HSC 2014 MX2 Marathon (archive) (1 Viewer)

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dunjaaa

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Re: HSC 2014 4U Marathon

(i)By inspection x=-1 is a factor of P(x), by polynomial division we get the required result
(ii) For 3 real roots, we set the discriminant >=0 and we get B<=-A
(iii) The region is bounded by lines B>=-1, A>=-1, B<=-A
 
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Sy123

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Re: HSC 2014 4U Marathon

(i)By inspection x=-1 is a factor of P(x), by polynomial division we get the required result
(ii) For 3 real roots, we set the discriminant >=0 and we get B<=-A
(iii) The region is bounded by lines B>=-1, A>=-1, B<=-A
In (ii) and (iii) you are missing another solution
 

dunjaaa

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Re: HSC 2014 4U Marathon

B>=3A?
 
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Sy123

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Re: HSC 2014 4U Marathon

Yep, however B=> 3A and -A => B are only the first half of hte solutions

I'm guessing you got to the point in that



So both brackets either have to be negative or both brackets have to be positive. You have only covered one of those 2 cases.
 

Sy123

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Re: HSC 2014 4U Marathon



Not sure if I'm overdoing it with the guiding thing now
 

dunjaaa

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Re: HSC 2014 4U Marathon

so careless tonight sigh..
(ii) First case: B>=3A, B<=-A (both positive)
Second Case: B>=-A, B<=3A (both negative)
(iii)First Case: A>=-1, B>=-1, B>=3A, B<=-A
Second Case: A<=1, B<=1, B>=-A, B<=3A
 

Sy123

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Re: HSC 2014 4U Marathon

so careless tonight sigh..
(ii) First case: B>=3A, B<=-A (both positive)
Second Case: B>=-A, B<=3A (both negative)
(iii)First Case: A>=-1, B>=-1, B>=3A, B<=-A
Second Case: A<=1, B<=1, B>=-A, B<=3A
Yea something like that

Even for the average 4U student, that would be too difficult.
Ah well, ok
 

Ikki

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Re: HSC 2014 4U Marathon

I'm probably going to sound stupid, but ok z=x+i.
Locus of z=x+i, is that just all of the plane? 0.0 Liek all of x values right, so adding i doesn't really do much...
Man i need to revise my complex.
K, sub any x
Then will lie in y=i. So just line y=i? Horizontal line at i?
 
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Sy123

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Re: HSC 2014 4U Marathon

I'm probably going to sound stupid, but ok z=x+i.
Locus of z=x+i, is that just all of the plane? 0.0 Liek all of x values right, so adding i doesn't really do much...
Man i need to revise my complex.
K, sub any x
Then will lie in y=i. So just line y=i? Horizontal line at i?
Yep the line y=1 (When using x,y we are talking about reala numbers so it doesn't make sense to say y=i)

Think of it this way

z is some number on the complex plane, where the real part can be anything (x is real), but its imaginary part has to always be 1, this is of course y=1


Try the second one
 

Sy123

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Re: HSC 2014 4U Marathon

 
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Ikki

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Re: HSC 2014 4U Marathon

Yep the line y=1 (When using x,y we are talking about reala numbers so it doesn't make sense to say y=i)

Think of it this way

z is some number on the complex plane, where the real part can be anything (x is real), but its imaginary part has to always be 1, this is of course y=1


Try the second one
Ok, z^2=(x+i)^2
Using the same logic, it is the line squared so a parabola with vertex at y=1?
 

Sy123

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Re: HSC 2014 4U Marathon

Ok, z^2=(x+i)^2
Using the same logic, it is the line squared so a parabola with vertex at y=1?
So

What kind of parabola would it be? How did you know it was a parabola?
 

Ikki

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Re: HSC 2014 4U Marathon

Facedesk*
I just recapped, I kept trying to think visually but i guess algebraic would be best...
Its just letting z=x+i that threw me off.
So i let z=x+iy.
for i) it is clear that locus is y=1 by equating coeff
ii) Similarly from your expansion y=1 should also be locus for z squared.
 
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