# HSC 2017 MX2 Marathon (archive) (1 Viewer)

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#### EngineeringHelp

##### New Member
Re: HSC 2017 MX2 Marathon

Just did my half yearly today... 30% of my internal mark... Calculating that im getting in the 70%'s
Why am i still in this course

#### Drsoccerball

##### Active Member
Re: HSC 2017 MX2 Marathon

$\bg_white \noindent Let n be a positive integer and let \zeta_1, \zeta_2, and \zeta_3 be the roots of the equation z^3 + 1 = 0. Here z \in \mathbb{C}.$

$\bg_white \noindent If a> 0 show that \frac{1}{3} \sum^3_{k = 1} \frac{1}{|\zeta_k - a|^2} = \frac{1 + a^2 + a^4}{(1 + a^3)^2}.$

#### davidgoes4wce

##### Well-Known Member
Re: HSC 2017 MX2 Marathon

HSC 2016 Q 10. I spent 5 minutes this morning on this question and couldn't work it out.

$\bg_white Suppose that x+\frac{1}{x}=-1$

$\bg_white What is the value of x^{2016}+\frac{1}{x^{2016}} ?$

Any tips? Advice? Short cuts? I assume 5 minutes would be too long for a question like this .

#### leehuan

##### Well-Known Member
Re: HSC 2017 MX2 Marathon

HSC 2016 Q 10. I spent 5 minutes this morning on this question and couldn't work it out.

$\bg_white Suppose that x+\frac{1}{x}=-1$

$\bg_white What is the value of x^{2016}+\frac{1}{x^{2016}} ?$

Any tips? Advice? Short cuts? I assume 5 minutes would be too long for a question like this .

#### InteGrand

##### Well-Known Member
Re: HSC 2017 MX2 Marathon

HSC 2016 Q 10. I spent 5 minutes this morning on this question and couldn't work it out.

$\bg_white Suppose that x+\frac{1}{x}=-1$

$\bg_white What is the value of x^{2016}+\frac{1}{x^{2016}} ?$

Any tips? Advice? Short cuts? I assume 5 minutes would be too long for a question like this .
$\bg_white \noindent Note that x must be non-real, since x+\frac{1}{x} is always at least 2 in absolute value for real x. Try substituting x = \mathrm{cis}\left(\theta\right). (If you initially substitute the more general x = r\,\mathrm{cis}\left(\theta\right), where r = |x|, you'll be able to show r must be 1.) Also recall the result that if x = \mathrm{cis}\left( \theta\right), then x^{n} + \frac{1}{x^{n}} = 2 \cos n\theta.$

#### Kingom

##### Member
Re: HSC 2017 MX2 Marathon

HSC 2016 Q 10. I spent 5 minutes this morning on this question and couldn't work it out.

$\bg_white Suppose that x+\frac{1}{x}=-1$

$\bg_white What is the value of x^{2016}+\frac{1}{x^{2016}} ?$

Any tips? Advice? Short cuts? I assume 5 minutes would be too long for a question like this .
we know that x^2+x+1=0. but this means that x is a cube root of unity ie x^3=1
therefore x^2016=1 and the solution is 2

##### -insert title here-
Re: HSC 2017 MX2 Marathon

$\bg_white \noindent Let n be a positive integer and let \zeta_1, \zeta_2, and \zeta_3 be the roots of the equation z^3 + 1 = 0. Here z \in \mathbb{C}.$

$\bg_white \noindent If a> 0 show that \frac{1}{3} \sum^3_{k = 1} \frac{1}{|\zeta_k - a|^2} = \frac{1 + a^2 + a^4}{(1 + a^3)^2}.$
Is this by any chance just you taking a special case of the AMM Problem 11947 of 2016?

##### Active Member
Re: HSC 2017 MX2 Marathon

$\bg_white \noindent Let n be a positive integer and let \zeta_1, \zeta_2, and \zeta_3 be the roots of the equation z^3 + 1 = 0. Here z \in \mathbb{C}.$

$\bg_white \noindent If a> 0 show that \frac{1}{3} \sum^3_{k = 1} \frac{1}{|\zeta_k - a|^2} = \frac{1 + a^2 + a^4}{(1 + a^3)^2}.$
Is this by any chance just you taking a special case of the AMM Problem 11947 of 2016?
Why indeed it is! As asked, the question makes for a very nice complex number question (and a bit of algebra) which is more than doable at the MX2 level.

#### si2136

##### Well-Known Member
Re: HSC 2017 MX2 Marathon

Hey guys, I have a question that I have never encountered before in 4U, how would I go about solving this? Thanks!

#### InteGrand

##### Well-Known Member
Re: HSC 2017 MX2 Marathon

Hey guys, I have a question that I have never encountered before in 4U, how would I go about solving this? Thanks!
View attachment 33841
$\bg_white \noindent It's of the form x = A\sec \theta, y = B\tan \theta, which as we know is a parametric representation of the hyperbola \frac{x^{2}}{A^{2}} - \frac{y^{2}}{B^{2}} = 1.$

#### si2136

##### Well-Known Member
Re: HSC 2017 MX2 Marathon

Just a curious question, is there a a faster way to prove the Recursive Integration Formulae without going through the long process?

#### pikachu975

Re: HSC 2017 MX2 Marathon

Just a curious question, is there a a faster way to prove the Recursive Integration Formulae without going through the long process?
What's the long process? Integration by parts?

If you mean integration by parts, then almost all questions involve IBPs, but some can be done by manipulating the integrand.

#### stupid_girl

##### Active Member
Re: HSC 2017 MX2 Marathon

Just a curious question, is there a a faster way to prove the Recursive Integration Formulae without going through the long process?
When you are asked to prove a given reduction formula, the efficient way is usually by differentiation.

##### -insert title here-
Re: HSC 2017 MX2 Marathon

Given:

x+y+z = 3a

xy + yz + zx = 3b

Where x,y,z are all real, and ab>0, find the greatest and least possible values of any of x,y,z

#### kawaiipotato

##### Well-Known Member
Re: HSC 2017 MX2 Marathon

$\bg_white \noindent A pack of sweets contains 25 sweets, a combination of mini-cupcakes and mini-macarons. If there are 18 varieties of mini-cupcakes, and 10 varieties of mini-macarons, how many ways can you fill a pack? (The order does not matter, just how many of each sweet type you have in the pack.)$

#### jathu123

##### Active Member
Re: HSC 2017 MX2 Marathon

$\bg_white \noindent Given S_n =\sum_{k=1}^{n}k^2. Prove by mathematical induction that: \\ \\ \qquad nS_{n}-\sum_{r=1}^{n-1}S_r= \sum_{r=1}^{n}r^3 \quad \forall n\in \mathbb{Z}: n>1$

#### EngineeringHelp

##### New Member
Re: HSC 2017 MX2 Marathon

Hey Guys!!

I am new around here, and not the best at this 4unit maths subject. When doing questions, I am not able to see the distinct ideas, and i get stuck, not knowing how to approach, or what the best and most efficient approach is. Any tips and help will be most appreciated! <3