An interesting question of functions
Let P ( x ) be a polynomial of degree 11 such that P ( x ) = 1 + x 1 , for x = 0 , 1 , 2 , . . . 1 1 . The value of P ( 1 2 ) is
The answer is 0.
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1 solution
Could you please explain why you consider x = − 1 ? I see that the product form of g ( x ) gives 0, but g ( x ) = ( x + 1 ) P ( x ) − 1 gives –1, so how does that give information about a ?
Thanks!
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Note that with g ( x ) = a ( x ) ( x − 1 ) ( x − 2 ) . . . . . . ( x − 1 1 ) we have that g ( − 1 ) = a × 1 2 ! = − 1 ⟹ a = − 1 2 ! 1 .
Then g ( 1 2 ) = a × 1 2 ! = − 1 ⟹ g ( 1 2 ) = ( 1 2 + 1 ) P ( 1 2 ) − 1 = − 1 ⟹ 1 3 P ( 1 2 ) = 0 ⟹ P ( 1 2 ) = 0 .
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Thanks for the explanation! Now I understand the solution.
Product form does not give 0. It give a(12!). And g(x) = -1 from the other form. So we get a(12!) = -1. The reason I put a = -1 Is that I can avoid the P(x). So I can find the value of a.
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Maybe I just didn't look carefully at the product form or you made a typo there, but now I understand. Great problen and solution!
We can consider a function g ( x ) such that it's roots are 0 , 1 , 2 , . . . 1 1
∴ g ( x ) = ( x + 1 ) P ( x ) − 1
g ( x ) = a ( x ) ( x − 1 ) ( x − 2 ) . . . . . . ( x − 1 1 ) ,where a is a constant since 0 , 1 , 2 , . . . . 1 1 are solution of g ( x )
If we put x = − 1 , we get a = − 1 2 ! 1
∴ P ( 1 2 ) = 0