HSC 2015 MX2 Marathon (archive) (1 Viewer)

Ekman · Mar 22, 2015

Re: HSC 2015 4U Marathon

Kaido said:
In hsc, they dont expect you to algebraically prove this locus type q.
this is one of those geometrical locus

In a geometrical sense, I drew up z+2 and z-2, and their intersection point. Thus utilizing the given formula I was able to prove that points z+2, z-2 and the intersection point are points on an isosceles triangle. Since z+2 and z-2 are equidistant from the origin the point of intersection is on the imaginary axis, thus as z varies, the point of intersection will have the locus of the imaginary axis, as long as it satisfies the formula given.

Sy123 · Mar 22, 2015

Re: HSC 2015 4U Marathon

$\\ $Prove the inequality$ \ x + x^3 - x^4 - x^6 < 1 \ $for all real$ \ x$

Sy123 · Mar 22, 2015

Re: HSC 2015 4U Marathon

$\\ $The real root of the equation$ \ 8x^3 - 3x^2 - 3x - 1 = 0 \ $has a real root of the form$ \ \frac{\sqrt[3]{a} + \sqrt[3]{b} + 1}{c} \ $where$ \ a,b,c \ $are positive integers$ \\ \\ $Find$ \ a + b + c$

Drsoccerball · Mar 22, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Prove the inequality$ \ x + x^3 - x^4 - x^6 < 1 \ $for all real$ \ x$

$Let F(x)= -x^6 -x^4 +x^3 +x$

$F'(x)=-6x^5 -4x^3+3x^2+1$

$$ Using newtons method of approximation the root is at $ x= 0.7097999841$

$$ Sub x value into f(x)$

$F(0.7097999841)= 0.6856940942$

$\therefore $ highest point on graph is 0.69 which is less than 1 $$

$$ Since its an even power with a negative coefficient always less than 1 (Starts from negative ends at negative)$

$\therefore x+x^3 -x^4 -x^6 <1$

Kaido · Mar 22, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
$Let F(x)= -x^6 -x^4 +x^3 +x$

$F'(x)=-6x^5 -4x^3+3x^2+1$

$$ Using newtons method of approximation the root is at $x= 0.7097999841$

$$ Sub x value into f(x)$

$F(0.7097999841)= 0.6856940942$

$\therefore $ highest point on graph is 0.69 which is less than 1 $$

$$ Since its an even power with a negative coefficient always less than 1 (Starts from negative ends at negative)$

$\therefore x+x^3 -x^4 -x^6 <1$

how did you get to that root lol

Drsoccerball · Mar 22, 2015

Re: HSC 2015 4U Marathon

Kaido said:
how did you get to that root lol

Im guessing you havnt done estimation of roots yet?
Newtons method of approximating roots
$a-\frac{f(a)}{f'(a)}$

if you sub in a random value (usually close to the root) you get an estimation of the root and you sub it in as a and keep doing this untill you get a value and that value was what i got as you see when you sub in the value into the equation

Carrotsticks · Mar 22, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
$Let F(x)= -x^6 -x^4 +x^3 +x$

$F'(x)=-6x^5 -4x^3+3x^2+1$

$$ Using newtons method of approximation the root is at $ x= 0.7097999841$

$$ Sub x value into f(x)$

$F(0.7097999841)= 0.6856940942$

$\therefore $ highest point on graph is 0.69 which is less than 1 $$

$$ Since its an even power with a negative coefficient always less than 1 (Starts from negative ends at negative)$

$\therefore x+x^3 -x^4 -x^6 <1$

How do you know for sure that the SP you approximated is the one that yields the maximal function value? The polynomial is of degree 6 so could have up to 5 stationary points, three of which could be maximum stationary points.

Kaido · Mar 22, 2015

Re: HSC 2015 4U Marathon

As Carrot stated.
Also, to get to that degree of accuracy, surely theres a more efficient method o.o

InteGrand · Mar 22, 2015

Re: HSC 2015 4U Marathon

Kaido said:
As Carrot stated.
Also, to get to that degree of accuracy, surely theres a more efficient method o.o

Newton's method is pretty efficient.

Drsoccerball · Mar 22, 2015

Re: HSC 2015 4U Marathon

Carrotsticks said:
How do you know for sure that the SP you approximated is the one that yields the maximal function value? The polynomial is of degree 6 so could have up to 5 stationary points, three of which could be maximum stationary points.

Testing points it never passes 0 again

Drsoccerball · Mar 22, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Testing points it never passes 0 again

Guess im not getting that free lunch ahahaha

Drsoccerball · Mar 22, 2015

Re: HSC 2015 4U Marathon

Carrotsticks said:
How do you know for sure that the SP you approximated is the one that yields the maximal function value? The polynomial is of degree 6 so could have up to 5 stationary points, three of which could be maximum stationary points.

Diferenciate F'(x) and find the turning points and prove that they're all below or above the axis (if there is turning points)

Sy123 · Mar 22, 2015

Re: HSC 2015 4U Marathon

$\\ x +x^3 - x^4 - x^6 = (x-x^4) + (x^3-x^6) = x(1-x^3) + x^3(1-x^3) = (x+x^3)(1-x^3)$

$\\ $For$ \ x \geq 1 \ $and for$ \ x \leq 0 \ $then$ \ (x+x^3)(1-x^3) \ $is negative, so is less than 1$ \\ \\ $For$ \ 0 < x < 1 \ $then both terms$ \ (x+x^3) \ $and$ \ (1-x^3) \ $are positive$ \\ \\ $So,$ \ (x+x^3)(1-x^3) < \frac{(x+x^3 - x^3 + 1)^2}{4} = \frac{1}{4}(x+1)^2 < 1 \ $for$ \ 0 < x< 1$

Drsoccerball · Mar 22, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ x +x^3 - x^4 - x^6 = (x-x^3) + (x^3-x^6) ?$

if you expand you dont get the LHS

Sy123 · Mar 22, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
if you expand you dont get the LHS

Typo the first x^3 is supposed to be x^4, fixed

Drsoccerball · Mar 22, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $The real root of the equation$ \ 8x^3 - 3x^2 - 3x + 1 = 0 \ $has a real root of the form$ \ \frac{\sqrt[3]{a} + \sqrt[3]{b} + 1}{c} \ $where$ \ a,b,c \ $are positive integers$ \\ \\ $Find$ \ a + b + c$

bump

Carrotsticks · Mar 22, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Testing points it never passes 0 again

You are an Extension 2 student, and you think that 'testing points' constitutes as a proof?

Drsoccerball said:
Diferenciate F'(x) and find the turning points and prove that they're all below or above the axis (if there is turning points)

How will the second derivative help with determining the location of the turning points?

Drsoccerball · Mar 22, 2015

Re: HSC 2015 4U Marathon

Carrotsticks said:
You are an Extension 2 student, and you think that 'testing points' constitutes as a proof?

How will the second derivative help with determining the location of the turning points?

I was joking about testing points:L
And for the double derivative if we can prove that all thr turning poinrs are under over the x axis it only intercepts once? Notsure

Carrotsticks · Mar 22, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
And for the double derivative if we can prove that all thr turning poinrs are under over the x axis it only intercepts once? Notsure

I understand what you're trying to say, but I don't understand how you intended to use the double derivative to acquire what you want to acquire.

Also, the turning points don't have to be under the x axis in order for the poly to have a maximal value less than 1. What if the function values were say 0.7, 0.5, 0.3 etc?

Sy123 · Mar 22, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $The real root of the equation$ \ 8x^3 - 3x^2 - 3x - 1 = 0 \ $has a real root of the form$ \ \frac{\sqrt[3]{a} + \sqrt[3]{b} + 1}{c} \ $where$ \ a,b,c \ $are positive integers$ \\ \\ $Find$ \ a + b + c$

$\\ 8x^3 - 3x^2 - 3x - 1 = 0 \\ 8x^3 = 3x^2 + 3x + 1 \\ 9x^3 = x^3 + 3x^2 + 3x + 1 \\ \Rightarrow \ 9x^3 = (x+1)^3 \\ \\ \therefore \ \left(\frac{x+1}{x} \right)^3 = 9$

$\\ $So,$ \ \frac{x+1}{x} = \sqrt[3]{9} \Rightarrow \ x = \frac{1}{\sqrt[3]{9} - 1} = \frac{\sqrt[3]{9^2} + \sqrt[3]{9} + 1}{9 -1} = \frac{\sqrt[3]{81} + \sqrt[]{9} + 1}{8} \\ \\ \therefore \ a + b + c = 81 + 9 + 8 = \fbox{98}$

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