polynomials q (1 Viewer)

[bleakarcher]

This is random, but can someone help me wif this question?

(a) Consider the following statements about a polynomial Q(x) .
(i) If Q(x) is even, then Q'(x) is odd.
(ii) If Q'(x) is even, then Q(x)is odd.
Indicate whether each of these statements is true or false. Give reasons for your
answers.

How would you answer this sort of a question since its telling you to give a proof in general?

 

[Shadowdude]

Re: BoS group study session? (Easter Edition)

This is random, but can someone help me wif this question?

(a) Consider the following statements about a polynomial Q(x) .
(i) If Q(x) is even, then Q'(x) is odd.
(ii) If Q'(x) is even, then Q(x)is odd.
Indicate whether each of these statements is true or false. Give reasons for your
answers.

How would you answer this sort of a question since its telling you to give a proof in general?

You mean of even degree? So just set up an example in general, I'd think.

 

[bleakarcher]

Re: BoS group study session? (Easter Edition)

You mean of even degree? So just set up an example in general, I'd think.

yeh degree

and yes thats what I did..not sure about it though

EDIT:
polynomial

 

[Carrotsticks]

Re: BoS group study session? (Easter Edition)

So even polynomial or even degree? There's a difference...

 

[RealiseNothing]

Re: BoS group study session? (Easter Edition)

So even polynomial or even degree? There's a difference...

Yer this, I was under the impression it was an even polynomial, and had no idea.

 

[Carrotsticks]

Re: BoS group study session? (Easter Edition)

I suppose you could say even polynomials must be even in degree...

 

[RealiseNothing]

Re: BoS group study session? (Easter Edition)

I suppose you could say even polynomials must be even in degree...

If it's of degree, say 3, it can still be even though.



This is even when x=1 for example.

Edit: Or did I misinterpret what you said.

 

[Carrotsticks]

Re: BoS group study session? (Easter Edition)

If it's of degree, say 3, it can still be even though.



This is even when x=1 for example.

Edit: Or did I misinterpret what you said.

I meant even/odd in terms of the definition

Q(x) = Q(-x) if even and -Q(x) = Q(-x) if odd etc.

 

[RealiseNothing]

Re: BoS group study session? (Easter Edition)

I meant even/odd in terms of the definition

Q(x) = Q(-x) if even and -Q(x) = Q(-x) if odd etc.

lols my bad.

 

[bleakarcher]

Re: BoS group study session? (Easter Edition)

This question was actually from the 1998 hsc paper but they didnt specify. It should be polynomial not degree*.

 

[AAEldar]

Re: BoS group study session? (Easter Edition)

I meant even/odd in terms of the definition

Q(x) = Q(-x) if even and -Q(x) = Q(-x) if odd etc.

Well if Q(x) = Q(-x) then all the powers must be even? So then just set up an equation like where n, k are even. Then likewise for an odd function and differentiate?

Haven't done that in ages D: Am I right? hahaha.

Also, good luck/have fun at the meet guys!

 

[Carrotsticks]

Re: BoS group study session? (Easter Edition)

Well if Q(x) = Q(-x) then all the powers must be even? So then just set up an equation like where n, k are even. Then likewise for an odd function and differentiate?

Haven't done that in ages D: Am I right? hahaha.

Also, good luck/have fun at the meet guys!

That is exactly what I am typing out right now, but all the sigmas are killing me.

 

[Trebla]

This is random, but can someone help me wif this question?

(a) Consider the following statements about a polynomial Q(x) .
(i) If Q(x) is even, then Q'(x) is odd.
(ii) If Q'(x) is even, then Q(x)is odd.
Indicate whether each of these statements is true or false. Give reasons for your
answers.

How would you answer this sort of a question since its telling you to give a proof in general?

If Q(x) is even then Q(-x) = Q(x)
Differentiate both sides wrt x, we have
- Q'(-x) = Q'(x)
=> Q'(-x) = -Q'(x)
Hence statement (i) is true.

If Q'(x) is even then Q'(-x) = Q'(x)
Now integrate both sides wrt x, we have
=> - Q(-x) = Q(x) + c
=> Q(-x) = - Q(x) - c
If c is non-zero then Q(-x) =/= - Q(x), hence statement (ii) is false

Note: I think this was in an Extension 2 paper as a harder Extension 1 type question.

 

[Drongoski]

Trebla - a masterly solution to a very good question.