Let P(x) be a polynomial with real coefficients such that ∫01 xmP(1−x)dx=0∀m∈N∪{0}, then

Let P(x) be a polynomial with real coefficients such that 01xmP(1x)dx=0mN{0}, then

  1. A

    P(x)=xn(1x)n for some nN

  2. B

    P(x)=(1x)2n for some nN

  3. C

    P(x)=1xm(1x)n for some m,nN

  4. D

    P(x) = 0

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    Solution:

    0=01xmP(1x)dx=01(1x)mP(x)

    Let P(x)=k=0nakxk where akR

    =k=0nak[1(1x)]k=k=0nakj=0k(1)k kCj(1x)j=k=0nbk(1x)k

    where bk=j=kn(1)k jCkaj

    Now 01(P(x))2dx=01P(x)k=0nbk(1x)kdx

    =k=0nbk01(1x)kP(x)dx=0

    As P(x) is a polynomial,

    P(x)0x[0,1] P(x)0

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