If $a_0,a_1,a_2 \cdots a_{99} \in R$ and $f(x) =x^{100}+a_{99}x^{99}+a_{98}x^{98} +\cdots +a_0$ be such that $|f(0)|=f(1)$..

Question

Problem :

If $a_0,a_1,a_2 \cdots a_{99} \in R$ and $f(x) =x^{100}+a_{99}x^{99}+a_{98}x^{98} +\cdots +a_0$ be such that $|f(0)|=f(1)$ and each root of f(x) =0 is real and between 0 to 1. If product of roots doesn't exceed $\frac{1}{(m)^{50}}$ then find the value of m.

Solution : Let $f(x) =(x-\alpha_1)(x-\alpha_2)\cdots (x-\alpha_{100})$

Now f(1) = $(x-\alpha_1)(x-\alpha_2)\cdots (x-\alpha_{100})$

How to proceed further please suggest don't have any clue. Thanks.

Answer 1

$f(0)$ and $f(1)$ must have the same sign because the polynomial has even degree, and approaches infinities of the same sign for large positive and negative $x$.
$m=1$ will do. Nothing larger will do because all the zeros can be squeezed up near 1.