i+1$ on one of the substitutions. Then you can pretty easily get to the desired result.$

$And for Q2 it is $n!2^n$ and $n!(-2)^n$ respectively right?$

dan964 · Apr 1, 2017

Re: HSC 2017 MX2 Marathon ADVANCED

fixed sorry.

dan964 · Apr 1, 2017

Re: HSC 2017 MX2 Marathon ADVANCED

calamebe said:
For Q1 instead of (n-1) at the top it should just be n-i.

$Anyway for that one in the prove for $k+1$ part (the induction step is it called?) get $(fg)^{(k+1)}$ which will be the derivative of $(fg)^{(k)}$, and this can value can be derived from the induction hypothesis. I derived by breaking the sum into two sums, and using a substitution $r=i+1$ on one of the substitutions. Then you can pretty easily get to the desired result.$

$And for Q2 it is $n!2^n$ and $n!(-2)^n$ respectively right?$

Q2 is correct. Split it into (x+1)(x-1)

jathu123 · Jun 12, 2017

Re: HSC 2017 MX2 Marathon ADVANCED

frog0101 said:
Find the Cartesian coordinates of a complex number that has a modulus of 3 and argument pi/8

$$ \noindent $ \cos \frac{\pi}{4}=\frac{\sqrt{2}}{2} \\ \\ 2 \cos^2 \frac{\pi}{8}-1=\frac{\sqrt{2}}{2} \\ \\ \Rightarrow \cos \frac{\pi}{8}=\frac{1}{2}\sqrt{2+\sqrt{2}} \\ \\$ Similarly, $ \sin \frac{\pi}{8}=\frac{1}{2}\sqrt{2-\sqrt{2}} \\ \\ \therefore 3 $cis$\frac{\pi}{8} = \frac{3}{2}\sqrt{2+\sqrt{2}}+\frac{3}{2}\sqrt{2-\sqrt{2}} $ $ i$

Paradoxica · Jun 21, 2017

Re: HSC 2017 MX2 Marathon ADVANCED

A square pyramid has height exactly half it's base side length, a. Find the volume of the pyramid.

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