Questions from 2003 CSSA (1 Viewer)

[ezzy85]

How do you sketch the cos-1? Its pi for x = 0, pi/2 for f(x) = 0, what next?

For the next polynomial q, how do you do i and the hence part of ii.?

Thanks

 

[underthesun]

For the polynomial q,

x^4 - x^3 + 2x^2 - 2x + 1 = 0

x (x^3 - x^2 + 2x - 2) = -1
x = -1 / (x^3 - x^2 + 2x - 2)

for x to be a whole number, x has to be one. however one is not a root of the equation. Any whole number value for x will give a fraction with a whole number denominator. If the expression has a whole number denominator other than one, then the value -1 / (x^3 - x^2 + 2x - 2) cannot be an integer. Hence, x cannot be an integer.

as for ii), i have no idea..

edit : am I allowed to put the answers out yet? Have all school done the CSSA 4U test?

 

[Newbie]

ii) x^4 + 3x^3 +5x^2 +3x + 1 =0 ??
something like that since im REALLY GOOD... at silly mistakes

and the other thing = 1?

oh shit my trials are tomorrow

anyway ezzy how did you get this paper?

 

[Constip8edSkunk]

for the graph when x -> +/- oo, f(x)->1, arccos f(x) -> 0

 

[felix_js]

this be the graph? hopefully it isnt wrong

 

[ezzy85]

Thanks guys for the help.

Originally posted by Newbie


anyway ezzy how did you get this paper?

look in the last few threads.

 

[ND]

a) ii)

Roots are in the form:
y = x-1
x = y+1
sub into x^4 - x^3 + 2x^2 - 2x + 1 = 0 ...[1]
(y+1)^4 - (y+1)^3 + 2(y+1)^2 - 2(y+1) + 1 = 0
this expands to:
y^4 + 3y^3 + 5y^2 + 5y + 1 = 0 ...[2]
now sum of roots of this eqn:
a-1+B-1+g-1+d-1 = -3
a+B+g+d=1
a+B+g=1-d
now 1-d is the same as the roots of [2] but negative, so the eqn with the roots in the form 1-x is:
x^4 - 3x^3 + 5x^2 - 5x + 1 = 0
so (a+B+g)(a+B+d)(a+d+g)(B+d+g) is the product of the roots of this eqn, which is 1.

I hope that works.

Just out of curiosity, which question/how many marks was that?

 

[ezzy85]

thats q5.a, part i. worth 2 marks and part ii. worth 4 marks.

 

[Newbie]

i still dont get part i

 

[deepulse]

i) Since the polynomial is monic, and the last term is 1, which is a prime number, the only integer roots can be + 1 or - 1

Subbing +1 and -1 You do not get 0. Hence no integer roots

 

[tonberry_kun]

that's a more clear and concise answer

 

[underthesun]

Originally posted by Toodulu
hmm, i think you're only meant to talk about it a week after you've done it.

drbuchanan posted the paper, so i guess its ok

Originally posted by deepulse
i) Since the polynomial is monic, and the last term is 1, which is a prime number, the only integer roots can be + 1 or - 1

Subbing +1 and -1 You do not get 0. Hence no integer roots

wuah man.. does that work? my long working would be useless then...

 

[deepulse]

well i did the paper and i got full marks for that question and thats wat i wrote hehe

 

[McLake]

Yes, that method is valid, but you would hae to be careful about how you word your answer ...

 

[underthesun]

i got it now. That's very nice deepulse

i think i just discovered the prime number formula..