HSC 2015 MX2 Marathon ADVANCED (archive) (3 Viewers)

Drsoccerball · Feb 1, 2015

Re: HSC 2015 4U Marathon - Advanced Level

dan964 said:
Next:
For 2015er's

1. Show that if w = 1/x + x
Then w^2 - 2 =x^2 + 1/x^2

2. Use part 1 to show that x^4+....+x+1 = x^2*(w^2+w+1)
And hence factorise the x^4+x^3+...+x+1

3. Solve x^5=1

4. Hence deduce using all parts above to find an exact value for cos (pi/5)) and cos (2pi/5)

5. Repeat the process this time expressing x and w in mod arg form.

$\text{number2 ?} x^4 +...+x+1 =x^2 (w^2 +w+1) \\ RHS= x^2 (x^2 +2 +\frac{1}{x^2} +x + \frac{1}{x}+1) \\= x^4 +2x^2 +1 +x^3 +x +x^2 \\ =x^4 +x^3 +3x^2 +x +1 ?...$

Drsoccerball · Feb 1, 2015

Re: HSC 2015 4U Marathon - Advanced Level

edit for number 2 $x^2 (w^2 +w -1)$

dan964 · Feb 2, 2015

Re: HSC 2015 4U Marathon - Advanced Level

whoops yep its is minus 1

leehuan · Feb 5, 2015

Re: HSC 2015 4U Marathon - Advanced Level

$1.\quad w=\frac { 1 }{ x } +x\\ Then\quad { w }^{ 2 }=\frac { 1 }{ { x }^{ 2 } } +\frac { 2x }{ x } +{ x }^{ 2 }\quad but\quad \frac { c }{ c } =1\\ Subtracting\quad by\quad 2:\quad { w }^{ 2 }-2=\frac { 1 }{ { x }^{ 2 } } +{ x }^{ 2 }\\ 2.\quad I\quad just\quad ignored\quad the\quad already\quad published\quad answer\quad and\quad went\quad ahead.\\ LHS=\quad { x }^{ 4 }+{ x }^{ 3 }+{ x }^{ 2 }+x+1\\ ={ x }^{ 2 }({ x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } +x+\frac { 1 }{ x } +1)\\ ={ x }^{ 2 }({ w }^{ 2 }-2+w+1)\\ ={ x }^{ 2 }({ w }^{ 2 }+w-1)\\ =RHS\\$
$Then,\quad { x }^{ 4 }+{ x }^{ 3 }+{ x }^{ 2 }+x+1\\ ={ x }^{ 2 }({ w }^{ 2 }+w+\frac { 1 }{ 4 } -\frac { 5 }{ 4 } )\\ ={ x }^{ 2 }[{ (w+\frac { 1 }{ 2 } ) }^{ 2 }-\frac { 5 }{ 4 } )]\\ ={ x }^{ 2 }(w+\frac { 1+\sqrt { 5 } }{ 2 } )(w+\frac { 1-\sqrt { 5 } }{ 2 } )\\ ={ x }^{ 2 }(\frac { 1 }{ x } +x+\frac { 1+\sqrt { 5 } }{ 2 } )(\frac { 1 }{ x } +x+\frac { 1-\sqrt { 5 } }{ 2 } )\\ =({ x }^{ 2 }+\frac { 1+\sqrt { 5 } }{ 2 } x+1)({ x }^{ 2 }+\frac { 1-\sqrt { 5 } }{ 2 } x+1)\quad irreducible\quad over\ \mathbb{R}.$
$3.\quad For\quad { x }^{ 5 }=1,\quad let\quad x=r\quad cis(\theta )\\ { \left( r\quad cis(\theta ) \right) }^{ 5 }=1\\ { r }^{ 5 }cis(5\theta )=1cis\left( 2k\pi \right) \quad (k\in \mathbb{Z})\\ Equating\quad moduli\quad and\quad arguments:\\ { r }^{ 5 }=1,\quad r=1\\ and\quad 5\theta =2k\pi \quad \quad (pick\quad k=-2,-1,0,1,2)\\ \theta =\frac { -4\pi }{ 5 } ,\frac { -2\pi }{ 5 } ,0,\frac { 2\pi }{ 5 } ,\frac { 4\pi }{ 5 } \\ Hence\quad the\quad solutions\quad are:\\ x=cis\left( \frac { -4\pi }{ 5 } \right) ,\quad cis\left( \frac { -2\pi }{ 5 } \right) ,\quad 1,\quad cis\left( \frac { 2\pi }{ 5 } \right) ,\quad cis\left( \frac { 4\pi }{ 5 } \right) \\$

leehuan · Feb 5, 2015

Re: HSC 2015 4U Marathon - Advanced Level

I suppose this answer doesn't really use Q2.
$4.\quad Alternatively\quad for\quad { x }^{ 5 }-1=0\\ \left( x-1 \right) \left( { x }^{ 4 }+{ x }^{ 3 }+{ x }^{ 2 }+x+1 \right) =0\\ As\quad x-1=0\quad solves\quad for\quad x=1,\quad the\quad other\quad 4\quad solutions\quad must\quad also\quad be\quad the\quad roots\quad of\\ { x }^{ 4 }+{ x }^{ 3 }+{ x }^{ 2 }+x+1=0\\ Substituting\quad x=cis\left( \frac { 2\pi }{ 5 } \right) \quad and\quad applying\quad De\quad Moivre's\quad Theorem\quad we\quad have:\\ cis\left( \frac { 8\pi }{ 5 } \right) +cis\left( \frac { 6\pi }{ 5 } \right) +cis\left( \frac { 4\pi }{ 5 } \right) +cis\left( \frac { 2\pi }{ 5 } \right) +1=0$
$\cos { \left( \frac { 8\pi }{ 5 } \right) } +i\sin { \left( \frac { 8\pi }{ 5 } \right) } +\cos { \left( \frac { 6\pi }{ 5 } \right) } +i\sin { \left( \frac { 6\pi }{ 5 } \right) } +\cos { \left( \frac { 4\pi }{ 5 } \right) } +i\sin { \left( \frac { 4\pi }{ 5 } \right) } +\cos { \left( \frac { 2\pi }{ 5 } \right) } +i\sin { \left( \frac { 2\pi }{ 5 } \right) } +1=0+0i\\ Equating\quad real\quad parts:\\ \\ \cos { \left( \frac { 8\pi }{ 5 } \right) } +\cos { \left( \frac { 6\pi }{ 5 } \right) } +\cos { \left( \frac { 4\pi }{ 5 } \right) } +\cos { \left( \frac { 2\pi }{ 5 } \right) } +1=0\\ 2\cos { \left( \frac { 4\pi }{ 5 } \right) } +2\cos { \left( \frac { 2\pi }{ 5 } \right) } +1=0\quad \quad \quad \quad since\quad \cos { \left( 2k\pi -\alpha \right) } =\cos { \alpha } \\ 4\cos ^{ 2 }{ \left( \frac { 2\pi }{ 5 } \right) } -2+2\cos { \left( \frac { 2\pi }{ 5 } \right) } +1=0\\$
$4\cos ^{ 2 }{ \left( \frac { 2\pi }{ 5 } \right) } +2\cos { \left( \frac { 2\pi }{ 5 } \right) } -1=0\\ \cos { \left( \frac { 2\pi }{ 5 } \right) } =\frac { -2\pm \sqrt { { \left( 2 \right) }^{ 2 }-4\left( 4 \right) \left( -1 \right) } }{ 2(4) } \\ \\ \cos { \left( \frac { 2\pi }{ 5 } \right) } =\frac { -2\pm \sqrt { 20 } }{ 2(4) } \\ \cos { \left( \frac { 2\pi }{ 5 } \right) } =\frac { -1\pm \sqrt { 5 } }{ 4 }$
$By\quad considering\quad the\quad quadrant\quad of\quad the\quad angle,\quad \cos { \left( \frac { 2\pi }{ 5 } \right) } =\frac { -1+\sqrt { 5 } }{ 4 } \\ Then,\quad \cos { \left( \frac { 2\pi }{ 5 } \right) } =2\cos ^{ 2 }{ \left( \frac { \pi }{ 5 } \right) } -1\\ \frac { -1+\sqrt { 5 } }{ 4 } +1=2\cos ^{ 2 }{ \left( \frac { \pi }{ 5 } \right) } \\ \frac { 3+\sqrt { 5 } }{ 8 } =\cos ^{ 2 }{ \left( \frac { \pi }{ 5 } \right) } \\ \cos { \left( \frac { \pi }{ 5 } \right) } =\frac { \sqrt { 3+\sqrt { 5 } } }{ 2\sqrt { 2 } }$
$=\frac { \sqrt { 6+2\sqrt { 5 } } }{ 4 } \\ =\frac { \sqrt { 1+2\sqrt { 5 } +5 } }{ 4 } \\ =\frac { 1+\sqrt { 5 } }{ 4 }$

dan964 · Feb 6, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Your answer is actually for Q5.

Q4 requires you to use at least part 2 (you may need part 1, depending on your approach).
In fact it may be easier or harder.

seanieg89 · Feb 6, 2015

Re: HSC 2015 4U Marathon - Advanced Level

seanieg89 said:
$Given that \\$p(z)=z^3+az^2+bz+c$\\ has all roots on the unit circle, show that \\$q(z)=z^3+|a|z^2+|b|z+|c|$ \\ also has all roots on the unit circle.$

bump.

FrankXie · Feb 6, 2015

Re: HSC 2015 4U Marathon - Advanced Level

seanieg89 said:
$Given that \\$p(z)=z^3+az^2+bz+c$\\ has all roots on the unit circle, show that \\$q(z)=z^3+|a|z^2+|b|z+|c|$ \\ also has all roots on the unit circle.$

$$ firstly, $ |c|=|z_1z_2z_3|=|z_1||z_2||z_3|=1. $ secondly $ |a|=|z_1+z_2+z_3|=|\bar{z_1}+\bar{z_2}+\bar{z_3}|=\left|\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3} \right|=\left|\frac{z_1z_2+z2_z_3+z_1z_3}{z_1z_2z_3}\right|=\frac{|z_1z_2+z_2z_3+z_1z_3|}{|z_1||z_2||z_3|}=|z_1z_2+z_2z_3+z_1z_3|=|b|$
$so $ q(z)=z^3+Az^2+Az+1, $ where $ 0\leq A\leq3 $ is a real number.$
$q(z)=(z^3+1)+(Az^2+Az)=(z+1)(z^2-z+1)+Az(z+1)=(z+1)[z^2+(A-1)z+1], $ so $ z=-1 $ is a root, and since $ -1\leq A-1\leq2, \Delta=(A-1)^2-4\leq0, $ where equality holds true when $ A=3, $ therefore $ z^2+(A-1)z+1 $ either has repeated real root $ z=-1 $ or has a pair of conjugate imaginary roots, and in both cases the modulii equal one since product of roots is one. $$
Q.E.D.

braintic · Feb 14, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Drsoccerball said:
Another self made one :

$If $ \sum_{r=1}^n bcis(r\theta) =Acis(5040\times \theta^7) $ Find n where $ \frac{A}{B} $ Is a real whole number. A and b are real numbers$

Hmmm .... are you sure about this question?
Just for starters, are there restrictions on the pronumerals?
And ... do you really mean to have a power inside the cis?
Don't get me wrong ... I'm not yet prepared to say it is all wrong .... but it just seems very weird to me.

Ekman · Feb 14, 2015

Re: HSC 2015 4U Marathon - Advanced Level

braintic said:
Hmmm .... are you sure about this question?
Just for starters, are there restrictions on the pronumerals?
And ... do you really mean to have a power inside the cis?
Don't get me wrong ... I'm not yet prepared to say it is all wrong .... but it just seems very weird to me.

The only correction he needs to make is to use $\prod^{n}_{r=1}$ rather than the sigma sign

After that, everything becomes straightforward, n=7

Drsoccerball · Feb 14, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Ekman said:
The only correction he needs to make is to use $\prod^{n}_{r=1}$ rather than the sigma sign

After that, everything becomes straightforward, n=7

Deleted the question because i didnt know how to code that multiplication sign and try the working out for it and ekman dont do it-.-.-

seanieg89 · Feb 15, 2015

Re: HSC 2015 4U Marathon - Advanced Level

A school offers 8 subjects in year 12, all of which are compulsory.

For each subject, each student either receives an A or a B.

We say student 1 beat student 2 if every grade of student 1 is at least equal to the corresponding grade of student 2 and student 1 got at least one HIGHER grade.

If there are n students, none of whom beat another student, and none of whom had identical grades, show that $n\leq 70.$

Sy123 · Feb 15, 2015

Re: HSC 2015 4U Marathon - Advanced Level

seanieg89 said:
A school offers 8 subjects in year 12, all of which are compulsory.

For each subject, each student either receives an A or a B.

We say student 1 beat student 2 if every grade of student 1 is at least equal to the corresponding grade of student 2 and student 1 got at least one HIGHER grade.

If there are n students, none of whom beat another student, and none of whom had identical grades, show that $n\leq 70.$

$\\ $Let the scores$ \ A \ $and$ \ B \ $be denoted by scores$ \ 1 \ $and$ \ 0 \ $respectively. So someone with$ \ 3 \ A $'s and$ \ 5 \ B $'s, means that they have a score of$ \ 1 + 1 + 1 + 5\cdot 0 = 3 \\ \\ $Therefore, for a student to beat another, means their score has to be higher, and the possible scores are$ \ 0,1,2,3,4,5,6,7,8 \\ \\ $For a selection of$ \ n\ $students, if they don't beat one another, this means that they have the same score, meaning there is only 1 unique score among the$ \ n \ $students$ \\ \\ $If the score that the collection of students got, was$ \ k \ $then$ \ \binom{8}{k} \ $is the number of possible unique grades with the same score (pick$ \ k \ $out of the$ \ 8 \ $grades to get an$ \ A \ $in)$ \\ \\ $So, the maximum possible number of students is identical to the maximum of$ \ \binom{8}{k} \ $which is$ \ \binom{8}{4} = 70 \\ \\ $Therefore$ \ n\leq 70$

seanieg89 · Feb 15, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Sy123 said:
$\\ $Let the scores$ \ A \ $and$ \ B \ $be denoted by scores$ \ 1 \ $and$ \ 0 \ $respectively. So someone with$ \ 3 \ A $'s and$ \ 5 \ B $'s, means that they have a score of$ \ 1 + 1 + 1 + 5\cdot 0 = 3 \\ \\ $Therefore, for a student to beat another, means their score has to be higher, and the possible scores are$ \ 0,1,2,3,4,5,6,7,8 \\ \\ $For a selection of$ \ n\ $students, if they don't beat one another, this means that they have the same score, meaning there is only 1 unique score among the$ \ n \ $students$ \\ \\ $If the score that the collection of students got, was$ \ k \ $then$ \ \binom{8}{k} \ $is the number of possible unique grades with the same score (pick$ \ k \ $out of the$ \ 8 \ $grades to get an$ \ A \ $in)$ \\ \\ $So, the maximum possible number of students is identical to the maximum of$ \ \binom{8}{k} \ $which is$ \ \binom{8}{4} = 70 \\ \\ $Therefore$ \ n\leq 70$

How did you deduce that in a group with no-one beating anyone else all the scores must be equal?

Eg AAAAB doesn't beat BBBBA despite having a wayy better score.

Sy123 · Feb 15, 2015

Re: HSC 2015 4U Marathon - Advanced Level

seanieg89 said:
How did you deduce that in a group with no-one beating anyone else all the scores must be equal?

Eg AAAAB doesn't beat BBBBA despite having a wayy better score.

I misinterpreted "beat", my bad

SilentWaters · Feb 15, 2015

Re: HSC 2015 4U Marathon - Advanced Level

I'm a bit rusty, but let's give this a go.

Consider each student's report in terms of an $8$ -element, ordered string of $A$ 's and $B$ 's. We know we must satisfy both the stated conditions, that $(1)$ no two strings are the same, and $(2)$ there cannot be a string that beats another in a cohort of reports. We examine the second condition, and note that in such a cohort of reports, we select any two strings and must find that there is an $A-B$ and a $B-A$ correspondence between the first and the second string respectively. This is the minimum difference ( $2$ grades) required to meet $(1)$ and equalize the two strings.

Given a fixed, arbitrary number of $A$ 's and $B$ 's, we generate different strings to meet $(1)$ , and find that in the process we satisfy $(2)$ since we only have $2$ possible elements, and hence must switch an $A$ and a $B$ in moving between unique strings.

In that case we have $\binom{8}{k}$ reports in a satisfactory cohort, where the integer $0\leq k\leq 8$ .

One could argue that moving on to a different number of $A$ 's and $B$ 's (changing $k$ ) satisfies $(1)$ by giving another combinatoric set of reports with totally different strings.

Consider one of these new strings, however. First, we find a string in the original cohort with the greatest number of identical grades. Then, we note that the difference would be either a number of $A-B$ correspondences or a number of $B-A$ correspondences. This is because the new string will have more/less of one grade than the original string when we are trying to match the two.

Such bias means one of these two strings would beat the other. Therefore, an exhaustive cohort of satisfactory reports must have a fixed number of $A$ 's and $B$ 's, and be of the above-stated combinatoric form. We know from the (ninth) row Pascal's triangle that to maximise this, we choose $k=4$ . Hence $n_{max}$ = $\binom{8}{4} = 70$ .

Pyjamees · Feb 15, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Heya. Could someone help me on a little bit of conics?
http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2000exams/hsc00_maths/95MAT4U.PDF
Question 8. Thanks. A little clueless.

braintic · Feb 15, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Pyjamees said:
Heya. Could someone help me on a little bit of conics?
http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2000exams/hsc00_maths/95MAT4U.PDF
Question 8. Thanks. A little clueless.

Have you at the solutions on BOSTES yet?

Drsoccerball · Feb 20, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Self made question... goodluck

:
$If n is not odd, find n and theta if :$

$4=cos(n\theta) +ncos((n-2))\theta+...+^nC_{\frac{n}{2}}cos(n-n)\theta +c$

FrankXie · Feb 20, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Drsoccerball said:
Self made question... goodluck:
$If n is not odd, find n if :$

$1=cos(n\theta) +ncos((n-2))\theta+...+^nC_{\frac{n}{2}} (n-n)\theta +c$

just clarify: if the last term is +c, what is c? and what is \theta? do u mean finding n given that equation is true for all \theta? or do u mean finding both n and \theta ?

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