HSC 2015 MX2 Marathon ADVANCED (archive) (1 Viewer)

Status
Not open for further replies.

Trebla

Administrator
Administrator
Joined
Feb 16, 2005
Messages
8,113
Gender
Male
HSC
2006
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/238/extracurricular-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.

I will get the ball rolling:

Prove that

 
Last edited:

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2015 4U Marathon - Advanced Level



-----------------------

 
Last edited:

FrankXie

Active Member
Joined
Oct 17, 2014
Messages
330
Location
Parramatta, NSW
Gender
Male
HSC
N/A
Uni Grad
2004

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2015 4U Marathon - Advanced Level



 

seanieg89

Well-Known Member
Joined
Aug 8, 2006
Messages
2,662
Gender
Male
HSC
2007
Re: HSC 2015 4U Marathon - Advanced Level

Here is my solution:

The "taking the limit" step is the real key. We can pass to the limit immediately from the question statement to get:



Which trivially has no real (or complex) solutions. (ie, we have the desired contradiction.)

This is all just streamlining though of course.
 

dan964

what
Joined
Jun 3, 2014
Messages
3,473
Location
South of here
Gender
Male
HSC
2014
Uni Grad
2019
Re: HSC 2015 4U Marathon - Advanced Level

** If f(nx) and g(nx) are to such that in terms of a polynomial; whereby
the sum of the roots in any amount (individually, in pairs, triples etc.) or the product of the roots is the same value - an integer
==> the coefficients of the polynomial are identical and integer values.


And also that g(x) = sqrt(1 - f(x)^2)
Show that n=5 is the first integer solution, apart from any trivial cases.

This question requires a bit of intuition.
edit: identical is wrong word maybe.
 
Last edited:

Axio

=o
Joined
Mar 20, 2014
Messages
484
Gender
Male
HSC
2015
Re: HSC 2015 4U Marathon - Advanced Level

I. E(x)-E'(x)=(x^n)/n! Which cannot equal zero. But for a double root, this must equal zero when the root is subbed in. Therefore a double root does not exist.

II. I'll try it tomorrow :p
 

FrankXie

Active Member
Joined
Oct 17, 2014
Messages
330
Location
Parramatta, NSW
Gender
Male
HSC
N/A
Uni Grad
2004
Re: HSC 2015 4U Marathon - Advanced Level

No it's not. It's asking for the value of the coefficient not the sum of everything.
Can you post your full working?




Now equating the coefficients of like powers of these two series, we have

 
Last edited:

FrankXie

Active Member
Joined
Oct 17, 2014
Messages
330
Location
Parramatta, NSW
Gender
Male
HSC
N/A
Uni Grad
2004
Re: HSC 2015 4U Marathon - Advanced Level

If f(nx) and g(nx) are identical polynomial of the form
A1^n + A2^(n-1) + ... + An
Ax are constants

And g(x) = sqrt(1 - f(x)^2)
Show that n=5 is the first integer solution.
I still don't understand, if g(x) = sqrt(1 + f(x)^2) and f(nx)=g(nx), isn't it that f(nx)^2=1/2 identically?
 
Last edited:

Axio

=o
Joined
Mar 20, 2014
Messages
484
Gender
Male
HSC
2015
Re: HSC 2015 4U Marathon - Advanced Level

I. E(x)-E'(x)=(x^n)/n! Which cannot equal zero. But for a double root, this must equal zero when the root is subbed in. Therefore a double root does not exist.

II. I'll try it tomorrow :p
II. E'(x)=1+x+x^2/2!...+x^(n-1)/(n-1)!
At stationary point, E'(x)=0, -1=x+x^2/2!...+x^(n-1)/(n-1)!
Therefore at stationary point E(x)=1-1+x^n/n!=x^n/n!, and as n is even, E(x) is positive at any stationary point and as the polynomial is of even degree and has a positive leading coefficient, this means that the graph is always above the x-axis and therefore has no real roots.
 
Last edited:

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2015 4U Marathon - Advanced Level

II. E'(x)=1+x+x^2/2!...+x^(n-1)/(n-1)!
At stationary point, E'(x)=0, -1=x+x^2/2!...+x^(n-1)/(n-1)!
Therefore at stationary point E(x)=1-1+x^n/n!=x^n/n!, and as n is even, E(x) is positive at any stationary point and as the polynomial is of even degree and has a positive leading coefficient, this means that the graph is always above the x-axis and therefore has no real roots.
Yep that's the right idea

A more rigorous way of doing so would be

"if E has no double roots, then each root is a single root, and so then there is some value 'c' So that E(c) < 0, but E(x) -> +inf as x grows large and E(x) -> +inf as x becomes negatively large, so, there must be another root, lets call these pair of roots A and B, since E(A) = E(B) = 0, then there is a point C where E'(C) = 0 And E(C) < 0, but as you have shown, E(C) > 0, so there is no root"
 

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2015 4U Marathon - Advanced Level

 

Chlee1998

Member
Joined
Oct 1, 2014
Messages
90
Gender
Male
HSC
2015
Re: HSC 2015 4U Marathon - Advanced Level

Rhs is (x-a2)(x-a3).....(x-an)+(x-a1)(x-a3)(x-a4)....(x-an).+...+ (x-a1)(x-a2).... (xn-1)
And its easy to see that this is p'(x) through extended product rule
 
Last edited:

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2015 4U Marathon - Advanced Level

Rhs is (x-a2)(x-a3).....(x-an)+(x-a1)(x-a3)(x-a4)....(x-an).+...+ (x-a1)(x-a2).... (xn-1)
And its easy to see that this is p'(x) through extended product rule
In HSC exams I would think that you'd need to justify the use of the "extended product rule", so how would you do it?
 

seanieg89

Well-Known Member
Joined
Aug 8, 2006
Messages
2,662
Gender
Male
HSC
2007
Re: HSC 2015 4U Marathon - Advanced Level

In HSC exams I would think that you'd need to justify the use of the "extended product rule", so how would you do it?
Where p is nonzero we have



The result extends to all x by continuity.
 

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Re: HSC 2015 4U Marathon - Advanced Level

 

SilentWaters

Member
Joined
Mar 20, 2014
Messages
55
Gender
Male
HSC
2014
Re: HSC 2015 4U Marathon - Advanced Level

For non-zero :

If has integer coefficients, then the product of its roots must be of a rational form (the quotient of two integers, the leading coefficient and the constant). To rationalize such a product with some irrational we must have its conjugate as a necessary root. This comes by way of the difference of two squares identity:



which is rational.

For :

Express as a mixed surd where and is coprime with the square numbers. For it is obvious that we need as a root for a rational sum of roots one at a time. For we could negate the irrational with the sum of some number of smaller surds greater than 1. We are then dealing with at least 3 potential roots. Any summation of roots an odd number at a time (from 3 onwards) would produce an irrational product retaining . The only way we can eliminate the irrational part in every case of summation would be to refrain from decomposing that root in the first place, and qualifying as a root. This is because taking cyclic sums will always either directly multiply the two simple surds to give the rational or multiply both by the same factors (comprising other roots), allowing one to factorise out and thus eliminate the irrational.
 
Last edited:

seanieg89

Well-Known Member
Joined
Aug 8, 2006
Messages
2,662
Gender
Male
HSC
2007
Re: HSC 2015 4U Marathon - Advanced Level

To rationalize such a product with some irrational we must have its conjugate as a necessary root.
You should probably provide more justification for this statement.
 
Status
Not open for further replies.

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top