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Dragon.Jade a posé la question dans Science & MathematicsMathematics · il y a 5 ans

Tricky math problem if you are bored.?

Hello,

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DISCLAIMER:

I am not trying to get my homework done. This quiz is provided as it is, and I do have the answer. So it is a challenge for anyone. Best answer will be awarded in approximatively two days. Have fun gals and guys!

= = = = = = = =

Let 𝑎 and 𝑏 be two real values.

Let 𝑃 be a polynomial such that:

   {   𝑃(𝑎) = 0

   {   𝑃(𝑏) = 0

A. What is the value of 𝑃'(𝑎)+𝑃'(𝑏) ?

B. What is the minimal degree 𝑚𝑖𝑛 of 𝑃?

C. If 𝑃 has a degree of 𝑚𝑖𝑛+1, what can you conclude?

Regards,

Dragon.Jade :-)

Mise à jour:

Mmm... After consideration. Question A is faulty and is hereby cancelled. It will not count toward Best Answer.

Questions B and C stand though. And they seem so trivial that no one wants to answer them correctly...?

Mise à jour 2:

= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

That was the trick in the question.

There's no doubt the Zero polynomial defined by:

   ∀𝑥∈ℝ,  Zero(𝑥) = 0

fits the given definition of 𝑃.

Mise à jour 3:

Now what is the degree of this polynomial?

It is indeed a convention that some think it should be undefined, 0, -1 or -∞.

If 𝑚𝑖𝑛 were undefined or 𝑚𝑖𝑛=-∞, then 𝑚𝑖𝑛+1 would have no meaning. Which means question C would be meaningless.

So those conventions need to be discarded in order to answer question C.

Mise à jour 4:

If 𝑚𝑖𝑛=-1, then 𝑚𝑖𝑛+1=0 and such 𝑃 would be a nonzero constant polynomial then 𝑃(𝑎)≠0 and 𝑃(𝑏)≠0 which is contrary to the problem.

The conclusion would then be there is no 𝑃 with degree 𝑚𝑖𝑛+1=0.

If 𝑚𝑖𝑛=0, then 𝑚𝑖𝑛+1=1 and such 𝑃 would be a linear polynomial which can only have an unique zero.

The conclusion would then be that 𝑎=𝑏.

► Thanks for participating. The BA is given to Barry G for being the only one seeing that the Zero polynomial did fit the definition of 𝑃.

4 réponses

Pertinence
  • il y a 5 ans
    Réponse favorite

    B. The minimum degree is zero :

    P(x) = x^0 - 1

    which is zero for all real values of x.

    C. If P has degree 1 then

    P(x) = Ax+B with A <>0.

    P(a) = Aa+B = 0 --> B = -Aa

    P(b) = Ab+B = 0 --> Ab-Aa = A(b-a) = 0

    so b=a.

  • il y a 5 ans

    A. Insufficient information given. There are infinity possible answers.

    B. Assuming a and b are different, then P must have at least degree 2. But if a and b can be the same, then P can have degree 1.

    C. There are min+1 roots (but not necessarily distinct roots; some or all may be equal), but some of those may be non-real roots.

  • il y a 5 ans

    Seems underspecified to me.

    p(x) = x^2 - x and q(x) = x^3 - 2x^2 + x are zero at a=0 and b=1, and nowhere else.

    p'(x) = 2x - 1 --> p'(0) = -1, p'(1) = 1, p'(0) + p'(1) = -1 + 1 = 0

    q'(x) = 3x^2 - 2x + 1 --> q'(0) = 1, p'(1) = 2, p'(0) + p'(1) = -1 + 1 = 3

    So, question A has no definite answer. The others two have pretty obvious answers.

  • ?
    Lv 7
    il y a 5 ans

    B. Assuming a and b are distinct, P must be of degree 2 or greater. The minimum degree is 2.

    C. A degree 3 polynomial with 2 real roots also has a third real root. Conclusion: there must be one more real root, though it isn't necessarily distinct from both a and b.

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